PrasannaStudents can Download Chemistry Part 2 Chapter 2 Non-Metals Questions and Answers, Summary, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 9th Standard Chemistry Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
You can Download Equal Triangles Questions and Answers, Activity, Notes, Kerala Syllabus 8th Standard Maths Solutions Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.
Textbook Page No. 11
Equal Triangles Class 8 Questions And Answers Kerala Syllabus Question 1.
In each pair of triangles below, find all pair of matching angles and write them down.
Solution:
(i) ∠A = ∠R (The angles opposite to 5cm sides)
∠B = ∠P (The angles opposite to the sides of length 4 cm)
∠C = ∠Q (The angles opposite to the sides of length 6 cm)
(ii) ∠L = ∠Y (The angles opposite to the sides of length 10cm)
∠M = ∠Z (The angles opposite to the side of length 4 cm)
∠N = ∠X (The angles opposite to the side of length 8cm)
Kerala Syllabus 8th Standard Maths Notes Question 2.
In the triangles below AB = QR, BC = RP, CA = PQ
Compute ∠C of ∆ABC and all angle of ∆PQR.
Solution: C = 80° (Use the property that the sum of three angles of a triangle is 180°)
AB = QR
∴ ∠C = ∠P
∠C = 80°
∴ ∠P = 80°
BC = RP
∴ ∠A = ∠Q
∠A = 40°
∴ ∠Q = 40°
CA = PQ
∴ ∠B = ∠R
∠B = 60°
∴ ∠R = 60°
(The angle opposite to equal sides are equal)
Equal Triangles – Class 8 Solutions Kerala Syllabus Question 3.
In the triangle below.
AB = QR BC = PQ CA = RP
Compute the remaining angles of both the triangles.
Solution:
AB = QR ∴ ∠C = ∠P
BC = PQ ∴ ∠A = ∠R
CA = RP ∴ ∠B= ∠Q
∠A = 60° ∴ ∠R = 6o°
∠Q = 70° ∴ ∠B = 70°
∠A = 60, ∠B = 70° then ∠C = 180 – (60° + 70°)
∴ ∠C = 50° ∴ ∠P = 50°
Equal Triangles Class 8 Notes Pdf Kerala Syllabus Question 4.
Are the angles of ∆ABC and ∆ABDequal in the figure above? Why?
Solution:
The side AB is common to both the triangles in the figure.
The side of ∆ABC are equal to the sides of ∆ABD. So the angles of ∆ABC are equal to the sides of ∆ABC.
8th Standard Maths Guide Kerala Syllabus Question 5.
In the quadrilateral ABCD shown below, AB = AD, BC = CD
Compute all the angles of the quadrilateral?
Solution:
AB = AD, BC = CD
AC is the common side
The sides of the triangles ABC and ADC are equal. So their angles are also equal. AB = DC
∴ ∠ACD = ∠ACB = 50° (Angles opposite to equal sides of a triangle are equal)
BC = CD
∴ ∠BAC = ∠DAC = 30° (Angles opposite to equal sides are equal in a triangle )
∴ ∠D = ∠B = 100°
Textbook Page No. 15
8th Class Maths Notes Kerala Syllabus Question 1.
In each pair of triangles below find the pairs of matching angles and write them down.
Solution:
(i) BC = PR (If two sides of a triangle and the angle made by them are equal to two sides of another triangle and the angle made by them, then the third sides of the triangle are also equal.)
∴ ∠B = ∠R
∴ ∠C = ∠P
∴ ∠A = ∠Q (The opposite angles of equal sides of two triangles are also equal)
(ii) MN = XY (If two sides of triangle and the angle made by then are equal to two sides of another triangle and the angle made by them, then the third sides of the triangle are also equal)
∴ ∠L = ∠Z
∠M = ∠Y
∠N = ∠X (If two sides of a triangle are equal, the angles opposite to these sides are also equal)
Class 8 Maths Chapter 1 Kerala Syllabus Question 2.
In the figure below, AC and BE are parallel lines.
(i) Are the lengths of BC and DE equal. Why?
(ii) Are BC and DE parallel? Why?
Solution:
(i) Given AC and BE are parallel lines.
∴ ∠CAB = ∠EBD
When we consider the triangles ∆CAB, ∆EBD (Corresponding angles)
BC = DE (The two sides of ∆ CAB and the angle made by them are equal to the two sides of ∆ EBD and the angle made by them. So the thirif side of triangle are also equal.)
(ii) Yes, they are parallel.
∠ABC = ∠BDE (The angles opposite to the equal sides of equal triangles are equal) But they are corresponding angles. BC and DE are parallel.
