Class 10

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SSLC Maths Chapter 2 Circles Textbook Questions and Answers

SCERT Class 10th Standard Maths Chapter 2 Circles Notes

Textbook Page No, 42

Circles Class 10 Kerala Syllabus Question 1.
Suppose we draw a circle with the bottom side of the triangles in the picture as diameter. Find out whether the top corner of each triangle is inside the circle, on the circle or outside the circle.

Answer:
Angle of the first triangle =110°
As 110° > 90° the top comer will be inside the circle
Angle of second triangle = 90°
∴ The top comer will be on the circle.
Angle of the third triangle = 70°
70° > 90°
∴ The top comer will be outside the circle

Sslc Maths Circles Questions And Answers Question 2.
For each diagonal of the quadrilateral shown, check whether the other two corners are inside, on or outside the circle with that diagonal as diameter

Answer:
The fourth angle in ABCD
= 360 – (110 + 105 + 55) = 360

Drawing diagonal AC and taking it as diameter of a circle As, ∠D = 90°
D will be on the circle. ∠B = 55° (90 > 55°)
∴ B will be outside the circle
Drawing diagonal BD and taking it as diameter of a circle.
∠A = 105°, ∠C = 110°
As both angles are greater than 90, they lie inside the circle.

Sslc Maths Chapter 2 Kerala Syllabus Question 3.
If circles are drawn with each side of a triangle of sides 5 centimetres, 12 centimetres and 13 centimetres, as diametres, then with respect to each circle, where would be the third vertex?
Answer:
As the sides are 5, 12, 13 cm and also
52 + 122 = 25 + 144 = 169 = 132
∴ ABC is a right triangle

Taking BC as diameter and drawing a circle, ∠A (<90°), A will be outside the circle.
Taking AB as diameter and drawing a circle ∠C (<90°) C will be outside the circle.
Taking AC as diameter and drawing a circle, ∠B = 90°, B will be on the circle.

Circles Class 10 Scert Solutions Kerala Syllabus Question 4.
In the picture, a circle is drawn with a line as diameter and a smaller circle with half the line as diameter. Prove that any chord of the larger circle through the point where the circles meet is bisected by the small circle.

Answer:
∠ADO = ∠APB = 90°
(angle subtended by diameter is always 90°)
⇒ OD\\PB

AO = OB (Radius of bigger circle)
(If in a triangle, the line drawn from midpoint of one side, is parallel to another side, then the line will bisect the third side)
Therefore AD = DP
(AB’s midpoint is ‘O’ and OD\\PB)

Sslc Maths Circles Questions And Answers Pdf Question 5.

Use a calculator to determine up to two decimal places, the perimeter and the area of the circle in the picture.
Answer:

Maths Chapter 2 Class 10 Kerala Syllabus Question 6.
The two circles in the picture cross each other at A and B. The points P and Q are the other ends of the diameters through A.

i. Prove that P, B, Q lie on a line.
ii. Prove that PQ is parallel to the line joining the centres of the circles and is twice as long as this line.
Answer:

i. ∆ PAB (angle subtended by semicircle)
∠PBA = 90°
∆ ABQ be a triangle on the semi¬circle of centre D.
∴ ∠ABQ = 90°
∠PBA + ∠ABQ = 180 (Linear pair)
As AP, AQ are diameters of the circle. PQ be the line drawn through B per¬pendicularly to AB. Therefore P, B, Q lies on the same line.

ii.

Sslc Maths Circles Notes Kerala Syllabus Question 7.
Prove that the two circles drawn on the two equal sides of an isosceles triangle as diameters pass through the midpoint of the third side.
Answer:

∠ADB= 90° (AABD angle subtended by semicircle)
∠CDA = 90°
∴ ∠ADB +∠CDA = 180° (linear pair)
∆ ABD ∆ ADC are right angled triangles.
In ∆ ABD
BD² = AB² – AD² ( AB = AC )
= AC² – AD² = DC²
BD = CD

Sslc Maths Chapter 2 Circles Questions And Answers Question 8.
Prove that all four circles drawn with the sides of a rhombus as diameters pass through a common point.

Prove that this is true for any quadrilat¬eral with adjacent sides equal, as in the picture.

Answer:

ABCD is a rhombus so diameter are perpendicular bisectors.
∠AOD = 90°
O be on the circle having diameter AD.
∠AOB = 90°, therefore
O be on the circle having diameter AB.
∠BOC= 90°, therefore
O be on the circle having diameter BC
∠DOC= 90°, therefore
O be on the circle having diameter DC
O be the common point on the circle.
∠A0D = ∠AOB and
∠COD = ∠BOC and,
AD = AB,
AO be the common side.

Δ AOD, Δ AOB are equal triangles.
OD = OB
Both the circles can passed through O.
A BCD is an isosceles triangle.
Those circles having diameters CD and BC are passing through midpoint of BD.
∴ O be common for the four circles. (Diameter)

Sslc Maths Chapter 2 Circles Notes Kerala Syllabus Question 9.
A triangle is drawn by joining a point on a semicircle to the ends of the diameter. Then semicircles are drawn with the other two sides as diameter.

Prove that the sum of the areas of the blue and red crescents in the second picture is equal to the area of the triangle.
Answer:

Area of triangle = \(\frac { 1 }{ 2 }\) × 2r × h = rh
Area of the semicircle = \(\frac{\pi r^{2}}{2}\)
Area of the rest of the figure

The diameters of the red and blue semi-circles are the sides of the two triangles.
∴ Their areas

Area of red and blue crescents

= area of the triangle

Textbook Page No. 53

Circles Class 10 Notes Pdf Kerala Syllabus  Question 1.
In all the pictures given below, O is the centre of the circle and A, B, C are points on it. Calculate all angles of Δ ABC and Δ OBC in each.

Answer:
a.

OA = OB (radii)
∴ ∠OAB = 20° ; (∠OAB = ∠OBA)
OC = OA (radii) ;
∠OAC = 30°
∠BAC = ∠OAB + ∠OAC
= 20 + 30 = 50°
∠BOC = 2x ∠BAC = 100°
OB = OC (radii)
∴ ∠OBC = ∠OCB = 40°

Angles of triangle ABC are ∠BAC = 50° ∠ABC = 60°, ∠ACB = 70°
Angles of ∆ BOC are
∠OBC = 40°, ∠OCB = 40°, ∠BOC = 100°

b.

Angles of ∆ ABC are ∠ABC = 50°, ∠BAC = 60°, ∠BCA = 70°
Angles of ∆ AOC are
∠OAC = 40°, ∠AOC = 100°, ∠OCA = 40°

c.

∠ACB = 180 – 55 = 125° ; OA = OC
∠CAO = ∠ACO = 70° ; ∠OBC = ∠BCO = 55° Angles of A OBC are
∠OBC = 55° ∠COB = 70° ∠BCO = 55°
Angles of ∆ ABC

Class 10 Maths Chapter 2 Circles Kerala Syllabus Question 2.
The numbers 1,4,8 on a clock’s face are joined to make a triangle.

Calculate the angles of this triangle.
How many equilateral triangles can we make by joining numbers on the clock’s face?
Answer:
The angle subtended by two adjacent numbers at the centre of the clock is 30°

We can make 4 equilateral triangles by joining the numbers on the clock (1. 5, 9), (2, 6, 10), (3. 7, 11), (4, 8, 12)

10th Class Maths Chapter 2 Circles Kerala Syllabus Question 3.
In each problem below, draw a circle and a chord to divide it into two parts such that the parts are as specified.
i. All angles on one part 80°.
ii. All angles on one part 110°.
iii. All angles on one part half of all angles on the other.
iv. All angles on one part, one and a half times the angles on the other.
Answer:
i. ∠AOB = 160°
Therefore all angles in the are ACB arc 80°.

ii. Draw angle as central angle 220° so angle on the small arc AB will be 110°.

iii. Draw angle as central angle 120° or 240° All angles on one part will be 120°, and other part be 60°.

iv. Draw a circle and draw central angle 144° All angles on the part APB will be 120° and All angles on the part AQB will be 108°.

Sslc Maths Chapter 2 Circles Kerala Syllabus Question 4.
A rod bent into an angle is placed with its corner at the centre of a circle and it is found that \(\frac { 1 }{ 10 }\) of the circle lies within it. If
it is placed with its corner on another circle, what part of the circle would be within it?

Answer:

Circles Class 10 Notes Kerala Syllabus Question 5.
In the picture, O is the centre of the circle and A, B, C are points on it. Prove that
∠OAC + ∠ABC = 90°

Answer:

Class 10 Maths Chapter 2 Kerala Syllabus Question 6.
Draw a triangle of circumradius 3 centimetres and two of the angles 32\(\frac { 1° }{ 2 }\) and 37\(\frac { 1° }{ 2 }\)
Answer:
Draw a circle with radius 3 cm and central angle 65°.
Half of 65° is 32\(\frac { 1° }{ 2 }\) Similarly we can draw 75°
Join the points A, B and C Halfof75°is 373\(\frac { 1° }{ 2 }\) Complete the triangle.

Circles Class 10 Notes State Syllabus Question 7.
In the picture, AB and CD are mutually perpendicular chords of the circle. Prove that the arcs APC and BQD joined together would make half the circle.

Answer:
If ∠ADC = x
∠AOC =2x
(Angle subtended on the alternate arc is half of the central angle of arc )
If ∠BAD = y
∠BOD= 2y
∠AOC + ∠BOD = 2x + 2y
= 2 (x + y) = 2 × 90 = 180°
∴ The arcs APC and BQD joined together will make half the circle.

Kerala Syllabus 10th Standard Maths Chapter 2  Question 8.
In the picture, A, B, C, D are points on a circle centred at O. The lines AC and BD are extended to meet at P. The line and BC intersect at Q. Prove that the angle which the small are AB makes at O is the sum of the angles it makes at P and Q.

Answer:

Textbook Page No. 59

Kerala Syllabus 10th Standard Maths Circles Question 1.
Calcula te the angles of the quadrilateral in the picture and also the angles between their diagonals:

Answer:
Since
∠ACD = 30°
∠ABD = 30°
(Angle in the same segment of a circle)
Since ZCBD = 45°
∠CAD = 45°
Since ZBDC = 50°
∠BAC = 50°
∠ABC + ∠ADC = 180 (cyclic quadrilateral)
∠ABC = 75°
∴ ∠ADC = 180 – 75 = 105°

∠ADB = 105 – 50 = 55°
As, ∠BAD = 95°
∠DCB = 180 – 95 = 85°
∴ ∠ACB = 85 – 30 = 55°

Circles Class 10 State Syllabus Kerala Syllabus Question 2.
Prove that any outer angle of a cyclic quadrilateral is equal to the inner angle at the opposite vertex.
Answer:

PQRS is cyclic Quadrilateral
∠PSR + ∠PQR = 180°
(sum of opposite angles)
∠PQR + ∠RQT = 180° (linear pair)
From this we get ∠PSR = ∠RQT

Maths Circles Class 10 State Syllabus Question 3.
Prove that a parallelogram which is not a rectangle is not cyclic.
Answer:
PQRS is cyclic quadrilateral.
∠P + ∠R = 180
Also in a parallelogram opposite angles will be equal.
∠P + ∠R = 180°
∠P = ∠R = 90°
This means that PQRS must be a rectangle, otherwise, it is not cyclic.

Sslc Maths Chapter 2 Notes Kerala Syllabus Question 4.
Prove that any non-isosceles trapezium is not cyclic.
Answer:

opposite angles are not supplementary.
∴ ABCD is not cyclic. Non-isosceles trapezium is not cyclic.

10th Class Maths Notes Kerala Syllabus Question 5.
In the picture, bisectors of adjacent angles of the quadrilateral ABCD intersect at P, Q, R, S.

Prove that PQRS is a cyclic quadrilateral.
Answer:

2x + 2y + 2z + 2w = 360°;
x + y + z + w = 180°
Δ DPC; ∠DPC = 180 – (w + z)
Δ ARB ; ∠ARB =180 – (x + y)
∠R + ∠P = 360 – (x + y + z + w) = 360 – 180 = 180°
In ΔBQC ZQ = m – (w + x)
In Δ ASD ∠S = 180 – (r + y)
Similarily ∠S +∠Q = 180.
PQRS is a cyclic quadrilateral.

Question 6.
i) The two circles below intersect at P, Q and lines through these points meet the circles at A, B, C, D. The lines AC and BD are not parallel. Prove that if these lines are of equal length, then ABDC is a cyclic quadrilateral.

ii) In the picture, the circle on the left and right intersect the middle circle at P, Q, R, S; the lines joining them meet the left and right circles at A, B, C, D. Prove that ABDC is a cyclic quadrilateral.

Answer:
i.

∴ ABCD is a cyclic quadrilateral.

ABCD is a cyclic quadrilateral.

Question 7.

In the picture, points P, Q, R are marked on the sides BC, CA, AB of AABC and the circumcircles of ΔAQR and ΔBRP are drawn. M is a point where these circles intersect.

Prove that the circumcircle of ΔCPQ also passes through M.
Answer:

Let M be a common point which three circles can passed.

Therefore the circumcircle of ACPQ also passes through M

Textbook Page No. 67

Question 1.
In the picture, chords AB and CD of the circle are extended to meet at P.

i. Prove that the angles of Δ APC and Δ PBD, formed by joining AC and BD, are the same.
ii. Prove that PA × PB = PC × PD.
iii. Prove that if PB = PD, then ABDC is an isosceles trapezium.
Answer:

i. As ABCD is a cyclic quadrilateral.
If ∠BAC = x° then
∠BDC = 180 – x ∠BDP = x°
If ∠ACD = y° then ∠PBD = y°
As ∠APC is common angle.
Angles of Δ APC and Δ PBD are same

iii. If PB = PD ; AP= PC
In ABCD
ABDC is a cyclic quadrilateral, so their opposite angles are supplementary.
If AP = PC, in ∆ APC
∠A =∠C

As AB = CD
AC || BD
Adjacent angles are supplementary
∴ ABCD will be an isosceles trapezium

Question 2.
Draw a rectangle of width 5 centimetres and height 3 centimetres.
i. Draw a rectangle of the same area with width 6 centimetres.
ii. Draw a square of the same area.
Answer:
i. Draw a rectangle of length 5 cm and width 3cm.

Extend AB up to 6cm.
(AE = 6cm) Draw an arc having radius as AE and A as centre. Extend DA and mark the point F.

Extend BA towards left. Mark G as AD = AG
Draw Δ GFB.

Circum circle of Δ GFB meets AD at D.
∴ AG × AB = AF × AH.
That is area of the rectangle having length AB and width AD is equal to the area of rectangle having length AE and width AH.

ii. Draw a rectangle of length 5cm and width 3cm. Area = 5 x 3 = 15 cm2.
Therefore side of a square will be √15.

Draw a semicircle of diameter AH.
Extend BC, and mark the point F.
AB × BH = 5 × 3 = 15
AB × BH = BF2 ; BF = √5 cm
Area of BEGF = √15 × √15 = 15 cm²

Question 3.
Draw a square of area 15 square centimetres.
Answer:
Draw a rectangle of length 5cm and width 3cm. Area = 3 × 5 cm² = 15 cm². Side of the square is √15.
Draw a semicircle of diameter AH.
BC can touch the point F.
AB × BH = BF2 = √152 = 15 cm².
Area of BEGF = 15 cm².

Question 4.
Draw a square of area 5 square centimetres in three different ways. (Recall Pythagoras theorem)
Answer:
i. Draw a rectangle of length 5cm and width 1 cm.

Draw a semicircle of diameter AE. Extend BC up to F. √5 is the side of the square BGHF.

ii. Draw a right-angled triangle of perpendicular sides 2 cm and 1cm.

Hypotenuse will be y is cm. The area of the square ACDE is 5 cm², because here we take the hypotenuse as sides of the square.

iii. Draw a right-angled triangle of hypotenuse 3 cm and one side 2cm.
Third side = \(\sqrt{3^{2}-2^{2}}=\sqrt{5} \mathrm{cm}\)
Draw a square having side BC, then area of the square BEDC will be 5cm²

Question 5.
In the picture, a line through the centre of a circle cuts a chord into two parts:

What is the radius of the circle?
Answer:

The intersection may be with in the circle.
Chords AB & CD intersect at P i. e.,
AP × PB = CP × PD
(The intersection will be inside the circle)
AP × PB = CP × PD
4 × 6 = CP × (OP + OD)
24 = CP × (OP + OC)
24 = CP × (OP + OP + CP)
24 = CP × (5 + 5 + CP)
24 = CP × (10 + CP)
CP =2
Radius = CP + OP = 2 + 5 = 7cm

Question 6.
In the picture, a line through the centre of a circle meets a chord of the circle:

What are the lengths of the two pieces of the chord?
Answer:


CP= 13 – PB
Therefore equation (1)
(13 – PB)PB = 40
PB = 5, 8
If PB = 5cm then PC = 8cm
If PB = 8cm then PC = 5cm
In figure PB > PC
PB = 8 cm, PC = 5 cm

Circles Orukkam Questions & Answers

Worksheet 1

Question 1.
In triangle ABC, AB = 8cm, BC = 6cm , AC = 10cm.
1. What kind of triangle is this?
2. What is the position of B based on the circle with AC as the diameter? Why?
3. What is the position of A based on the circle with BC as the diameter? Why?
4. What is the position of the point C based on the circle with diameter AB?
Answer:
In Δ ABC
AB2 + BC2 = 82 + 62 = 64 + 36 = 100 = AC2
Δ ABC is a right angled triangle.
If we draw a circle taken in a AC as diameter, ∠B = 90°, Therefore the point B on the circle.
Δ ABC is a right-angled triangle.
If we draw a circle taking BC as diameter, ∠A < 90°, Therefore the position of point A will be outside the circle.
If we draw a circle taking AB as diameter, ∠C < 90°, Therefore the position of point C will be outside the circle.

Question 2.
Three vertices of a parallelogram are on a circle and the fourth vertex is at the center. Find the angles of the parallelogram.

Mark a point P on the top of the figure on the circle, join AP and CP. If angle AP C = x then write? AOC
Write ABC?
Write ∠ABC + ∠APC ?
What is APC?
Find the angles
Answer:

The angles of a parallelogram are 60, 120, 60 and 120.

Question 3.
In triangle ABC ,AB = AC, angle BAC = 30, BC = 5cm Find the radius of ABC
Draw the figure
Mark the center, BO and CO
Find the measure of angle BOC
Write the angles of triangle OBC
What kind of angle is triangle OBC
Write the radius of the circumcircle
Answer:

∠OBC= 75 – 15 = 60 = OCB (∵OB = OC)
Angle of Δ OBC are: 60, 60, 60
ΔOBC is an equilateral triangle.
Radius of circum circle of Δ OBC = OB = OC = BC = 5 cm.

Question 4.
P QRS is cyclic.
∠P = 3x, ∠Q = y, ∠R = x, ∠ = 5 v
Find the angles
Draw circle , mark P, Q, R, S on it, complete PQRS Enter the given angles.
What is 3x + x? Find x
What is y + 5y? Find y
Find the angles
Answer:

3x + x = 180° (Sum of the opposite angles of a cyclic quadrilateral is 180″)
=> 4x = 180° => 4x = 45°
y + 5y = 180 (Sum of the opposite angles of a cyclic quadrilateral is 180°)

Question 5.
In the figure ABC D is a trapezium. If the vertices are on a circle, prove that it is an isosceles trapezium draw figure
What is ∠A + ∠C?

What is ∠B + ∠C?
Write the relation between A and B. Write the conclusion.
Answer:

∠A + ∠C = 180° (Sum of the opposite angles of a cyclic quadrilateral is 180°)
∠B + ∠C = 180° (AB || CD)
(Sum of the alternate angles of a cyclic quadrilateral is 180″)
∠A + ∠C = ∠B + ∠C
∴ ∠A = ∠B
∴ ABC’D is an isoceles trapezium

Question 6.
In ABC AB = AC . P is the midpoint of AB and Q is the midpoint of AC. Prove that BPQC is cyclic?
Draw figure. Mark PQ and complete BP QC?
Is PQ parallel to BC?
Note that ∠B = ∠C?
What is ∠C + ∠Q?
What is ∠B + ∠Q?
Write conclusion.
Answer:

PQ || BC
(The line joining midpoints of two sides of a circle will be parallel to the third side).
∠B = ∠C (1)
(∵ AB = AC are given )
∠C + ∠Q = 180° (2)
(∵ PQ || BC, QC is the bisector, so the sum of alternate angles are 180°)
∠B + ∠Q = 180° .
From (1) and (2) we conclude that BPQC is an cyclic trapezium

Worksheet 2

Question 7.
Prove that ABCD given in the figure is cyclic

Draw figure and mark PQ
If ∠BAP =xthen
what is ∠BQP?
Find ∠PQD
Find ∠PDC? Why?
What is ∠A + ∠C?
Answer:
i. Quadrilateral ABPQ is cyclic
If ∠A = x, then ∠BQP = 1 80 – x
If ∠B = y, then ∠APQ = 180 – y
Quadrilateral PQCD is cyclic, SO
∠QCD = 180 – x (∠DPQ = x )
∠PDC = 180 – x (∠PQC = y)
∴ ABCD is a cyclic quadrilateral.

Worksheet 3

Question 8.
In the figure AB, C Dare extended and intersect at P. If AB = 5, BP = 3, P D = 2 then find CD? Draw the figure.

Write the relation between PA, PB, PC, PD
Find C D
Answer:
PA x PB = PCxPD
If CD = x, then
⇒ 8 × 3 = (2 + x)2 ⇒ 24 = 4 + 2x
⇒ 2x = 20 ⇒ x = \(\frac { 20 }{ 2 }\) = 10
∴ CD= 10

Question 9.
In the figure AB is the diameter and CD is parallel to the diameter. AB = 8cm,BD = 2cm, find CD

Answer:
If we draw a perpendicular DP to AB from D. Then PAxPB = PD2.
Here PB = x, then PA = 8 – x.

x(8 – x) = PD², 2² = x² + PD²
x(8 – x) = 4 – x², 8x – x² = 4 – x²
8x = 4, x = \(\frac { 1 }{ 2 }\)
Similarly, let us draw a perpendicular CQ to AB from C
AQ = ,PQ = 8– \(\left(\frac{1}{2}+\frac{1}{2}\right)\) = 7
CD = 7cm

Question 10.
Draw a rectangle of length 6cm and width 4cm. Draw another rectangle whose area equal to area of the first rectangle and one of the sides is 8cm.
Ans: Draw ABCD as in the given measurement. Mark E by extending AB 2cm more. AE = 8 will be 8cm. WithAas centre and AE radius draw an arc. This arc cut DA produced at F. Extend BA such that AD = AG and mark G. Draw triangle GF B and construct a circumcircle. The circle meets AD at H . Complete the rectangle AHIE

Worksheet 4

Question 11.
Draw an equilateral triangle of height 3cm. What is the length of a side?

Write the principle of construction. Students are advised to construct as in the steps given below.
Answer:
Draw a circle of radius 2cm and mark a diameter AB which is 4cm. Mark a point P from one end A is 3 cm apart on the diameter.
Draw a chord C D perpendicular to AB. Complete triangle C AD.
Using PA x PB = PD², PD= √3.
Now we get AD = AC = C D = 2√3
Height AP = 3cm.

Worksheet 5

Question 12.
In the figure, PA is a tangent and O is the centre of the circle. P A = 17, ∠OPA = 30° then calculate the radius of the circle and distance from centre to the point P Triangle OAP with 30°, 60°, 90° is right triangle. Using the property of this special right triangle find the radius and the distance.

Answer:
Δ OAP is a right-angled triangle having sides 30°, 60°and 90°.
Length of side which is opposite to the angle 90°, is twice the side which is opposite to the angle 30°.
Length of side which is opposite to the
angle 60°, is √3 times of the side which is opposite to the angle 30°
That is radius of the circle , OA = \(\frac { 17 }{ √3 }\)
Distance from centre to the point P

Worksheet 7

Question 13.
Draw a circle and construct 30°, 150° angles on it.
Answer:

Question 14.
Draw a circle and construct \(22 \frac{1}{2}^{0}\) on it.
Answer:

Question 15.
In triangle ABC the radius of the circumcircle is 6 cm, ∠A = 70°, ∠B = 80°. Construct the triangle
Answer:

Question 16.
Draw a rectangle of length 7cm, and width 5cm and construct a square whose area is same as the area of this rectangle.
Answer:
Draw a rectangle of length 7cm, and width 5cm. Area = 7 × 5 = 35 cm².
Therefore length of one side of the square is √35.

Draw a semicircle taking AH as diameter Extend BC, and mark a point F.
AB × BH = 7× 5 = 35
AB × BH = BF2 ;
BF = √35cm
Area of BEGF= √35 × √35 = 35 cm²

Question 17.
Draw a rectangle of one side 5cm, width 7cm. Construct another rectangle whose one side is 8cm and area equal to the area of the first rectangle.
Answer:
Draw a rectangle of length 7cm and width 5cm.

Extend AB up to 8 cm (AE = 8cm) Draw an arc taking A as centre and AE as radius. Extend DA and mark the point F.

Elongate BA towards left, and mark G such that AD = AG.
Draw Δ GFB

Circumcircle of A GFB will meet side AD on H.
AG × AB = AF × AH.
That is area of the rectangle having length AB and width AD is equal to the area of rectangle having length AE and width AH.

Question 18.
Draw a square of side 5cm and construct a rectangle having one side 7cm and area equal to area of the square.
Answer:
Draw a square of side 5cm.

Extend AB to 7 cm
(AE = 7cm) Draw an arc taking A as centre and AE a radius. Extend DA and mark the point F.

Elongate BA towards left, and mark G such that AD = AG.
Draw Δ GFB.

Circumcircle of A GFB will meet side AD on H.
AG × AB = AF × AH.
That is area of the rectangle having length AB and width AD is equal to the area of rectangle having length AE a width AH.

Question 19.
What is the position of the vertex of an equilateral triangle with opposite side as the diameter?
Answer:

Angles of an equilateral triangle is 60° each. Position of the vertex of triangle with opposite side as the diameter is outside of the circle, because the angle is less than 90°.

Circles SCERT Questions & Answers

Question 20.
In the figure “ ABC is a right triangle
a. If a circle is drawn with AC as diameter find the position of B.
b. If a circle is drawn with BC as diameter, find the position of A. [ Score: 3 Time: 5 minutes]

Answer:
a. On the circle (1)
∠B = 90°

b. Outside the circle (1)
Position of the vertex of an triangle with opposite side as the diameter is outside the circle, because. ∠A <9o°.

Question 21.
A circle is drawn with AB as diameter. Find the positions of the points C, D, E related to the circle.

[ Score: 3 Time: 5 minute]
Answer:
C inside the circle (1)
∠C > 90°.
D on the circle (1)
∠D = 90°.
E outside the circle (i)
∠E <90°.

Question 22.
In Δ ABC and Δ PQR, BC = QR, ∠ A = ∠P, ∠Q = 90°, QR = 5 cm, PQ = 12 cm.

Find the diameter of the circumcircle of Δ ABC. [ Score: 4 Time: 6 minute]
Answer:
QR = BC (1)
∠A = ∠P (1)
PR Diameter of the circumcircle of Δ ABC (1)
Diameter = PR= \(\sqrt{12^{2}+5^{2}}=\sqrt{169}=13 \mathrm{cm}\) .(1)

Question 23.
PQ and RS are two mutually prependicular chords of a circle. < QPR=50° find< PQS. [ Score: 3, Time: 6 minute]
Answer:
∠PRS = 90 – 50 = 40° (1)
∠PQS = 40° (1)

Question 24.
O is the centre of the circle. If ∠BOC = 130° and ∠AOB = 110°. What is ∠AOC?
Find all angles of Δ ABC
[ Score: 3, Time: 3 minute]
Answer:

Question 25.
Find all angles of the hexagon ABCDEF
[ Score: 4 Time:5 minute]
Answer:
∠ EFD = ∠EAD = 30°
∠ FE A = ∠FDA = 40°
∠FDE = ∠FAE = 35° (1)
∠BAC= ∠BDC = 45°
∠ABD= ∠ACD = 62°
∠ACB = ∠ADB= 35° (1)
∠A = 1480, ∠B = 100°
∠C = 97° ∠D= 155° (1)
∠E= 115° ∠F= 105° (1)

Question 26.
O is the centre of the circle and AB is a chord. AC is the bisector of ∠OAB. ∠OAB = 56°.
a. Prove that OC and AB are parallel,
b. Find ∠ABC and ∠OBE.

