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Students can Download Kerala SSLC Maths Model Question Paper 1 English Medium, Kerala SSLC Maths Model Question Papers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala SSLC Maths Model Question Paper 1 English Medium

Time: 2½ Hours
Total Score: 80 Marks

Instructions

• Read each question carefully before writing the answer.
• Give explanations wherever necessary.
• First 15 minutes is Cool-off time. You may use the time to read the questions and plan your answers.
• No need to simplify irrationals like √2, √3, π etc., using approximations unless you are asked to do so.

Answer any three questions from 1 to 4. Each question carries 2 scores. (3 × 2 = 6)

Wandoor Ganitham Sslc Model Question Paper Question 1.
Consider the arithmetic sequence 13, 23, 33, …..
a) What is its common difference?
b) What is the first three-digit term of this sequence?
Answer:
(a) 10
(b) 103

Wandoor Ganitham Model Question Paper Question 2.
In the figure, O is the centre of the circle ∠A = 60°.

a) ∠BOD = __________
b) ∠C = _________
Answer:
(a) ∠BOD = 120°
(b) ∠C = 120°

Wandoor Ganitham Answer Key Polynomials Question 3.
If x – 1 is a factor of the polynomial 5x³ – 4x² + x – k, what number is k?
Answer:
P(x) = 5x³ – 4x² + x – k
P(1) = 5 × 1³ – 4 × 1² + 1 – k = 0
5 – 4 + 1 – k = 0
k = 2

Wandoor Ganitham Question Paper Question 4.
A circle is drawn with the origin as centre. It passes through the point (3, 3).
a) What is the radius of the circle?
b) Write the coordinates of a point where the circle meets the x-axis.
Answer:
(a) Radius = 3√2 cm
(b) (-3√2, 0)

Answer any five questions from 5 to 11. Each question carries 3 scores. (5 × 3 = 15)

Wandoor Ganitham Answer Key 2021 Question 5.
a) If a square is inscribed in a circle of diameter 4cm, what will be the length of a side of the square?
b) What is the base edge of a square pyramid of maximum size that can be carved out from a hemisphere of radius 5 cm?
Answer:
(a) Base edge of the square pyramid = 2√2 cm
(b) Base edge = \(\frac{10}{\sqrt{2}}=\frac{5 \times 2}{\sqrt{2}}\)
\(\begin{aligned}
&=\frac{5 \times \sqrt{2} \times \sqrt{2}}{\sqrt{2}}\\
&=5 \sqrt{2} \mathrm{cm}
\end{aligned}\)

Wandoor Ganitham Model Question Paper 2021 Question 6.
Draw the co-ordinate axes and mark the point (4, 0). Draw an isosceles right angled triangle with this point as one of its vertices.
Answer:

Wandoor Ganitham Answer Key Arithmetic Sequence Question 7.
Draw a circle of radius 3.5 centimetres. Draw a triangle of angles 50°, 60°, 70°with its vertices as points on the circle.
Answer:

Wandoor Ganitham Model Question Paper Answer Key Question 8.
The algebraic form of an arithmetic sequence is 5n + 5
(a) What is its first term?
(b) What is the difference pf its 10th and 20th terms?
(c) Can the difference between any two terms of this sequence be 368? Justify.
Answer:
(a) x₁ = 5 × 1 + 4 = 9
(b) Difference: x₂₀ – x₁₀ = 10d = 10 × 5 = 50
(c) The difference between any two terms of an arjthmetic sequence is divisible by the common difference.
368 is not divisible by 5, the common difference that is, 368 is not the difference btween two terms.

Sslc Model Question Paper Maths English Medium Question 9.
The length of a rectangle is 6 centrimetres more than its breadth. Its area is 1216 square centimetres. Find its length.
Answer:
Breadth = x
Length = x + 6
x(x + 6) = 1216
⇒ x² + 6x = 1216
⇒ x² + 6x + 9 = 1216 + 9 = 1225
⇒ (x + 3)² = 1225
⇒ x + 3 = √1225 = ±35
⇒ x + 3 = 35 or x + 3 = -35
⇒ x = 35 – 3 = 32 or x = -35 – 3 = -38
i.e Breadth = 32
Length = 32 + 6 = 38

Sslc Maths Question Paper 2020 Kerala English Medium Question 10.
In the figure, C, D are points on the circle AD is a diameter of the circle. ∠C = 30°, AB = 4 centrimetres.

a) ∠D = ________
b) ∠ABD = _________
c) What is the length of the diameter?
Answer:
(a) ∠D = 30° (Angle in the same arc)
(b) ∠ABD = 90° (Angle in the semicircle is 90°)
(c) 8 cm

Sslc Maths Model Question Paper English Medium Question 11.

In the figure C is the centre of the circle, PA and PB are tangents. PC = 5 centimetres and radius of the circle is 3 centimetres.
a) Find the length of PA
b) What is the area of the quadrilateral PACB?
Answer:
(a) PA = 4 cm
(b) Area of PAC = \(\frac { 1 }{ 2 }\) × 4 × 3 = 6 cm²
Area of PACB = 2 × 6 = 12 cm²

Answer any 7 questions from 12 to 21. Each question carries 4 scores. (7 × 4 = 28)

Wandoor Ganitham Sslc Model Question Paper Answer Key Question 12.
The coordinates of two opposite vertices of a rect-angle are (7, 8) and (1, 3).
a) Without drawing coordinate axes, mark these points as the vertices of a rectangle with left-right, top-bottom positions correct.
b) Find the co-ordinates of other two vertices.
c) What is the length of its diagonals?
Answer:

Wandoor Ganitham Sslc Model Answer Key 2021 Question 13.
Consider the polynomial P(x) = ax³ – x² – bx – 1
a) Find P(1).
b) What is the relation between a and b if x – 1 is a factor of P(x)?
c) What is the relation between a and b if x + 1 is a factor of P(x)?
d) Will, P(x) have both (x + 1) and (x – 1) as factors for any numbers a and b? Justify.
Answer:
P(x) = ax³ – x² – bx – 1
(a) P(1) = a × 1³ – 1² – b × 1 – 1 = a – 1 – b – 1 = a – b – 2
(b) If (x – 1) is a factor, P(1) = 0
ie. a – b – 2 = 0
ie. a – b = 2
(c) If x + 1 is a factor, P(-1) = 0
ie. a (-1)³ – (-1)² – b(-1) – 1 =0
ie. -a – 1 + b – 1 = 0
ie. b – a = 2
(d) It is not possible to find a and b such that a – b = b – a = 2
Therefore we cannot find a and b.