Kerala Syllabus Class 8 Maths Solutions Question 3.
Is ABCD in the figure, a parallelo¬gram? Why?
Solution:
AC = BD
AB is the common side.
The angles between the sides AC, AB and BD, AB are equal.
∴ BC = AD
The opposite sides of quadrilateral ∆CBD are equal. The angles opposite to equal sides of triangles ∆ABC and ∆ABD are equal. So the opposite angles in quadrilateral ACBD are also equal.
∴ ACBD is a parallelogram.
Kerala Syllabus 8th Standard Maths Notes Pdf Question 4.
In the figure below, M is the midpoint of the line AB. Compute the other two angles of ∆ABC
Solution:
AM = BM (Given M is the mid point of AB)
CM = CM (common)
∠AMC = 90° = ∠BMC
∴ The two sides in ∆AMC and ∆BMC and the angle made by them are equal.
So the third side and other angles are equal.
∠A = 50° ∴ ∠B = 50°
∠ACM = 40° ∴ ∠BCM = 40°
∴ ∠C = 80°
Class 8 Maths Chapter 1 Notes Kerala Syllabus Question 5.
In the figure below, the lines AB and CD are parallel and M is the mid point of AB.
(i) Compute the angle of ∆AMD, ∆MBC and ∆DCM?
(ii) What is special about the quadrilateral AMCD and MBCD?
Solution:
Given AB = 12 cm and M is the mid-point of AB.
∴ AM = MB = 6 cm
In quadrilateral AMCD,
AM = CD
AB||CD ∴ AM||CD
∴ AMCD is a parallelogram.
∴ ∠AMD = ∠CDM (Alternate interior angles)
∠ADM = ∠CMD (Alternate interior angles)
∠A = ∠DCM = 40° = ∠CMB
∴ ∠MCB = 80° [180 – (60 + 40)]
(i) The angles of ∆AMD, ∆MBC and ∆DCM are 40°, 60° and 80° respectively.
(ii) Both of them are parallelograms.
Textbook Page No. 21
Class 8 Maths Scert Solutions Kerala Syllabus Question 1.
In each pair of triangles below, find matching pairs of sides and write their names.
Solution:
(i) BC = PQ
AC = PR (The two angles and the side in between them in ∆ ABC are equal to the two angles and the side in between them in ∆PQR. So the third angles of the triangles ∠C and ∠R are also equal. Also BC and PQ, opposite to the 50° angle are also equal. The sides AC and PR opposite to the 70° angle are also equal.
(ii) ∠N = 70°
∠Z= 80°
MN = XZ
∠M = ∠Z(Sides opposite to equal angles are also equal)
8th Maths Notes Kerala Syllabus Question 2.
In the figure, AP and BQ equal and parallel are lines drawn at the ends of the line AB. The point of inter section of PQ and AB is marked as M.
(i) Are the sides of ∆AMP equal to the sides of ∆BMQ? Why?
(ii) What is special about the position of M on AB.
(iii) Draw a line 5.5 cm long. Using a set square, locate the midpoint of this line.
Solution:
(i) Yes, they are equal
∠P = ∠Q
∠A = ∠B (alternate angles formed by cutting the parallel lines AP and QB by PQ and AB.)
AP = QB
∴ The third angle of ∆APM and ∆BMQ and opposite sides of equal angles are equal.
(ii) AM = BM. So M is the midpoint of AB.
(iii) Draw a line segment of length 5.5 cm. Draw perpendiculars of equal lengths upward at one end of the line and downwards at the other end. Join the ends. This line divides the first one equally.
Hsslive Guru Maths 8th Standard Kerala Syllabus Question 3.
In the figure, ABCDE is a pentagon with all sides of the same length and all angles of the same size. The sides AB and AE extended, meet the side CD extended at Px and Q.
(i) Are the sides of ∆BPC equal to the sides of ∆EQD? Why?
(ii) Are the sides of AP and AQ of ∆ APQ equal? Why?
Solution:
(i) Yes, this are equal the sides and angles of a pentagon are equal.
∴ BC = DE
∠PBC = ∠PCB (Exterior angles of a regular pentagon)
∠QDE = ∠QED (Exterior angles of a regular pentagon)
∆QDE = ∆QED (If two angles and side of one triangle are equal are equal to two angles and corresponding side of the other triangle then their sides are equal.
BP = EQ and PC = DQ
(ii) AB = AE sides of regular pentagon.
BP = EQ
∴ AP = AQ [AB + BP = AC + EQ]
8th Standard Maths Notes State Syllabus Question 4.
In ∆ABC and ∆PQR shown below.