Answer:

Question 27.
O is the centre of the circle. AD and BC are perpendicular to XY. CB cuts the circle at E. Prove that –CE = AD.

Answer:
∠AEB = 90° (Angle in a semi circle) ( 1)
∠AEC=90°
∴ AECD is a rectangle
∴ AD = CE (2)

Question 28.
ABCD is a parallelogram. A, B, E, F are the points on a circle. ∠DEF = 80° Find out the angles of the quadrilateral AEFB.
[ Score: 4, Time: 4 minute]
Answer:
∠AEF = 180 – 80 = 100° (1)
∠ABF = 180 – 100 = 80° (1)
∠A = 180 – 80 = 100° (1)
∠EFB = 180 – 100 = 80° (1)

Question 29.

O is the centre of the circle. ? ∠OCA = x °.
a. Find ∠OAC
b. Prove that ∠OCA + ∠ABC = 90°
c. Prove that ∠ADC – ∠OCA = 90° [Score: 4, Time: 4 minute]
Answer:

Circles Exam Oriented Questions & Answers

Short Answer Type Questions (Score 2)

Question 30.
In the figure AB is the diameter. PC is perpendicular to AB. PC = 6cm, PB = 3cm. Find the radius of the semi-circle.

Answer:
AP × PB = PC²
AP × 3 = 62
AP = 36/3 = 12 .
AB = 12 + 3 = 15
ie Radius = 15/2 = 7.5 cm

Question 31.
In Δ PQR, ZP = 60°, ∠R = 30° find whether the vertex Q on the circle with PR as diameter.
Answer:
∠P + ∠R = 60 + 30 = 90°.
∴ ∠Q = 180 – 90 = 90°.
So point Q on the circle.

Question 32.
In Δ ABC, ∠A = 60°, ∠B = 70°. Find whether the vertex C is inside or outside the circle with AB as diameter.
Answer:
The vertex C is outside the circle Since ∠C = 180 – (60 + 70) = 50° ∠90°

Question 33.
In the diagram, the central angle of arc ABC is 100° and ∠OAD is 30°. Find ∠OCD.
Answer:

∠D = 50°
As ∠OAD
is an isosceles triangle.
∠ODA = 30°
∠ODC = 20°
∴∠OCD = 20°

Question 34.
In the figure, find ∠PQB, O is the centre.

Answer:

Short Answer Type Questions (Score 3)

Question 35.
The central angle of arc ABC is 60°, then find out the following,
i) ∠D
ii) Central angle of arc AEC,
iii) ∠B

Answer:
i. ∠D = 30°
ii. Then central angle of arc AEC = 300°.
iii. ∠B = 150°

Question 36.
Show that the arcs APC, BQD when joined make a semicircle?

Answer:
∠ADC is the half of the central angle of arc APC
The central angle of are APC = 2 ∠ADC
The central angle of arc BQD = 2 ∠DAB
In triangle AOD, CD ⊥ AB, ∠AOD = 90°,
∠DAO + ∠ADO = 180 – 90° =90°,
ie, ∠DAB + ∠ADC = 90°
2( ∠DAB + ∠ADC ) = 90° × 2 = 180°

Question 37.
The central angle of the complementary arc of a circle is 40° more than 3 times the central angle of the arc. Find out the central angles of each arc?

Answer:
x + 3x + 40 = 360
4x + 40 = 360
4x = 360 – 40 = 320
x = \(\frac { 320 }{ 4 }\) = 80
∴ central angle of arc
ABC = 80
central angle of arc ADC = 360 – 80 = 280°

Question 38.
ABCD in the diagram is a rectangle. Then find out the area of the circle.

Answer:
ABCD is a rectangle ∠B = ∠D = 90°
AC in the diameter of the circle
AC = \(\sqrt{8^{2}+6^{2}}=\sqrt{64+36}=\sqrt{100}=10\)
∴ radius = 5cm
∴ Area of the circle = πr² = π × 5² = 25πcm²

Question 39.
In the figure, AD = 16cm, BD = 6cm, CD = 2cm. Find the length EF.

Answer:

Question 40.
In the given figure, O is the centre of the circle. If ∠OAP = 35° and ∠OBP = 40°, find the value of ∠x.

Answer:
Join OP
Since OA = OP and
∠APO = ∠OAP=35°
Similarly, OB = OP and ∠OPB = ∠OBP = 40°
∠APB = 35°+ 40° = 75°
∠AOB = 2 × 750 = 150°

Long Answer Type Questions (Score 4)

Question 41.
In the figure find ∠APB,∠ABQ ; where O is the centre of the circle ∠OAP = 32° and  ∠OBP = 47° .

Answer:
JoinOP.
In OAP, OA = OP = Radius
∠OAP = ∠OPA = 32°
In OPR, OB = OP = radius
∠OBP = ∠OPB =47°
∠APB = 32°+ 47°= 79°
∠AQB = 180° – 79°= 10°

Question 42.
Draw a line of √7 cm.
Answer:

Question 43.
Ois the centre of the circle as shown in the figure.
Find ∠CBD.

Answer:

Takeapoint E on the circle, join AE and CE.
∠AEC= \(\frac { 100 }{ 2 }\) = 50°
∠AEC + ∠ABC = 180° (Opposite angles of a cyclic quadrilaterals)
∠ABC = 130°
∠ABC + ∠CBD = 180° (linearpair)
130°+ ∠CBD = 180°
∠CBD = 50°

Long Answer Type Questions (Score 5)

Question 44.
In the figure O is the centre of the circle. Central angle of arc AXB is 60°, arc CYD is 80°. Then find all the angles of ΔAPD.

Answer:
Central angle of arc AXB = 60°
i.e., ∠AOB = 60°
i.e., ∠ADP = 30°
Central angle of arc CYD = 80°
i.e., ∠COD = 80°
i e., ∠DPA = 40°
i e., ∠APD = 180° – (30 + 40) = 110°
Angles of = 30°, 40°, 110°

Question 45.
‘O’ is the centre of the circle ∠D = 80°, find the following measurements.

a. ∠C
b. ∠ABC
c. ∠BAC
d. ∠F
Answer:
∠D = 80°
a. ∠C = 80° (∠D and ∠C are angled on a same arc So, both are equal)

b. ∠ABC = 90°
(AC is diameter, Angle of a hemisphere is right)

c. ∠BAC = 180 – (80 + 90)
= 180 – 170 = 100

d. ∠F= 180 – 80 = 100° (Opposite angles of a cyclic quadrilateral are equal)

Circles Memory Map

Angle in a semicircle is right:
The angle made by any arc of a circle on the alternate arc is half the angle made at the centre.

All angles in an arc is equal:

If AB, CD are two chords, then
PA × PB = PC × PD

The area of the rectangle formed of parts into which a diameter of a circle is cut by a perpendicular chord is equal to the area of the square formed by half the chord.
PA × PB = PC²

Students can Download Chemistry Chapter 2 Gas Laws Mole Concept Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Chemistry Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

SSLC Chemistry Chapter 2 Gas Laws Mole Concept Questions and Answers Malayalam Medium

SCERT 10th Standard Chemistry Textbook Chapter 2 Solutions Malayalam Medium

You can Download Reactivity Series and Electrochemistry Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemsitry Solutions Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Chemistry Chapter 3 Reactivity Series and Electrochemistry Textbook Questions and Answers

SCERT Class 10th Standard Chemistry Chapter 3 Reactivity Series and Electrochemistry Solutions

Reactivity Series And Electrochemistry Kerala Syllabus Text Book Page No: 48

→ Which metal reacts vigorously?
Answer:
Sodium.

→ Which gas is formed as a result of this reaction?
Answer:
Hydrogen.

→ Write down its chemical equation.
Answer:
2Na + 2H₂O → 2NaOH + H₂

→ Complete the table (3.1) given below.

Answer:

→ Based on your observation, arrange these metals in the decreasing order of their reactivity.
Answer:
Sodium > Magnesium > Copper
→ 2Mg + O₂ →
Answer:
2Mg + O₂ → 2MgO

Sslc Chemistry Chapter 3 Kerala Syllabus Text Book Page No: 49

→ Which metal among magnesium, copper, gold, sodium and aluminium, loses its lustre at a faster rate?
Answer:
Sodium

→ List the above metals in the decreasing order of their reactivity with air and thereby losing lustre
Answer:
Sodium > Magnesium > Aluminium > Copper > Gold.

Text Book Page No: 50

→ What happened to the Zn rod?
Answer:
Before the experiment the Zn rod was colourless. After the experiment Zn rod became blue due to the deposition of copper.

→ What is the reason for this?
Answer:
When the Zn rod is dipped in CuSO₄ solution, the Cu²⁺ ions in the solution get deposited at the Zn rod as Cu atoms.

→ What is the reason for the change in intensity of the colour of CuSO₄ solution?
Answer:
The blue colour of CuSO₄ solution is due to the presence of Cu²⁺ ions. The change in intensity of the colour of CuSO₄ solution because when the Zn rod is dipped in CuSO₄ solution, the Cu²⁺ ions in the solution get, deposited at the Zn rod as Cu atoms.

→ Which is the metal that gets displaced here?
Answer:
Copper

→ Which is more reactive Zn or Cu?
Answer:
Zn

→ On the basis of the position of Zn and Cu in the reactivity series, can you explain why Cu had been displaced?
Answer:
Zn is placed above Cu in the reactivity series because Zn has a higher reactivity than Cu.

→ Isn’t it due to the higher reactivity of zinc (Zn) when compared to copper (Cu)?
Answer:
Yes.

Chemistry Class 10 Chapter 3 Kerala Syllabus Text Book Page No: 51

→ Is this reaction oxidation or reduction? Why?
Answer:
Oxidation. Because the losing of electrons is called oxidation.

→ The change that happened to Cu²⁺
Answer:
Cu^–2+ + 2e^– → Cu

→ What is the name of this reaction? Why?
Answer:
Reduction. The gaining of electrons is called reduction.
→

Complete this chemical equation by assigning oxidation numbers.
Answer:
2 Ag⁺¹ NO₃^1–+ Cu°→ Cu²⁺ (NO₃)^–1₂ + 2Ag⁰

→ Which metal was oxidised in this case? Which metal was reduced?
Answer:
Metal which was oxidised: Cu
Metal ion which was reduced: Ag⁺

→ Write equations showing oxidation and reduction.
Answer:
Oxidation : Cu⁰ → Cu²⁺ + 2e^–
Reduction : Ag⁺+ le^– → Ag⁰

Reactivity Series And Electrochemistry Sslc NotesText Book Page No: 52

→ Complete the table 3.3.

Answer:

Sslc Chemistry Chapter 3 Notes Pdf Kerala Syllabus Text Book Page No: 53

→ Which electrode has the ability to donate electrons in a cell constructed using these metals?
Answer:
Zn

→ Which one can gain electrons?
Answer:
Cu

→ Identify the chemical reaction that takes place at the Zn electrode. Tick ✓ the right one.
Answer:
Zn(s) → Zn²⁺ (aq) + 2e^–  (✓)
Zn²⁺(aq) + 2e^– → Zn(s)   (✘)

→ What is the reaction taking place here?
Answer:
Oxidation.

→ Write the chemical equation for the reaction taking place at the Cu electrode.
Answer:
Cu²⁺ (aq) + 2e^– → Cu(s)

→ Sketch the cell constructed.
Answer:

→ Note down the reaction of the Galvanic cell.
Answer:
Cu(s) + 2Ag⁺(aq) → Cu²⁺ (aq) + 2Ag(s)
Cu(s) → Cu²⁺(aq) + 2e^– (Anode)
Ag⁺(aq) + le^– → Ag(s) (Cathode)

→ Direction of flow of electrons From Cu to Ag
Answer:
Mark the direction of electron how in the cell illustrated.

→ write the reactions taking place at cathode and anode.
Answer:
At cathode : Ag⁺ + le^– → Ag
At anode : Cu → Cu²⁺ + 2e^–

Hss Live Guru 10th Chemistry Kerala Syllabus Text Book Page No: 55

→ You have used three metals Zn, Cu and Ag. How many cells can be produced using these?
Answer:
Three.

→ Complete the Table 3.4 by writing anode and cathode in each.

Answer:

→ What are the substances obtained when electricity is passed through acidified water?
Answer:
Hydrogen, Oxygen.

→ Do such type of chemical changes happen when electricity is passed through metals?
Answer:
Yes.

Sslc Chemistry Chapter 3 Questions And Answers Kerala Syllabus Text Book Page No: 56

→ To which electrodes are the positive ions attracted during electrolysis?
Answer:
Towards negative electrodes(Cathode)

→ To which electrodes are the negative ions attracted?
Answer:
Towards positive electrodes(Anode),

→ What changes happen to the ions which N are attracted to cathode?
Answer:
Reduction

→ What about the changes happening to the ions attracted to anode?
Answer:
Oxidation

→ Which ion is attracted to the positive electrode (anode)?
Answer:
Chloride ion (Cl^–)

→ What is the chemical reaction taking place there?
Answer:
2Cl^– (aq) → Cl₂(g) + 2e^–

Text Book Page No: 57

→ Which is the gas liberated at the anode?
Answer:
Chlorine(Cl₂)

→ Which is the ion attracted to the negative electrode (cathode)? Write the change happening to it?
Answer:
Na⁺ ions. These ions accept one electron and changes to sodium atom. That is sodium ions are reduced.

→ Which is the metal deposited at the cathode?
Answer:
Sodium (Na)

→ Which are the ions attracted to the positive electrode?
Ans.
Cl^–,OH^–

→ Which are the ions attracted to the negative electrode?
Answer:
Na⁺,
H₃O⁺,
H₂O.

Text Book Page No: 59

→ Which metal is connected to the negative terminal of the battery?
Answer:
Iron.

→ Which metal is connected to the positive terminal of the battery?
Answer:
Copper.

→ Which solution is used as the electrolyte?
Answer:
Copper sulphate solution.

→ What happens to Cu²⁺ ions at the cathode? Complete the equation.
Answer:
Cu²⁺ + 2e → Cu

→ What happened to the copper ions? Oxidation/Reduction?
Answer:
Reduction.

→ Complete the equation given below.
Answer:
Cu → Cu²⁺ + 2e^–

Text Book Page No: 60

→ Find out more examples and extend the list.
Answer:

• Chromium plating is used in motor car etc.
• To make metal coating easily corroding metals to prevent corrosion.
• In ICs (Integrated Circuits) coating of gold /silver is made by electroplating.

Reactivity Series and Electrochemistry Let Us Assess

Kerala Genetics Question 1.
The solutions of ZnSO₄, FeSO₄, CuSO₄ and AgNO₃ are taken in four different test tubes. Suppose, an iron nail is kept immersed in each one
In which test tube the iron nail undergoes a colour change?
What is the reaction taking place here?
Justify your answer. (Refer reactivity series of metals).
Answer:
Iron nail immersed in solution of CuSO₄ and AgNO₃ undergoes a colour change.
i. Fe(s) + CuSO₄(aq) →
FeSO₄ (aq) + Cu (s).
ii. Fe(s) + 2AgNO₃ (aq) →
Fe(NO₃)₂(aq) + 2Ag(s).
Iron displaces Cu from CuSO₄ and Ag from AgNO₃ because Fe has higher reactivity than Cu and Ag.

Genetic Engineering Syllabus Question 2.
Compare the electrolysis of molten potassium chloride and solution of potassium chloride. What are the processes taking place at the cathode and the anode?
Answer:
Molten KCl
KCl (s) → K⁺ + Cl^–
At the negative electrode:
K⁺ + le^– → K (reduction – cathode)
At the positive electrode:
2Cl^– → Cl₂ + 2e^– (oxidation- anode)
Solution of potasium chloride.
At the negative electrode:
2H₂O + 2e^– → H₂ + 2OH^– (cathode).
At the positive electrode:
2Cl^– → Cl₂ + 2e^– (anode).

Future Diary 10th Question 3.
You are given a solution of AgNO₃, a solution of MgSO₄, a Ag rod and a Mg ribbon. How can you arrange a Galvanic cell using these? Write down the reactions taking place at the cathode and the anode.
Answer:
At anode,
Mg (s) → Mg²⁺ (aq) + 2e^–
At cathode,
Ag (aq) + le^– → Ag (s)

Reactivity Series and Electrochemistry Extended Activities

The Reactivity Question 1.
1. Keep two carbon rods immersed in copper sulphate solution. Then pass electricity through the solution.
i. At which electrode does colour change occur anode or cathode?
ii. Is there any change in the blue colour of the copper sulphate solution?
iii. Write down chemical equations for the changes occurring here.
Answer:
i. At cathode
ii. Colour fades
iii.At cathode: Cu²⁺ (aq) + 2e^– → Cu (s)
At anode: 2H₂O → O₂(g) + 4H⁺ (aq) + 2e^–

SSLC Chemistry Chapter 4 Question 2.
When acidified copper sulphate solution is electrolysed oxygen is obtained at the anode. What arrangements are to be made for this? Find the element deposited at the cathode.
Answer:

Element deposited at the cathode: Copper.

Question 3.
a. When Galvanic cells are made using the metals like Mg, Cu, Zn and Ag, what will be the nature of reactions in each cell?
(Reactivity: Mg > Zn > Cu >Ag)
b. How many Galvanic cells can be made by using the metals Ag, Cu, Zn and Mg?
Chapter 7 Biology test Answers:
a. i. Cu-Ag cell
Anode: Cu (s) → Cu²⁺ (aq) + 2e^–
Cathode: Ag⁺ (aq) + le^– → Ag(s)
ii.Zn-Ag cell
Anode: Zn (s) → Zn²⁺ (aq) + 2e^–
Cathode: Ag⁺ (aq) + le^– → Ag (s)
(iii) Mg-Ag cell
Anode: Mg (S) → Mg²⁺ (aq) + 2e^–
Cathode: Ag⁺ (aq) + le^– → Ag (s)
(iv) Zn – Cu cell
Anode: Zn (s) → Zn²⁺ (aq) + 2e^–
Cathode: Cu²⁺ (aq)+ 2e^– → Cu(s)
(v) Mg-Cu cell
Anode: Mg (s) → Mg²⁺ (aq) + 2e^–
Cathode: Cu²⁺ (aq) + 2e^– → Cu (s)
(vi) Mg-Zn cell
Anode: Mg (s) → Mg²⁺ (aq) + 2e^–
Cathode: Zn²⁺(aq) + 2e^– → Zn(s)
b. 6 cells

Reactivity Series and Electrochemistry Orukkam Questions and Answers

Scope of Genetic Engineering Question 1.
1. Take cold water and Hot water in two test tubes, Add one or two drops of phenolphtha lein in it. Drop equally sized Mg ribbon in it.
a. In which test tube pink colour occured sharply?
b.Why did pink colour appear in that test tube so early?
c. Which gas evolved out from both test tubes?
d. Write balanced equation for the above mentioned reaction.
Answer:
a. Pink colour occurred sharply in the test tube with hot water.
b. Temperature is a factor that affects the rate of a reaction. When heated the kinetic energy of molecules increases and hence the rate of chemical reaction also increases which causes the pink colour to appear early.
c. Hydrogen
d. Mg(s) + 2H₂O(l) → Mg(OH)₂(aq) + H₂(g)

SCERT Question Pool 2017 Question 2.
Cut a small sodium metal piece into two, watch it.
a. What change occurred on the surface of sodium metal?
b.Write one word for the process of this type of decomposition.
c. Write down the equations for this.
Answer:
a. After some time the cut piece of sodium will turn dull.
b.Corrosion.
c. 4Na(s) + O₂(g) → 2Na₂O
2Na(s) + 2H₂O(l) → 2NaOH(s) + H₂(g)
2NaOH (s) + CO₂(g) → Na₂CO₃(s) + H₂O(l)

Question 3.
Take equal quantities of dil HCl in five test tubes. Drop Mg, Zn, Fe, Cu in each test tube. Watch carefully.
a. Arrange metals in decreasing order of reactivity.
b.Write balanced equations for each reaction.
SCERT Question Pool Answer:
a. Mg > Zn > Fe > Cu
b. Zn + 2HCl → ZnCl₂ + H₂
Fe + 2HCl → FeCl₂ + H₂
Cu + HCl → No reaction
Mg + 2HCl → Mg Cl₂ + H₂

SCERT Question Pool Question 4.
Some metals and metallic compounds are given in the table. If the metal substitute the metal in the compound put a tick mark in the corresponding column and otherwise a cross mark in the column. Write down correct answer based on the table given below.

a. Correct the table if necessary.
b. Is it possible to substitute lower positioned metals by top positioned metals in the reactivity series?
c. What type of reaction is this?
d. Write down balanced equations for all the true sign given in the table.
Answer:
a.

b.Yes. It is possible.
c. Substitution reactions.
d.CuSO₄ + Mg → Cu + MgSO₄
CuSO₄ + Zn → ZnSO₄+ Cu
CuSO₄ + Fe → FeSO₄ + Cu
ZnSO₄ + Mg → MgSO₄ + Zn
AgNO₃ + Mg → Ag + MgNO₃
AgNO₃ + Cu → CuNO₃ + Ag
AgNO₃ + Zn → ZnNO₃ + Ag
AgNO₃ + Fe → FeNO₃ + Ag

Human Insulin Gene Question 5.
Draw maximum number of Galvanic cell using substances given in the table.
Salt bridge, Zinc rod, Copper rod, Voltmeter, Aluminium chloride, Copper sulphate, Zinc sulphate, Silver nitrate, Silver rod, Calcium chloride
a. Complete the table based on the figures drawn.

B. Write down the general names used for an electrode which gives electrons,
c. Metals in that electrode in the reactivity series is (in the Top, Bottom)
d. General name of the Electrode which accepts electron.
e. Process of giving electron is
f. Process of Accepting electron is
g.Direction of the flow of Electron
h.Write down the balanced equation taking place in both electrodes.

Answer:

Zn-Ag cell is also possible.
a.

b. Anode.
c. In the top.
d. Cathode.
e. Oxidation.
f. Reduction.
g. From Anode to Cathode.

Chemistry Reactivity Series Question 6.
Take Cupric chloride (CuCl₂) solution in a beaker. Dip two graphic rod in it. Pass 5V electricity through it.
a. Why electricity passes through cupric chloride solution?
b.Which gas evolved out through positive electrode? How did you identify that gas?
c. Which product formed in negative electrode?
d. In which electrode oxidation and reduction take place?
e. Write one word for the process of chemical change happening in a Electrolyte while passing Electricity?
Answer:
a. Electrolytes are substances which conduct electricity in molten states or in aqueous solutions. In molten state ions of CuCl₂ can more freely. These ions are responsible for the Conduction of electricity by electrolytes.
b.Chlorine.
c. Copper.
d.Oxidation takes place at the positive electrode and Reduction takes place at the negative electrode.
e. Electrolysis.

Reactivity Series Class 10 Question 7.
Take 25 ml water in a beaker and the pass electricity through it. Then add little sulphuric acid in it.
a. Why electricity didn’t pass through pure water?
b. What happens when dil H₂SO₄ is added?
c. Which type of ion formed more when sulphuric acid is added in water?
d. Complete the equation of the Ionization 0f H₂SO₄
H₂SO₄ → 2H⁺ +
Based on the equation given below write down the correct answers.
2H⁺ +  + 2H₂O → 2H₃O +  + SO₄^2–
e. Complete the equation.
f. Write down the name of H3O⁺ ion?
g.Which ion is moving towards negative ion?
h.Complete the reaction taking place in the negative electrode.
2H₃O⁺(aq) + 2e^– →  +
i. Whicfrion is having highest oxidation potential?
j. Complete the reaction taking place in positive electrode.
2H₂O →  + 4H⁺
k. Ions remain in the beaker after the electrolysis are
I. What product form when these two combined together
Answer:
a. Since the number of ions is less, pure water does not allow the passage of electricity.
b. When dil H₂SO₄ is added the water becomes a good electrolyte. Hence electricity passes through it. When a little dilute sulphuric acid is mixed with water large quantity of hydronium ions are formed.
c. Hydronium ion (H₃O⁺)
d. H₂SO₄ → 2H⁺ + SO₄^2–
e. 2H⁺ + S0₄^2– + 2H₂O → 2H₃O⁺ + SO₄^2–
f. Hydronium ion
g. H₃O⁺
h. 2H₃O⁺(aq) + 2e^– → H_2(g) + 2H₂O
i. H₂O has the highest oxidation potential when compared to SO₄^2–
j. 2H₂O → O₂(g) + 4H⁺ + 4e^–
k. 4H⁺, SO₄^2–ions.
l. H₂SO₄ is formed when these 2 combine together.

Reactivity Series Chemistry Question 8.
a. Complete the table based on the Electrolysis of molten sodium chloride.

b. Write down the reaction taking place in each electrodes and products formed in the electrolysis of sodium chloride solution.

c. Why is hydrogen formed in the cathode instead of sodium?
d. Write one word for a solution undergoes chemical change when electricity passes through it.
e. Write the name of the above process.
f. Write down the uses of above type of reaction.
Answer:

c. When Na+ ion and water are compared reduction occurs to water. Hence H₂ is liberated at cathode.
H⁺+ e → H,
H + H → H2.
d. Electrolyte.
e. Electrolysis.
f. Electroplating, Production of chemicals, Purification of metals.

Reactivity Series and Electrochemistry Evaluation Questions

Take little water in a test tube add two drops of phenolphthalein in it Same quantity of Kerosene is added to the mixture and a small piece of sodium is dipped in it.

Insulin Gene Question 1.
What kind of colour formed in the test tube? Why?
Answer:
Water become pink in colour. Because when phenolphthalein is added to water it becomes alkaline.

Question 2.
Which gas is bubbled on the surface of sodium metal?
Answer:
Hydrogen

Question 3.
Write balanced equation of the reaction between sodium and water.
Answer:
2Na(s) + 2H₂O(l) → 2NaOH (aq) + H₂(g).

Question 4.
What products occurs when Iron reacts with water vapour?
Answer:
Fe₃O₄ (Iron Oxide) and Hydrogen gas.

Question 5.
Lustre of magnesium disappeared fast when it placed in open space why?
Answer:
When Magnesium is kept in open space it reacts with atmosphere air and a light coat of magnesium oxide is formed. This is the reason for Magnesium to lose its lustre.

Question 6.
Verdigris formed on copper utensils disappeared after some days why?
Answer:
The copper in copper utensils react with atmospheric air and forms copper oxide. Due to this verdigris are formed on Copper utensils.

Question 7.
Lustre of aluminium utensils disappear after some days. Why?
Answer:
Aluminium reacts with atmospheric air and Aluminium oxide is formed. This process takes place slowly. Hence Aluminium utensils loose their lustre.

Question 8.
Write down the equation for the reaction between CuSO₄ and iron nail? What type of reaction is this?
Answer:
CuSO₄ + Fe → FeSO₄ + Cu.
Redox reaction.

Reactivity Series and Electrochemistry SCERT Questions and Answers

Question 1.
5ml water is taken in 3 test tubes. Copper, sodium and magnesium of equal mass are dropped in different test tubes. Test tubes having copper and magnesium are heated.
a. Write the observations in the heated test tubes.
b. Write the equation for the reaction in the test tube in which sodium is dropped,
c. Arrange these metals in the decreasing order of their reactivity.
Answer:
a. Mg reacts with hot water liberating hydrogen, Copper does not react with hot water.
b. 2Na + 2H₂O → 2NaOH + H₂.
c. Na > Mg > Cu.