Hss Live Guru Model Question Paper Question 14.
The radius and height of a cone are 12 centimetres and 6 centimetres respectively.
a) What is its volume?
b) If this cone is cut parallel to its base, along the midpoint of its height, what is the radius of the small cone obtained?
c) What is the volume of the small cone?
d) Find the ratio of the volumes of the small cone and the first cone.
Answer:
r = 12 cm, h = 6 cm
(a) Volume = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\) × π × 12² × 6 = 288π cm³
(b) R = 12 cm, H = 6 cm, h = 3 cm

\(\begin{aligned}
&\frac{r}{R}=\frac{h}{H}\\
&\frac{r}{12}=3
\end{aligned}\)
r = 6 cm
(c) Volume = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\) × π × 6² × 3 = 36π cm³
(d) Ratio = 36π : 288π = 1 : 8

Wandoor Ganitham Sslc Unit Test 2021 Answer Key Question 15.
In the figure ABCD is a parallelogram. ∠E = 90°. A(3, 5) B(8, 5) are two vertices BE = 3 units, CE = 4 units.

a) Write the coordinates of C.
b) What are the coordinates of D?
c) Find the coordinates of meeting point of the di-agonals of the parallelogram.
Answer:
(a) B(8, 5), BE = 2
E(11, 5), C (11, 9)
(b) D (6, 9)

(c) The point of intersection of the diagonals is the midpoint of a diagonal.
The mid point of AC is \(=\left(\frac{11+3}{2}, \frac{9+5}{2}\right)\) = (7, 7)

Wandoor Ganitham Pre Model Answer Key Question 16.
Draw a circle of radius 3 centrimetres. mark a point 7 centimetres away from its centre. Draw the tan¬gents to the circle from that point.
Answer:

Hsslive Guru Question Paper Question 17.
In triangle PQR, ∠Q = 90°, ∠R = x°. Lengths of the sides PQ, QR and PR are a, b, c respectively.

a) Which among the following is tan x°?
\(\left(\frac{a}{c}, \frac{b}{a}, \frac{a}{b}, \frac{b}{c}\right)\)
b) Similarly write sin x° and cos x° from this triangle.
c) Prove that \(\frac{\sin x^{0}}{\cos x^{0}}=\tan x^{0}\)
Answer:

Question 18.
Consider two fractions having numerator 1. The denominator of one fraction is 2 more than the denominator of the other. Sum of these fractions is \(\frac{5}{12}\).
a) Write the above fact as an algebraic equation.
b) Find the fractions.
Answer:
Let x be the denominator of the first fraction.
The denominator of the other fraction is x + 2

Question 19.
A bag contains some red and green balls. If we take a ball from it, without looking, the probability of getting a red ball is \(\frac{1}{4}\)
a) What is the total number of balls, if there are 8 red balls?
b) What is the probability that a ball taken is green?
c) Find the sum of both the probabilities.
d) From a box containing some red balls and some blue balls th& probability of getting a red ball is \(\frac{a}{b}\). What is the probability of getting a blue ball?
Answer:

Question 20.
In the figure, ∠P = 90°. Sides of triangle APC are extended to B and D.

a) If a circle is drawn with AC as its diameter, where will be the position of P with respect to that circle?
b) What about the position of P, if the circle is drawn with AD as diameter?
c) Prove that, the circles drawn with the sides of a quadrilateral with perpendicular diagonals, will meet at a common point.
Answer:
(a) On the circle
(b) On the circle
(c) Since AB and CD are perpendicular to each other, the circle with diameter AB, diameter AD, diameters DB and diameter GB passes through P.

Question 21.
a) Write the sequence of odd numbers greater than 1.
b) What is the algebraic form of this sequence?
c) What is the algebraic form of the arithmetic sequence \(\frac{3}{6}, \frac{5}{6}, \frac{7}{6} \ldots .\)
d) Prove that this sequence does not contain any natural number.
Answer:
(a) 3, 5, 7, …
(b) x_n = 2n + 1
(c) x_n = \(\frac{2 n+1}{6}\)
(d) To get integer terms, the numerator should be a multiple of 6.
2n + 1 is odd. It is not a multiple of 6 for carry n.
The sequence does, not contain integers.

Answer any five questions from 22 to 28. Each question carries 5 score. (5 × 5 = 25)

Question 22.
The table below shows the students of a class sorted according to their scores in a test

a) If the students are arranged in the increasing order of their scores, the score of the student at what position is taken as the median?
b) What may be assumed as the score of the 15th student in the arrangement?
c) Find the median score.
Answer:

(a) n = 41 (odd number)
\(\frac{41+1}{2}=\frac{42}{2}\) = 21st score comes in the middle.
It is in between 20 and 30
(b) When 10 scores we equally divided among 10 children, each one’s share is \(\frac { 10 }{ 10 }\) = 1
Score of 15th child = 20 + \(\frac { 1 }{ 2 }\) = 20.5
(c) Consider the arithmetic sequence with first term 20.5 and common difference 1
7th term will be the median Median = 20.5 + 6 × 1 = 20.5 + 6 = 26.5

Question 23.
A solid is made by fixing a hemisphere of same radius on the flat face of a cone. The height of the cone is 12 centimetres and its slant height is 13 centimetres.

a) What is the radius of the cone?
b) What is the curved surface area of the hemisphere?
c) What is the total surface area of the solid?
Answer:
l = 13 cm, h = 12 cm

Question 24.
Draw a triangle of sides 5 dentimetres, 6 centimetres and 7 centimetres. Draw its incircle.
Answer:
Draw triangle, Draw angle bisectors. They meet at a point. That point will be the incenter.

Question 25.
A boy is standing between two buildings of equal height. The boy and the buildings are in a straight line. He sees the tops of those buildings at elevations of 45° and 30°. The nearest building is 20 metres away from him.
a) Drawa rough figure.
b) Find the height of the buildings and distance between the buildings.
Answer:
(a)

(b) BC = 20m, ∠CBD = 45°, CD = 20 m
Height of the building = 20 m
AE = 20 m
ie. AB = 20√3 m
Distance between the building 20 + 20√3 = 20(1 + √3)m

Question 26.
In the figure chords AB and CD intersect at P. PA = 8 centrimetres, PB = 6 centimetres, PC = 4 centimetres, BC = 4 centimetres.

a) Which angle is equal to ∠A?
b) Write one more pair of equal angles.
c) Find the length of PD.
d) What is the length of AD?
Answer:
(a) ∠A = ∠C
(b) ∠D = ∠B
(c) PD × PC = PA × PB
PD × 4 = 8 × 6
PD = 12 cm
(d) ΔAPD , ΔCPB are similar
\(\begin{aligned}
&\frac{A P}{P C}=\frac{P D}{P B}=\frac{A D}{B C}\\
&\frac{8}{4}=\frac{A D}{4}
\end{aligned}\)
AD = 8 cm