AB = QR BC = RP CA = PQ
(i) Are CD and PS equal? Why?
(ii) What is the relation between the areas of ∆ABC and ∆PQR?
Solution:
(i) AB = QR
BC = RP
∠A = ∠Q
∴ ∆ABC and ∆QRP are equal triangles. Given all sides of ∆ABC are equal to the sides of ∆QRP
∴ CD and PS are equal. (Opposite sides of equal angles
(ii) AB = QR and CD = PS
⇒ 1/2 AB × CD = 1/2QR × PS
∴ The areas of ∆ABC and ∆PQR are equal.
Question 5.
In the quadrilateral ABCD shown below the sides AB and CD are parallel. M is midpoint of the side BC. The lines DM and AB extended meet at N.
(i) Are the areas of ∆DCM and ∆BMN equal? Why?
(ii) What is the relation between the areas of the quadrilateral ABCD and the triangle ADN.
Solution:
(i) M is the midpoint of the line BC.
∴ CM = MB; BN || DC
∴ ∠DCM = ∠NBM (Alternate angles)
∠DMC = ∠NMB (Vertically opposite angles)
∴ ∆DCM and ∆BMN are equal triangles. So their areas are equal.
(ii) The areas of ∆DCM and ∆BMN are equal and quadrilateral AB, MD common
∴ The area of the quadrilateral
ABCD and the area of ∆ADN are equal.
Question 6.
Are the two diagonals of a rectangle equal. Why?
Solution:
ABCD is a rectangle.
Consider the ∆ABD and ∆ABC
AB = AB, common AD = BC (opposite sides of rectangle); ∠A = ∠B =90°
Both sides of ∆ABD and ∆ABC and the angle formed by them are equal. So the third sides BD and AC are equal. So the diagonals of the rectangle are equal.
Textbook Page No. 26, 27
Question 1.
Some are equal isosceles triangles are drawn below, In each, one angle is given. Find the other angles.
Solution:
(i) 30°, 75°, 75°
(ii) 40°, 70°, 70°
(iii) 20°, 8o°, 8o°
(iv) 100°, 40°, 40°
Question 2.
One angle of an isosceles triangle is 90°. What are the other two angles?
Solution:
The other two angles are equal. So they are 45°, 45°
Question 3.
One angle of an isosceles triangle is 6o°. What are the other two angles.
Solution:
The other two angles are equal. So they are 60°, 60°
Question 4.
In the figure below, O is the centre of the circle and A, B are points on the circle.
Compute ∠A and ∠B?
Solution:
OA = OB (radius of circle)
∆AOB is a isosceles triangle; ∴ ∠A = ∠B
∠O = 60°
∴ ∠A = ∠B = 60°
Question 5.
In the figure below, O is the centre of the circle and A, B, C are points on the circle.
What are the angles of ∆ABC?
Solution:
∆AOB, ∆AOC, ∆BOC are isosceles triangles. Each triangles are with angles 120°, 30° and 30°.
∴ ∠A = ∠B = ∠C = 30° + 30° = 60°
Text Book Page No. 29
Question 1.
Draw a line of 6.5 centimetres long and draw its perpendicular bisector.
Solution:
Draw a line segment AB of length 6.5 c.m with A and B as centres draw arcs on both sides of AB with equal radii. The radius of each of these arcs must be more the half the length of AB. Let these arcs cut each other at points C and D. Join CD which cuts AB at M
Then AM = BM. Also ∠AMC = 90°
Hence, the line segment CD is the perpendicular bisector of AB as it bisects AB at M and is also perpendicular to AB.
Question 2.
Draw a square, each side 3.75 centimetres long?
Solution:
Draw AB = 3. 75 cm at. A Construct ∠PAB = 90° from AP, cut AD = 3.75 cm
Taking D as centre, draw an arc of radius 3.75 cm and taking B as centre, draw one more are of radius 3.75 cm.
Let the two arcs intersect at point C. Join BC and DC.
Then ABCD is the required square.
Question 3.
Draw an angle of 750 and draw its bisector?
Solution:
Draw a line segment AB of any suitable length with A as centre. Draw an arc of any size to cut AB at D. With D as centre. Draw another arcs of some size to cut the previous arc at C.
Now ∠CAD = 60°. Draw ∠EAB = 90° and bisect ∠EAC.
∴∠PAC = 150 ∠DAC + ∠CAP = 60 + 15 = 75°
∴ ∠BAP = 75°
Then bisect
∠BAP AQ to the bisectors of ∠PAB
Question 4.
Draw a circle of radius 2.25 centimetres.