Question 2.
a. Which metal among copper, aluminium and gold loose its metallic lustre at a faster rate? Write the equation of the reaction.
b. Sodium is kept in Kerosene. Why?
Answer:
a. Aluminium, Al + 3O₂ → 2Al₂O₃
b. Na reacts with air (oxygen) and water.

Question 3.
An experimental setup is made to compare the reactions of Mg, Zn and Cu with dilute hydrochloric acid.
a. Write the procedure and observation of the reaction.
b. Which is the gas evolved when zinc reacts with dilute hydrochloric acid?
Answer:
a. Take Mg, Zn and Cu in different test tubes and add dilute HCl to each.
Observation: Magnesium and Zinc reacts with dilute hydrochloric acid copper does not react with the acid.
b.Hydrogen.

Question 4.

a. What are the changes that can be observed with the iron rod and the colour of copper sulphate solution?
b. Write the equations of the oxidation and reduction reactions.
c. What will be the change if silver rod is used instead of iron rod? What is the reason?
Answer:
a. Copper is deposited on iron and the blue.colour of copper sulphate solution decreases.
b. Cathode – Cu²⁺(aq) + 2e^– → Cu (s) (Reduction)
Anode – Fe (s) → Fe²⁺(aq) + 2e^– (Oxidation)
c. No change occurs, Reactivity of silver is less than that of copper. In the reactivity series, the position of silver is below copper.

Question 5.
Sodium reacts with water.
a. Identify the gas evolved in the reaction h If two drops of phenol[ihthaloin is added to the water, what will be colour change of the resultant solution? Explain the reason?
Answer:
a. Hydrogen.
b. Colour changes to pink. Due to the presence of sodium hydroxide (alkaline nature).

Question 6.
Three Galvanic cells are given.

a. Find out the most reactive metal and least reactive metal among them.
b. In cell, which electrode undergoes oxidation why?
c. Write the equation of the redox reaction occurring in cell 3 (Valency of A, B are 2.)
Answer:
a. Most reactive metal A, Least reactive metal B.
b. A Reactivity of A is higher than B.
c. A+ 2e^– → A²⁺,
C²⁺ + 2e^– → C ,
A + C²⁺ → A²⁺ C .

Question 7.
Some metals and salt solutions are given (Cu, Zn, Ag, ZnSO₄, AgNO₃, MgCl₂)
a. Draw the diagram of a Galvanic cell that can be made using these substances.
b. Find out the anode and cathode of this cell and write the chemical equation for the reaction at cathode.
Answer:

b. Anode – Zn,
Cathode – Ag
2Ag⁺+ 2e^–→ 2Ag

Question 8.
Give reasons for the following.
a. Iron vessels are not used a boilers that are used to boil water.
b. Blue vitriol solution is not kept in iron vessels.
Answer:
a. Iron reacted with steam-heated to high temperature

Question 9.
Examine the given electrolytic cell.

a. Which gas is evolved at the positive electrode?
b. Write the oxidation and reduction reactions of this cell.
c. What is the difference in the energy transformation of a Galvanic cell and an electrolytic cell?
Answer:
a. Chlorine / Cl₂
b. 2Cl^– → Cl₂ + 2e^–
Cu²⁺ + 2e^– → Cu
c. Galvanic cell – Chemical energy is converted to electrical energy.
Electrical Cell – Electrical energy is changed to chemical energy.

Question 10.
The solutions in the given table electrolyzed.

List any two areas in which electrolysis is made use of?
Answer:
a. i. Hydrogen.
ii. Chlorine.
iii.Chlorine.
iv. Hydrogen.
b.

• Purification of metals.
• Electroplating.
• Production of chemicals.

Question 11.
The position of iron is below that of zinc in reactivity series. The cell formed by them is given. Correct the mistakes and redraw.

Answer:

Question 12.
Sodium chloride solution is electrolysed using platinum electrodes.
a. Write the chemical equation of the reaction at cathode.
b. What happens when phenolphthalein is added to the solution? State the reason?
Answer:
a. 2H₂O + 2e^– → H₂ + 2OH
b. Colour turns pink. Presence of sodium hydroxide in the solution.

Question 13.
The anode and cathode of two Galvanic cells are given.

A. Mg → Mg²⁺ + 2e^–
B. Zn²⁺+2e^– → Zn
C. Ag⁺ +le^– → Ag
D. Zn → Zn²⁺ + 2e^–
E. Ag → Ag⁺ + le^–
F. Mg²⁺ +2e^– → Mg
a. Find out the reactions at the anode and cathode for each cell from the above.
b. Which metal can act only as cathode? Why?
Answer:
a. Cell 1: Anode Mg → Mg²⁺ + 2e^–
Cathode Zn²⁺ + 2e^– → Zn
Cell 2: Anode Zn → Zn²⁺ + 2e^–
Cathode Ag⁺ + le^– → Ag
b. Ag. Lesser tendency to give up electron that is it is a less reactive metal.

Question 14.
The chemical reactions of various Galvanic cells are given as incomplete in the table. Complete them.

Answer:
a. Zn → Zn²⁺+ 2e^–
b. Zn+Cu²⁺ → Zn²⁺ + Cu
c. Fe → Fe²⁺ + 2e^–
d. 2Ag⁺ + 2e- → 2Ag
e. Pb²⁺+ 2e^– → Pb
f. Mg + Pb²⁺ → Mg²⁺ + Pb

Question 15.
The picture of a Galvanic Cell is given below.

a. Identify A and B.
b. Give the direction of electron flow?
c. Write the chemical equation at the anode and cathode.
Answer:
a. A – Zn,
B – FeSO₄Solution
b. From Zn to Fe
c. Anode Zn → Zn²⁺ + 2e^–
Cathode Fe²⁺ + 2e^– → Fe

Question 16.
An incomplete table about the electrolysis of different electrolytes are given below. Complete it.

Answer:
a. Cu²⁺,
Cl^–,
H₂O.
b. 2H₂O → O₂ + 4H⁺ + 4e^–
c. Na⁺,
Cl^–
d. Na⁺ + le^– → Na
e. 2Cl → Cl₂ + 2e^–
f. 2H₂O + 2e^– → H₂ + 2OH^–

Question 17.
5mI AgNO₃ is taken in a test tube and a copper rod is dipped into
a. Identify the changes occurring with the copper rod and the solution?
b. Complete the equation of the reaction.
Cu + 2AgNO₃ →  +
c. Write the equations of the oxidation and reduction reactions.
Answer:
a. Silver is deposited in copper rod. Colour of solution changes to blue.
b. Cu + 2 AgNO₃ → Cu(NO₃)₂ + 2Ag
c. 2Ag⁺ + 2e^– → 2Ag (Reduction)
Cu → Cu²⁺ + 2e^– (Oxidation)

Reactivity Series and Electrochemistry Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 1.
The symbols of certain metals are given below. Arrange them as they are given in the reactivity series.
Mg, Pb, Ag, Cu, Zn, Fe, Au, Sn.
Answer:
Mg, Zn, Fe, Sn, Pb, Cu, Ag, Au.

Question 2.
Analyse the table given below and answer the questions.

a. Find out the metals which are likely to be A, B and C from the box given below.

b. Write down the chemical equation between metal ‘B’ and water.
Answer:
a. A – Mg,
B – Cu,
C – Ca
b. Cu (s) + 2H₂O (l) → Cu OH₂ (aq) + H₂ (g)

Very Short Answer Type Questions (Score 2)

Question 3.
Certain metals are given below:
Ag, Zn, Pb, Sn, Fe
a. When a galvanic cell is constructed using these metals, which one acts only as anode? Give the reason.
b. Draw Zn-Fe cell. Mark the direction of electron flow and write the chemical equation anode.
Answer:
a. Zn. Because Zn has a higher reactivity than the other 4 metals.

Direction of electron flow from Zn to Fe Chemical reaction at anode:
Zn(s) → Zn²⁺(aq) + 2e^–

Question 4.
Zn (s) + 2AgNO₃ (aq) →
Zn(NO₃)₂ (aq) + 2Ag
a. Write the oxidation state of each element in this displacement reaction.
b. Write the chemical equation for oxidation and reduction.
Answer:
a. Zn° (s) + 2Ag¹⁺N⁵⁺ O₃^2– (aq)
→ Zn²⁺(N⁵⁺O^2–₃)₂ (aq) + 2Ag°

b.Oxidation:
Zn° (s) → Zn²⁺ (aq) + 2e^–
(Oxidation means losing of electrons)

Reduction:
Ag¹⁺ (aq) + le^– → Ag (s)
(Reduction means gaining of electrons)

Question 5.
Based on the reactions given below, answer the following questions.
i. Aqueous solution of CuCl₂ undergoes electrolysis using graphite rods.
ii. Molten KCl undergoes electrolysis.
iii. Aqueous solution of NaCI undergoes electrolysis.
a. In which all reactions Cl₂ gas is formed? At which electrode is Cl₂ gas formed?
b. In which reaction H₂ gas is formed? Write the chemical equation of this reaction.
Answer:
a. Cl₂ gas is formed in all the three reactions. Chlorine gas is formed at anode.

b. When aqueous solution of NaCI undergoes electrolysis, hydrogen gas is formed at cathode.
Equation: 2H₂O + 2e^– → H₂ + 2OH^–

Very Short Answer Type Questions (Score 3)

Question 6.
Take water in 4 different beakers and add a small piece of sodium, lead, iron and copper in each.
a. In which all solutions gas bubbles will be formed? Which gas is formed?
b. Which solution will turn pink on adding phenolphthalein? Why?
c. Write the chemical equation between this metal and water.
Answer:
a. Gas bubbles are formed in the beaker containing sodium. Hydrogen is tlie gas formed.

b. Beaker containing sodium turns pink because sodium reacts with water and forms an alkali, sodium hydroxide. Alkalies turn phenolphthalein pink.

c. 2Na (s) + 2H₂O (l) → 2NaOH (aq) + H₂ (g)

Question 7.
Analyse the picture given below.

a. Identify ‘A’.
b. Write the chemical equation at ‘B’.
c. Add few drops of phenolphthalein to the remaining solution after electrolysis. What change will take place? Why?
Answer:
a. A – cathode.

b. Chemical equation at ‘B’:
2Cl^– → Cl₂ + 2e^–

c. Solution turns pink. Because, after electrolysis K⁺ and OH^– ions are present in the remaining solution. That means KOH, an alkali is formed.

Question 8.
The flow of electron in certain galvanic cells are given below:
i. Cu → Ag
ii. Ag → Zn
iii.Na → Mg
iv.Fe → K
a. Choose the incorrect ones.
b. Explain your answer.
Answer:
a. ii. Ag → Zn,
iv. Fe → K are the incorrect ones.

b. Zn has a higher reactivity than Ag. Hence Zn undergoes oxidation (loses electrons). So the direction of electron flow is from Zn to Ag. Similarly, K has higher reactivity than Fe. Hence direction of electron flow is from K → Fe.

Question 9.
Which of the chemical reactions given below are wrong? Explain the reason.
a. Cu (s) + 2HCl (aq) →
CuCl₂(aq) + H₂ (g)
b. Mg(s) + Pb(NO₃)₂ (aq) →
Mg(N0₃)₂ (aq) + Mg (s)
c. 3Fe (s) + 4H₂O (l) →
Fe₃O₄(s) + 4H₂(g)
Answer:
Equations (a) and (c) are wrong.
a. Copper cannot displace hydrogen from acids because it is placed below hydrogen in the reactivity series.

c. Fe reacts only with superheated steam. Fe does not react with water in liquid state.

Question 10.
Write the chemical equation of the electrolysis of water to which little sulphuric acid (H₂SO₄) is added.
Answer:
H₂SO₄ (l) + 2H₂O (l) → 2H₃O (aq) + SO₂^2–(aq)
at cathode:
2H₃O+ (aq) + 2e^– → H₂(g) + 2H₂O (l) (reduction)
at anode:
2H₂O (l) → O₂(g) + 4H^– (aq) + 4e^– (oxidation)

Long Answer Type Questions (Score 4)

Question 11.
Analyse the reactions and answer the following questions:

a. Which among the following test tubes will undergo a chemical reaction?
b. What are these chemical reactions called?
c. Explain the oxidation and reduction reactions taking place here including the chemical equation?
Answer:
a. Chemical reaction takes place in test tube (ii) only.
b. These chemical reactions are called displacement reactions. .
c. Chemical equation: (including oxidation states)
Mg° (s) + Fe²⁺S⁶⁺O4^2– (aq) →
Mg²⁺S⁶⁺O^2–₄ (aq) + Fe° (s)
At Mg : Mg°(s) → Mg²⁺(aq) + 2e^– (oxidation)
At Fe²⁺, Fe²⁺ (aq) + 2e^– →  Fe°(s) (reductipn)

Question 12.
A students observation is given below:
i. When Zn is put in salt solution, Na gets deposited over Zn.
ii. Au reacts with water vapour and hydrogen gas is formed.
iii. Al reacts with acid and forms hydrogen gas.
iv. Mg reacts with hot water and forms hydrogen gas.
a. Which statements are incorrect?
b. Give reason for your answer.
Answer:
a. Statements (i) and (ii) are wrong.

b. i. Zn cannot displace sodium. Because sodium has a higher reactivity than Zn.
ii. Au does not react with water vapour.

Question 13.
“Sodium cannot be kept open in atmospheric air, and cannot be stored in water. So it is stored in kerosene.” Give explanation for the above statement with its chemical equation.
Answer:
Sodium which is placed above in the reactivity series reacts vigorously with water and oxygen.
4Na (s) + O₂ (g) → 2Na₂O (s)
2Na (s) + 2H₂O (l) →
2NaOH (aq) + H₂(g)
This NaOH reacts with CO₂ in the atmospheric air and forms sodium carbonate.
2NaOH (s) + CO₂ (g) → Na₂CO₃ + H₂O (l)
To avoid the reaction with O₂, H₂O and CO₂, sodium is stored in kerosene.

Question 14.
The direction of electron flow in certain galvanic cells are given below. (Symbols are not real)
i. B → A,
ii. E → C,
iii.D → E,
iv.A → D.
a. Arrange the metals A, B, C, D and E in the decreasing order of their reactivity.
b. Choose the reaction taking place at ‘C’ in cell (ii) E → C. Give the reason.
i. C⁺ (aq) + le^– → C(s)
ii. C (s) → C⁺ (aq) + le^–
iii.C (s) → C⁺ (aq) + le^–
Answer:
a. B,
A,
D,
E,
C.
b.C⁺ (aq) + le^– → C (s) is the correct equation. Because ‘E’ has higher reactivity than. So ‘C’undergoes reduction.

Question 15.
Give reason for the following.
a. CuSO₄ solution is not stored in iron vessels.
b. Buttermilk is not stored in aluminium vessels.
Ans.
a. Fe displaces copper from copper sulphate solution and forms FeSO₄.
b. Aluminium reacts with acid in the buttermilk.

Students can Download Maths Chapter 1 Arithmetic Sequences Questions and Answers, Notes PDF, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Maths Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

SSLC Maths Chapter 1 Arithmetic Sequence Questions and Answers Malayalam Medium

SCERT 10th Standard Maths Textbook Chapter 1 Solutions Malayalam Medium

You can Download Trigonometry Questions and Answers, Activity, Notes PDF, Kerala Syllabus 10th Standard Maths Solutions Chapter 5 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Maths Chapter 5 Trigonometry Textbook Questions and Answers

SCERT Class 10th Standard Maths Chapter 5 Trigonometry Notes

Textbook Page No. 103

Trigonometry Class 10 Kerala Syllabus Question 1.
In the triangle shown, what is the perpendicular distance from the top vertex to the bottom side? What is the area of the triangle?

Answer:

The sides of the right angle triangle with angles 30°, 60°, 90° are proportional to the number 1 : √3 : 2
AD : BD : AB = 1 : √3 : 2
AB = 2x = 4cm
x = 2 cm
AD = 2 cm
BD = 2√3cm
Perpendicular distance from the top vertex to the bottom side = AD = 2 cm
BC = 2BD = 4√3 cm
Area of triangle = 1/2 bh
1/2 × 4√3 × 2 = 4√3 cm²

Sslc Trigonometry Solutions Kerala Syllabus Question 2.
In each of the following parallelograms, find the distance between the top and bottom side? Calculate the area of parallelogram.

Answer:

Sslc Maths Chapter 5 Kerala Syllabus Question 3.
A rectangular board is to be cut along the diagonal and the pieces rearranged to form an equilateral triangle as shown below. The sides of the triangle must be 50 centimetres. What should be the length and breadth of the rectangle?

Answer:
The sides of the right angle triangle with angles 30°, 60°, 90° are proportional to the number 1 :√3: 2

Sslc Maths Trigonometry Kerala Syllabus Question 4.
Two rectangles are cut along the diagonal and the triangles got are to be joined to an-other rectangle to make a regular hexagon as shown below:

If the sides of the hexagon are to be 30 centimetres, what would be the length and breadth of the rectangles?
Answer:

The sides of the right angle triangle APF with angles 30°, 60°, 90° are proportional to the number 1 :√3: 2

Length of the smaller rectangle = 25.98 cm
Breadth of the smaller rectangle = 15 cm
Length of the bigger rectangle = 51.96 cm
Breadth of the bigger rectangle = 30 cm

Trigonometry Sslc Kerala Syllabus Question 5.
Calculate the area of the triangle shown.

Answer:
Angles of the first triangle are in ratio 45 : 45: 90. So sides are in ratio x : x: √2 x
Angles of the second triangle are in ratio 30 : 60: 90, sides are in ratio, y :√3y: 2y

Textbook Page No. 109

Trigonometry Problems For Class 10 State Syllabus Question 1.
The lengths of two sides of a triangle are 8 centimetres and 10 centimetres and the angle between them is 40°. Calculate its area. What is the area of the triangle with sides of the same length, but angle between them 140°?
Answer:

If the angle between the sides is 140° sin 140 = sin(180 – 40) = sin 40 The area of the triangles will be same.

Trigonometry Questions For Class 10 Kerala Syllabus Question 2.
The sides of a rhombus are 5 centimetres long and one of its angles is 100°, Compute its area.
Answer:

Sslc Maths Trigonometry Notes Kerala Syllabus Question 3.
The sides of a parallelogram are 8 centimetres and 12 centimetres and the angle between them is 50°. Calculate its area.
Answer:

Trigonometry Class 10 Scert Kerala Syllabus Question 4.
Angles of 50° and 65″ are drawn at the ends of a 5 centimetres long line to make a triangle. Calculate its area.
Answer:
If diameter is d

Sslc Trigonometry Kerala Syllabus Question 5.
A triangle is to be drawn with one side 8 centimetres and an angle on it 40°. What should be the minimum length of the side opposite this angle?
Trigonometry Questions and Answer: The side opposite to 40° will be at least

Textbook Page No. 114

Class 10 Maths Chapter 5 Kerala Syllabus Question 1.
The figure shows a triangle and its circumcircle: What is the radius of the circle?

Answer:
∠BAC = 60°
∠BOC = 120°
The angle made by any arc of a circle on the alternate arc is half the angle made at the centre.

The sides of the right angle triangle ΔCOD with angles 30°, 60°, 90° are proportional to the number 1 : √3: 2 .

Sslc Maths Chapter 5 Solutions Kerala Syllabus Question 2.
What is the circumradius of an equilateral triangle of sides 8 centimetres?
Answer:
The sides of the right angle triangle OBD with angles 30°, 60°, 90° are proportional to

Hsslive Trigonometry Kerala Syllabus Question 3.
The figure shows a triangle and its circum.

i. Computer the diameter of the circle.
ii. Compute the lengths of the other two sides of the triangle.
Answer:

Maths Chapter 5 Class 10 Kerala Syllabus Question 4.
A circle is to be drawn, passing through the ends of a line, 5 centimetres long; and the angle on the circle on one side of the line should be 80°. What should be the radius of the circle?
Answer:

Maths Trigonometry Class 10 State Syllabus Question 5.
The picture below shows part of a circle:

What is the radius of the circle?
Answer:

First, complete the circle. Draw BD and join one of its end D to C. ∠BAC + ∠BDC= 180° ∠D = 40°, ∠BCD (angle on semicircle). So ABCD is right-angled.

Trigonometry Questions For Class 10 Scert Question 6.
A regular pentagon is drawn with all its vertices on a circle of radius 15 centimetres. Calculate the length of the sides of this pentagon.
Answer:
Sum of angles of pentagon
= (n – 2)180
= (5 – 2)180
= 540°
One angle of regular pentagon = \(\frac { 540 }{ 5 }\) = 180°

Textbook Page No. 117

Kerala Syllabus 10th Standard Maths Chapter 5 Question 1.
One angle of a rhombus is 50° and the larger diagonal is 5 centimetres. What is its area?
Answer:
In rhombus ABCD, One angle of a rhombus is 50° and one diagonal is 5 centimetres.

Sslc Maths Chapter 5 Trigonometry Kerala Syllabus Question 2.
A ladder leans against a wall, with its foot 2 metres away from the wall and the angle with the floor 40°. How high is the top end of the ladder from the ground?
Answer:
tan 40 = \(\frac { QR }{ 2 }\)
QR = 2 × tan 40
= 2 × 0.8391 = 1.6782
Height of the ladder from ground = 1.68m

Hss Live Guru 10th Maths Kerala Syllabus Question 3.
Three rectangles are to be cut along the diagonals and the triangles so got rearranged to form a regular pentagon, as shown in the picture. If the sides of the pentagon are to be 30 centimetres, what should be the length and breadth of the rectangles?

Answer:

36°, 54°, 90°
Sin 54 = \(\frac { DG }{ 30 }\)
DG = 30 × 0.8090
= 24.27 cm
Cos 54° = \(\frac { EG }{ 30 }\)
EG = 30 × 0.5878 = 17.63 cm
Length of larger rectangle = 46.17 cm.
Breadth = 15 cm
Length of smaller rectangle = 24.27 cm
Breadth = 17.63 cm

10th Standard Maths Trigonometry Kerala Syllabus Question 4.
In the picture, the vertical lines are equally spaced. Prove that their heights are in arithmetic sequence. What is the common difference?

Answer:

10th Trigonometry Questions And Answers Kerala Syllabus Question 5.
One side of a triangle is 6 centimetres and the angles at its ends are 40° and 65°. Calculate its area.
Answer:
∠C = 180 – (40 + 65) = 75°
Draw a perpendicular BD from B to AC

Textbook Page No. 122

Trigonometry Hsslive Kerala Syllabus Question 1.
When the sun is at an elevation of 40°, the length of the shadow of a tree is 18 metres. What is the height of the tree?
Answer:
In the right triangle ΔPQR
tan 40° = \(\frac { QR }{ 18 }\)
QR = 18 × tan40° =18 × 0.8391
Height of the tree = 15.1 m

Class 10 Maths Chapter 5 Trigonometry Kerala Syllabus Question 2.
When the sun is at an elevation of 35°, the shadow of a tree is 10 metres. What would be the length of the shadow of the same tree, when the sun is at an elevation of 25°?
Answer:

Question 3.
From the top of an electric post, two wires are stretched to either side and fixed to the ground, 25 metres apart. The wires make angles 55° and 40° with the ground. What is the height of the post?
Answer:

Question 4.
A 1.5-metre tall boy saw the top of a building under construction at an elevation of 30°. The completed building was 10 metres higher and the boy saw its top at an elevation of 60° from the same spot. What is the height of the building?
Answer:
In the right triangle BDE,

Question 5.
A 1.75-metre tall man, standing at the foot of a tower, sees the top of a hill 40 metres away at an elevation of 60°. Climbing to the top of the tower, he sees it at an elevation of 50°. Calculate the heights of the tower and the hill.
Answer:
In the right triangle CEF
BD = CE = AF = HG = 40m

AB = tower
DF = hill
Height of hill = 69.28 + 1.75 = 71.03 m
In the right triangle HGF,
\(\tan 50=\frac{G F}{H G}=\frac{G F}{40}\)
GF = 40 × tan 50= 40 × 1.1918 = 47.67 m
Height of tower = 71.03 – (47.67 + 1.75)
= 71.03 – 49.42 = 21.61 m

Question 6.
A man 1.8 metre tall standing at the top of a telephone tower, saw the top of a 10-metre high building at a depression of 40° and the base of the building at a depression of 60°. What is the height of the tower? How far is it from the building?
Answer:

Height of the building = 10m
Height of the towerAG
Height of the man GF = 1.8m
AB = x
In the right triangle CHF
tan 40 = \(\frac { HF }{ x }\)
HF = x tan 40
= x × 0.8391
= 0.8391x
In the right ABF
\(\tan 60=\frac{A F}{A B}=\frac{A F}{x} \Rightarrow\)
AF = x tan 60 = 1.732x
BC = AH = AF – HF
= 1.732x – 0.8391x = 10
= o.8929x = 10

Height of tower = 19.4 – 1.8 =17.6 m
Distance from building to tower = 11.2m

Trigonometry Orukkam Questions and Answers

Worksheet 1

Question 1.
Complete the table given below.

Answer:
∠C = 90°
The sides of the right angle triangle with angles 30°, 60°, 90° are proportional
1 : √3: 2

Question 2.
Complete the table given below.


Answer:
The sides of the isosceles right-angle triangle with angles 45: 45 :90 will have sides proportional to [opposite to corresponding angles] 1: 1: √2

Question 3.
Calculate the perimeter of the triangle.

Answer:
i. BC = 10
∴ AB = 10

Question 4.
ABC Dis a square AC = 10cm . Find B, BACFind the length of AB. Find the perimeter of the square.
Answer:
∠B = 90°
∠BAC = 45° = ∠BCA
\(\mathrm{AB}=\frac{10}{\sqrt{2}}=5 \sqrt{2}\)
Perimeter of the square D
= 4 × 5√2 = 28.28 cm

Question 5.
PQRS is a rectangle. Find angle SP R? Find angle PRQ. If PR = 30 then find P Qand QR. Calculate the perimeter of the rectangle.

Answer:

∠SPQ = 90° ( ∴ PQRS is a square, so angles are 90° each.)
∴ ∠SPR = 90 – 30 = 60°

Question 6.
If C D = 5 then find ∠ACD,∠BCD . Find AB, AD, BD, BC. Find the angles of triangle ABC. If the angles are 45°, 60°, 75° find the ratio of the sides?

Answer:

Worksheet 2

Question 7.
In the figure BC = 12 ∠D = 90°,Find ∠CBD, ∠ACD, ∠ABC . Find BD, C D, AD, AC, AB. Find the ratio of the sides of the triangle having the angles 30°, 15°, 135° C

Answer;
∠ACD = 180 – (30 + 90) = 60
∠BCD = 60 – 15 = 45 ∠CBD = 45
∠ABC = 135 [∵ 180 – (30 + 15)]

Question 8.
In the figure AD = 7, CD = 8, BD = 5, ∠ADP = 50° then find ADB ?

Answer:

Question 9.
Find the measure of the remaining part of the triangle from the figure given below

Question 10.
∠AOB = 2x, radius of the circle R. Find ∠AOC? Find sin x, AC and AB
Answer:

Worksheet 3

Question 11.
Using the figure find AB

Question 12.
In the figure BD = 10,

Find ∠BAD and ∠ADB.
AD, CD and AC
Answer:
∠ADB = 180 – 60 = 120°
∠BAD = 180 – (120 + 30) = 30

The sides of the right angle triangle with angles 30°, 60°, 90° are proportional to
1 :√3: 2
AD = 2 × 5√3 = 10√3
AC = 5 √3 × √3 = 15

Question 13.
In the figure QR = 7, find ∠QRP, ∠QPR Find the length of PR, PS and RS.