Question 27.
The sum of first 9 terms of an arithmetic sequence is 45 and the sum of first 18 terms is 171.
a) What is the sum of its 10th to 18th terms?
b) What is tis 5th term?
c) Find its 14th term.
d) Find the sum of 5th to 14th terms.
Answer:
S₉ = 45, S₁₈ = 171
(a) If S_n denotes the sum of first n terms = S₁₈ – S₉ = 171 – 45 = 126
(b) \(x_{5}=\frac{S_{9}}{9}=\frac{45}{9}=5\)
(c) \(x_{14}=\frac{126}{9}=14\)
(d) Sum of 5th to 14th terms = \(\frac { 10 }{ 2 }\) (X₅ + X₁₄)
= 5(5 + 14)
= 5 × 19
= 95

Question 28.
In the figure ABCD is a square. The axes are drawn throgh the midpoints of its sides. The length of its side is 6 units.

a) Write the coordinates of its vertices.
b) Write the equation of its diagonal BD.
Answer:
(a) A (-3, -3) B(3, -3) C (3, 3) D (-3, 3)
(b) Slope of BD = \(\frac{Y_{2}-Y_{1}}{X_{2}-X_{1}}=\frac{3–3}{-3-3}=\frac{6}{-6}=-1\)
Let (x, y) be a point on BD
\(\frac{y-3}{x-3}=-1\)
y + 3 = -1(x – 3)
y + 3 = -x + 3
x + y = 0

Read the following. Understand the Mathematical Idea expressed in it and answer the questions that follow. (6 × 1 = 6)

Question 29.
Consider the number 2751. The sum of its digits is 2 + 7 + 5 + 1 = 15. Adding the digits of 15 we get 1 + 5 = 6. This number 6 is called the ‘digital root’ of the number 2751. That is, to find the digital root of a number, find the sum of its digits (Dont forget to find the sum of the digits again if the first sum has more than one digit)
Let us see one more example.
The sum of the digits of the number 679412 is 6 + 7 + 9 + 4 + 1 + 2 = 29
Sum of digits of 29 = 2 + 9 = 11
Sum of digits of 11 = 1 + 1 = 2
Therefore the digital root of 679412 is 2.
Digital roots have an interesting property.
To see this, consider the product 43 × 27 = 1161.
The digital roots of the numbers 43 and 27 are 4 + 3 = 7 and 2 + 7 = 9.
Product of the digital roots = 7 × 9 = 63.
Digita root of 63 = 6 + 3 = 9.
The digital root of 1161 is also 9 (1 + 1 + 6 + 1 = 9)
That is the digital root of 1161 = The digital root fo 63, where 63 is the product of the digital roots of 43 and 27.
This property is true for all other natural numbers.
a) What is the digital root of 345?
b) What is the digital root of 927?
c) What is the digital root fo 345 × 927?
d) The digital root of the number 63 □ 5 is 8 (□ represents a missing digit). Find the missing digit.
e) 121 × 92 = 11□32. Find the missing digit.
f) If the digital root of a is 5 and the digital root of b is 2. Then what is the digital root of ab?
Answer:
(a) 3 + 4 + 5 = 12 = 1 + 2 = 3
(b) 9 + 2 + 7 = 18 = 1 + 8 = 9
(c) 9 × 3 = 27 = 2 + 7 = 9
(d) 6 + 3 + x + 5 = 8
14 + x = 8
1 + 4 + x = 8
5 + x = 8
x = 8 – 5 = 3
(e) (1 + 2 + 1) (9 + 2) = 1 + 1 + X + 3 + 2
4(1 + 1) = 7 + x
4 × 2 = 7 + x
8 = 7 + x
x = 1
(f) Digital root f a × b = 5 × digital root of 2
5 × 2 = 10
Digital root of 10 is 1.

You can Download Equal Triangles Questions and Answers, Activity, Notes, Kerala Syllabus 8th Standard Maths Solutions Chapter 8 help you to revise complete Syllabus and score more marks in your examinations.

Kerala Syllabus 8th Standard Maths Solutions Chapter 8 Area of Quadrilaterals

Area of Quadrilaterals Text Book Questions and Answers

Textbook Page No 149

Area Of Quadrilateral Class 8 Scert Chapter 8 Question 1.
Draw a parallelogram of sides 5 centimetres, 6 centimetres and area 25 square centimetres.
Solution:

• Area of parallelogram = one side × distance to the opposite side.
• Since 5x distance to the opposite side is 25 sq.cm. Distance to the opposite side is 5 cm.
• Draw a square ABCD of side 5cm.
• Draw an arc of radius 6cm with centre A to cut DC at E.
• Length of radius 6cm with centre B and arc of radius 5 cm with centre E meet at F.

Class 8 Mathematics Area Of Quadrilaterals Kerala Scert Solutions Question 2.
Draw a parallelogram of area 25 square centimetres and perimeter 24 centimetres.
Solution:
Area of the parallelogram to be 25 cm². The product of one side and distance to the opposite side to be 25 cm².
One side and distance to the opposite side can be 5 cm each.
The other side is = 12 – 5 = 7 cm.
Draw a square with side cm and draw an arc of radius 7 cm with A as centre such that it intersects the side CD. Mark the point E and join AE. Draw an arc of radius 7 cm with B as centre and draw another arc of radius 5 cm with E as centre, Two arcs meet at the point
P. JoinBF and CF.

Area Of Quadrilaterals Class 8 Kerala Syllabus Question 3.
In the figure, the two bottom corners of a parallelogram are joined to a point on the top side.

Solution:
Area of dark triangle = \(\frac{1}{2} b h\) cm²
Area of parallelogram = bh
= 2 × area of triangle 10 cm²

Class 8 Mathematics Construction Of Quadrilaterals Kerala Scert Solutions Question 4.
The picture below shows the parallelogram formed by the intersection of two pairs of parallel lines?

What is the area of this parallelogram? And the perimeter?
Solution:
One side of the parallelogram = 4 cm
Distance to the opposite side = 3 cm
Area of the parallelogram = 4 × 3 = 12 sq.cm
BC × DE = 12
BC × 2 = 12
BC = 6, AD = 6

Perimeter = 4 + 6 + 4 + 6 = 20 cm.

Area Of Quadrilateral Questions For Class 8 Kerala Syllabus Question 5.
Compute the area of the parallelogram below:

Solution:

Textbook Page No 153

Hss Live Guru 8 Maths Kerala Syllabus Chapter 8 Question 6.
Draw a square of area 4½ square centimetres.
Solution:


Draw a circle of radius 3 cm. Draw the diameter AC and construct the perpendicular bisector of AC which meets the circle at B and D. Join AC, AD, CD and BD to get the square ABCD.