Solution:
Draw a line of length 4.5 cm. Draw its perpendicular bisector it meet at point ‘O’.
‘O’ is the centre of the circle and radius = 2.25 cm. Then complete the circle.
Question 5.
Draw ∆ ABC, with AB = 6 cm,
∠A = 22\(\frac{1}{2}\)°, ∠B = 67\(\frac{1}{2}\)°
Solution:
Draw the line AB in 6 cm length. Draw angle A at 45° and draw its bisector. Draw angle 135° at B and draw its bisector. Mark the point as C where bi-sectors meet.
Question 6.
Draw a triangle and perpendicular bisectors of all three sides. Do all these three bisectors intersect at the same point?
Solution:
Draw a triangle ABC. By using compass mark the arcs on both sides from each ends.
Draw the same for all sides. They intersect at same point
Question 7.
Draw a triangle and the bisectors of the three angles. Do all three bisectors intersect at the same point.
Solution:
Draw a triangle PQR and by using com-pass draw the bisectors of angles. All three bisectors meet at the same point.
Question 8.
Prove that if both pairs of opposite sides of a quadrilateral are equal, then it is a parallelogram.
Solution:
When the diagonal of a quadrilateral with equal opposite sides is drawn, we get two equal triangles. The angels opposite to the diagonal in the triangles are equal. That is the opposite sides and angles in the quadrilateral are equal. So the quadrilateral is a parallelogram.
Question 9.
In the figure, ABCD is parallelogram and AP = CQ
Prove that PBQD is a parallelogram
Solution:
DC = AB ………. (1)
AD = CB
QC = AP ……. (2) (as ABCD is a parallelogram)
(1) – (2) ⇒ DC – QC = AB – AP; ∴ DQ = PB
When, ∆ APD, ∆ CQB are considered.
AD = CB
AP = QC
∠A = ∠C, The two sides and angle formed by them in these triangles are equal. So the third sides PD and BQ are equal.
∴ Two pairs of opposite sides in the quadrilateral PBQD are equal. So PBQD is a parallelogram.
Question 10.
Prove that if all sides of a parallelogram are equal, them each diagonal is the perpendicular bisector of the other.
Solution:
The diagonal BD divides the parallelogram into two isosceles triangles. [The angles opposite to the equal sides in an isosceles triangles are equal.] So the diagonal DB bisect ∠D and ∠B.
Similarly the diagonal AC bisect A and ∠C. 4x +4y = 360° ⇒ x + y = 90°
The four triangles formed by intersec¬ting the diagonals are equal triangles. Each one 90 angle. So each diagonal is the perpendicular bisector of the other. In ∆AMD ∠AMD = 180 – (x – y) = 180 – 90 = 90° ⇒ BD ⊥ AC
Question 11.
In the figure below O is the centre of the circle and AB is the diameter. C is the point on the circle.
(i) Compute ∠CAB
(ii) Draw another figure like this with a different number for the size of ∠COB. Calculate ∠CAB
Solution:
(i) ∠BOC = 50°
∴ ∠COA = 180° – 50°= 130° (straight angle)
OA = OC (radii)
∴ AOC is an isosceles triangle.
∴ ∠A = ∠C = \(\frac{180-130}{2}\) = 25°
(ii) ∠O = 70°
∴ ∠COA = 180 – 30 = 150°
OA = OC
∴ ∠CAB = ∠ACO =\(\frac{180-150}{2}=\frac{30}{2}\) = 15°
Question 12.
In the figure below, O is the centre of the circle and AB is a diameter. C is a point on the circle.
(i) Compute ∠ACB
(ii) Draw another figure like this, changing the size of ∠COB and calculate ∠ACB.
Solution:
(i) ∠BOC = 50°
∠AOC = 130°
∆ AOC and ∆ BOC are isosceles triangles.
∠OAC = ∠OCA = \(\frac{180-130}{2}\) = 25°
∠OBC = ∠OCR = \(\frac{180-50}{2}\) = 65°
∠ACB = ∠OCA+ ∠OCB
25° + 65° = 90°
(ii) ∠AOC = 180 – 80 = 100°
∠OAC + ∠OCA = 180 – 100 = 8o°
∴ ∠OAC = ∠OCA = 80 ÷ 2 = 40°
∆ OBC
∠OBC + ∠OCB = 180 – 80 = 100°
∠OBC =∠OCB = 100 ÷ 2 = 50°
∠ACB = ∠OCA + ∠OCB
40° + 50° = 90°
Question 13.
How many different isosceles triangles be drawn with one angle 50° and any one side 7 centimetres.