Answer:


Question 14.
In the figure BD = 10, CD = x, find the length of BC. Using tan 40, tan 50 find the length of AC.
Answer:

Worksheet 4

Question 15.
In triangle ABC , AB = 7, BC = 12, ∠B = 40 Find the area of the triangle. Calculate the length of AC.
Answer:
Draw a perpendicular line
AD from A to BC.
AD = AB sin40
= 7 × 0.6428 = 4.4996 = 4.5
Area = \(\frac { 1 }{ 2 }\) × BC × AD
= \(\frac { 1 }{ 2 }\) × 12 × 4.5 = 27m2
BD = AB cos 40 = 7 x 0.7660 = 5.36
∴ CD = 12 – 5.36 = 6.64

From right angled ΔADC

∴ Length of AC = 8.02 units

Question 16.
In triangle ABC , AB = 7, BC = 12, ∠B = 40 Find the area of the triangle.
Answer:
See the above question

Question 17.
In the figure AD = BD = C D = 5 ∠ADC = 50° . find the area of triangle AC D, triangle ABD and triangle ABC.

Answer:
AD = BD = CD, So ΔABD and ΔACD are isosceles triangles. AE is the perpendicular from D to AC

Area of ΔABD = AF × DF
= 4.5 × 2.1 = 9.45
Area of ΔABC= \(\frac { 1 }{ 2 }\) × AB × AC
\(\frac { 1 }{ 2 }\) × 9 × 4.2 = 18.9m²

Question 18.
ABC D is a parallelogram, angle D = 120°, AB = 10, AC = 12. Calculate the area of the parallelogram.

Answer:
Area of the parallelogram= \(\frac { 1 }{ 2 }\) × AC × BD
In ΔABD, ∠A = 60°
ΔABD is an equilateral triangle.
BD = 10
Area of the parallelogram ABCD = \(\frac { 1 }{ 2 }\) × 10 × 12 = 60m²

Question 19.
One angle of a triangle is 30°, prove that radius of the circumcircle is equal to the side opposite to 30°
Answer:
For a right-angled triangle one of the angles is 30° then other one is 60°.
Side which is opposite to the angle of 90° is twice of the side which is opposite to the angle of 30°
Center of circumcircle is the midpoint of the side, which is opposite to the angle 90° that means half.
∴ Radius of the circumcircle is equal to the side opposite to 30°.

Question 20.
O is the centre of a circle having a chord
AB. AB= 12, angle AOB = 120°. Find the radius
Answer:
AC = BC = 6
A perpendicular is drawn through center which can bisect the perpendicular line AB into half.
In ΔAOC , ∠AOC = 60°, ∠ACO = 90°


Question 21.
Above viewed the top of a tree at an angle of elevation 30°. He moved 10 m towards the tree and saw the top of the tree ant the angle 60° Find the height of the tree
Answer:

Question 22.
In the figure BC =14, ∠5 = 40°, ∠C = 50° Find the area of triangle ABC.
Answer:

Area of ΔABC =\(\frac { 1 }{ 2 }\) × AB × AC
AC = BC sin 40 = 14 × 0.6428 = 8.99
AB = BC cos 50 = 14 × 0.7660 = 10.72
Area of ΔABC = \(\frac { 1 }{ 2 }\) × 8.99 × 10.72 = 48.2 m²

Worksheet 5

Question 23.
A child observed an aeroplane flying horizontally at the height 1km at ah angle of elevation 60°at an instant. After ten seconds he saw the plane at the angle 30°. Calculate the speed of the plane.
Answer:

speed = distance/time = 1150m/ 10sec = 115 m/s

Question 24.
In the figure BC = a, CD = b. Then prove that a = 3b.
Answer:

Worksheet 6

Question 25.
A man observed the top of a tower at a distance a from its base at an angle of elevation 60°. He saw the top of the tower at an angle of elevation 30° from a point at the distance b from the base. Prove that height of the tower h = √ab
Answer:

Trigonometry SCERT Questions and Answers

Question 26.
The diagonal of a square is 4cm long. Find its perimeter and area. [Score: 2, Time: 3 minute]

Question 27.
AC and BC are two equal chords of a circle with diameter AB. If the equal chords have lengths 10cm find the area of the circle. [Score: 3, Time: 3 minute]

10 Diameter AB = 10√2 cm. (1)
Radius = 5√2 cm (1)
Area = πr² = π × (5√2)² = 5π sq. (1)

Question 28.
In ΔABC, AB = 10 cm. AC 8 cm, ∠A = 45°
a. Find the perpendicular distance from C to AB.
b. Find the area of the triangle. [Score: 2, Time: 3 minute]

Answer:

Question 29.
One side of a rhombus is 12 cm and one angle is 135°.
a. Find the distance between the parallel sides?
b. Find the area of the rhombus. [Score:3,Time:3minute]

Answer:
a. ABCD is a rhombus
AB = AD = 12 cm, ∠B = 135° ∠A = 180 – 135 = 45°
Angles of ΔADE are 45°, 45°, 90°

Question 30.
In ΔABC ∠A = 45°, BC = 6 c m. Find the diameter of the circumcircle. [Score: 4, Time:4 minute]

Answer:
Draw diameter BD and Join CD. Angles of ΔBCD is 45°, 45°, 90° (1)

Question 31.
12 centimetre long diagonal of a rectangle makes an angle 30° With its one side. Find its perimeter and area. [Score: 4, Time:4 minute]

Answer:
ABCD is a square
AC = 12cm , ∠BAC = 30° Angles of ΔABC are 30°, 60°, 90°

Question 32.
In ΔABC, AB = 20 cm ∠A = 30°, AC = 12 cm
a. Find the length of the perpendicular from C to AB.
b. Find the area of the triangle. [Score: 3, Time: 4 minute]

Answer:
a. Draw CD perpendicular to AB. Angles of ΔADC are 30°, 60°, 90° (1)

Question 33.
In ΔABC, AB = 8 cm , BC = 10 cm and ∠B = 60°
a. Find the area of the triangle ΔABC.
b. Find AC. [Score: 4, Time:6 minute]

Answer:

Question 34.
One angle of a triangle is 150° and its opposite side 3 centimetre. Find the diameter of its circumcircle. [Score: 3, Time:5 minute]

Answer:
In DABC, ∠B = 150°, AC = 3 cm
Draw diameter AD and join CD (1)
∠ADC = 180 – 150 = 30°, ∠ACD = 90°
Angles of DADC are 30°, 60°, 90°
30° 60° 90°
1 : √3 : 2
↓    ↓     ↓
3 3√3 6 (1)

Diameter, AD = 6 cm (1)

Question 35.
In ΔABC AB = 12 cm, ∠A=45° and ∠B = 30°
a. Find the area of the triangle ABC.
b. Find the ratio of the sides of the triangle having angles 30°,45°,105° [Score: 5, Time:8 minute]

Answer:
a. Angles of ΔABC are 45°, 45°, 90°
Angles of ΔBDC are 30°, 60°, 90°. (1)

Question 36.
The diagonal of a rectangle is 12 cm and it makes an angle 35° With one side. Find the perimeter of the rectangle. [sin 35° = 0.57, cos 35° = 0.82] [Score, 3, Time: 5 minute]

Answer:
Let the breadth of the rectangle = x and length = y
sin 35° = \(\frac { x }{ 12 }\) (1)
x = 12 × sin 35 = 12 × 0.57 = 6.84 cm.
cos 35° = \(\frac { y }{ 12 }\)
y = 12 × cos 35 = 12 × 0.82 = 9.84 cm. (1)
Perimeter = 2(6.84 + 9.84) = 2 × 16.68 = 33.36 cm (1)

Question 37.
In ΔABC, ∠A = 125°,BC = 8 cm. Find the diameter of the circumcircle. [sin 55 = 82] [Score: 3, Time:6 minute]

Answer:
Draw diameter BD and jon CD. (1)
In ΔBCD, sin55° = \(\frac { BC }{ BD }\) (1)

Question 38.
Can one cut out a triangle of one side 7 cm and its opposite angle 40° from a circular sheet of diameter 10 cm. Justify your answer. [sin 40° = 0.64]
[Score: 4, Time:7 minute]
Answer:
The diameter of the circumcircle of a triangle with one angle 40° and it’s opposite
side 7 cm = \(\frac { 7 }{ sin 40 }\) (1)
\(\frac { 7 }{ 0.64 }\) = 10.93 cm
Diameter of the paper is 10 cm, which is less than 10.93 cm.
Hence triangle cannot be cutout. (2)

Question 39.
Find the area of a triangle Whose sides are a and b and the angle between those sides is C. [Score: 2,Time:3 minute]
Answer:
In ΔABC Draw AD perpendicular to BC.
sin C = \(\frac { h }{ b }\)
h = b sin C
Area = \(\frac { 1 }{ 2 }\) ah
= \(\frac { 1 }{ 2 }\) ab sin C (1)

Question 40.
Find the sides of a triangle whose angles are A, B and C and its circumdiameter d. [Score: 3, Time: 5 minute]
Answer:
Draw a diameter BD and join CD

∠BDC = A : BC = a (1)
sm A = \(\frac { a }{ BD }\) = \(\frac { a }{ d }\)
Similarly
b = d sin B° = AC (1)
AB = c = d sin C°. (1)

Trigonometry Exam Oriented Questions and Answers

Short Answer Type Questions (Score 2)

Question 41.
The angle of a right triangle is 30° and its hypotenuse is 4 cm. What is its area?

Answer:
Triangle side ratio is 1 : : 2
Altitude = hypotenuse \(× \frac{\sqrt{3}}{2}=4 \times \frac{\sqrt{3}}{2}\)
Area = \(\frac { 1 }{ 2 }\) × 23.46 = 3.46 cm²

Question 42.
In the figure, find the length of the side represented by x.

Answer:
\(\tan 50^{\circ}=\frac{\text { opposite side }}{\text { adjacent side }}=\frac{x}{24}\)
x = 2.4 × tan 50° = 2.4 × 1.1918 = 2.86

Question 43.
Calculate the area of a right-angled triangle whose one angle is 45° and hypotenuse 20 cm.
Answer:
Angle of the right 45°, 45°, 90°
Ratio of sides = 1 : 1 : √2
hypotenuse = 20cm
∴ The other two sides are \(\frac { 20 }{ √2 }\) cm each
∴ Area of the right angled triangle
\(=\frac{1}{2} \times \frac{20}{\sqrt{2}} \times \frac{20}{\sqrt{2}}=\frac{1}{2} \times \frac{20 \times 20}{2}=100 \mathrm{cm}^{2}\)

Question 44.
Different sizes of isosceles triangle are given. In the table given below some of its sides are given. Fill the table.

Answer:
a. 4. 4, √32
b. 2, 2, √8
c. 3, 3, √18
d. 10, 10, √200
e. 1, 1, √2

Question 45.
The area of a parallelogram with one side 8cm and an angle 30° is 80 cm².
Find out the length of the other side.

Answer:

Area of the parallelogram = 80cm2
8h = 80
h = \(\frac { 80 }{ 8 }\)= 10cm
The angles of ΔAPD are 30°, 60° and 90°
The sides are in the ratio 1 : √3 : 2
DP = 10cm
∴ AD = 20cm

Short Answer Type Questions (Score 3)

Question 46.
In the figure, ∠BAC = 90°, AD = 6cm, CD = 9cm, ∠ACD = x.
a. What is tan x?
b. How much is ∠BA D
c. What is the length of BD?

Answer:

Question 47.
In the quadrilateral ABCD shown below,
∠A = ∠C = 90% ∠ABD = 45° ∠CDB = 60° and AB = 6cm.
Find out the lengths of the other sides of the quadrilateral.

Answer:
In ΔABD,
The angles are 45°, 45°, 90°
As the triangle is equilateral, sides are in the ratio, 1: 1: √2
AB = 6cm
∴AD = 6cm, BD = 6√2cm

Question 48.
P be a point on a circle having diameter AB. ∠ABP = 30°, BP = 6cm. Draw a rough figure.
Find the length of AP and area of circle ?
Answer:
∠P = 90°, ∠A = 60°,
AP: PB: AB = 1: √3 :2
AP = 2√3 cm,
AB = 4√3 cm
Radius of circle = 2√3 cm
Area of circle = π(2√3)2
cm = 12 π cm²

Long Answer Type ? Questions (Score 4)

Question 49.
In ΔABC, ∠A = 110° and BC = 8cm Find out the radius of the circumcircle.

Answer:
Draw diameter BD and Join D and C. The opposite angles of cyclic quadrilaterals are supplementary ∠D = 70°, BCD is a semicircle. ∠BCD is the angle in a semicircle, ∠BCD =90°

Question 50.
Length of two sides of a triangle are 20cm and 16cm and the angle between them is 135°.
a. Draw a rough figure and mark the measurements.
b. Find the perpendicular distance of the vertices to the side of length 20cm.
c. Find the area of the triangle.
Answer:

b. Now AADB is a right triangle with angles 45°, 45°, 90°. Since the side opposite to 90° angle is 16cm, the other two sides are 8√2 each.

Perpendicular distance of the vertex to the side of length 20cm is 8√2 cm.

Long Answer Type Questions (Score 5)

Question 51.
A girl standing on a lighthouse built on a cliff near the seashore, observes two boats due East of the lighthouse. The angles of depression of the two boats are 300 and 600. The distance between the boats is 300m.
a. Draw a rough figure based on the given details.
b. Find the distance of the top of the lighthouse from the sea level. (Boats and foot of the lighthouse are in a straight line).
Answer:


Distance of the top of the lighthouse from the sea level = 259.8m

Question 52.
In the figure, OR is perpendicular to OP and OP = 12cm. A, B and C are points on OR. If ∠OPA = 30°? ∠APB = 15°,and ∠BPC = 15°. Find OA, OB and OC. Also find AB: BC.

Answer:
ΔOPA is a right triangle with angles 30°, 60°, 90°. Since the side opposite to 30° angle is 4√3


Now consider ΔOPB. It is a right triangle with angles 45°, 45°, 9Q° Since the side OP = 12cm, side OB is also 12cm.

Also ΔOPC is a right triangle with angles 30°, 60°, 90°. Since the side opposite to 30° angle is 12cm, the side opposite to the 60° angle ie OC is 12√3.
Thus OA = 4√3 cm,

Trigonometry Memory Map

You can Download Coordinates Questions and Answers, Activity, Notes PDF, Kerala Syllabus 10th Standard Maths Solutions Chapter 6 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Maths Chapter 6 Coordinates Textbook Questions and Answers

SCERT Class 10th Standard Maths Chapter 6 Coordinates Notes

Textbook Page No. 134

Coordinates Class 10 Kerala Syllabus Chapter 6 Question 1.
Find the following.
i. The y -coordinate of any point on the x-axis.
ii. The x -coordinate of any point on the y-axis.
iii. The coordinates of the origin
iv. The y-coordinate of any point on the line through (0,1), parallel to the x-axis.
v. The x- coordinate of any point on the line through (1,0), parallel to the y-axis.
Answer:
i. y-coordinate of any point on the x-axis is zero.
ii. x-coordinate of any point on the y-axis is zero.
iii. Origin is (0, 0).
iv. y = 1
v. x =1

Sslc Maths Coordinates Kerala Syllabus Chapter 6 Question 2.
Find the coordinates of the other three vertices of the rectangle below:

Answer:

A(0,0), B (4,0), D(0,3)

Sslc Maths Chapter 6 Kerala Syllabus Question 3.
In the rectangle shown below, the sides are parallel to the axes and origin is the midpoint:

What are the coordinates of the other three vertices?
Answer:

B (–3, 2), C(–3, -2). D (3, –2)

Sslc Coordinates Kerala Syllabus Chapter 6 Question 4.
The triangle shown below is equilateral:

Find the coordinates of its vertices.
Answer:
ΔOAB is an equilateral triangle. So angles of AOCB are 30°, 60°, 90°.
∴ Its sides are in the ratio of 1: √3: 2

OC = 2, BC = 2 √3
Hence the coordinates are B (2, 2, √3), A (4, 0), O(0, 0)

Coordinates Questions And Answers Kerala Syllabus Question 5.
A large trapezium made up of four equal trapeziums:

Find the coordinates of the vertices of all these trapeziums. Draw this picture in GeoGebra.
Answer:
In trapezium (1). Coordinates of the vertices are (0, 0), (2, 2), (4, 2), (4, 0)
In trapezium (2). Coordinates of the vertices are (4,0), (4,2), (6, 2), (8, 0)
In trapezium (3), Coordinates of the vertices are (2, 2), (4, 4), (6, 4), (6, 2)
In trapezium (4). Coordinates of the vertices are (8, 0), (6, 2), (6, 4), (8, 4)
For drawing figure in GeoGebra, Put input in the sequence of input bar as [(a, a + 1), a, 0.5,0.5].

Sslc Coordinates Questions And Answers Kerala Syllabus Question 6.
In the picture the centre O of the circle is the origin and A, B are points on the circle. Calculate the coordinates of A and B.

Answer:
As the angles are in the ratio 30 : 60: 90 the sides will be in the ratio 1: √3: 2
that is A(√3, 1), B(–1, √3)

Textbook Page No. 139

Coordinates 10th Class Kerala Syllabus Chapter 6 Question 1.
All rectangles given below have sides parallel to the axes. Find the coordinates of the remaining vertices of each.

Answer:

Sslc Maths Coordinates Questions And Answers Chapter 6 Question 2.
Without drawing coordinate axes, mark each pair of points below with left-right, top-bottom position correct. Find the other coordinates of the rectangles drawn with these as opposite vertices and sides parallel to the axes.
i. (3, 5), (7, 8)
ii. (6, 2), (5, 4)
iii. (-3, 5), (-7, 1)
iv. (-1, -2), (-5, -4)
Answer:
i. If A (3, 5) and C (7, 8) then other coordinates are, B (7, 5), D (3, 8)

ii. If B(6, 2) and D (5, 4) D<5,4)other coordinates are A (5, 2) & C (6, 4)

iii. If A(-7, 1) and C (-3, 5) other coordinates are B(-3, 1), D (-7, 5)

iv. If A(-5, -4) and C(-1, -2) other coordinates are B(-1, -4), D (-5, -2)

Textbook Page No. 146

Coordinate Geometry Class 10 State Syllabus Chapter 6 Question 1.
Calculate the length of the sides and diagonals of the quadrilateral on the right.

Answer:
in quadrilateral ABCD

Let AC, BD be the diagonals,

Length of AC = Distance between the points (-3, -2) and (0, 0)

Length of BD = Distance between the points (1, -2) and (-3, 1)

Sslc Maths Coordinate Geometry Kerala Syllabus Chapter 6 Question 2.
Prove that by joining the points (2, 1), (3, 4), (-3, 6) we get a right triangle.
Answer:
Let, A(2, 1), B(3, 4), C(-3, 6) be the vertices of the triangle ABC.

Using Pythogoras theorem AC2 = AB2 + BC2
50 = (√10)² + (√40)² = 50
∴ We obtain a right triangle.

Sslc Maths Coordinate Geometry Notes Kerala Syllabus Chapter 6 Question 3.
A circle of radius 10 is drawn with the origin as centre.
i. Check whether each of the points with coordinates (6, 9), (5, 9), (6, 8) is inside, outside or on the circle.
ii. Write the coordinates of 8 points on this circle.
Answer:

i. Distance of the point (5, 9) from the centre of the circle = \(\sqrt{25+81}=\sqrt{106}>10\)
Distance of the point (6, 9) from the centre of the circle = \(\sqrt{36+81}=\sqrt{117}>10\)
Distance of the point (6, 8) from the centre of the circle = \(\sqrt{36+64}=\sqrt{100}=10\)
(5, 9), (6, 9) are outside the circle (6, 8) is on the circle

ii. (10, 0), (0, 10), (–10, 0), (0, –10)
(6, 8), (–6, 8), (6, –8), (–6, –8)

Sslc Maths Chapter 6 Solutions Kerala Syllabus Question 4.
Find the coordinates of the points where a circle of radius √2, centred on the point with coordinates (1, 1) cuts the axes.
Answer:

It cuts at (0, 0), (2, 0) on the x-axis and at (0, 0), (0, 2) on the y-axis.

Hss Live Guru 10th Maths Kerala Syllabus Chapter 6 Question 5.
The coordinates of the vertices of a triangle are (1, 2), (2, 3), (3, 1), Find the coordinates of the centre of its circumcircle and the circumradius.
Answer:

Coordinates Orukkam Questions and Answers

Worksheet 1

Coordinates Class 10 Kerala Syllabus Chapter 6 Question 1.
Draw x, y axis and mark the points A(0, 5), B(0, –2), C(4, 0), D(–3, 0), E(4, 5), F(–3,–2), G(4, –2)
Answer:

Points on x axis = – D(–3, 0), C(4, 0)
Points on y axis = A(0, 5), B(0, –2)

Question 2.
What are the points on x-axis, on y-axis?
Answer:

Points on x axis = – D(–3, 0), C(4, 0)
Points on y axis = A(0, 5), B(0, -2)

Question 3.
Write coordinates of two more points on AE
Answer:
(1, 5), (2, 5)

Question 4.
Write the coordinates of two more points onCE.
Answer:
(4, 1), (4, 2)

Worksheet 2

Question 5.
Given A(2, 3), B(5, 4), C(6, 7), D(3, 6). Find the lengths AB, BC, CD, AD.
Answer:

Question 6.
Check whether AC = BD or not.
Answer:

Question 7.
Prove that P(4,5) the point on AC and BD.
Answer:

AP +PC = 4√2 = AC
∴ P (4, 5) is a point on AC .

BP + PD = √2 + √2 = 2√2 = BD
∴ P (4, 5) is also a point on BD.

Question 8.
Find a point on x-axis equidistant from A and B.
Answer:
Let (x, 0) be a point on the x axis which is at equal distance from A and B .
(2 – x)² + (3 – 0)²= (5 – x)² + (4 – 0)²
(2 – x)² + 9 = (5 – x)² + 16
(2 – x)² + (5 – x)² = 16 – 9
4 – 4x + x² – 25 + 10x – x² = 7
6x = 28
x = \(\frac { 28 }{ 6 }\) = \(\frac { 14 }{ 3 }\)
∴ Point on x axis is \(\left(\frac{14}{3}, 0\right)\)

Worksheet 3

Question 9.
The sides of ABCD are parallel to the coordinate axes and A(3, 7), C(7, 9) are the opposite vertices. Write the coordinates of B and D.
Answer:

Question 10.
Find the lengths of AB and BC.
Answer:
AB = |7 – 3| = 4 BC = |9 – 7| = 2

Question 11.
Calculate the area of the rectangle ABCD
Answer:
Area of the rectangle ABC D = AB × BC
4 × 2 = 8 m²

Question 12.
If P, Q, R, S are the midpoints of the sides write the coordinates of P, Q, R, S
Answer:

Question 13.
Calculate the sides of PQRS.
Answer:

Question 14.
Suggest a name suitable to PQRS
Answer:
Rhombus (SQ = 4, RP = 2, diagonals are not equal, sides are equal hence it is a rhombus)

Worksheet 4

Question 15.
If A(4, 3), B(-4, 3) are two points on the line AB write two more points on this line.
Answer:
A(4, 3), B(-4, 3)
Hence other points on AB = (-2, 3) (1, 3)

Question 16.
Write the coordinates of two more points on the line perpendicular to AB and passing through (4, 3)
Answer:
(4, 1) (4, 5)

Question 17.
Find the length AB.
Answer:

Question 18.
Write the coordinates of the mid point of AB.
Answer:
Coordinates of the mid point of AB = \(\left(\frac{4+-4}{2}, 3\right)=(0,3)\)

Worksheet 5

Question 19.
Without drawing coordinate axes mention the positions of A(2, 1), B(6, 1), C (6, 5) as left-right, above-below
Answer:

Question 20.
Draw coordinate axes, mark the points and complete triangle ABC.

Question 21.
Which side of the triangle is parallel to x-axis?
Answer:
AB, parallel to x-axis.

Question 22.
Which side is parallel to y-axis?
Answer:
BC, parallel to y-axis.

Question 23.
Write the coordinates of the mid point of AB
Answer:
Coordinates of the mid point of AB = \(\left(\frac{2+6}{2}, 1\right)=(4,1)\)

Question 24.
Write the coordinates of the midpoints of BC.
Answer:
Midpoints of BC = \(\left(6, \frac{5+1}{2}\right)=(6,3)\)

Question 25.
Write the coordinates of the midpoints of AC.
Answer:
Midpoints of AC = \(\left(\frac{2+6}{2}, \frac{1+5}{2}\right)=(4,3)\)

Worksheet 6

Question 26.
A (6, 0) is a point on a circle with centre (0,0). Find the radius of the circle.
Answer:
Radius of the circle = 6

Question 27.
Show that B (–3, 3√3) ,C(–3, –3√3) are the points on this circle
Answer:
Distance of B from origin \(\sqrt{(-3-0)^{2}+(3 \sqrt{3}-0)^{2}}\)

Distance of C from origin

B(–3, 3√3) and C(–3, –3√3) are on the circle.