Area Of Quadrilateral Formula Class 8 Kerala Syllabus  Question 7.
Draw a non-square rhombus of area 9 square centimetres.
Solution:

• Draw AC = 1 cm
• Find midpoint of AC by drawing per perpendicular bisector
• Find B and D, such that OB = OD = 9 cm

Similarly we can draw the other two rhombus
AC = 9 cm,
OB = OD = 1 cm
AC = 3 cm, OD = OB = 3 cm

8th Standard Maths Notes Kerala Syllabus Chapter 8 Question 8.
The area of the dark triangle in the figure is 5 square centimetres. What is the area of the parallelogram?
Solution:

iii. Perimeter = 4 × 15 = 60 m
iv. Rhombus is a parallelogram. So,
Area of parallelogram = one side × distance to the opposite side
15 × h = 216
h = \(\frac{216}{15}=14.4 \mathrm{cm}\)

8th Standard Maths Notes PdfKerala Syllabus Chapter 8 Question 9.
A 68 centimetre long rope is used to make a rhombus on the ground. The distance between a pair of opposite corners is 16 metres.
i. What is the distance between the other two corners?
ii. What is the area of the ground bounded by the rope?
Solution:

Length of one side of a parallelogram

Diagonals are perpendicular bisector

Kerala Syllabus 8th Standard Maths Notes Pdf Chapter 8 Question 10.
In the figure, the midpoints of the diagonals of a rhombus are joined to form a small quadrilateral:

i. Prove that this quadrilateral is a rhombus.
ii.The area of the small rhombus is 3 square centimetres. What is the area of the large rhombus?
Solution:

The diagonals of the rhombus intersect at O and they bisect each other at right angles.
OA = OC
\(\frac{\mathrm{OA}}{2}=\frac{\mathrm{OC}}{2}\)
OQ = OS
Similarly, since OB = OD
ie., OR = OP
The diagonals of PQRS bisect each other. (Since OQ = OS and OR = OP)
Since the diagonal AC is perpendicular to BD, the diagonals of PQRS are also mutually perpendicular bisectors.
therefore PQRS is a rhombus

ii. Area of quadrilateral PQRS = 3 sq. cm.

8th Area Of Quadrilateral Formula Class 8 Kerala Syllabus Question 11.
What is the area of the largest rhombus that can be drawn inside a rectangle of sides 6 centimetres and 4 centimetres?
Solution:
If we join the midpoints of the opposite sides of rectangle, we get the longest diagonal of the rhombus.

Textbook Page No 156

Hsslive 8th Maths Kerala Syllabus Chapter 8 Question 12.
Draw a rectangle of sides 7 centimetres and 4 centimetres. Draw isosceles trapeziums of the same area, with the following specifications.
i. Lengths of parallel sides 9 centimetres, 5 centimetres.
ii. Lengths of non-parallel sides 5 centimetres.
Solution:
i. Draw a rectangle of length 7 cm and breadth 4 cm.

Extent AB either side by 1 cm and mark E and F. Also mark the points G and H on CD at a distance 1 cm from C and D respectively. Draw the rhombus EFGH.

ii. Draw the rectangle ABCD.The arc of radius 5cm drawn with A as centre intersects DC at P. The arc of 5cm drawn with C as the centre intersect extented AB at Q.

Std 8 Maths Kerala Syllabus Kerala Syllabus Chapter 8 Question 13.
Calculate the area of the isosceles trapezium drawn below:

Solution:

8th Std Maths Notes Kerala Syllabus Chapter 8 Question 14.
The parallel sides of an isosceles trapezium are 8 centimetres and 4 centimetres long; and non – parallel sides are 5 centimetre long. What is its area?
Solution:
Divide the isosceles trapezium into a rectangle and two triangles.

Textbook Page No 158

Area Of Quadrilateral Class 8 Kerala Syllabus Chapter 8 Question 15.
The lengths of the parallel sides of a trapezium are 30 centimetres, 10 centimetres and the distance between them is 20 centimetres. What is its area?
Solution:

Question 16.
Compute the area of the trapezium shown below:

Solution:
Consider the ∆ PQS, it is a right angled triangle

Textbook Page No 159

Question 17.
Compute the area of the hexagon below.

Solution:

Total area = 108 + 108 + 168 = 384 cm²

Question 18.
This is a picture drawn in the lesson Construction of Quadrilaterals.

What is the area of the large trapezium made up of the four smaller ones?
Solution:

Question 19.
What is the area of the quadrilateral shown below?

Solution:

Question 20.
Prove that for any quadrilateral with diagonals perpendicular, the area is half the product of the diagonals.
Solution:

Question 21.
Compute the area of the quadrilateral shown below:

Solution:

Question 22.
Compute the area of the parallelogram shown below:

Solution:
The diagonal divides the parallelogram into two equal triangles.

Area of the parallelogram = 2 × area of triangle = 2 × 96 = 192 cm²

Question 23.
The three blue lines in the picture below are parallel:

Prove that the areas of the quadrilaterals ABCD and PQRS are in the ratio of the lengths of the diagonals AC and PR.
i. How should the diagonals be related for the quadrilaterals to have equal area?
ii. Draw two quadrilaterals, neither parallelograms nor trapeziums, of area 15 square centimetres
Solution:


i. For the quadrilaterals to have equal area, the diagonals AC and PR should be equal.
ii. Draw three parallel lines, distance between them is 5 cm each. Also draw two quadrilaterals with diagonals PQ and RS 6 cm each.

Additional Questions And Answers

Question 1.
Which of the following has greater area; a parallelogram with one side 12 cm and distance between parallel sides is 6 cm or a square having diagonals 12 cm each?
Solution:

Question 2.
Area of a trapezium is 128 sq. cm and the distance between its parallel sides is 8 cm. Length of one parallel side is 28 cm. Find the length of the other parallel side.
Solution:

Length of the other Parallel side = 4 cm

Question 3.
Find the area of a rhombus of one side 6 cm and one diagonal 6 cm.
Solution:

Question 4.
It was decided to lay tiles in the room of Hari’s house with tiles in the shape of trapezium. The parallel sides of a tile is 40 cm and 20 cm and the distance between them is 10 cm. If 200 tiles are needed to lay tiles in the hall, find the area of the hall in square metres.
Solution:

Number of tiles required = 200
Area of the hall = 200 × 300
= 60000 cm²
= 6 m²

Question 5.
In the pictures given below, which is has more area?

Solution:
Rectangle has the maximum area among the parallelograms with same sides.

Question 6.
The ratio of the two adjacent sides of a parallelogram is 3 : 2. Distance between longer sides is 10 cm. If area is 900 cm², find the sides of the parallelogram?
Solution:
Let the sides be 3x and 2x

Question 7.
If one side of a parallelogram is ‘a’ and height of that side is h, prove that area = ah.
Solution:
In the figure, ABCD is a parallelogram. AB = a

The perpendicular distance from D to AB = h
By drawing the diagonal BD, we can divide the parallelogram into two equal triangles. ∆ ABD and ∆ BCD are equal triangles. ∴ their areas are equal, ie; area of the parallelogram is two times of the area of ∆ ABD.