Coordinates SCERT Questions and Answers

Question 28.
Consider an equilateral triangle ABC with A (0,3) AD is the height. If G is the centroid and D is the origin, Find the coordinates of B, C, D and G. [Score: 3, Time: 6 minutes]

Answer:
D is the origin coordinates is (0, 0) (1)
G divides AD in the ratio 2 : 1
∴ Coordinates of G is (0,1) (1)
Triangle ADB is a right triangle with 30°, 60°, 90°
∴ BD = \(\frac { 3 }{ √3 }\) = √3
∴ Coordinates of B is (–√3, 0) (1)
∴ Coordinates of C is (√3, 0) (1)

Question 29.
Find the coordinates of a point on the x -axis which is at a distance of 5 units from (4, -5). Find the coordinates on the y-axis [Score: 3, Time: 6 minutes]
Answer:
Thc point on x-axis which is at a distance of 5 units from (4,-5) is (4, 0).
(x – 4)² + 52 = 25,
∴ x = 4
Point on x axis is (4, 0) (1)
Point on y axis is (0, y).
(0 – 4)² + (y + 5)² = 25,
∴ y = –8, –2
The point on y -axis which is at a distance of 5 units from (4, –5) are (0, –2), (0, –8)

Question 30.
Consider a rhombus ABCD whose diago¬nals meet at origin. The length of diagonals are 8 units and 6 units. Find the coordinates of the vertices. [Score: 4, Time: 4 minutes]
Answer:
Find the coordinates of the vertices by drawing the figure.
(3, 0), (–3, 0), (0, 4), (0, –4)

Question 31.
Find the coordinates of the vertices of a square of side 10 units whose diagonals meet at origin. [Score: 4, Time: 4 minutes]
Answer:

Question 32.
Consider a regular hexagon of side 6 units whose diagonals meet at the origin. Find the coordinates of the vertices. [Score: 5, Time: 8 minutes]

Answer:
(6, 0) (0, 6)
Find the coordinates of the vertices using the concept of right-angle triangles 30°, 60°, 90° (1)
(3, 3√3), (–3, 3√3), (3, –3√3), (–3, –3√3) (3)

Question 33.
Find the coordinates of other points.. [Score: 4, Time: 8 minutes]

Answer:
Points are (–2, 0), (0,2), (0, –2) (2)
Points are (3, –3), (–3, –3), (–3,3) (2)

Question 34.
Find the coordinates of the fourth vertices of the parallelogram. Find the length of the sides of the parallelogram. Write the length of diagonals. [Score: 5, Time: 8 minutes]

Answer:
The difference of x-coordinate of the points A, C = 2
The difference of x-coordinate of the points B, D = 2
Hence the x-coordinate of the point x = 11
The difference of y-coordinate of the points A, C = 7 – 5 = 2
The difference of y-coordinate of the points B, D = 2
Hence the y-coordinate of the point x = 10 + 2 = 12

Question 35.
Consider a rhombus of side 10 units whose diagonals are coordinate axis. If an angle is 120°, Find the coordinates of the other vertices. [Score: 4, Time: 6 minutes]
Answer:
A rhombus is formed by joining two equilateral triangles. Using the right triangle of 30°, 60°, 90°, find the diagonals as 10,10√3 units. Hence the coordinates of the vertices of the rhombus are(5, 0) (-5, 0)(0,5√3), (0,-5√3). (4)

Question 36.
Find the coordinates ofthe points on x-axis which are equidistant from the points (-5, 8) and (6, -4).[Score: 4,Time: 8 minutes]
Answer:
The y-coordinate of points on x-axis will be zero. Hence assume that a point on x-axis is (x, 0)
Distance between (x, 0) and (-5, 8) = \(\sqrt{(x-5)^{2}+8^{2}}\)
Distancebetween (x, 0) and (6, 4) = \(\sqrt{(x-6)^{2}+(-4)^{2}}\)
(x + 5)² + 8² = (x – 6)² + (–4)²
x² + 10x + 25 + 64 = x² – 12x + 36 + 16
22x = 36 + 16 – 25 – 64 = 52 – 89 = –37
x = \(\frac {–37 }{ 22 }\)
Coordinate \(\left(\frac{-37}{22}, 0\right)\) (4)

Question 37.
Prove that A (4,5), B (4,2), C(8,2) represents the vertices of a right angled triangle. Find the coordinate of the circumcentre. What is the radius of the circumcircle? [Score: 4, Time: 8 minutes]
Answer:
Distance between (4, 5) and (4, 2) = 5 – 2 = 3
Distance between (4, 2) and (8, 2) = 8 – 4 = 4
Distance between (4, 5) and (8, 2)

3, 4, 5 are sides of a right angled triangle,
since 3² + 4² = 5² (1)
The coordinates of circumcentre is (6 ,3, 5) (1)
Radius of circumcircle = 2.5 unit (1)

Question 38.
P is a point on the perpendicular bisector of the line joining (2, 5) and (6, 5). If the x coordinate and y coordinate of P are equal write the coordinate of P. [Score: 5, Time: 8 minutes]
Answer:
Understanding the concept that the distance of a point on the perpendicular bisector is equidistant from the endpoints. (1)
Assume that the coordinate of P is (a, a)
Then; Distance between (a, a) and (2, 5)

Coordinate of P (4, 4) (1)

Question 39.
Write the coordinate of the centre of a circle passing through the points (9, 3), (7,-1) (1,-1). Find the radius of the circle. [Score: 5, Time: 8 minutes]
Answer:
Assume that (x, y) is centre of the circle Distance between (9,3) and (x,y)

The coordinate of the centre of the circle is (4, 3)
∴ Radius = Distance between (4, 3) and (1,-1)
= \(\sqrt{(4-1)^{2}+(3+1)^{2}}=\sqrt{3^{2}+4^{2}}=5\) (1)

Question 40.
If (x, y) be apoint equidistant from the points (7, 5), (4, 3).then show that 6x + 4y = 49 [Score: 4, Time: 7 minutes]
Answer:
Distance between (x, y) and (7, 5)
\(\sqrt{(x-7)^{2}+(y-5)^{2}}\) (1)
Distancebetween (x, y) and (4, 3)
\(\sqrt{(x-4)^{2}+(y-3)^{2}}\) (1)
They are equal
(x – 7)² + (y – 5)² = (x – 4)² + (y-3)²
by solving
6x + 4y = 49 (2)

Question 41.
Three vertices of a square are given. If the fourth vertex is (2, P) find the ratio of P. Find the area of the square. [Score: 4, Time: 7 minutes]

Answer:

Question 42.
In the figure if P (36, 48), then find the coordinates of A, B, M [Score: 4, Time: 7 minutes]
Answer:

Coordinates of M (36, 0) (1)
Coordinates of A (0, 75) (1)
Coordinates of B (100, 0) (1)

Coordinates Exam Oriented Questions and Answers

Short Answer Type Questions (Score 2)

Question 43.

Find the coordinates of die other three verti¬ces of the rectangle in the figure.
Answer:

The coordinates of are (0, 0), ( 5, 0), (5, 4)

Question 44.
Check whether the points P(0, 7), Q(5, 12), R(-10, -3) lie on a single line.
Answer:

∴ The three points are on the same line.

Question 45.
The origin in the diagram is o. The endpoints of the diameter of the circle are A and B. Find the coordinates of P. OQ is perpendicular to OP. Find out the coordinates of Q.

Answer:

30: 60: 90 = 1: √3: 2
Coordinates of p is (√3, 1)
The Coordinates of Q = (–1, √3)

Question 46.
ABCDEF is a regular hexagon. Of these points C and D are on the number line. C is three unit to the left of zero and D is 5 unit to the right. What is the length of the side EF? What is the measure of an angle of the regular hexagon?
Answer:
CD = |5– –3| = 8
∴ CD = 8 unit
EF = 8 unit
Angle of a regular hexagon = 720 ÷ 6 = 120°

Question 47.
A circle is drawn with centre at (-1, 0) and radius 5 units in a coordinate system. What are the coordinates of the points at which it cuts the x-axis and the points where it cuts the y-axis?
Answer:

Question 48.
A circle is drawn with centre at (0, 0) and radius 6 units in a cooredinate system. What are the coordinates of the points which it cuts the x-axis? And the points where it cuts the y-axis?
Answer:
x axis (6,0) and (–6,0)
y axis (0, 6 ) and (0, –6)

Short Answer Type Questions (Score 3)

Question 49.
Find the length of the sides of the rectangle given in the figure.

Answer:

Question 50.
A small rectangle is cut from the centre of the big for rectangle. Based on the perpendiculars drawn at the centre of the big rectangle, find out the coordinates of the vertices of the small rectangle cut.

Answer:
P(–3, –2), Q(3, –2), R(3, 2), S(–3, 2)

Question 51.
Each side of the square OPQR given in the diagram is 6 unit. Write the coordinates of the vertices of the square.

Answer:
P(6, 0), Q(6, 6), R(0, 6), O(0, 0)

Question 52.
a. In the figure ABCD is a square. If A is (2, 0) then find the coordinates of B,C, D. Also find the coordinates of the centre.

b. Classify the points (5, 0), (0, –2), (–2, 0), (2, 3), (10, 9), (0, 8) and (6, 0) according to:
i. Points on the x-axis
ii. Points on the y-axis
iii. Does not belong to the axis
Answer:
a. B(0, 2), C(–2, 0), D(0, –2) Coordinate of vertex (0,0)

b, i. (5, 0), (–2, 0), (6, 0)
ii. (0, –2), (0, 8)
iii. (2, 3), (10, 9)

Question 53.
Selecting the given pair of coordinates given below, mark the opposite corners of a rectangle whose sides are parallel to the axis. Measure the lengths of the sides. Find out the coordinates of other points.
a (4, –4), (–4, 4)
b. (2, 1), (6, 5)
Answer:

Long Answer Type Questions (Score 4)

Question 54.
a. Write the coordinates of the points of intersection of the circle with the axis, where circle having radius 10cm and centre at the origin.
b. Write the co-ordinates of a point which are not on the circle.
Answer:
a. (10, 0), (0, 10); (–10, 0), (0, –10)
b. (9, 6); ie, all the point which satisfies.
x² + y²= 10² are on the circle, others are inside or outside the circle

Question 55.
Calculate the length of sides of quadrilateral.

Answer:

Question 56.
P is a point inside the rectangle ABCD. Prove that PA² + PC² = PB² + PD²

Answer:
Take AB as x axis and AD as y axis, a and b are the sides of the rectangle. The coordinates of the points are marked on the picture.
PA² + PC² = x² + y² + (x – a)² + (y – b)²
PB² + PD² = (x – a)² + y² + x² + (y – b)²
PA² + PC² = PB² + PD²

Long Answer Type Questions (Score 5)

Question 57.
In ΔABC, drawn above the x axis with ∠A=90° the coordinates of A (2,0) and that of B (8,0). The length of AC is 5 units and the area of the triangle is 12 square units. Draw a rough sketch of coordinate axis and the triangle. Find the coordinates of C.
Answer:
\(\frac { 1 }{ 2 }\) × AB × h = 12
\(\frac { 1 }{ 2 }\) × 6 × h = 12
h = 4 units
AC = 5
PC = 4 then
AP = 3
∴ OP = 5, PC = 4, C(5,4)

Question 58.
Find out the coordinates for the other vertices of the rectangle ABCD in the diagram. If the unit used to measure the length is 3/4cm, Find out the area of the rectangle ABCD.

Answer:
C(6, 2); D(2, 2)
Length of the rectangle = 4 unit,
Breadth = 2 unit
Unit= 3/4 cm,
Length = 4 × 3/4 = 3cm
Breadth = 2 × 3/4 = 3/2 = 1 1/2cm
Area = 3 × 1.5 = 4.5 cm²

Question 59.
a. From the points given below, find the pair which are on a line parallel to the x – axis and the pair which are on a line parallel to the y-axis.
A(4, 3), B(3, 5), C (–6, 3),
D(3, –2), E(5, 4)
b. Find the distance between the points given below
i. (–3, 2) and (4, 2)
ii. (5, 8) and (6, 9)
Answer:
a. Parallel to the x-axis A(4, 3) and C(–6, 3)Parallel to the y-axis B(3, 5) and D(3, –2)
b. i. Distance = |–3 –4| = 7

Coordinates Memory Map

Horizontal line is x-axis and vertical line is y-axis.

Point of intersection of the x- axis and y axis is called the origin O. Numbers which are above and to the right side of the origin O will be always positive while numbers on left and below the origin will be negative.

All points on the x – axis have y – coordinates zero and all points on y-axis have x- coordinates zero.

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SSLC Physics Chapter 1 Effects Of Electric Current Textbook Questions and Answers

SCERT Class 10th Standard Physics Chapter 1 Effects Of Electric Current Solutions

Textbook Page No. 7

SSLC Physics Chapter 1 Question 1.
Some electrical devices are shown in the house of the child. What are they?
Answer:

• Electric bulb
• Electric fan
• Mixi
• Induction cooker
• Microwave oven
• Storage batten,
• Inverter

Textbook Page No. 8

HssLive Physics Chapter 1 Question 2.
Write down the energy changes in them with respect to their use.

Answer:

Heating Effect of Electric Current Class 10 Question 3.
which are the devices that give heating effect of electric current?
Answer:

• Electric iron
• Electric stove
• Microwave oven
• Heating coil
• Induction cooker

Textbook Page No. 9

Heat Chapter Class 10 Questions and Answers Question 4.

→ How does the ni-chrome wire become red hot while passing electricity through the circuit?
Answer:
Due to the resistance of Ni-chrome wire.

→ In this case which form of energy was converted into heat energy?
Answer:
Electrical energy

→ How does this energy change occur?
Answer:
This based on the concept that energy can neither be created nor destroyed. It can only be converted from one form to another (Law of Conservation of Energy) As the resistance of Ni-chrome is higher it produced more heat.

Lighting Effect of Electric Current Question 5.

If the ammeter shows a current I ampere on applying a potential difference V across the resistor of resistance R Ω
Current \(I=\frac{Q}{t}\)
Then, the charge that flows through the conductor in t second,
Q = ……………coulomb.
Answer:
Q = I × t coulomb

Heat Chapter Class 10 Question 6.
factors influencing the heat developed when a current passes through a conductor.
Answer:

• Electric current
• Resistance of the conductor
• Time of current flow

Textbook Page No. 11

Sslc Physics Chapter 1 Questions And Answers Question 7.
Complete the following table on the basis of Joule’s Law. (Page no.11).

Answer:

Kerala Syllabus 10th Standard Physics Chapter 1 Question 8.
Analyse the table and find out the factor that influences heat the most.
Answer:
Intensity of current

Sslc Physics Chapter 1 Questions And Answers Pdf Question 9.
Experiment (Textbook Page no. 11)

→ Of the water in beakers A and B which one got heated more? Why?
Answer:
since nichrome has high resistance, more heat is produced. Hence temperature of water is increased.

→ What change is observed in the temperature of water in both the beakers when the current is increased using the rheostat?
Answer:
As the current increases, the heat produced is increases.

→ What was the change that happened to the temperature of water in both the beakers on increasing the time?
Answer:
Temperature is increased more

Textbook Page No. 12

10th Physics Chapter 1 Kerala Syllabus Question 10.
Heat generated = 2400 J
If 4.2 J is one calorie then H = ……… calorie
Answer:
H = 2400 J = \(\frac{2400}{4.2}\) = 571.42 calorie

Physics Chapter 1 Class 10 Kerala Syllabus Question 11.
Let’s find out the heat developed in 3 minutes by a device of resistance 920 Ω working under 230 V.
Answer:
V = 230V,
R = 920W,
t = 3 × 60 sec
I = \(\frac { V }{ R }\) = \(\frac { 230 }{ 920 }\) = 0.25
H = I²Rt = 0.25² × 920 × 3 × 60 = 10350 J

→ Is there any difference in the amount of heat energy thus obtained?
Answer:
No

Textbook Page No. 13

→ How this problem can be solved using the relation, H = VIt.
Answer:
V = 230 V,
R=920 W,
t = 3 × 60 sec
I = \(\frac { V }{ R }\) = \(\frac { 230 }{ 920 }\) = 0.25
H = VIt = 230 × 0.25 × 3 × 60 = 10350 J

Effects Of Electric Current Class 10 Kerala Syllabus Question 12.
Let’s calculate the heat developed when 3 A current flows through an electric iron box designed to work under 230 V.
Answer:
V=230 V,
I = 3 A
t = 30 m = 30 × 60 = 1800 s
H = VIt =230 × 3 × 1800 = 1242000 J = 1242 kJ

Sslc Physics Chapter 1 Notes Kerala Syllabus Question 13.
Table 1.3

→ Why does the heater having low resistance get heated more?
Answer:
Intensity of electric current is high.

→ In which way does the change in resistance influence the heat developed?
Answer:
As resistance increases heat decreases.

→ Find out the current in the heaters A and B and compare the heat developed.
Answer:
I = \(\frac{V}{R}\)
Heater A: I = \(\frac { 230 }{ 1150 }\) = 0.2
Heater B: I = \(\frac { 230 }{ 460 }\) = 0.5
In heater A and B, I increase H increases.

→ How do the resistors bring about a change in the current in the circuit?
Answer:
As resistance increases, I decrease.

Textbook Page No. 14

Physics 10th Class Chapter 1 Kerala Syllabus Question 14.
Consider the fig 1.4(a) and 1.4(b).

→ In which circuit does the bulb glow with high intensity?
Answer:
Circuit A [figure 1.4(a)]

→ Remove one bulb from each circuit. What do you observe?
In figl.4 (a)
In figl.4 (b)
Answer:
In fig 1.4 (a) : bulb glows
In figl .4 (a) : bulb does not glow

→ Why do the bulbs in Fig 1.4 (a) glow with maximum brightness?
Ans: Low resistance

Textbook Page No. 15

10 Physics Chapter 1 Kerala Syllabus Question 15.
Table 1.4.

Answer:

Physics Class 10 Chapter 1 Kerala Syllabus Question 16.
Analyse the table and tick (✓) the best suited

Answer:

Textbook Page No. 17

Sslc Physics Chapter 1 Questions Kerala Syllabus Question 17.
Complete Table 1.6 by analysing Tables 1.4 and 1.5.

Answer:

Textbook Page No. 18

Class 10 Physics Chapter 1 Kerala Syllabus Question 18.
10 resistors of 2 Ω each are connected in parallel. Calculate the effective resistance.
Answer:
R = \(\frac { 2Ω }{ 10 }\) = 0.2 Ω

Textbook Page No. 19

Sslc Physics Chapter 1 Pdf Kerala Syllabus Question 19.
A few heating appliances are shown in the figure(1.8). Examine any one of them and answer the following questions. Record the answers in the science diary.

→ Name the part in which electrical energy changes into heat energy.
Answer:
Heatingcoil

→ Which material is used to make this part?
Answer:
Nichrome:
Heating coils are made of nichrome. It is an alloy of nickel, chromium and iron.

→ What are the peculiarities of such substances?
Answer:

• High resistivity
• High melting points
• Ability to remain in red hot condition for a long time without getting oxidized.
• Thermal expansion is negligible.

Textbook Page No. 20

Physics Class 10 Chapter 1 Kerala Syllabus Question 20.
Which are the circumstances that cause high electric current, leading to the melting of fuse wire?
Answer:
Overloading or short circuit

10th Standard Physics Chapter 1 Kerala Syllabus Question 21.
How is the fuse wire connected to a circuit? In series/ parallel?
Answer:
in series

10th Physics 1st Chapter Kerala Syllabus Question 22.
You know that according to Joule’s Law, more heat will be produced when electric current is increased. What happens to the fuse wire due to this?
Answer:
Fuse wire melts

Class 10 Physics Chapter 1 Kerala Syllabus Question 23.
When heat is generated, why does the fuse wire melt?
Answer:
The melting point of fuse wire is lower than that of other metals

10th Physics Chapter 1 Notes Kerala Syllabus Question 24.
When the fuse wire melts, the circuit is broken. What happens to the current in the circuit?
Answer:
The flow of current in the circuit stops simultaneously with the melting of fuse wire.

Class 10 Physics Chapter 1 Question Answer Kerala Syllabus Question 25.
Why is the? fuse used in a circuit called safety fuse? Explain.
Answer:
The current in the circuit may increase due to reasons such as short circuit, overload, excess flow of current or any problems in the insulation. As the higher temperature produced in the circuit due to these reasons makes the fuse wire to melt and the flow of current stops. Thus the circuit and the appliances are protected.

Kerala Syllabus 10th Physics Chapter 1 Kerala Syllabus Question 26.
When a fuse wire is included in a household wiring. What are the precautions to be taken?
Answer:

• The ends of the fuse wire must be connected firmly at appropriate points.
• The fuse wire should not project out of the carrier base.
• Use the fuse wire of proper amperage.

Physics Class 10 Chapter 1 Notes Kerala Syllabus  Question 27.
You might have noticed the marking of 500 W on an electrical appliance. What does it indicate?
Answer:
It indicates the power of the instrument. The amount of energy consumed by an electrical appliance in unit time is its power.

10th Class Physics Chapter 1 Kerala Syllabus Question 28.
What is the unit of power?
Answer:
Watt (W)

Textbook Page No. 22

Effect Of Electric Current Class 10 Questions And Answers Question 29.
If R= V/I What will be P ?
P = IR = I × ……… = ……….
Answer:
P = I²R = I² × \(\frac { V }{ I }\) = VI

10th Standard Physics Kerala Syllabus Question 30.
A current of 0.4 A flows through an electric bulb working at 230 V. What is the power of the bulb?
Answer:
V = 230 V
I = 4 A
P = VI
P = 230 × 4 = 92 W

Textbook Page No. 23

Question 31.
What happens if the interior of the bulb is not evacuated?
Answer:
The interior of the bulb is evacuated to pr-event the oxidation of filament. When the tungsten comes in contact with the air in the heated condition
then it undergoes oxidation.

Question 32.
Why is the bulb filled with an inert gas or nitrogen?
Answer:
To prevent the vaporization of filament.

Question 33.
What are the advantages of using tungsten as a filament?
Answer:

• High resistivity
• High melting point
• High ductility
• Ability to emit white light in the white-hot conditions

Textbook Page No. 24

Question 34.
Nichrome is not used as filament in incandescent lamps. Why?
Answer:
It can remain only in red hot condition but it can’t give light.

Question 35.
A filament lamp is lit for a short period of time. Touch it. What do you feel?
Answer:
Heat is experienced.

Question 36.
What are the other types of lamps working on electricity? List them.
Answer:

• Discharge lamp
• Fluorescent lamp
• Arc lamp
• CFL
• LED

Textbook Page No. 25

Question 37.
What are the advantages of using discharge lamps instead of incandescent lamps?
Answer:
Loss of electricity in the form of heat is less, more life span, no shadow formation, more light is obtained, less consumption of electricity.

Question 38.
What are the factors to be considered when you select a bulb?
Answer:
Efficiency, energy consumption, low energy loss, less environmental pollution.

Question 39.
Which are the lamps that are mostly used? Why?
Answer:
LED: Low energy consumption, low energy 1 loss, less environmental pollution.

Question 40.
LED.
Answer:

• LED’s are Light Emitting Diodes.
• As there is no filament, there is no loss j of energy in the form of heat.
• Since there is no mercury in it, it is not harmful to environment
• Very small
• It requires only small amount of power.
• No filaments

Effects of Electric Current Let Us Assess

Question 1.
Fuse wire is to be used by understanding the amperage correctly. Write down the amperage of the fuse wires that are currently available in the market.
Answer:
Fuse wire of suitable amperage should be selected. Amperage is the ratio of the power of an equipment to the voltage applied. Amperage increases with the thickness of the conductor. Thick wire takes more time to melt. Due to high current flow, the circuit may be damaged. If thin wire is used resistance is increased and hence current is reduced.
Amperage of available fuse : 0.1 A, 0.2 A, 0.5 A, 1.5 A, 3 A, 5 A.

Question 2.
0.5 A current flows through an electric heating device connected to 230 V supply.
a. the quantity of charge that flows through the circuit in 5 minutes is
i. 5 C
ii. 15 C
iii. 150 C
iv. 1500 C
b. How much is the resistance of the circuit?
c. Calculate the quantity of heat generated when current flows in the circuit for 5 minutes.
d. How much is the power of the heating device connected to the circuit if we ignore the resistance of the circuit wire?
Answer:

Question 3.
According to Joule’s Law the heat generated due to the flow of current is H = I² Rt. Will the heat developed increase on increasing the resistance without changing the voltage? Explain.
Answer:
Heat decreases.
Reason H α \(\frac { 1 }{ R }\)

Question 4.
The table shows details of an electric heating device designed to work in 230 V. Complete the table by calculating the change in the heat and power on changing the voltage and resistance of the device. Analyze the table and answer the following questions.

a How does the voltage under which a device works affect its functioning?
b.What change happens to power on increasing the resistance without changing the voltage?
c. What change is to be brought about in the construction of household heating devices in order to increase their power?
Answer:
a. As voltage increases, power increases
b. Power decreases
c. Use thick wire, use conductor of suitable material, use conductor of small length.

Question 5.
a. Complete the table based on the amperage of the fuse wire.

b. The amperage of the fuse wire used in a circuit that works on 230 V is 2.2 A. If so the power of the device is
i. less than 300 W
ii. 300 W to 500 W
iii. between 500 Wand 510 W
iv. more than 510 W
Answer:
iii. between 500 W and 510 W

Question 6.
A 230 V, 115 W filament lamp works in a circuit for 10 minutes,
a. What is the current flowing through the bulb?
b. How much is the quantity of charge that flows through the bulb in 10 minutes?
Answer:
a. I = \(\frac { P }{ V }\) = \(\frac { 115 }{ 230 }\) = 0.5 A

b.  Q = I × t = 0.5 × 10 × 60 = 300 c
Question 7.
An electric heater conducts 4 A current when 60 V is applied across its terminals. What will be the current if the potential difference is 120 V?
Answer:
R = \(\frac{V}{I}=\frac{60}{4}=15 Ω
If the voltage is changed to 120 V
I = [latex]\frac{V}{R}=\frac{120}{15}\) = 8 A

Question 8.
Three resistors of 2 Ω,3 Ω and 6 Ω are given in the class.
a. What is the highest resistance that you can get using all of them?
b. What is the least resistance that you can get using all of them?
c. Can you make a resistance 4.5 Ω using these three? Draw the circuit.
Answer:
a. Highest resistance
R = R + R + R = 2 + 3 + 6 = 11 Ω

b. Least resistance

c. Yes

Question 9.
A girl has many resistors of 2 Q each. She needs a circuit of 9 Q resistance. For this draw a circuit with the minimum number of resistors.
Answer:

Question 10.

If a bulb is lit after rejoining the parts of a broken filament, what change will occur in the intensity of the light from the lamp? What will be the change in the power of the bulb?
Answer:

• Intensity of bulb
• Power increases.

Question 11.
Which of the following does not indicate the power of a circuit?
a. I²8
b. VI
c. 1R²
d. V²/R
Answer:
c. 1R²

Question 12.
How much will be the power of a 220 V, 100 W electric bulb working at 110 V?
a. 100 W
b. 75 W
c. 50 W
d. 25 W
Answer:
d. 25 W

Question 13.
Which of the following should be connected in parallel to a device in a circuit?
a. voltmeter
b. ammeter
c. galvanometer
Answer:
a. voltmeter

Question 14.
When a 12 V battery is connected to resistor, 2.5 mA current flows through the circuit. If so what is the resistance of the resistor?
Answer:

Question 15.
If 0.2Ω, 0.3Ω, 0.4 Ω, 0.5 Ω, and 12Ω resistors are connected to a 9 V battery in parallel, what will be the current through the 12 Ω resistor?
Answer:
I = \(\frac{V}{R}=\frac{9}{12}\) = 0.75 A

Question 16.
How many resistors of 176 Ω should be connected in parallel to get 5A current from 220 V supply?
a. 2
b. 3
c. 6
d. 4
Answer:
4

Question 17.
Depict a figure showing the arrangement of three resistors in a circuit to get an effective resistance of
(i) 9 Ω
(ii) 4 Ω
Answer:
i.

ii.

Effects of Electric Current Extended Activities

Question 1.
Analyse and describe the working of a microwave oven.
Answer:
Microwave oven is a device which is used for heating effect of electric current. It produces heat without a heating coil. Eddy current is used in the working of microwave oven. The heat is generated as a result of microwave radiations. The water molecules are stimulated using ‘magnet one’ and thus attains high temperature. Metal utensils are avoided and the substances without water content are not heated by this.

Question 2.
How does an arc lamp help in rescue operations?
Answer:
The intense light of arc lamps are used in search lights and rescue operations during night time. It is used to rescue the victims of natural calamities, boat accidents.

Question 3.
With the help of teachers and the Internet find out the following
a. What is the percentage of nickel, chromium and iron in Nichrome?
b. How much is the melting point of nichrome in degree Celsius?
c. How much is the resistivity of Nichrome?
d Does the result of your observation justify the use of nichrome as a heating element?
Answer:
a. 61 % Ni, 15% Cr, 24% Fe
b. 1350°C
c. (1.0 – 1.5)x 10 – 6 Ωm
d. 1. High resistivity
2. High melting points
3. Ability to remain in red hot condition for a long time without getting oxidized. :
4. Thermal expansion is negligible.

Question 4.
Analyse the merits and demerits of the following lamps and find out which is best in the group. Justify your answers.
a. filament lamp
b. fluorescent lamp
c. arc lamp dCFL
e. LED bulb
Answer:
a. Filament lamp:
Demerits: Loss of energy as heat, low light, low light.
Merits: Work in both DC and AC.

b. Fluorescent lamp:
Demerits: Cause environmental pollution,
Merit: Produce more light

c. Arc lamp:
Demerits: Carbon rods are to be changed frequently.
Merit: Intense light is produced.

d. CFL:
Merit: Produce more light at low power.
Demerits: Cause environmental pollution,

e. LED Bulb :
Merit: No environmental pollution, low energy consumption, long life. The best lamp is LED.

Effects of Electric Current Orukkam Questions and Answers

Question 1.
Complete the table based on the effects of electric current and energy change.

Answer:
a. Light energy
b. Mechanical effect
c. Heat energy
d. Chemical effect

→ Given below questions are related with heating of an electric iron box. Answer them.
a. Which is the part that produces heat in an electric iron?
b. Which nature of this part is made use in the above situation?
c. What is the relation between intensity of electric current and heat energy generated?
d. What are the factors that affect the heat generated in such heating appliances?
e. What is the relation connecting these factors with the heat generated?
f. What is this law known as?
g. Name a device that works on this law used for ensuring safety in electric circuits?
Answer:
a. Heating coil
b. High resistivity, High melting point
c. As electric current increases, Heat energy increases.
d. Electric current, Resistance, time
e. H = I² Rt
f. Joules law
g. Safety fuse

Question 2.
Incandescent lamp, discharge lamp, C.F.L, LED lamp, arc lamp are given for observation (Otherwise, make use of their pictures)
Answer the following questions after observing them.
a. Which metal is used to make filament of an incandescent lamp? What are the advantages of using this metal as a filament?
b. Name the lamps which belong to the group of discharge lamp.
c. The color of the light depends on the gas inside the discharge lamp. Which gases are to be filled for getting white light and yellow light?
d. Which lamps are harmful to environment because of the presence of mercury in it?
e. Which lamp is used in rescue work during night time and used in searchlights?
Answer:
a. Tungsten. Advantages of tungsten are high ductility, high resistivity, high melting point
b. Fluorescent lamp, CFL, LED, Arc lamp.
c. For white light – Mercury
For yellow light- Sodium Vapour
d. Fluorescent lamps
e. Arc Lamp

Effects of Electric Current SCERT Questions and Answers

Question 1.
Observe the circuit diagram given below and answer the following questions.

a. Which are the instruments labeled as P and Q in the diagram?
b. If you replace the copper wire AB with a Nichrome wire of same length and area of cross-section.
i. What change would you notice in the reading on the device Q? Why?
ii. What will happen to the heat produced in the conductor? Explain with reference to Joule’s Law.
Answer:
a. P- Rheostat.
Q- Ammeter.

b. i. Reading will decrease. Due to higher resistance current decreases
ii. Because of Nichrome’s high resistivity, the current in the circuit will decrease. According to Joules law H= I² Rt, decrease in the amount of current will reduce the amount of heat.

Question 2.
Write down the names of four types of lamps working on the lighting effects of electricity.
Answer:
Discharge lamp, Fluorescent lamp, LED, Arc lamp.

Question 3.
Two types of lamps are given below.
1. Discharge lamp
2. Filament lamp.
a. If nitrogen gas is filled in each lamp, what change will happen to their working?
b. Why is it said that the use of filament lamp must be controlled?
Answer:
a. Nitrogen filled discharge lamp will give red light, but in a filament lamp nitrogen is filled to reduce the evaporation of the filament.

b. In a filament lamp, major part of the electrical energy we supply is converted into heat energy.

Question 4.

a. Identify the device shown in the picture.
b. How are such devices used in rescue ope-rations?
Answer:
a. Arc lamp

b. Light intensity in an Arc lamp is very high compared to other lamps, so Arc lamps are helpful in rescue operations even in adverse climatic conditions.

Question 5.
Observe the diagram and answer the questions below.

a. Which bulbs will glow when S₁ switched on?
b. Which bulbs will low when S₁ and S₁ are switched on?
c. What change will happen to the circuit when S₃ is switched on?
d. Calculate the amperage of the fuse to be used in the circuit,
e. Describe short circuit and overloading with the help of the given circuit
Answer:
a. B₁
b. Bulb will not glow, (fuse burns out due to overloading)
c. The circuit breaks as fuse burns out due to short circuit.
d. Amperage
\(\frac{200}{100}=2 A(\text { more than } 2 \mathrm{A})\)
e. Short circuit happens when two wires in the mains in contact without the presen-ce of a resistance in between. Over loading is connecting appliances with more power in the circuit, that it can bear.

Question 6.
A 800W electrical device is designed to work in 200V. What will be its power if the device is working in 100 V?
Answer:
Resistance of the appliance

Question 7.
Bulbs marked 200V and 500W are shown in the picture.

a. Calculate the resistance of each bulb in the circuit.
b. What is the power with which the bulb in circuit 1 glows?
c. What is the power of the bulb in circuit 2? Why?
d. How is the power and resistance of an electrical device related to each other when the voltage is the same?
Answer:
Resistance of B₁ = \(\frac { V² }{ P }\) = \(\frac { 200 × 200 }{ 50 }\) = 800 Ω
Resistance of B₂=800 Ω

b. Will work in 50 W

c. Since the 50 W bulbs are in series current will be equal and voltage will behalf. Since P=VI, Power also will become half.

d. P = \(\frac { V² }{ R }\) (No change in resistance, Power will change)

Question 8.
Electric bulbs are connected in a 240V supply line are shown in the figure.

a. What is the total wattage of appliances used in the circuit?
b. Calculate the amperage of the fuse to be used in the circuit.
Answer:
a. P = 20 W+ 20 W + 20 W = 60W

Question 9.
Given below are the steps in the working of a fluorescent lamp. Arrange them in the proper order.
a. Ultraviolet rays are produced
b. Visible light is emitted.
c. Fastly moving electrons collide with the unionized molecules of mercury.
d. Due to electric current thorium oxide coated heating element becomes red hot.
Answer:
(d),
(c),
(a),
(b)

Question 10.
Correct the mistakes, if any:
a. Amperage decreases in proportion to the decrease in the area of cross-section of the conductor.
b. Connecting appliances in a circuit beyond its power capacity is short circuit.
c. It is to reduce the heat loss that electric lamps are filled with inert gases.
Answer:
b. Overloading is connecting appliances with more power in the circuit, than it can bear.
c. Inert gases are filled in filament lamps to reduce the rate of evaporation.

Question 11.
Power of an electric heater working in 230 V is 1000 W. Calculate the heat produced if the current passes for 5 minutes through the circuit.
Answer:
Heat = P × t
= 1000 × 5 × 60
= 300000 J

Question 12.
An electric bulb has marking 110 V, 100 W on it.
a. How much energy is used per second by the circuit?
b. What is the resistance of electric bulb?
Answer:

Question 13.
Fill up the blanks by finding the suitable relationship.

Answer:
A= Fuse wire
B= High melting point

Question 14.
If the resistance of two soldering irons working in 250 V are 500 Ω and 750 Ω.
a. Calculate which one of these will carry more current.
b. Find out which soldering iron has more power.
c. Calculate the heat produced in 5 minutes in the soldering iron having resistance 750 Ω.
Answer:

b. Soldering iron has less resistance

Question 15.
Two bulbs of 500 W and 100 W are connected parallel in a circuit of 250 V.
a. Which bulb will have more brightness?
b. Through which bulb is current greater?
c. Find out the resistance of filaments in each bulb.
d. Which of these two bulbs will glow with more brightness if they are connected in series? Explain the reason.
Answer:
a. 500 W Bulb

b. Bulb which has power 500 W (P = VI).

c. 500 W bulb
R = \(\frac { V² }{ P }\) = \(\frac { 250 × 250 }{ 500 }\) = 125 Ω
100 W Bulb
R = \(\frac { V² }{ P }\) = \(\frac { 100 × 100 }{ 250 }\) = 40 Ω

d. 100 W bulb — Current will be the same in series, High resistance bulb will shine more brightly.

Question 16.
Fill in the blanks suitably using the relationship in the first pair,
a. Filament: High melting point
Fuse : ………………
b. Nitrogen filled discharge lamp : Red
………………………… : Blue
Answer:
a. Low melting point

b. Hydrogen used in discharge lamp

Question 17.
Choose and write the facts related to LED lamps from the given list.
a. Require only a small quantity of power.
b. UV rays are produced due to electric discharge.
c. Not harmful to environment since there is no mercury.
d. Intense light is produced when high voltage is applied.
Answer:
a, c

Effects of Electric Current Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 1.
Using the relation from the first pair, complete the other.
Filament: Tungsten Heating coil:
Answer:
Nichrome

Question 2.
Which of the given material has the highest resistivity?
a. Nichrome
b.Copper
c. Aluminium
d. Silver
Answer:
Nichrome

Question 3.
What is the color of the light emitted by discharge lamp filled with nitrogen when it works ?
Answer:
Red

Question 4.
Using the relation from the first pair, complete the other.
Fuse wire : low melting point
Tungsten: …………..
Answer:
High melting point

Question 5.
Which among the following is the special characteristics of a material to be used as a heating coil in a electric heating device?
a. Low melting point
b. High resistivity
c. Large area of cross section
d. Low resistance
Answer:
High resistivity

Question 6.
Which material is used as filament of bulb?
Answer:
Tungsten

Question 7.
Find out the odd one from the following also write the reason.
[long glass tube, Mercury vapor, Fluorescent coating, Thin tungsten filament]
Answer:
Thin tungsten filament. Others are parts of a fluorescent lamp.

Question 8.
If we apply 230 V for a device having 230 V and 400 W, what will happen the power of the device ?
(increases, decreases, doesn’t change)
Answer:
decreases

Question 9.
In incandescent lamp nichrome is not used as filament. Choose the reason from the following
a. Glows brightly
b. It emit white light when it is in hot condition. ,
c. It can remain only in red hot condition but it can’t give light.
Answer:
c

Question 10.
Find out the odd one which is not suitable for the advantages of LED.
a. It requires only a small quantity of power.
b. It is not harmful to environment.
c. Ultraviolet fays are formed due to discharge.
Answer:
c

Short Answer Type Questions (Score 2)

Question 11.
A 400 W electrical device is designed to work in 100 V. What will be its power if the device is working in 100 V?
Answer:
Resistance of the appliance
R = \(\frac { V² }{ P }\) = \(\frac { 100 × 100 }{ 400 }\) = 25 Ω
Power when connected in 100 V
P = \(\frac { V² }{ R }\) = \(\frac { 100 × 100 }{ 25 }\) = 400 Ω
Answer:
b,
d,

Question 12.
An electric bulb marked 500 W, 250 V is connected to a 250 V supply.
a. Find the electric current through the circuit.
b. Calculate the resistance of the bulb.
Answer:

Question 13.
What are the advantages of fluorescent lamps over incandescent lamps.
Answer:

1. Saves electrical energy to a greater extent.
2. The inconvenience caused by shadow is minimized.
3. The life of fluorescent lamp is about 5 times that of incandescent lamps.

Question 14.
Complete the table suitably.

Answer:
a. Blue
b. Neon
c. Red
d. Chlorine

Short Answer Type Questions (Score 3)

Question 15.
In figure, Beakers A and B contain 100 ml water. PQ is a nichrome wire and RS is a copper wire of same length and diameter.

a. Water in which beaker will be at higher temperatures? What is the reason?
b. If the current is doubled using the rheostat, what happens to the quantity of heat produced in the wire PQ?
Answer:
a. Beaker A, since the combination is series, the current remain same. As nichrome has high resistance, more heat is produced in it. (H = I² Rt)

b. Heat produced becomes 4 times.
Heat H = I² Rt, Heat increases as current is squared.

Question 16.
Classify the following as those suitable for fluorescent lamps and for arc lamps.
a. The main part is carbon rods.
b. The heating coil is coated with thorium oxide.
c. used in searchlights.
d. more harmful for the environment.
e. more intense light.
f. ultraviolet rays are produced.
Answer:
Arc lamps: a, c, e
Fluorescent lamps : b, d, f

Question 17.
Find out the relation and complete the missing parts.
i. tungsten: ………….. (A) ……………. : high melting point
ii. alloy of suitable : fuse wire: ……….(B)……….. metals
iii (C)…….. heating coils: high melting point
Answer:
A. Filament
B. Low melting point
C. Nichrome

Question 18.
The filament of a filament lamp is broken. It is rejoined and is lit again.
a. What happens to the intensity of light from it? Describe the reason behind.
b. Incandescent lamps are filled with nitrogen at low pressure. What is the advantage of doing so?
c. You are given two filament lamps of resistance 75012 and 100012. Of these, which has more power?
Answer:
a. Increases. Because resistance decreases and current increases when length decreasing.

b. Nitrogen doesn’t expand over less change in temperature. At normal temperature it behave as a inert gas. It is plenty in atmosphere.

c. 750 Ω

Questions 19.
Choose the appropriate items from the box.
Nichrome, Tungsten, Fuse Wire, Nitrogen
a. Which is used as heating coil ?
b. Which is an alloy of tin and lead?
c. Which is used as filament ?
Answer:
a. Nichrome
b. Fuse wire
c. Tungsten

Short Answer Type Questions (Score 4)

Question 20.
a. What is meant by amperage ?
b. An appliance of power 690 W is used in a branch circuit. If the voltage is 230 V,what is its amperage?
Answer:
a. Amperage is the ratio of the power of an equipment to the voltage applied.
b. A = \(\frac { P }{ V }\) = \(\frac { 690 }{ 230 }\) = 3 A

Question 21.
Analyse the table and fill it suitably.

Answer:
(1) 9 times,
(2) Series,
(3) 1/16 times

Question 22.
An electric heater working at 230 V produces 1000 J energy in one second.
a. What is the power of the lamp ?
b. Calculate the resistance of the heating coil used in it.
c. Calculate the heat generated by it when it works for 5 minute.
Answer:

Question 23.
Write down the reasons for the following statements.
a. Don’t throw unwanted fluorescent tubes.
b. Nichrome is not used as filament in incandescent lamps.
c. Fuse wire is melt when electric current increased.
d. Why is the incandescent filled with an inert gas or nitrogen?
Answer:
a. It contains mercury which causes several environmental problems.
b. It can remain only in red hot condition but it can’t give light.
c. When current flows as the higher temperature produced in the circuit. Fuse wire has low melting point compared to other, so it melts.
d. To prevent the vaporization of filament.

Memory Map:

You can Download Tangents Questions and Answers, Activity, Notes PDF, Kerala Syllabus 10th Standard Maths Solutions Chapter 7 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Maths Chapter 7 Tangents Textbook Questions and Answers

SCERT Class 10th Standard Maths Chapter 7 Tangents Notes

Textbook Page No. 163

Tangents Class 10 Chapter 7 Kerala Syllabus Tangents Class 10 Kerala Syllabus Question 1.
In each of the two pictures below, a triangle is formed by a tangent to a circle, the radius through the point of contact and a line through the center:


Draw these in your notebook.
Answer:
Draw a circle with radius 2.5cm and center as O. Mark a point P at a distance 5cm from O. Taking OP as the diameter, draw a circle.

Mark A at the point where the two circles intersect each other. Join PA.
Radius of the circle = \(\sqrt{4^{2}-2^{2}}\)
= \(\sqrt{16-4}\) = √12 = 2√3 cm
= 2 × 1.73 = 3.46 cm ~ 3.5 cm
Draw a circle with radius 3.5 cm. OA is the radius. Draw AP = 2 cm, perpendicular to OA. Join O and P, such that PA is the tangent

Draw this picture in your notebook.

Answer:
Draw a circle of radius 4 cm and center O. Sdesof a rhombus are the tangents of the circa Let the angle between the tangents be 40°, then the center angle of arc between them = 180 – 40 = 140°.
Draw a rhombus by given measures.

Sslc Maths Chapter 7 Kerala Syllabus Question 3.
Prove that the tangents drawn to a circle at the two ends of a diameter are parallel.
Answer:

A tangent through P is perpendicular to PQ, PQ ⊥RP. A tangent through Q is perpendicular to PQ, PQ ⊥ SQ ∴∠SQP = ZRPQ = 90°. But they are co-interior angles. RP || SQ This means that RP is parallel to SQ. Hence proved.

Sslc Maths Chapter Tangents Kerala Syllabus Chapter 7 Question 4.
What sort of a quadrilateral is formed by the tangents at the ends of two perpendicular diameters of a circle?
Answer:
Let centre of the circle be O. Draw 2 mutually perpendicular diameters, PQ and SR.
Tangents drawn from a point outside to circle have equal length.

AP = AS, PD = PR, CQ = CR, QB = BS
AD + BC = AP + PD + BQ + CQ = AS + DR + SB + CR
= AS + SB + DR + CR
= AB + CD
Opposite sides are equal.
Hence ABCD is a square.

Textbook Page No. 166

Class 10 Maths Tangents Kerala Syllabus Chapter 7 Question 1.
Draw a circle of radius 2.5 centimeters. Draw a triangle of angles 40°, 60°, 80° with all its sides touching the circle.
Answer:
First, draw a circle with radius 2.5cm. PCOB is a quadrilateral, ∠COB = 360 – (90 + 90 + 40) = 140°. Divide the circle into three as 100°, 120°, 140°. Draw an arc equal distance from A and B find the point R. Similarly, find P and Q. Complete the figure,

Class 10 Tangents Kerala Syllabus Chapter 7 Question 2.
In the picture, the small (blue) triangle is equilateral. The sides of the large(red) triangle are tangents to the circumcircle of the small triangle at its vertices.
i. Prove that the large triangle is also equilateral and its sides are double those of the small triangle.
ii. Draw this picture, with sides of the smaller triangle 3 centimeters.

Answer:
i. Let O be the center of the radius
In ΔAOB
OA = OB (Radius)
∴ ∠OAB = ZOBA =30°
∴ ∠AOB = 120° ⇒ ∠APB = 60°
Similarly
∠AOC = 120° ⇒ ∠ARC = 60°
∠BOC = 120° ⇒ ∠CQB = 60°
Angle in the A PQR 60° each.
ΔPQRis an equilateral triangle.
In Δ APB, AP = PB
∠APB = 60°
∠PAB = ∠PBA = 60°
∴ Δ PAB is a equilateral triangle.
PA = PB=AB Similarly
AR = RC = AC and CQ = BQ =BC
Δ ABC is a equilateral triangle. Hence
PR = PA + AR = AB + AC = 2 AC
∴ The large triangle is also equilateral and its sides are double that of the small triangle.

Tangent Class 10 Solutions Kerala Syllabus Chapter 7 Question 3.
The picture shows the tangents at two points on a circle and the radii through the points of contact

i. Prove that the tangents have the same length.
ii. Prove that the line joining the center and the point where the tangents meet bisects the angle between the radii.

iii. Prove that this line is the perpendicular bisector of the chord joining the points of contact.

Answer:
i. Δ OAP, Δ OBP are similar triangles.
In.ΔOAP
OP²=OA²+AP²
AP²=OP^{2 –} OA²

Length of the tangents are equal,

ii. . Consider Δ AOP, ΔBOP
PA= PB
(tangents)
OA = OB (radii)
PO = PO (common side)
∴ ΔAO ~ ΔBOP (Since all the sides are same)
Hence triangles are similar.
∴ ∠OPA = ∠OPB
OP bisects ∠ APB

iii. As ΔOAP ~ ΔBOP
∠POA = ∠POB
Considering
ΔAOM, ΔBOM
OA =OB(radii)
OM = OM (common side)
∠AOM = ∠BOM
ΔAOM ~ ΔBOM (By SAS property).
∴ AM = BM, OP bisects AB

Sslc Maths Chapter 7 Notes Kerala Syllabus Question 4.
Prove that the quadrilateral with sides as the tangents at the ends of a pair of perpendicular chords of a circle is cyclic.
What sort of a quadrilateral do we get if one chord is a diameter? And if both chords are diameters?

Answer:
LetAB, AD be tangents
∠PON + ∠PCN = 180°
∠PON = ∠MOQ
∠MAQ + ∠PCN = 180°
(PQ ⊥ MN)
∠MOQ = 90°
∠MAQ = 180° – 90° = 90°,
Similarly
∠NCP = 90°, ∠AMQ || ∠NCP = 180°,
∠A+ ∠ C = 180°
ABCD is a cyclic quadrilateral. If one chord is the diameter then the quadrilateral is a trapezium. If two chords are diameters then the quadrilateral is a square.

Textbook Page No. 172

Tangents 10th Class Kerala Syllabus Chapter 7 Question 1.
In the picture, the sides of the large triangle are tangents to the circumcircle of the small triangle, through its vertices.

Calculate the angles of the large triangle
Answer:
In a circle, the angle which a chord makes with the tangents at its ends on any side are equal to the angle which it makes on the part of the circle on the other side.

In Δ ABC, ∠B = 80°, ∠C = 60°
∠R = 180 – (80 + 60) = 40
∠CAR = 80°= ∠ACR
∠QBC = ∠QCB = 40°
∠P = 60°, ∠Q = 100°
∠NCP = 20°
Then the angles of bigger triangle are 60°, 100°, 20°.

Sslc Tangent Questions And Answers Kerala Syllabus Chapter 7 Question 2.
In the picture, the sides of the large triangle are tangents of the circumcircle of the smaller triangle, through its vertices.

Calculate the angles of the smaller triangle
Answer:

∠ACB =180-(60+ 40) = 80°
In a circle, the angle which a chord makes with the tangents at its ends on any side are equal to the angle which it makes on the part of the circle on the other side.
AP = AR ∴ ∠APR = ∠ARP = 60°
CR = CQ ∴ ∠CRQ =∠CQR = 50°
PB = BQ ∴ ∠BPQ = ∠BQP = 70°
Angles of the smaller triangle are 50°, 60°, 70°

Hss Live Maths 10th Kerala Syllabus Chapter 7 Question 3.
In the picture, PQ, RS, TU are tangents to the circumcircle of ΔABC.
Sort out the equal angles in the picture

Answer:
∠QAB = ∠ACB = ∠ABS
∠RBC = ∠BAC = ∠BCU
∠PAC = ∠ABC = ∠TCA

10th Maths Tangents Kerala Syllabus Chapter 7 Question 4.
In the picture, the tangent the circumcle of a regular pentagon through a vertex is shown.
calculate the angle which the tangent makes with the two sides of the pentagon through the point of contact.

Answer:
Sides of the pentagon are equal.


Textbook Page No. 179

Tangent Class 10 Kerala Syllabus Chapter 7 Question 1.
In the picture, a triangle is formed by two mutually perpendicular tangents to a circle and a third tangent. Prove that the perimeter of the triangle is equal to the diameter of the circle.

Answer:

O is the center and OS perpendicular to PS
(Radius is perpendicular to tangent)
OQ perpendicular PQ
(Radius is perpendicular to tangent)
OS = PS = PQ = OQ
(PQRS is a square)
PS= PA + AS
AS = AM (Tangents from A is equal) PS= PB + BQ
Hence BM= BQ (Tangents from B is equal)
PQ = PB + BM
Therefore
PS+ PQ= PA+AM + PB + BM
= PA+PB+AM + BM
= PA+PB+AB = Perimeter of right angled triangle PS + PQ is diameter of the circle.
So, the perimeter of the triangle is equal to the diameter of the circle The picture shows a

Tangent Questions And Answers Kerala Syllabus Chapter 7  Question 2.
The picture shows a triangle formed by three tangents to a circle.

Calculate the length of each tangent from the corner of the triangle to the point of contact
Answer:

Sslc Maths Tangents Model Questions Kerala Syllabus Chapter 7 Question 3.
In the picture, two circles touch at a point and the common tangent at this point is drawn

i. Prove that this tangent bisects! another common tangent of these circles.

ii. Prove that the points of contact of these two tangents from the vertices of a right triangle.

iii. Draw the picture on the right in your notebook, using convenient lengths. What is special about the quadrilateral formed by joining the points of contact of the circles?

Answer:
i.

PB = PQ and AP = PQ (Since they are tangents from P). PQ is common side.
∴ PA = PB, Therefore, the first tangent bisects the second tangent

ii. AP = QP
=> ∠ PAQ = ∠ PQA = x
BP = QP
=> ∠BQP = ∠PBQ = y
In AAQB
∠QAP + ∠ABQ + ∠AQB = 180°
x + y + x + y = 180°
2(x + y) = 180°
x + y = 90°
∠AQB = 90°
Δ ABQ is a right angled triangle.

iii. It is a rectangle

Sslc Maths Tangents Notes Kerala Syllabus Chapter 7 Question 4.
In the picture below, AB is a diameter and p is a point on AB extended. A tangent from P touches the circle at Q.
What is the radius of the circle?

Answer:
AP = 8 cm PQ = 4 cm
QP² = AP × BP

AB = AP – BP
= 8 – 2 = 6
Diameter = 6
∴ Radius = \(\frac { 6 }{ 2 }\) = 3 cm

Kerala Syllabus 10th Standard Maths Guide Pdf Download Chapter 7 Question 5.
In the first picture below, the line joining two points on a circle is extended by 4 centimeters and a tangent is drawn from this point. Its length is 6 centimeters, as shown:

The second picture shows the same line extended by 1 centimeter more and a tangent drawn from this point. What is the length of this tangent?
Answer:
In the first circle
PC² = PB × PA
62 = 4 x PA
PA = \(\frac { 36 }{ 4 }\) = 9

AB = AP – PB = 9 – 4 = 5 cm In the second
PC² = PA × PB
PC²= 10 × 5 = 50
PC= √50 = 5 √2cm
= 7.05 cm

Textbook Rage No. 185

Kerala Syllabus 10th Standard Maths Guide Malayalam Medium Chapter 7 Question 1.
Draw a triangle of sides 4 centimetres, 5 centimetres, 6 centimetres and draw its incircle. Calculate its radius.
Answer:
Draw a line BC = 6 cm. Draw an arc with a radius of 4 cm with the B as center. Also, draw an arc with a radius of 5 cm taken C as center. Let the bisector of both arcs drawn and they meet at ‘A’. Complete Δ ABC.
Let the bisectors of ∠A and∠B be drawn and they meet at ‘O’. Draw an incircle to the triangle with center O.

10th Class Maths Malayalam Medium Kerala Syllabus Chapter 7 Question 2.
Draw a rhombus of sides 5 centimeters and one angle 50° and draw its incircle.
Answer:
Draw AB with length 5 cm. Draw a ray from A making angle 50° with AB. Mark D at 5 cm from A. Draw a perpendicular bisector to ∠A. That is meet BD atO.
Extend line AO up to C such that AO = OC. Join BC and DC. Draw a perpendicular OP to AB through O. Draw a circle with centre as O and OP as radius.

Question 3.
Draw an equilateral triangle and a semi-circle touching its two sides, as in the picture.

Answer:
Draw an equilateral triangle with side 4cm.
The bisector of ∠A is passing through the midpoint of BC. Take the midpoint of AB as P. Join OP. Let draw a circle with OP as radius, we get the required one.

Question 4.
What is the radius of the incircle of a right triangle having perpendicular sides of length 5 centimeters and 12 centimeters?
Answer:

Question 5.
Prove that if the hypotenuse of a right triangle is h and the radius of its incircle is r, then its area is r(h+r).
Answer:

Let the perpendicular sides of right-angled triangle be a, d and hypotenuse be h then

Question 6.
Prove the radius of the incircle of an equilateral triangle is half the radius of its circumcircle.
Answer:
The center of the circumcircle and the incircle are the same.

r = \(\frac { r }{ R }\), Hence the radius of the incircle of an equilateral triangle is half the radius of its circumcircle.

Tangents Orukkam Questions & Answers

Question 1.
ΔABC is an equilateral triangle. A circumcircle is drawn to it Prove that the triangle formed by the tangents to the circle at the vertices of ABC is another equilateral triangle.
Answer:

∴ ∠AOB = 120° ∠BOC = 120°
∠A = ∠B = ∠c = 60° (v Equilateral triangle)
∴ ∠AQC = 180 – 120° = 60°
(Sum of opposite sides of a quadrilateral is 180°)
∠BPC = 180 – ∠BOC = 60°
∠ARB = 180 –∠AOB = 60°
Three angles of the ΔPQR is 60° each.
So ΔPQR is an equilateral triangle.

Question 2.
Δ lf the perimeter of ΔABC is 10cm, calculate the perimeter of triangle PQR?
Answer:
Perimeter of ΔPQR = 2 × 10 = 20
One side =20/3
Area of ΔPQR =

Question 3.
What is the relation between the perimeters of ΔABC and ΔPQR?
Answer:
Area of Δ ABC =
Area of ΔPQR is four times of the area of ΔABC.

Question 4.
Draw the figure. Mark the circumcenter of AABC.
Answer:

Question 5.
Join OA, OB, OC. Note the cyclic quadrilaterals in it.
Answer:
OBPC, OAQC, OARB are the cyclic quadrilaterals by joining tangents and radius.

Question 6.
Since ∠B = 60°. ∠AOC will be 120° Write is ∠AOC?
Answer:
∠B = 60° hence ∠AOC = 120°
Angle made by an arc on the center which is twice the angle made by it on the alternate arc.