Question 8.
PQRS is a rhombus. If the diagonals are 8 cm and 9 cm each, compute the area.

Solution:
Let the diagonals of the rhombus are d₁ and d₂

Question 9.
The perimeter of a rhombus is 40 cm. If the length of one diagonal is 16 cm. What is the length of the other diagonal. Find the area?
Solution:
Perimeter = 40 cm
One side = 10 cm .
∆ POQ is a right angled triangle Since d₁ = 16 cm, OP = 8 cm
∴ 8² + OQ² = 10²

Question 10.
Draw a square of area 12.5 cm² and write the geometric principles of the construction.
Solution:

Draw a line of length 5 cm and its perpendicular bisector.
Draw a circle with the midpoint of this line as centre and the distance to one end as radius. This line will be the diameter of the circle. Mark the point of intersection of the perpendicular bisector as the circle and complete the square.

Question 11.
Compute the area of the trapezium shown below.

Solution:

Question 12.
Compute the area of the quadrilateral ABCD

Solution:

Question 13.
In the figure AB || CD. AD = BC = 13 cm
Distance between the parallel side is 12 cm. If CD = 20 cm. Find the area of ABCD.

Solution:
To compute AB

Question 14.
Draw a square and mark the mid-points. How is the area of the first square with the second square.
Solution:
Draw the square ABCD.
The perpendicular bisector of the side AB bisects CD also at right angles.

Similarly draw the perpendicular bisectors of AD and BC. Which intersect the sides. Join the points of intersection to complete the square.

We see, 8 triangles in the figure. The square PQRS is formed by joining 4 among these. The area of the square PQRS is half of the area of the square ABCD.

Question 15.
What is the maximum area of parallelogram of sides 8 cm and 5 cm? What is the speciality of the parallelogram of maximum area?
Solution:
The area will be maximum for a rect¬angle and the maximum area is 40 cm².

Question 16.
See the quadrilateral Ravi drew. In it AC = 12 cm and perpendiculars to AC are 6 cm and 12 cm.

a. Find the area of the quadrilateral.
b. Raju has to draw a parallelogram of this area. If its base is 12 cm what should be its height?
Solution:

b. If the height of the parallelogram is h its area is bh = 12 × h
This is 54 sq. cm

Question 17.
The area of a rhombus is 112 sq. cm and one of its diagonal is 16 cm long. Find the length of the other diagonal.
Solution:
Area of the rhombus \(=\frac{1}{2} \mathrm{d}_{1} \times \mathrm{d}_{2}\)
= 112 sq. cm

Question 18.
Draw a parallelogram of sides 6 cm, 4 cm and area 18 cm².
Solution:
Draw a line AB of length 6 cm. Draw a circle with centre A and radius 4 cm

In the circle (as shown in the picture) mark a point D and join AD. Draw an arc of radius 4 cm with centre B as centre and radius 4 cm and draw another arc with D as centre and radius 6 cm. Two arcs meet at C. Join CD and BC.

ABCD will be a parallelogram.

Question 19.
In the figure ABCD, AB parallel to CD and the distance between them is 8 cm. ?

AB = 12 cm, CD = 10 cm. Compute the area of quadrilateral (trapezium) ABCD?
Solution:

Question 20.
Compute the area of the quadrilateral ABCD in the figure.

Solution:

You can Download सुख-दुख Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 1 सुख-दुख (कविता)

सुख-दुख पाठ्यपुस्तक के प्रश्न और उत्तर

सुख दुख कविता का सारांश Kerala Syllabus 8th प्रश्ना 1.
‘सुख-दुख की खेल मिचौनी खोले जीवन अपना मुख।’ इन पंक्तियों का आशय क्या है?

उत्तर:
सुख और दुख एक सिक्के के दो पहलू हैं। जीवन में सुख और दुख का होना स्वाभाविक है। जीवन का असली मुख सुख-दुख के मिलन में खुलता है।

Sukh Dukh Poem Summary In Hindi Kerala Syllabus 8th प्रश्ना 2.
‘फिर घन में ओझल हो शशि फिर शशि में ओझल हो घन।’ यहाँ घन और शशि किन-किन के प्रतीक हैं?

उत्तर:
यहाँ घन दुख का प्रतीक है और शशि सुख का प्रतीक है।

सुख-दुख कविता का सारांश Kerala Syllabus 8th प्रश्ना 3.
‘अविरत सुख भी उत्पीड़न’ क्या आप इससे सहमत है? क्यों?

उत्तर:
मैं इस कथन से सहमत हूँ। क्योंकि जीवन का असली मुख सुख-दुख पूर्ण है। हमेशा सुख हो तो जीवन की असलियत की पहचान नहीं होगी। सदा सुखी रहनेवालों के जीवन में दुख के आने पर वे उसके सामना करने में असफल हो जाते हैं।

सुख-दुख Textbook Activities

Sukh Dukh Kavita Ka Saransh Kerala Syllabus 8th प्रश्ना 1.
कविता से शब्द युग्मों का चयन करके लिखें।

उत्तर:
1. सुख-दुख
2. निशा-दिवा
3. सोता-जागता
4. साँझ-उषा
5. विरह-मिलन
6. हास-अश्रु

Sukh Dukh Poem Kerala Syllabus 8th प्रश्ना 2.
वर्गपहेली की पूर्ति करें।


दाईं ओर:

1. ‘मुख’ का समानार्थी शब्द।
5. इसका अर्थ है- ‘आकाश’।
6. सुख-दुख के मिलन से यह परिपूर्ण हो जाता है।
9. यह ‘निरंतर’ का समानार्थी है।

नीचे की ओर:

2. सफलता पाने के लिए इसकी ज़रूरत है।
3. ‘मेघ’ के लिए इस कविता में प्रयुक्त शब्द।
4. कविता में ‘संसार’ का प्रतीकात्मक शब्द।
7. ‘उषा’ का अर्थ।
8. ‘सुख-दुख’ किस विधा की रचना है?
10. यह कभी नहीं बोलना चाहिए।
उत्तर:
दाईं ओर:

1. आनन,
5. गगन,
6. जीवन,
9. अविरत
नीचे की ओर:

2. मेहनत,
3. घन,
4. जग,
7. प्रभात,
8. कविता,
10. झूठ

सुख दुख कविता का आशय Kerala Syllabus 8th प्रश्ना 3.
सुख-दुख कविता का आशय लिखें।

उत्तर:
सुख-दुख कविता श्री. सुमित्रानंदन पंत की है। इसमें कवि जीवन की असलियत को समझाने का प्रयास किया है। वे कहते हैं कि जीवन सुख-दुखपूर्ण है। सुख-दुख की खेलमिचौनी से जीवन अपना वास्तविक मुख खोल देता है। सुख-दुख के मधुर मिलन से जीवन पूर्ण हो जाता है। जीवन के दुख कभी सुख में बदलता है तो कभी सुख दुख में परिणित हो जाता है। संसार अति सुख और दुख से पीड़ित है। सुख-दुख मानव जीवन में बराबर रहें। निरंतर दुख और निरंतर सुख दोनों पीड़ा देनेवाले हैं। सुख-दुख रूपी दिन-रात में संसार का जीवन सोता और जागता है। इस संध्या और उषा के आँगन में विरह और मिलन का आलिंगन हो रहा है। मानव जीवन का मुख सदा हँसी और आँसू से भरा है।

सुख-दुख Summary in Malayalam and Translation

सुख-दुख शब्दार्थ Word meanings

You can Download Equations Questions and Answers, Activity, Notes, Kerala Syllabus 8th Standard Maths Solutions Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Maths Solutions Chapter 2 Equations

Equations Text Book Questions and Answers

Textbook Page No. 34

Kerala Syllabus 8th Standard Maths Solution Chapter 2 Question 1.
“Six more marks and I would’ve got full hundred marks in the maths test.” Rajan was sad. How much mark did he actually get? the num¬ber.
Solution:
The marks Rajan got is six less from 100
∴ 100 – 6 = 94

Kerala Syllabus Class 8 Maths Solutions Chapter 2 Question 2.
Mother gave Rs.6o to Lissy for buying hooks. She gave back the 13 rupees left. For how much money did she buy books?
Solution:
Money Lissy got from mother = Rs. 6o
Amount she returned = Rs. 13
Amount Lissy used to buy books = 60 – 13 = Rs. 47

Kerala Syllabus 8th Standard Maths Notes Chapter 2 Question 3.
Gopalan bought a bunch of bananas. 7 of them were rotten which he threw away. Now there are 46. How many bananas were there in the bench?
Solution:
No. of rotten bananas = 7
No. of bananas left =46
∴ Total number of bananas
= 46 + 7 = 53

Hss Live Guru 8th Maths Chapter 2  Question 4.
Vimala spent 163 rupees shopping and now she has 217 rupees. How much money did she have at first?
Solution:
Amount spent by Vimala for shopping = Rs. 163
Amount left with her after shopping Rs. 217
Total amount = 163 + 217 = Rs. 380

Hsslive Guru 8th Class Maths Chapter 2 Question 5.
264 added to a number makes it 452. What is the number?
Solution:
Number = 452 – 264 = 188

Kerala Syllabus 8th Standard Notes Maths Chapter 2 Question 6.
198 subtracted from a number makes it 163. What is the number.
Solution:
Number = 198 + 163 = 361

Textbook Page No. 35

8th Scert Maths Solutions Chapter 2 Question 1.
In a company the manager’s salary in five times that of a peon. The manager gets Rs, 40,000 a month. How much does a peon get a month?
Solution:
Salary of the manager = Rs.40,000.
\(\frac{1}{5}\) th salary of the manager in the salary of the peon.
∴ Salary of the peon = \(\frac{40000}{5}\)
= 8000

Kerala Syllabus 8th Standard Maths Textbook Solutions Question 2.
The travellers of a picnic split equally the 5200 rupees spent. Each gave 1300 rupees. How many travellers were there?
Solution:
Total amount spent = Rs. 5200
Share of one = Rs.1300
∴ Members in the group = \(\frac{5200}{1300}\) = 4

Kerala Syllabus 8th Maths Solutions Chapter 2 Question 3.
A number multiplied by 12 gives 756. What is the number?
Solution:
Number = \(\frac{756}{12}\) = 63

Class 8 Maths Chapter 2 Kerala Syllabus Question 4.
A number divided by 21 gives 756. What is the number
Solution:
Number 756 × 21 = 15,876

Textbook Page No. 37

8th Standard Maths Guide Kerala Syllabus Chapter 2 Question 1.
Anita and her friends bought pens. For five pens bought toge¬ther, they got a discount of three rupees and it cost them 32 rupees. Had they bought the pens separately, how much would each have to spend?
Solution:
Cost for 5 pens = Rs. 32
discount = Rs.3
Real cost = 32 + 3 = 35
∴ Real cost of 1 pen = \(\frac{35}{5}\) = 7

8th Class Maths Notes Kerala Syllabus Chapter 2 Question 2.
The perimeter of a rectangle is 25 metres and one of its side is 5m. How many metres is the other side?
Solution:
Perimeter of the rectangle = 25m
One side = 5m
Double of the sum of length and breadth is perimeter = \(\frac{25}{2}\) – 5 = 7.5

Maths Class 8 Kerala Syllabus Chapter 2 Question 3.
In each of the problems below, the result of doing some operations on a number is given. Find the number?
(i) Three added to double is 101
(ii) Two added to triple is 101
(iii) Three substracted from double is 101
(iv) Two substracted from triple is 101
Solution:

Hsslive 8th Class Maths Chapter 2 Question 4.
Half a number added to the number gived III. What is the number?
Solution:
1\(\frac{1}{2}\) of the numbers = 111
ie \(\frac{3}{2}\) of the number = 111
number = 111 × \(\frac{2}{3}\) = 74

Hss Live Guru 8 Maths Chapter 2 Question 5.
A piece of folk maths. A child asked a flock of birds, ‘How many are you”? A bird replied
“We and us again
with half of us
And half of that
with one more.
would make hundred” How many birds were there?
Solution:
We and us ⇒ double (2 times)

Textbook Page No. 41

Hsslive Guru Maths 8th Standard Chapter 2 Question 1.
The perimeter of a rectangle is 80 metre and its length is one metre more than twice the breadth. What are its length and breadth?
Solution:
If x be the breadth, Then length
= 2x + 1
2 (x + 2x + 1) = 80
2(3x + 1) = 80
6x + 2 = 80
6x = 80 – 2
x = \(\frac{80-2}{6}\) = 13.
breadth = 13 metre
Length = 2 × 13 + 1 = 27 metre

Question 2.
From a point on a line another line is to be drawn such that the angle on one side is 50° more than the angle on the other side. How much is the smaller angle?
Solution:
Let one angle is x, the second angle = x + 50
Sum of two angles on a line = 180°
ie x + x + 50 = 180°
2x + 50 = 180°, 2x = 180 – 50
x = \(\frac{180-50}{2}\) = 65
The angles are 65°, 115°

Question 3.
The price of a book is 4 rupees more than the price of a pen. The price of a pencil is 2 rupees less than the price of the pen. The total price of 5 books, 2 pens and 3 pencils is 74 rupees. What is the price of each?
Solution:
Let the cost of a pen = x
cost of a book = x + 4
Cost of a pencil = x – 2
ie.5(x + 4) + 2x + 3(x – 2) = 74
5x + 20 + 2x + 3x – 6 = 74
10 x + 14 = 74, 10x = 74 – 14
x = \(\frac{74-14}{10}\) = 6
Cost of pen = Rs. 6
Cost of book = Rs. 10
Cost of pencil = Rs. 4