Question 7.
Find ∠P, ∠R . Write conclusions
Answer:
∠P = 60°, ∠R = 60°, both are equal

Question 8.
See three parallelograms like ABCQ. Find the perimeter of the outer triangle by the equality of opposite sides.
Answer:
ABCQ, CABP, CARB are three parallelograms.
In parallelogram. ABCQ , AB = CQ and BC = AQ . AB = BC
QA = QC (tangents from a outside point) = AB + BC = 2BC
PC = PB (tangents from a outside point) = 2 BC
RA = RB (tangents from a outside point) = 2 AC
∴ PQ + QR + PR =QC + PC + QA + RA + RB + PB
= 2 AB + 2 BC + 2 AC = 2(AB + BC + AC) = 2 × 10 (Perimeter of ΔABC is 10) = 20
Perimeter of ΔPQR =20

Question 9.
AC is the diagonal which divides the parallelogram by two equal triangles.
Answer:
The diagonal AC divides parallelogram ABCQ into two equal triangles.
Area of ΔACQ = Area of ΔABC
Similarly, area of ΔBPC
= Area of ΔABC
Area of ΔPQR
= Area (ΔACQ + ΔBPC + ΔABR + ΔABC)
= (4 × Area ΔABC)
∴ The perimeter of the outer triangle is twice of the perimeter of inner triangle.

Question 10.
Draw a circle and mark a point on it. Construct tangent to the circle at this point without using center.
Draw the circle and mark the point(P) DrawachordAB and joinAP and BP.
See the chord in the figure that you have drawn. This chord made the angle ∠PAB on one side. An equal angle will be formed on the other side of the chord at P with P B as one arm.(Use compass and scale method)
Answer:

All angles in the arc PB are equal.

Worksheet 6

Question 11.
In the figure, a circle touches the sides of?
ABC at P, Q, R.
If AB =AC then prove that BR = C R
Why is AP=AQ?

Answer:
AP = AQ
(Tangent drawn from a point A to circle have equal length)
Establish BP = C Q.
Establish BR = C R.
If AB = AC
AP + PB = AQ + QC
AP = AQ (Tangent drawn from a point A to circle have equal length)
∴ PB = QC
BP = BR (Tangent drawn from a point B to circle have equal length)
CR = CQ (Tangent drawn from a point C to circle have equal length)
∴ BR = CR

Question 12.
In the figure AP, BQ, P Qare tangents to the circle. The line AP is parallel to BQ. Find ∠POQ

Draw figure, mark OA, OB, OC.
Establish angle OAP, angle OCP equal.
Take ∠AOP, ∠COP as x.
Triangles BOQ, and COQ are equal.
∠BOQ, ∠COQ = Y
2x + 2y = 180. Write x + y and ∠POQ
Answer:
Δ OAP, Δ OCP are equal triangles.
So, ∠AOP = ∠COP = x
ΔBOQ, ΔCOQ are equal triangles.

So,
∠BOQ= ∠COQ = y
2x + 2y = 180°
Angle on semicircle.
So, x + y = 90°, ∠POQ = x + y = 90°

Question 13.
If a circle can be drawn by touching the sides of a parallelogram inside it will be a rhombus. Prove.
Answer:

Draw the figure AP = AS, BP = BQ, DR= DS, CR = CQ.
Using these equations prove the statement given as the 11^th point in the basic concepts.
2 × AB = 2 × AD.
Answer:
AP = AS, BP = BQ,
DR = DS, CR = CQ
(Tangent drawn from a point to circle have equal length)
Sum of opposite sides of a quadrilateral formed by joining the tangents on four points of a circle are equal.
So, 2 × AB = 3 × AD.
Therefore ABCD is a rhombus.

Question 14.
If r is the radius of the incircle of a right triangle prove that \(r=\frac{a+c-b}{2}\)

BP = BQ = r
AP = AR = c – r
CR = CQ = b – r
b = c – r + a – r
Answer:
In figure BP = BQ = r
∴ AP = AR = c – r
(Tangent drawn from a poijt A to circle have equal length)
CR = CQ = a – r
(Tangent drawn from a point C to circle have equal length)
AC = b = c – r + a – r
= c + a – 2r,
c + a – 2r = b
∴ \(r=\frac{a+c-b}{2}\)

Question 15.
Find the inradius of an equilateral triangle of side 10cm. User r = \(\frac { A }{ s }\)
Answer:

Tangents SCERT Questions and Answers

Question 16.
PQ is a tangent to the circle with center O.
a. Find ∠p?
b. If ∠O = 42° ,what is ∠Q? [Score: 3, Time: 5 minutes]

Answer:
∠P = 90°, ∠Q = 90 – 42 = 48°

Question 17.
Draw this figure using the given measurements. [Score: 3, Time: 4 minutes]
Answer:
Draw a circle of radius 4 centimeters. (1)
Draw a radius and a perpendicular to it. (1)
Complete the triangle after marking angle 60° at the center.

Question 18.
In the figure, AC and BC are tangents to the circle from C. Centre of the circle O.
i. Find ∠A
ii. If ∠C is 2 times ∠O, then what is ∠C ? [Score: 3, Time: 4 minutes]

Answer:
i. ∠A= 90° (1)
ii. ∠C + ∠O = 180° (0)
∠C= 60° (0)

Question 19.
The radius of a circle touching all sides of an equilateral triangle is 3centimetres. Draw this triangle. [Score: 3, Time: 4 minutes]
Answer:
Draw a circle of radius 3 cm. (1)
Mark 120° at the center of circle. (2)
Complete the equilateral triangle. (2)

Question 20.
Radius of an incircle to a triangle is 3 centimeters. Two angles of this triangle are 55° and 63°. Draw this triangle. [Score: 5, Time: 8 minutes]
Answer:
Draw a circle of radius 3 cm. (1)
Mark the angles 180 – 55 = 125°, 180 – 63 = 117°, at the centre (1)
Draw the tangents.
Complete the triangle. (3)

Question 21.
In the figure PA, PB are tangents through A and B of a circle with center O. If the radius of the circle is r, then prove that OP × OQ = r². [Score: 3, Time: 5 minutes]

Answer:
Δ OQA, ΔOPA are triangle with equal angles.
The ratio of sides opposite to the equal angles. (1)

Question 22.
In the figure, angles formed by the radius segment of the meeting points of the tan-gent to incircle are given. Find all angles of the triangle. [Score: 3, Time: 5 minutes]

Answer:
Angles 180 – 120 = 60° (1)
180-130 = 50° (1)
Third angle 180 – (60 + 50) = 70° (1)

Question 23.
The incircle triangle ABC touches the tri-angle sides at P, Q& R. as shown in the figure. Find all angles of PQR [Score: 3, Time: 5 minutes]

Answer:
The angles at the center of the circles are 180 – 70 = 110°,
180 – 80 = 100 and
360 + ( 110 + 100) = 150° (2)
The angles of triangle PQR are

Question 24.
Find all angles of triangle AOP and OPT. [Score: 4, Time: 7 minutes]

Answer:
The angles of ΔAOP are 32°, 32°, 116° (2)
The angles of ΔOPT are 64°, 26°, 90° (2)

Question 25.
QP is a tangent of the circle with center O. AR is a diameter. Find all angles of triangle PQR. [Score: 4, Time: 6 minutes]

Answer:
∠PQR = ∠QAR = 30° (1)
∠PRQ = 180 – 60 = 120° (1)
∠P = 180 – (120 + 30) = 30°

Question 26.
In the figure, PQ is a diameter and O is the center of the circle. ∠R = ∠T= 90°
1. Prove that ∠PSR – ∠OSQ.
2. Prove that ΔPSR and ΔSQT are similar. [Score: 3, Time: 5 minutes]
Answer:
1. ∠PSR = ∠PQS (1)
Q is a diameter ZPSQ = 90° (1)
∠PSR = 90 – ∠OSP = ∠OSQ

2. ∠PSR = ∠SQT (1)
∴ Triangle are similar (2 angles are same) (1)

Question 27.
Draw a circle of radiu$3 3 cm. Draw a chord AB = 4 cm of this circle. Draw tangents through A and B. [Score: 3, Time: 5 minutes]
Answer:

Question 28.
In the figure, O is the center. C is a point on the semicircle with diameter OA.
BC is a tangent through B, If OB = 1 cm AB = 3 cm then what is BC f Find all angles of triangle OBC ? [Score: 4, Time: 6 minutes]

Answer:
OB × AB = BC² (1)
1 × 3 = BC² = 3 (1)
BC = √3
The angles of ΔOBC 30°, 60°, 90°. (2)

Question 29.
In the figure, radius of the circle centered at O is 9 cm.
OA = 15 cm. Semicircle with diameter O A cuts the circle with center O at D and BC is a tangent through B.
1. What is the length of BC?
2. If the line PD is perpendicular to OA, then what is the length of PD? [Score: 4, Time: 7 minutes]

Answer:
1. BC² = OB × BA = 9 × 6 = 54
BC = √54cm (1)
2. OP × OA = r²

Question 30.
Hypotenuse of a right triangle is 18 cm and its inradius 3 cm. What is its perimeter? What is its area? . [Score: 3, Time: 5 minutes]
Answer:

Perimeter = 3 + x + x +y +y + 3
= 3 + 3 + 18 + 18 = 42 cm (1)
Area = \(\frac { 42 }{ 2 }\) × 3 = 21 × 3 = 63 sq.cm (1)

Question 31.
Area of a right triangle is 60 sq. centimetres and its inradius 3 cm. What is the length of its hypotenuse? [Score: 3, Time: 5 minutes]
Answer:
Area = 60 sq.cm, radius = 3

Tangents Exam Oriented Questions and Answers

Short Answer Type Questions (Score 2)

Question 32.
From a point P, the length of the tangent to a circle is 24 cm and distance of P from the center is 25cm. Find radius.
Answer:
OP² = OQ² + QP²

25² = OQ² + 24²
OQ² = 25² – 24²
OQ²= 625 – 576 = 49
OQ = 7 cm

Question 33.

Let PQ be a tangent to a circle at A and AB be a chord. Let C be a point on the circle such that ∠BAC = 54° and ∠BAQ= 62°. Find ∠ABC.
Answer:
∠ABC = l80° – ( ∠BAC + ∠ACB )
∠ABC = i8o° – (54° + 62°) = 64°

Question 34.
Draw a circle of radius 3.5cm and construct an equilateral triangle intersecting all sides with this circle. What is the radius of a circumcircle?
Answer:

Radius of circumcircle = 7 cm

Question 35.
In the figure, AB is the tangent at B of the circle centered at O. How much is ∠OBA? How much is ∠AOB?

Answer:
∠OBA = 90°
∠AOB = 55°

Question 36.
In the figure O is the center of the circle and P, Q,R are points on it. Find the angles of the triangle formed by the tangent at P, Q,R.

Answer:
Angle in a triangle and opposite angle in its center is supplementary. One angle is 40°, other angles are 70° each.

Question 37.
ABCD is a quadrilateral such that all of its sides touch a circle. If AB = 6cm, BC = 6.5cm and CD = 7cm, then find the length of AD.

Answer:
We have, AP = AS
BP = BQ
CR = CQ
DR = DS
Hence,
AP + BP+CR + DR = AS + BQ + CQ + DS
AB +CD = AD + BC
AD = AB + CD – BC = 6 + 7 – 6.5 = 6.5 cm

Short Answer Type Questions (Score 3)

Question 38.
Draw a circle of radius 2cm. Also, construct a regular hexagon intersecting all sides with this circle.
Answer:

ABCDEF is a regular hexagon.

Question 39.
AB 4cm and BC = 6cm. Find the length of AD? Draw a line segment of length √12 cm by using the same idea?

Answer:
AB × AC = AD²
AB = 4cm
AC = AB + BC = 4 + 6= 10cm
4 × 10 = AD2
∴ AD = √40
Inorder to draw √12 , draw AC = 6cm by taking BC = 4 cm and AB = 2 cm.
When a tangent is drawn from A, its length is √12 cm.

Question 40.
The radius of a circle with center O is 8cm. P is a point outside the circle. PQ and PR are the tangents drawn from P to the circle. If ∠QOR = 60° then find the lengths of PQ, PR, and OP.
Answer:
In Δ OPQ the angles are 30°, 60°, and 90°.
∴ The sides are in the ratio 1 : √3: 2

Question 41.
In the figure PA, PB and RS are the three tangents to the circle. Prove that the perimeter of ΔPRS is the sum of the lengths PAandPB.

Answer:
PA = PB (Tangents from an outside point are equal)
RA = RT
ST = SB
Perimeter of ΔPRS = PR + RS + PS
= PR + (RT + ST) + PS
= (PR + RA) + (SB + PS)
(Since RT = RA, ST= SB)
= PA + PB

Question 42.
In the figure, tangents PA and PB are drawn to a circle with center O from an external point P. If CD is a tangent to the circle at E and AP=25cm, find the perimeter of ∠PCD,
Answer:
We know that the lengths of the two tangents from an exterior point to a circle are equal.

CA = CE, DB = DE and PA = PB Now,
the perimeter of = PC + CD + DP
= PC + CE + ED + DP
= PC + CA + DB + DP
= PA + PB = 2PA(PB = PA)
Thus, the perimeter of ∠PCD = 2 × 25 = 50cm

Question 43.
In the figure, PQ is the diameter and RS is a tangent to the circle.

If ∠SPQ = 34°,fmd
a) ∠SQP
b) ∠RSQ
c) ∠SRP
Answer:
∠PSQ = 90° (Angle in a semicircle)
∴ ∠ SQP
= 180 – (90 + 34)
= 180 – 124 = 56°
b. ∠RSQ = ∠SPQ = 34° (Angle between chord and tangent = Angle in the segment on the other side)
c. ∠SRP= 180 – (124 + 34) s = 180 – 158 = 22°

Question 44.
Prove that the angles formed by the tangents from the endpoints of a chord are equal.
Answer:
Each angle between a chord and the tangent at one of its ends is equal to the angle in the segment on the other side of the chord. [Angles in the alternate segments are equal].
∴ ∠P = x ie ∠RBA = x.
ie ∠QAB = ∠RBA

Long Answer Type Questions (Score 4)

Question 45.
In the following figure, PQ is the tangent and PB is the scent. Prove that PQ2 = PA × PB

Answer:
Considering the triangles PAQ and PQB.
∠APQ = ∠BPQ (Common angle)
∠PQA = ∠QBP (Angle between tangent and chord = the angle made by the chord on its . complimentary arc)
ΔPAQ ~ ΔPQB
[A.A Similarly]

[Ratio of the similar sides are equal]

PQ² = PA × PB

Question 46.
AB, BC, and AC are the tangents of the circle. If AR = 5cm, BP = 4cm and BC = 10cm, find the perimeter of the triangle.
Answer:
AR = AQ = 5 cm
BP = BR = 4 cm
BC – BP = PC
∴ PC = 10 – 4 = 6cm,
PC = PQ = 6cm
∴Perimeter = 2 × 5 + 2 × 4 + 2 × 6
= 2 (5 + 4 + 6 ) = 30 cm

Long Answer Type Questions (Score 5)

Question 47.
AB and AC are the tangents of the circle with center O.

Show that the quadrilateral ABOC is a cyclic quadrilateral.
Answer:
AB and AC are tangents
∠B = ∠C = 90°
∠B + ∠C = 180°
The sum of the four angles of a quadrilateral is 360°. ie, ∠A + ∠O = 180°
The opposite angles of quadrilateral ABOC are complementary.
∴ ABOC is a cyclic quadrilateral.

Question 48.
AB, BC, CD and AD are the tangents of the circle. If AP = x, BP = y, CR = z, SD =w then show that the perimeter of the quadrilateral ABCD is 2(x + y + z + w).

Answer:
AP = AS = x;
BP = BQ = y
CR = CQ = z;
SD = DR = w
ie, AB = x + y
BC = y + z;
CD = z + w
AD = x + z
∴ Perimeter = 2x + 2y + 2z + 2w = 2(x + y + z + w)

Question 49.
In the picture PQ, RS, and TU are the tan-gents of the circle with center O. Find the pairs of angles with same measures

Answer:
∠QAB = ∠RBA ; ∠QAB = ∠ACB
∠RBA =∠ACB ; ∠SBC = ∠UCB
∠SBC = ∠BAC ; ∠UCB = ∠BAC
∠ACT = ∠PAC ; ∠ACT = ∠ABC
∠PAC = ∠ABC

Tangents Memory Map

Tangent to the circle is perpendicular to the radius through the point of tangency.

The tangents from an exterior point to a circle and radii to the points of tangency form a cyclic quadrilateral. In figure PAOB is a cyclic quadrilateral.
Tangents from an exterior point to a circle are equal. If PA, PB are the tangents then PA=PB.

The angle between a chord of a circle and the tangent at one end of the chord is equal to angle formed by the chord in the other side of the circle.

If a circle touches the sides of a quadrilateral, that circle will be the incircle of that quadrilateral. Sum of the opposite sides of such quadrilateral are equal. In the figure, ABCD is a quadrilateral having incircle AB + CD = AD + BD.

If P is an exterior point to a circle, a line from P touches the circle at T on the circle and a line intersects the circle at A and B then PA × PB = PT2.

The center of the circle which touches two lines will be a point on the bisector of the angle between the lines. The bisectors of the angles of a triangle passes through a point. That point will be the incenter of the triangle.

The circle drawn inside a triangle which touches the sides of the triangle is called incircle. The circle drawn outside the triangle which touches the sides of the triangle are excircles.

The radius of the incircle of a triangle is obtained by dividing area of the circle by its semi perimeter.

If a, b, c are the sides of a triangle then the area of the triangle
A= \(\sqrt{\mathrm{S}(\mathrm{S}-\mathrm{a})(\mathrm{S}-\mathrm{b})(\mathrm{S}-\mathrm{c})}, s=\frac{a+b+c}{2}\)
This is popularly known as Hero’s formula.

Students can Download Physics Chapter 5 Refraction of Light Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Physics Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

SSLC Physics Chapter 5 Refraction of Light Questions and Answers Malayalam Medium

SCERT 10th Standard Physics Textbook Chapter 5 Solutions Malayalam Medium

You can Download Energy Management Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Physics Solutions Chapter 7 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Physics Chapter 7 Energy Management Textbook Questions and Answers

SCERT Class 10th Standard Physics Chapter 7 Energy Management Solutions

Textbook Page No. 147

Sslc Physics Chapter 7 Kerala Syllabus Question 1.
It is said that energy can neither be created nor destroyed; then how does energy crisis occur?
Answer:
When energy is transformed from one form to another, some part of it gets lost in other forms. Such a loss is the main cause for energy crisis.

Textbook Page No. 148

Energy Management Class 10 Kerala Syllabus Question 2.
List down the different forms of energy you use for various purposes from the time you wake up till you reach your school.
Answer:

• Muscular energy – for different physical activities
• Chemical energy – for cooking
• Mechanical energy – for moving
• Sound energy – to call friends.

Sslc Physics Chapter 7 Notes Kerala Syllabus Question 3.
From which sources are you getting these forms of energy?
Answer:
From different sources like sun, fuels and power stations.

Textbook Page No. 149

Sslc Physics Chapter 7 Energy Management Kerala Syllabus Question 4.
Complete the table 7.1

Answer:

Physics Chapter 7 Class 10 Kerala Syllabus Question 5.
Take three papers of the same size. Keep one stretched. Crumble the next. Make the third paper wet using water. Burn each of them over a candle flame using pincers. Compare the burning of each.
Complete the table 7.2
.
Physics Chapter 7 Answer:

Textbook Page No. 150

Hss Live Guru 10th Physics Kerala Syllabus Question 6.
Write down the situations/specialties for partial combustion.
Answer:

• Partial dryness
• Insufficient availability of O₂
• Lack of facilities for the removal of oxygen.

Sslc Physics Chapter Wise Questions And Answers Kerala Syllabus Question 7.
What are the drawbacks of partial combustion?
Answer:

• Fuel loss
• Wastage of time
• Economic loss
• Atmospheric pollution
• More smoke is produced
• CO₂, CO (carbon monoxide) are produced as byproducts.

Sslc Physics Textbook Solutions Kerala Syllabus Question 8.
What are the advantages of using smokeless choolahs at home? Note them down in the science diary.
Answer:
Makes home neat, lung disease can be reduced, does not affect the oxygen-carrying capacity of blood. Reduces wastage of time and fuel loss.

Class 10 Physics Kerala Syllabus  Question 9.
Visit a nearby pollution testing center, interact with the staff there and prepare a note on the permissible pollution rate.
Answer:
The carbon monoxide produced as a result of combustion of fuels causes environmental pollution. More carbon monoxide may be produced if the vehicles are not working properly. Smoke testing is conducted to know what quantity of carbon monoxide is present in the smoke coming out of vehicle.

Kerala Syllabus 10th Standard Physics Question 10.
Which are the fuels that are used in vehicles and industries?
Answer:
Petrol, Diesel, LPG, CNG etc

Hss Live 10th Physics Kerala Syllabus Question 11.
Tabulate the category to which these fuels belongs to.

Answer:

Textbook Page No. 151

Hsslive Physics 10th Kerala Syllabus Question 12.
Which are the products obtained from fractional distillation of petroleum?
Answer:
Petrol (or gasoline), naptha, kerosene, diesel oil, lubricating oil, fuel oil, grease wax, and some residue.

Physics Class 10 Chapter 7 Kerala Syllabus Question 13.
Which is the cooking gas that we get in cylinders for domestic use?
Answer:
LPG

10th Scert Physics Solutions Kerala Syllabus Question 14.
How will you know if there is leakage in a LPG cylinder?
Answer:
The smell of LPG is felt

Physics Kerala Syllabus 10th Standard Question 15.
Never switch on or switch off electricity when there is a leakage of LPG Why?
Answer:
It is usually advised not to switch on or off any of the electric switches if you detect a gas leakage. It is because the fumes of gas are highly flammable and even smallest of sparks can ignite a huge fire.

Class 10 Kerala Syllabus Physics Solutions Question 16.
If there is a leakage of LPG does it rise up or come down in the atmosphere? Why?
Answer:
Another reason why care should be taken during storage is that LPG vapor is heavier than air, so any leakage will sink to the ground and accumulate in low lying areas and may be difficult to disperse. LPG expands rapidly when its temperature rises

Class 10 Physics Chapter 7 Kerala Syllabus Question 17.
If there is leakage of LPG it is mandatory to open the doors and windows. Why?
Answer:
Immediately open all the doors and windows to your house so that the gas can escape. Never open electrical fans or even an exhaust fan. Let the gas escape naturally. Once you do that, go outside the house and isolate the main electric supply. Be sure to evacuate yourself and others from the area.

Kerala Syllabus 10 Physics  Question 18.
What precautions are to be taken to avoid accidents due to LPG leakage? Discuss and record in the science diary.
Answer:

• Examine the rubber tube at regular intervals and ensure that it does not have a leakage.
• Turn on the knob of stove only after the regulator is turned on.
• Always buy LPG cylinders from authorized franchisees only
• Check that the cylinder has been delivered with the company seal and safety cap intact, do not accept the cylinder if the seal is broken.
• Please look for the due date of test, which is marked on the inner side of the cylinder stay plate and if this date is over, do not accept the cylinder
• Disconnect LPG regulator and affix safety cap on the cylinder when your gas stove is not in use for prolonged period.
• Always store the LPG cylinder in an upright position and away from other combustible and flammable materials.
• Check for gas leaks regularly by applying soap solution on cylinder joints and Surakshapipes.

Textbook Page No. 153

Physics Class 10 Kerala Syllabus Question 19.
If a gas leak is suspected or if the fire spreads on a cylinder, what else could be done? Think it over.
Answer:
If you are convinced that there is a gas leak, disconnect electricity from outside the home (switch off the main switches). Switch off the regulator and shift the cylinder to an empty space. Keep the windows and doors open. Request help from the Fire Force by calling in the toll-free number 108.

Well trained rescue operators can put out the fire by covering the top end of the cylinder with wet sack to prevent the contact with oxygen. If the fire is in flat or the top story, then one should not try to escape using lifts. Only staircase should be used. Cover the nose and the mouth with soft cloth to avoid the intake of smoke or gases.

Kerala Syllabus 10th Standard Physics Notes Question 20.
Haven’t you seen bio-wastes dumped in public places? Don’t you experience a putrid smell, when you pass them by? Which are the gases responsible for this smell?
Answer:
Methane gas is responsible for the putrid smell.

Physics 10th Class Notes Kerala Syllabus Question 21.
Besides air pollution, what are the problems that may arise when garbage is heaped? Discuss and record.
Answer:

• Soil contamination
• Air contamination
• Water contamination
• Bad impact on human health
• Impact on animals and marine life.
• Disease-carrying pests
• Adversely affect the local economy
• Missed recycling opportunities.

Textbook Page No. 154

Question 22.
Some of you may be using LPG in your houses. What is the weight of the LPG filled in the cylinders supplied to your homes?
Answer:
14.2 kg

Question 23.
Using this quantity of LPG for how many days can you cook?
Answer:
This must be sufficient for a month.

Question 24.
How many days can you cook using firewood of the same weight?
Answer:
14.2 kg firewood would be sufficient for a few a days only.

Question 25.
What difference do you feel in the efficiency of these two fuels?
Answer:
The LPG with more calorific value has more fuel efficiency.

Textbook Page No. 155

Question 26.
Based on its calorific value, which fuel can be considered as the most efficient?
Answer:
Hydrogen

Question 27.
Which are the instances when hydrogen is used as fuel?
Answer:
Hydrogen used as fuel in rockets.

Question 28.
Why is hydrogen not used as a domestic fuel?
Answer:
The combustion rate is very high for hydrogen, the possibility of explosion is also more. And it is also difficult to store hydrogen.

Textbook Page No. 156

Question 29.
What is the energy conversion in a generator?
Answer:
Mechanical energy → Electrical energy

Question 30.
From where do we get energy required for the working of a generator?
Answer:
The mechanical energy required can be provided by engines operating on fuels such as diesel, petrol, natural gas, etc. or via renewable energy sources such as a wind turbine, water turbine, solar-powered turbine, etc

Question 31.
Power stations can be classified based on the nature of the source providing the energy required to operate the generator.
Answer:
Flowing water -Hydroelectric power station.
Nuclear energy – Nuclear power station
Coal – Thermal power station

Textbook Page No. 157

Question 32.
On the basis of notes and discussions, complete the table :

Answer:

Textbook Page No. 158

Question 33.
We get different forms of energy from the Sun. Attempts are underway now to utilize solar energy of its maximum. What are the devices used for this?
Answer:

• Solar panel
• Solar water heater
• Solar cooker
• Solar water heater
• Solar thermal power plant

Textbook Page No. 159

Question 34.
What is the energy transformation that takes place in a solar cell?
Answer:
In a solar panel light energy is converted into electrical energy.

Question 35.
There are certain situations in which a solar panel cannot be put to use. Which are they?
Answer:
When the sky is cloudy, during the rain and night time we can not use solar panels.

Question 36.
What are the situations where solar panel alone is depended on?
Answer:
In space stations, satellite, and in remote islands where there is no electricity, etc solar panels are used.

Question 37.
Take two conical flasks. Paint the outer surface of one with black paint and the other, with white paint. Fill both of them with water and expose them to direct sunlight for the same period of time. Which one gets heated first? What may be the reason?
Answer:
The conical flask which is painted black seems to be more hot when kept under the sun for an hour because the black body absorbs more heat and do not let it out from it. It is the property of black body to absorb heat and there is no chance of heat to escape.

Textbook Page .159

Question 38.
List down the specialties of a solar cooker by examing Fig. 7.7.

1. A box with blackened interior
2. A glass cover for the box
3. A mirror outside the each? What is the function of each?
Answer:
1. A box with blackened interior:
The black color of the vessel lets it to absorb more heat as dark colors absorb more heat.
2. A glass cover for the box:
The glass lid will allow the light and heat energy of the sun to come inside but the light energy will go out but the heat will be trapped inside the cooker to cook food.
3. A mirror outside the each?:
The plane mirror will reflect the maximum light of the Sun inside the cooker.