Question 4.
(i) The sum of three consecutive natural numbers is 36. What are the numbers
(ii) The sum of three consecutive even numbes is 36. What are the numbers?
(iii) Can the sum of three consecutive odd numbers be 36. Why?
(iv) The sum of three consecutive odd numbers is What are the numbers?
(v) The sum of three consecutive natural numbers is 33. What are the numbers?
Solution:
x + x + 1 + x + 2 = 36
3x + 3 = 36
x = \(\frac{36-3}{3}\) = 11
∴ Numbers are 11, 12, 13
x – 1 + x + x + 1 = 36
3x = 36
x = 12
Or
∴ Numbers are 11, 12, 13

(ii) (x – 2) + x + (x + 2) = 36
3x = 36
x = 12
∴ The numbers are 10, 12, 14

(iii) The sum of three odd numbers is odd.
∴ It is not possible to get 36 as the sum of three odd numbers.

(iv) (x – 2) + x + (x + 2) = 33
3x = 33
x = 11
∴ Numbers are 9, 11, 13

(v) (x – 1)+ x + (x + 1) = 33
3x = 33
x = 11
The numbers are 10, 11, 12

Question 5.
(i) In a calender, a square of four numbers is marked. The sum of the numbers is 80. What are the numbers?
(ii) A square of nine numbers is marked in a calendar. The sum of all there numbers is 90. What are the numbers?
Solution:
(i) Let,
x x + 1
x + 7 x + 8
four numbers in the square of 4 numbers.
Sum of the numbers = x + (x + 1) + (x + 7) + (x + 8) = 80
4x + 16 = 80, 4x = 80 – 16

Textbook Page No. 44

Question 1.
Ticket rate for the science exhibition is rupees 10 for a child and 25 rupees for the adult. Rs. 740 was got from 50 persons. How many children among them?
Solution:
Let the numbers of children be x
Then the number of adult = 50 – x
∴ 10 x + 25 (50 – x) = 740
10 x + 1250 – 25 x = 740
1250 – 15 x = 740
15 x = 1250 – 740
x = \(\frac{1250-740}{15}\) = 34
Number of adults = 50 – 34 = 16

Question 2.
A class has the same numbers of girls and boys. Only 8 boys were absent on a particular day and then the number of girls was double the number of boys and girls?
Solution:
Let the number of boys =number of girls = x
2 (x – 8) = x
2x – 16 = x, 2x – x = 16
x – 16 = 0
∴ x = 16
∴ The number of boys = No. of girls = 16

Question 3.
Ajayan is ten years older than Vijayan. Next year Ajayan’s age would be double that of Vijayan. What are their ages now?
Solution:
Let the age of Vijayan = x
Age of Ajayan = x + 10
Age of Vijayan after an year = x + 1
Age of Ajayan after an year = x + 11
2 (x + 1) = x + 11
2x + 2 = x + 11
2x – x = 11 – 2
x = 9
∴ Age of Vijayan = 9
Age of Ajayan = 19

Question 4.
Five times a number is equal to three times the sum of the number and 4. What is the number?
Solution:
Let the number = x
Five times number = 5x
4 more than the number = x + 4
Three times of it = 3(x + 4)
ie. 5 x = 3 (x + 4)
5 x = 3x + 12
5 x – 3 x = 12
2 x = 12
x = 6

Question 5.
In a co-operative society, the number of men is thrice the number of women 29 women and 16 men more joined the society and now the number of men is double the number of women. How many women were there in the society at first?
Solution:
Let the number of women = x
No of men = 3 x
No. of women when 29 more were joined = x + 29
No. of men when 16 more were joined = 3x + 16
3 x + 16 = 2 (x + 29)
3x + 16 = 2x + 58
3x – 2x = 58 – 16
x = 42
No. of women = 42;
No. of men = 3 × 42 = 126

Equations Additional Questions & Answers

Question 1.
“If you give rupees 5 to me both of us have equal amounts with us”. Ajay told to Vineeth. Then Vineeth told to Ajay “ If you give rupees 5 to me I will have 5 times more money than yours. Find out the amount both of them have?
Solution:
Let Ajay has Rs.x with him and when Vineeth gives Rs. 5, both of them have x + 5 rupees with them. If Vineeth gets RS.5 he has 5 times more than Ajay.
x + 5 + 5 = 5x
4 x = 10
x = \(\frac{10}{4}=\frac{5}{2}\) = 2.5 Rupees
Amount with Ajay = Rs. 2.50
Amount with Vineeth = Rs.12.50

Question 2.
The perimeter of a triangle is 49 cm. One side is 7 cm more than the second side and 5 cm less than the third side. Find out the lengths of three sides?
Solution:
Let the second side = xcm
First side = x + 7 cm
Third side = x + 12 cm
Perimeter = x + x + 7 + x + 12 = 49
= 3x = 49 – 19 = 30
3x = 20, x = 10 cm
The sides are = 10 cm, 17 cm, 22 cm

Question 3.
The sum of two numbers is 9, 8 times the smaller number is 2 more than 6 times the bigger number. Write an equation to find the numbers?
Solution:
Let the smaller number = x
Bigger number = 9 – x
Equation is, 8 x = 6 (9 – x) + 2

Question 4.
Anitha and friends bought pens. When they bought a packet of 5 pens they got a discount of rupees 2. They paid rupees 18. Find the cost of each one buys separately?
Solution:
Total cost in including discount = Rs. 20
Cost of 5 pens = Rs.20
Cost of 1 pen = \(\frac{20}{5}\) = 4

Question 5.
The cost of a chair and a table is Rs. 1500. The cost of table is 4 times the cost of chair. Find the expense of each?
Solution:
Let the cost of chair = x
Cost of table = 4x
x + 4 x = 1500
5x = 1500
x = 300
cost of chair = Rs. 300
Cost of table = Rs. 1200

Question 6.
There were 25 questions in the examination written by Jafar. Each correct answer gets 2 marks. There is a loss of mark for each wrong answer. Jafar answered all the questions. He got 35 marks. Find^ut the number of correct answers?
Solution:
Let the number of correct answers = x
Total number of questions = 25
No. of wrong answers = 25 – x
Marks for correct answers = 2x
marks losses for wrong answers = 25
ie. 2x – (25 – x) = 25
2x – 25 + x = 35
3x = 60
x = \(\frac{60}{3}\) = 20, ∴ number of correct answers = 20.

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Kerala SSLC Biology Model Question Paper 2 Malayalam Medium

Instructions:

• The first 15 minutes is the cool-off time.
• You may use the time to read the questions and plan your answers.
• Answer only on the basis of instructions and questions given.
• Consider score and time while answering.

Time: 1½ Hours
Total Score: 40 Marks

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Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 20 Static Electricity in Malayalam

Static Electricity Text Book Questions and Answers

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Kerala SSLC Social Science Model Question Paper 3 Malayalam Medium

Instructions:

• The first 15 minutes is the cool-off time. You may use the time to read the questions and plan your answers.
• Answer all questions in PART – A. Answer any one from the questions given under each question number in PART – B.

Time: 2½ Hours
Total Score: 80 Marks

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Kerala SSLC Physics Model Question Paper 3 Malayalam Medium

General Instructions:

1. The first 15 minutes is the cool off time. You may – use the time to read and plan your answers.
2. Answer the questions only after reading the instructions and questions thoroughly.
3. Questions with marks series 1, 2, 3 and 4 are categorized as sections A, B, C and D respectively.
4. Five questions are given in each section. Answer any four from each section.
5. Answer each question by keeping the time.

Time: 1½ Hours
Total Score: 40 Marks

You can Download गांधीजी गांधीजी कैसे बने Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Hindi Solutions Unit 2 Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Hindi Solutions Unit 2 Chapter 2 गांधीजी गांधीजी कैसे बने (लेख)

गांधीजी गांधीजी कैसे बने Textual Questions and Answers

गांधीजी गांधीजी कैसे बने विश्लेषणात्मक प्रश्न

Gandhiji Gandhiji Kaise Bane Kerala Syllabus 9th प्रश्ना 1.
वे धीरे-धीरे गांधी बने’ इसका मतलब क्या है?

उत्तर:
अपने कर्मों से ही एक व्यक्ति महान बनता है। कोई भी एकाएक अपने आप महान नहीं बनते। गांधीजी पहले एक साधारण आदमी था। लेकिन उन्होंने अपने अनुभवों से अपने को गढ़ा और वे एक समाज सेवी बन गए। ऐसा परिवर्तन उनके जिंदगी में धीरे-धीरे आया था।

Kerala Syllabus 9th Standard Hindi Solutions Unit 2 Chapter 2 प्रश्ना 2.
औरत की हालत देखकर गांधीजी ने एक ही धोती पहनने का फैसला कर लिया। ऐसा फैसला लेने का उद्देश्य क्या था?

उत्तर:
तत्कालीन समाज गरीबी से त्रस्त था। यह दृश्य तत्कालीन भारत की गरीबी का चित्र उनके सामने पेश किया। भारत की आम जनता गरीबी एवं अभावों से विवश थी.। यह समझकर गांधीजी ने उनके समान जीने का निश्चय किया। यह गांधीजी की ज़िंदगी का एक अहम मोड़ था।

गांधीजी गांधीजी कैसे बने Text Book Activities

गांधीजी गांधीजी कैसे बने विधात्मक प्रश्न

Gandhiji Gandhiji Kaise Bane Notes Kerala Syllabus 9th प्रश्ना 1.
भाषण तैयार करें। ‘मेरा जीवन ही मेरा संदेश है’ यह गांधीजी का कथन है। पाठभाग के आधार पर इसका विश्लेषण करके भाषण तैयार करें।

उत्तर:

गांधीजी गांधीजी कैसे बने Additional Questions and Answers

गांधीजी गांधीजी कैसे बने आशयग्रहण के प्रश्न

Gandhiji Gandhiji Kaise Bane Malayalam Kerala Syllabus 9th प्रश्ना 1.
गांधीजी की पहले की वेश-भूषा कैसी थी?

उत्तर:
गांधीजी कुर्ता पाजामा पहन रखा था। पाँव में चप्पल थे। सर पर गांधी टोपी थी। बच्चों के स्कूल बैग की तरह गले में झोला टाँगता था।

Gandhiji Gandhiji Kaise Bane Question Answer Kerala Syllabus 9th प्रश्ना 2.
बचपन में गांधीजी के चरित्र की विशेषता क्या थी?

उत्तर:
बचपन में गांधीजी को अंधेरे से डर लगता था। उन्हें लगता था कोई भूत आकर उन्हें पकड़ लेगा।

9th Class Hindi Notes Kerala Syllabus प्रश्ना 3.
किस घटना गांधीजी की वेशभूषा में परिवर्तन लाया?

उत्तर:
एक बार गांधीजी भाषण देने मदुरै गए। वहाँ एक औरत को देखा जो तालाब में अपनी धोती धो रही थी। इसप्रकार धो रही थी कि आधी पहनती थी और बाकी आधी धोती थी। फिर धुली हुई पहन लेती थी और शेष को धोती थी। इस घटना ने गांधीजी को गरीबी पर सोचने को विवश किया। इससे उन्होंने एक धोती ही पहनने का फैसला ले लिया।

Kerala Syllabus 9th Standard Hindi Notes प्रश्ना 4.
गांधीजी सादगी से जीवन बिताते थे। पाठ भाग से एक उदाहरण पेश करें।

उत्तर:
गांधीजी का बढ़िया पेन एक दिन चोरी चला गया। तब से एक बच्चे से दी गई पेंसिल से लिखना शुरू किया। लिखते-लिखते पेंसिल छोटी हो गई। तो उन्होंने कागज़ की भोंगली लगाकर धागे से बाँधकर लिखते थे।

गांधीजी गांधीजी कैसे बने Grammar

गांधीजी गांधीजी कैसे बने व्याकरण के प्रश्न

9th Hindi Notes Kerala Syllabus प्रश्ना 1.
ये वाक्य पढ़ें :

प्रत्येक वाक्य में रेखांकित शब्दों का आपसी संबंध पहचानें।

i. औरत तालाब में अपनी धोती धो रही थी।
ii. परीक्षा में गांधीजी चौंतीस बच्चों में से बत्तीसवें स्थान पर रहे।
iii. ये लोग कविता करते है
iv. मेरे पड़ोस में एक गरीब और मेहरून्नीसा रहती है।

Hss Live Guru Class 9 Hindi Kerala Syllabus 9th प्रश्ना 2.
तालिका के शब्दों से वाक्य बनाएँ:

उत्तर:
1. धरती पानी के लिए तरस रही है।
2. रीता अभी -अभी घर आई है
3. गाती आती है
4. सलमान क्रिकेट खेलता है
5. गौरव पत्र लिखता हे.
6. सलमान परसो आएगा

गांधीजी गांधीजी कैसे बने Summary in Malayalam and Translation

गांधीजी गांधीजी कैसे बने शब्दार्थ

Students can Download Maths Chapter 10 Statistics Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 8th Standard Maths Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

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Statistics Text Book Questions and Answers

Students can Download Maths Chapter 9 Negative Numbers Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 8th Standard Maths Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

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Students can Download Basic Science Chapter 2 Cell Clusters Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 8th Standard Basic Science Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

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