Question 39.
Find out the working of solar cookers other than the box type and record them in the science diary.
Answer:
Panel solar cookers are inexpensive solar cookers that use reflective panels to direct sunlight to a cooking pot that is enclosed in a clear plastic bag.

Textbook Page .160

Question 40.

Observe the figure. Discuss and record how hot water is formed in the tank of solar heater.
Answer:
Solar water heating (SWH) is the conversion of sunlight into heat for water heating using a solar thermal collector. A sun-facing collector heats a working fluid that passes into a storage system for later use. SWH are active (pumped) and passive (convection-driven). They use water only, or both water and a working fluid. They are heated directly or via light-concentrating mirrors. They operate independently or as hybrids with electric or gas heaters. In large- scale installations, mirrors may concentrate sunlight into a smaller collector.

Textbook Page .162

Question 41.
‘Ocean as a source of energy: its possibilities and limitations’ Prepare a seminar paper on this.
Answer:
Hints: Oceans cover 70 percent of the earth’s surface and represent an enormous amount of energy. Although currently under-utilized, Ocean energy is mostly exploited by just a few technologies: Wave, Tidal, Current Energy and Ocean Thermal Energy.
Limitations:

• Few limited sites/places where this wave, tidal or ocean thermal energy can be obtained.
• The cost of construction of plants is very; high
• The efficiency of producing energy is also low.

Question 42.
Why is it said that geothermal power plants are not possible in Kerala? Discuss and record.
Answer:
There are some minor environmental is-sues associated with geothermal power. Geothermal power plants can in extreme cases cause earthquakes. There are heavy upfront costs associated with both geothermal power plants and geothermal heating/cooling systems. Very location-specific (most resources are simply not cost-competitive). Some countries have been blessed with great resources – Iceland and Philippines meet nearly one-third of their electricity demand with geothermal energy. If geothermal energy is transported long distances by the means of hot water (not electricity), significant energy losses has to be taken into account. Geothermal power is only sustainable (renewable) if the reservoirs are properly managed.

Textbook Page .164

Question 43.
What are the different methods by which energy is produced from the nucleus?
Answer:
Nuclear fusion, nuclear fission.

Question 44.
Even if the matter converted is very small, the energy produced is very large. What is the reason?
Answer:
According to Einstein’s equation E = mc². The amount of energy produced is high, even though the transformed mass is very less. Here’ represents the converted mass, c is the speed of light (3 x 10⁸ m/s) and E is the amount of energy obtained.

Question 45.
What is the reason for an uncontrolled fission reation ending in an explosion?
Answer:
Nuclear fission is the process in which the nuclei of greater atomic mass are split into lighter nuclei using neutrons. The mass of nucleus produced in such a process is less than of its parent nuclei. Thus there will be loss of matter in fission process. The mass lost during fission converts into energy. The two or three neutrons produced during the process, bombards with other nucleus and fission process continue rapidly and ends in big explosion.

Textbook Page . 165

Question 46.
Complete the Table 7.5

Answer:

Textbook Page .166

Question 47.
Classify the energy from the follow-ing sources as green energy and bown energy. Solar cells, atomic reactors, tidal energy, hydroelectric power, diesel engines, windmills, thermal power stations.
Answer:

Question 48.
What must be done to ensure maximum utilization of green energy while constructing a house?
Answer:

• Sufficient sunlight should be available in the rooms during day time.
• Comfortable warmth, coolness and air circulation must be available without the help of electricity.
• Using the sun for heating through south facing windows during the winter lowers heating costs. Shading those same windows in summer lowers cooling costs.
• Grid-tied solar photovoltaic (PV) panels currently provide the most cost- effective form of renewable energy for a zero energy home.
• Fresh filtered air and moisture control are critical to its success. This need for ventilation has a silver lining:

Let Us Assess

Question 1.
In a way most of the important energy sources of today can be said to be solar. Which of the following does not belong to the solar energy?
a. fossil fuel
b. energy from the mind
c. nuclear energy
d. biomass
Answer:
Nuclear energy

Question 2.
Which of the following is a green energy?
a. coal
b. naptha
c. biogas
d.petroleum gas
Answer:
Biogas

Question 3.
Write down the advantages and limitations of solar cooker.
Answer:
Advantages:

• Eco-friendliness
• Maintain better air quality indoors, reduce carbon monoxide emissions
• Less expensive
• Don’t cause any environmental pollution

Limitation:

• Cooking with solar cookers obviously requires sunlight, which makes it difficult to use during winter months and on rainy days.
• Cooking also takes a significantly longer time compared to conventional methods.
• Solar cookers are not as efficient at retaining heat as conventional cooking devices. Factors such as wind,  rain, and snow can seriously hinder operation, and in such weather conditions, even after the food is cooked, it  will lose its warmth very quickly.

Question 4.
Kerala has a long coastal land. Still, ocean is not considered as a major source of energy. Why?
Answer:
In Kerala usually, waves are of low. So they have less energy. Also during tides Kerala sea coasts, the sea attitude does not increasingly rise.

Question 5.
The graph regarding the calorific value of certain fuels is given below. Analyze the graph and answer the following Questions.

a. Which fuel has the highest calorific value? Which has the lowest?
b. How many kilogram of dried cow- dung is to be burnt to obtain the same amount of heat produced when 1 kg of LPG burns?
c. From the graph find out the most suitable fuel for household pur-pose. Justify your answer.
Answer:
a. Highest – Hydrogen
Lowest – Cowdung (Dried)
b. Energy released when 1kg LPG bums = 54000 kJ.
Dried cow-dung is to be burnt to obtain the same amount of heat \(\frac { 54000 }{ 6000 }\) = 9 kg
c. Biogas, because it is a renewable energy. Comparatively high calorific value.

Extended Activities

Question 1.
Find out the scope of hydrogen as a fuel with a high calorific value and prepare an essay.
Answer:
The amount of heat liberated by the complete combustion of 1kg of fuel is its calorific value. Hydrogen has high calorific value and also possess high combustion rate too. Hydrogen does not create much atmospheric pollution like other gases. Therefore seek method to combust it in moderate rates and use fuel cells.

Question 2.
Visit a hydroelectric power station and
try to understand different stages of the production of electricity. Make use of this principle and find out the scope of mini hydroelectric power project.
Answer:
There are only some primary expenses to build a hydrogen-electric power station. Hydroelectric power stations must be constructed only in areas of heavy rainfall and good streamflow. If these required conditions are satisfied, the production of electricity from such power station is very profitable one.

Question 3.
Visit a biogas plant and explore the possibility of establishing a community biogas plant in your region.
Answer:
Generally, the biomass required to construct a biogas plant may not be available from a single house and it is not profitable to construct separate biogas for each house. We know that the social cleanliness is equally important as personal cleanliness, due to several reasons, like.economic advantages, it is necessary to run a social biogas plant. And it will be useful to more people.

Question 4.
Write a short play to make the public aware of the need for making use of solar energy.
Answer:
Sun is the major source of energy available in earth, during photosynthesis the light energy is converted into chemical energy and is stored in plant cells. Since this plant reaches in animals as food, the stored energy reaches in animals too. The remainings of the animals hurried in earth transforms under high pressure and temperature and becomes fuels. Write drama using these hints.

Question 5.
Solar energy has an incredible future in the field of transportation. We are in its infant stage. Write an essay on the topic “Prospects of solar energy”.
Answer:
Solar energy is the ultimate energy source. All energy sources are related and connected with solar energy in one or the other way. Expand these points and complete the essay.

Question 6.
Find out the advantages and disadvantages of main energy sources and tabulate them.
Answer:

Question 7.
A nuclear reactor is about to be established in Kerala. What is your reaction to this proposal? Justify.
Answer:
Kerala is a highly populated state in India. Therefore there are not much vacant places available for the construction a nuclear reactors here. Since the radiations given out from such power plant are very harmful to human life, it may cause unpredictable effects. I strongly disagree with this opinion.

Question 8.
A man pointing at a car running on petrol says. “This car is running on solar energy”. Write down your responses about this matter.
Answer:
We use fossil fuels like petrol or diesel in cars. The plants which lived millions of year ago have also used solar energy for photosynthesis to make food. Humans and animals grew up by taking these plants and fruits. Then this humans and animals are burned in earth after their death and became fossil fuels. Fuels are formed out of them. This fuel is being used in cars so the ultimate source of energy here is the solar energy itself. The vehicles connected with solar panels uses solar energy in the direct form.

Energy Management Orukkam Questions and Answers

Question 1.
Observe the fast and complete combustion of a flat paper. Understand that combustion of a crumbled paper make more smoke.
a. List the conditions for the complete combustion.
b. Name the gases produced during complete combustion.
c. Name the gases produced during partial combustion.
d. How does the partial combustion causes environmental pollution?
e. How are the fossil fuels formed?
f. Why are they considered as non-renewable?
g. Write the full form of LNG and CNG
h. Which is the main constituent of LNG? What are the uses of CNG?
i. Name the main constituent of CNG.
Answer:
a. Conditions for complete combustion. Dryness, Fast evaporation, Materials should reach at a certain temperature that is required for combustion.
b. Carbon dioxide, Steam, Heat, and light.
c. Carbon monoxide, soot and a little of car-bondioxide.
d. Besides the fuel loss and time loss, the carbon monoxide produced during partial combustion
e. Fossil fuels are formed by the transformation of plants and animals that went under the earths crust millions of years ago. Eg: Coal, petroleum and natural gases.
f. They are not replenished or renewed in proportion to their consumption.
g. LNG – Liquefied Natural Gas CNG – Compressed Natural Gas
h. Methane. CNG is used as fuel in vehicles, thermal power stations.
i. Methane

Question 2.
1. Write the full form of LPG, what is its main constituent?
2. Why is ethyl mercaptan added to LPG?
3. Which element is the main component of coal?
4. Name the different types of coal?
5. Name the process by which the components of coal are separated?
6. Write the products obtained by the distillation of coal.
Answer:
1. Liquefied Petroleum Gas, Butane
2. To identify the leak of LPG.
3. Carbon
4. Peat, Lignite, Anthracite and Bituminous coal.
5. Distillation
6. Ammonia, Coal gas, Coal tar, and Coke.

Work Sheet:

Question 1.
Identify the relation and fill in the blanks
a. LPG: Butane
LNG ………….
b. LNG: Fuel in Vehicles
LPG …………..
Answer:
a. Methane
b. For domestic use.

Question 2.
Name the chemical used to identify the leakage of cooking gas.
Answer:
Ethyl mercaptan

Question 3.
Find the odd one out and explain the reason
(Peat, Anthracite, Bauxite, Lignite)
Answer:
Bauxite, Others are types of coal.

Question 3.
1. Is the heat produced by the combustion of different fuels the same? Discuss
2. What do you mean by fuel efficiency?
3. What is its unit?
4. List the qualities of a good fuel.
5. Compare biomass and biogas
6. Name the device that converts solar energy into electrical energy.
7. What is the reason for electric current in this device?
8. In a Solar Panel Solar energy is converted into electrical energy. What do you mean by a Solar panel?
9. Write down the situation where solar panels are used.
10. What is the energy change in a solar heater?
11. What is the difference between a solar voltaic power plant and solar thermal power plant?
Answer:
1. No, the heat produced by combustion of different fuels is not the same because their calorific value is not the same.
2. We use firewoods, kerosene, LPG etc in our homes as fuels. Each of them liberates different amounts of heat. This is indicated as fuel efficiency.
3. Kilo Joule/ Kilogram
4. Should be easily available Should be of low cost
Should cause minimum atmospheric pollution on combustion.
5. Biogas: Biogas refers to a mixture of different gases produced by the breakdown of organic mater in the absence of oxygen. Biomass: The body parts of plants and animals are known as Biomass.
6. Solar panel
7. The flow of electrons from p – region to n – region.
8. A large number of solar cells are suitably assembled to form a solar panel.
9. In lighting street lamps, In artificial satellites
10. Solar energy to heat energy
11. Both photovoltaic and solar thermal are the two established solar power technologies. Photovoltaics use semi conductor technology to directly convert sunlight into electricity. Solar thermal works by using mirrors to concentrate sunlight.

Question 4.
Watch the animation video of nuclear fission and fusion
a. Which nuclear reaction happens in an atom bomb?
b. E=mc²
E = energy, then m = ………., c = ………….
c. What will you call the nuclear reaction in which small atoms combine?
d. Which nuclear reaction is carried out in stars?
e. What is energy change in a nuclear reactor?
f. Classify the given energy sources into conventional and non-conventional energy.
(Fossil fuels, Solar energy, Nuclear energy, Biomass, Hydro-Electric Power)
g. What will call the energy sources that lead to global warming?
h. What do you mean by green energy?
i. Why is it instructed to control the use of brown energy sources?
j. List the names of renewable energy sources.
k. Write down the causes an remedies of energy crisis.
Answer:
a. Nuclear Fission
b. m = mass, c = velocity of light
c. Nuclear fusion
d. Nuclear fusion
e. Nuclear energy to electical energy
f. Conventional energy sources – Fossil fuels, Biomass, Hydroelectric power Nonconventional energy sources – Solar energy, Nuclear energy
g. Brown energy
h. Green energy is the energy produced from natural sources which does not cause environmental pollution.
i. Brown energy are the sources which cause environmental problems including global warming, so it is essential to control the use of brown energy sources.
j. Solar energy, Wind energy, energy from biomass.
k. Judicious utilization of energy, Maximum utilization of energy Timely repairing of machines

Work Sheet

Question 1.
Classify the given energy sources into renewable and nonrenewable.
(Solar energy, Petroleum, Nuclear energy, Coal, Geothermal energy)
Answer:
Renewable energy resources – Solar energy, Geothermal energy, Coal

Question 2.
Identify the relation and fill in the blanks.
a. Hydrogen bomb: Nuclear fusion
Atom bomb: …………
b. Solar energy: Green energy
Nuclear energy: ……….
Answer:
a. Nuclear fission
b. Brown energy

Question 3.
Why green energy is called clean energy?
Answer:
Green energy is the energy produced from natural sources which does not cause environmental pollution. Hence it is known as clean energy.

Question 4.
What is the fuel used in a nuclear reactor?.
Answer:
Enriched Uranium

Energy Management SCERT Question Pool Questions and Answers

Question 5.
Though energy is available in many forms we depend mostly on electrical energy.
a. Write down two forms of energy, other than electrical energy, used in daily life.
b. Electrical energy is used much in daily life. Why?
c. Does increase in population and mechanization lead to energy crisis? How?
Answer:
a. Heat energy, Light energy.
b. Electrical energy can be easily converted into many other forms.
c. Small increase in population cause large increase in the use of energy consumption, mechanization will lead to excessive use of sources of energy and leads to energy crisis

Question 6.
The heat energy obtained on burning 2 kg of hydrogen, coal and petrol are given below:
Petrol – 9 × 107 J
Hydrogen- 3 × 10⁸ J
Coal – 6 × 10⁷ J
a. Which of these is the most efficient fuel?
b. How much is the calorific value of hydrogen?
c. Arrange these fuels in the ascending order of their calorific values.
d. Which of the above will you select as a good fuel? What is the basis of your selection?
Answer:
a. Hydrogen
b. 150000 kj/kg
c. Coal, petrol, hydrogen.
d. Petrol.
Easy to handle: less atmospheric pollution, Calorific value is high.

Question 7.
It is advisable to keep stirring heap of wastes while burning them.
a. Write down two essential situations for the complete burning of fuels.
b. How does the stirring help the combustion? Explain.
c. Write down two disadvantages, of partial combustion.
Answer:
a. To increase the availability of oxygen.
i. Increase the surface area exposed to air
ii. Make available the temperature needed for burning.
b. Sufficient oxygen is made available which increases the rate of combustion.
c. Partial combustion pollutes the air by giving excess of smoke, soot, carbon monoxide an carbon dioxide.

Question 8.
a. What is meant by nonrenewable sources of energy?
b. How did such fuels originate in nature?
c. Write down any two examples for it.
Answer:
a. sources that cannot be renewed.
b. The biowastes (bio residue) that went into the earth change into petroleum by undergoing chemical changes due to high temperature, in the absence of air.
c. Petrol, coal.

Question 9.
a. What are fuels? Write down the names of two fuels each from the various category of fuels.
b. How does the excessive use of fuels in-fluence global warming? Explain.
c. English the need for prohibiting diesel vehicles in our country in the context of environmental pollution.
Answer:
a. Fuels are substances that release large quantities of heat on burning; solid: logs: liquid: kerosene.
b. Excessive of fuels produce greenhouse gases which increase the temperature of atmosphere and cause global warming.
c. The amount of CO² and SO² released by diesel vehicles is very large. This accelerates global warming.

Question 10.
a. How does the biomass change into biogas in a biogas plant?
b. Of these which is the fuel that is advantageous?
c. What are the advantages of having community biogas plants?
Answer:
a. Biogas is formed in the biogas plants by the action of bacteria on biomass in the absence of oxygen.
b. Biogas has higher calorific value; environmental pollution is less.
c. Community biogas plants help in con-trolling environment pollution, centralised energy source minimizes expense.

Question 11.
a. On what basis are the sources of energy classified as green energy and brown energy?
b. Classify the following into green energy and brown energy Solar cell, nuclear reactor, tidal energy, hydroelectric, diesel engine, windmill, thermal power station
c. Give any two suggestions to use the green energy to the maximum level.
Answer:
a. Green energy – green energy is the energy produced from natural resources that do not cause environmental pollution Brown energy – Energy from petroleum, coal, etc., and nuclear energy.
b. Green energy: solar cell, tidal energy, electricity from water, windmill.
Brown energy: nuclear reactor, diesel engine, thermal power station.
c. i. Install biogas plants
ii. Use solar panels

Question 12.

The figure indicates the fission reaction of a heavy nucleus of Uranium.
a. Write down an activity to make energy available by using lighter nucleii.
b. Calculate the energy obtained if Ig matter changes into energy during nuclear fission reaction (c = 3 x 10⁸ m/s).
c. What will be the result if the reaction shown in the above figure continues?
d. Why are we establishing nuclear reactors despite of the realization that nuclear reactors are controlled nuclear bombs?
Answer:
a. Nuclear fusion
b. E = mc² m = lg

c. Continuous fission reaction ends in an explosion (atom bomb).
d. Energy crisis and industrial growth promote production of energy.

Question 13.
Coal is the most abundant fossil fuel on the earth.
a. How did the fossil fuels originate?
b. Which is the main constituent of coal?
c. What are the substances obtained when coal is allowed to undergo distillation in the absence of air?
d. How does the excessive use of fossil fuels cause global warming?
Answer:
a. Plants and animals that went under the earth many years ago changed into fossil fuels by undergoing changes under high pressure, high temperature, and absence of oxygen.
b. Carbon
c. Ammonia, coal gas, coaltar, coke
d. The greenhouse gases given out by the burning of fossil fuels increase the temperature of atmosphere and causes global warming

Question 14.
LNG and CNG are made from natural gases.
a. What is the advantage of LNG over CNG?
b. Compare LNG and CNG from the point of view of fuel efficiency.
Answer:
a. LNG – liquefied natural gases can be conveniently transported by liquefaction. Can be changed into gaseous state at the same temperature and can be distributed through pipes.

Question 15.
Hydrogen is a fuel with a high calorific value.
a. What is the limitation of hydrogen as a fuel?
b. Which is the substance added to hydro-gen to make a hydrogen fuel cell? ‘
Answer:
a. Hydrogen easily catches fire and has a high chance of explosion. It is difficult to store and transport.
b. Oxygen.

Energy Management Exam Oriented Questions and Answers

Very Short Answer Type Questions (Score 1)

Question 16.
Which of the following is a renewable energy source?
(Petroleum, Solar energy, Coal)
Answer:
Solar energy

Question 17.
What is the fuel used in a nuclear reactor?
Answer:
Enriched Uranium

Question 18.
Using the relation from the first pair, complete the other.
LNG – Methane
LPG –
Answer:
Butane

Question 19.
Write the full expansion of LPG.
Answer:
Liquified Petroleum Gas

Question 20.
Which nuclear reaction is carried out in stars?
Answer:
Nuclear Fusion

Question 21.
Which material is used along with hydro-gen to make Hydrogen fuel cells?
Answer:
Oxygen

Question 22.
Using the relation from the first pair, complete the other. Solar cell – Green energy Atomic reactor –
Answer:
Brown energy

Question 23.
Find the odd one in the group and write the reason.
[Coal gas, Coaltar, Butane, Coke]
Answer:
Butane. Others are those materials got from distillation of coal.

Question 24.
Which nuclear reaction takes place in an atom bomb?
a.Nuclear fission
b. Nuclear fusion
c. Radioactivity
d. Combustion
Answer:
a.Nuclear fission

Question 25.
Using the relation from the first pair, complete the other.
Atomic bomb – Nuclear fission
Hydrogen bomb – …………….
Answer:
Nuclear fusion

Short Answer Type Questions (Score 2)

Question 26.
What is meant by Biomass? Give examples for biomass.
Answer:
The body parts of plants and animals are known as biomass. Examples for biomass are Wood, Excreta, etc.

Question 27.
L.P.G is one of the common fuel for domestic use.
a. Which is the main constituent of LPG?
b. Write full form of LPG.
Answer:
a. Butane b. Liquified Petroleum Gas

Question 28.
Calorific value of Hydrogen is 1,50,000 kj / kg. What does it mean?
Answer:
The amount of heat liberated by the complete combustion of 1 kg of hydrogen is 1,50,000 kj/kg.

Question 29.
a. Write any two qualities of hydrogen as a fuel.
b. What is the limitation of hydrogen as a domestic fuel?
Answer:
a. High calorific value, High availability b. Hydrogen easily catches fire and has a high chance of explosion. It is difficult to store and transport.

Question 30.
Complete the flowchart.

Answer:
a. Methane
b. LNG

Short Answer Type Questions (Score 3)

Question 31.
Write down the remedies of energy crisis.
Answer:

• Judicious utilization of energy.
• Maximum utilization of solar energy
• Minimizing the wastage of water
• Making use of public transportation as far as possible
• Construction and beautifying of houses and roads in a scientific manner.
• Controlling of the street lamps with LDR.

Question 32.
Classify the following suitably.
Solar cell, Atomic reactor, Tidal energy, Hydroelectric power, Diesel engine, Thermal power station, Windmill, Biogas

Question 33.

Various situations which uses solar energy are given in the diagram. What are the advantages & disadvantages of solar energy list them
Answer:

Question 34.
a. What are fossil fuels?
b. Why it is called as fossil fuels?
Answer:
a. Coal, Petroleum, Natural gas
b. Fossil fuels are formed by the transformation of animals and plants that went under the earth millions of years ago.

Question 35.
High calorific value is one among the properties that a substance must possess to be considered as a good fuel.
a. What do you mean by calorific value?
b. What are the other properties a substance must have in order to be considered as a good fuel?
Answer:
a. The amount of heat liberated by the complete combustion of 1kg of fuel is its calorific value.
b. Low price, more availability, less atmospheric pollution, portability, and easiness to store, etc. are the properties of a good fuel.

Long Answer Type Questions (Score 4)

Question 36.
Firewood, Cowdung cake, Kerosene, Petrol, LPG, etc are fuels.
a. Which among these has high calorific value?
b. In which state it exists?
c. What are the qualities of a good fuel?
d. Complete the table.

Answer:
a. LPG
b. Gaseous state
c. Fuel with high calorific value is considered as good fuel
d. 1. Butane
2. Methane
3. Methane

Question 37.
A few news related to energy crisis are given below. State two reasons of energy crisis and write how we can overcome them.

Answer:
Reasons: Increase in population, Overexploitation of nonrenewable energy resources, Industrialisation, Urbanisation. Remedial measures:- Population control, Use of renewable energy sources, Industrialisation after finding required energy sources.

Question 38.
Choose the correct answer from the bracket
Fossil fuel, Hydrogen, electrical energy
methane, Butane, Petroleum, Solar panal
a. The most abundant element in sun is ………
b. A solar cell converts solar energy into ………..
c. ……….. is the main constituent of biogas.
d. ……….. provides the power required for artificial satellite.
Answer:
a. Hydrogen
b. Electrical energy
c. Methane
d. Solar panel

Question 39.
a. Among the energy sources identify which will give green energy.
Energy from fossil fuel. Nuclear energy, Solar energy, Energy from wind
b. What are the precautions to be taken for green energy in household buildings.
c. When fuels v are partially burned it can cause certain hazards. List out some.
Answer:
a. Solar energy, Energy from wind
b. Use solar panels, Use biogas plants.
c. Energy loss, Pollution

Question 40.
a. What are the properties that a good fuel must-have?
b. What is the full form of LNG? List out its characteristics.
Answer:
a. 1. Should be of low cost
2. Should be easily available
3. Should cause minimum atmospheric pollution on combustion.
b. Liquefied Natural gas. LNG is a natural gas that can be liquefied and transported to long distances conveniently. It can be again converted into gaseous format atmospheric temperature and distributed through pipelines.

Question 41.
a. What is the full form of LNG? List out its characteristics.
b. Compare Nuclear fission and Nuclear Fusion.
Answer:
a. Liquefied Natural gas. LNG is a natural gas that can be liquefied and transported to long distances conveniently. It can be again converted into gaseous form atmospheric temperature and distributed through pipelines.
b. Nuclear fusion: The process in which lighter nuclei are combined to form heavier ones. Nuclear Fission: The process by which the nuclei of greater mass are split into lighter nuclei, using neutrons. The mass of small nuclei is less than that of parent nucleus.

You can Download Arithmetic Sequences Questions and Answers, Activity, Notes PDF, Kerala Syllabus 10th Standard Maths Solutions Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

SSLC Maths Chapter 1 Arithmetic Sequence Textbook Questions and Answers

SCERT Class 10th Standard Maths Chapter 1 Arithmetic Sequences Notes

Textbook Page No. 10

Arithmetic Sequence Questions And Answers 10th Standard Question 1.
Make the following number sequences, from the sequence of equilateral triangles, squares, regular pentagons and so on, of regular polygons:
Number of sides 3, 4, 5, ………
Sum of inner angles
Sum of outer angles
One inner angle
One outer angle
Arithmetic Sequence worksheet with Answer:
No. of triangles of a regular polygon having no. of sides are 3, 4, 5 ………… n respectively is given as 1, 2, 3, 4. Sum of interior angles of a triangle is 180°.

Then the sequence of Sum of interior angles
= 180, 180 × 2, 180 × 3 …. = 180, 360, 540, ……………….
Sum of exterior angles of any geometric structure having any number of sides is always 360°.
Sum of exterior angles = 360, 360, …………..
One interior angle
= \(\frac { 360 }{ 3 }\), \(\frac { 360 }{ 4 }\), \(\frac { 360 }{ 5 }\), ……………. = 120, 90, 72, ………….

Find the sequence calculator can determine the terms (as well as the sum of all terms) of the arithmetic, geometric, or Fibonacci sequence.

Arithmetic Sequence Class 10 Kerala Syllabus Question 2.
Look at these triangles made with dots. How many dots are there in each ?

Compute the number of dots needed to make the next three triangles.
Arithmetic Sequence worksheet with Answer:
3, 6, 10
The number of dots needed to make the next three triangles will be:
10 + 5 = 15
15 + 6 = 21
21 + 7 = 28
15, 21, 28 Dots

Arithmetic Sequence Question 3.
Write down the sequence of natural numbers leaving remainder 1 on division by 3 and the sequence of natural numbers leaving remainder 2 on division by 3.
Answer:
The numbers that leave 1 as remainder when divided by 3 are 1, 4, 7, 10, 13, ………..
(Sequence each with difference of 3 and starting from 1)
The numbers that leave 2 as remainder when divided by 3 are 2, 5, 8, 11, 14, ………….
( Sequence each with difference of 3 and starting from 2)