Solution:
An isosceles triangle can be drawn with one angle 50° as angles either 50°, 50°, 8o° or 50°, 65°, 65°. In both the cases, 7 cm can be taken as equal sides or can be without 7 cm one as side. So there can be 4 ways of drawing diagram.
Question 14.
Draw ∆ ABC with AB = 7 cm, ∠A = 67\(\frac{1}{2}\), ∠B = 15° without using protector.
Solution:
Draw AB with length 7 cm. Extend both the sides. Draw the perpendicular from A. Draw the bisector through the left 90° angle among the 90° angles obtained. Draw an angle as 90° + 45° = 135°, Draw its bisector.
Now ∠A = 67\(\frac{1}{2}\). Draw an angle 60° in B to construct an equilateral triangle. Draw the bisector of its bisector. Then ∠B = 15°. We get ∆ ABC.
Question 1.
O is the centre of the circle in the diagram. If AB = BC,
(a) Then prove that ∠AOB = ∠BOC
(b) If OA = AB = BC, then find the values of ∠AOB and ∠BOC?
(c) Find out how many equilate¬ral triangles can be drawn in a circle with length of its side is radius.
Solution:
(a) OA = OB = OC, AB = BC
∆ OAB and ∆ OBC are equal triangles.
∴ ∠AOB and ∠BOC are equal which are opposite to the equal sides AB and BC.
(b) If OA = AB then ∆ OAB is an equilateral triangle.
If OB = BC, ∆ OBC is equilateral triangle.
∴ ∠AOB = ∠BOC = 60°
(c) Each angle at O is 60°. The angle at the centimeter is O is 360° and 6 triangles can be drawn.
Question 2.
If AB = AD, BC = CD in the diagram, then prove that ∠ABC = ∠ADC
Solution:
The three sides of triangles ∆ ABC are equal. The angles opposite to the sides are also equal.
AB = AD, BC = DC, AC = AC
AC is the common side. So the angles opposite to this side ∠ADC and ∠ABC are also equal,
i e ∠ABC = ∠ADC
Question 3.
Draw a rhombus with sides and a diagonal as 5 cm.
Solution:
Draw a line of length 5 cm. Draw equilateral triangles on both ends of the line with length 5 cm and line as one side.
Question 4.
In the figure, AB = DE, BC = EF and AC = DF. Can ∠BPD = ∠C? Prove it?
Solution:
The sides of ∆ DEF and ∆ ABC are equal. The angles opposite to equal sides are equal.
∠E = ∠B, ∠F = ∠C, ∠D = ∠A
But ∠C = 180 – (∠B + ∠A)
∠P = 180 – (∠B + ∠D)
∴ ∠D = ∠A
∴ ∠C = ∠P = 180 – (∠B + ∠A)
∴ ∠C = ∠P
Question 5.
In the figure, AB = PQ, AC = PR, BC = QR. PQ is parallel to AB.
(a) Then show that BC is parallel to QR.
(b) Also show that PR is parallel to AC
Solution:
AB = PQ, AC = PR, BC = QR
∴ The angles of ∆ ABC and ∆ PQR are equal.
∠A = ∠P, ∠B =∠Q, ∠C = ∠R
(a) AB||PQ, ∴ ∠B = ∠PMN (corresponding angle);
∴ ∠PMN = ∠Q
∴ MN||QR
∴ BC||QR
(b) BC||QR
∠R = ∠MNP = ∠C
∴ NP||AC, PR||AC
Question 6.
Diagonals of three parallelograms with equal areas are given. Draw the parallelograms.
(i) length of diagonal 7 cm
(ii) length of diagonal 6 cm
(iii) length of diagonal 5 cm.
Solution:
(i) Draw a line of length 7 cm. Draw triangles of sides 7 cm, 6 cm, 5 cm at both the ends of the line to get a parallelogram by joining both the triangles.
(ii) Draw a line of length 6 cm and follow the above method.
(iii) Draw a line of length 5 cm and follow the above method.
Question 7.
O is the centre of circle and AC, BD are diameters in the figure. Prove that AB = CD
Solution:
Consider ∆ODC and ∆OAB.
OD = OC = OA = OB (radi ∠AOB = ∠DOC; Two triangles are equal So the third sides of the triangles AB and CD are equal.)
Question 8.
ABCD in the figure is a parallelogram. P, Q, R and S are the mid points of the sides of the parallelogram. The prove that PQ = RS, and QR = PS.
Solution:
Consider the triangles ∆APS and ∆CRQ
AP = CR, (half of the equal lines AB and CD)
AS = CQ (half of the lines with equal lengths AD and BC.)
∠A = ∠C (opposite angles of the parallelogram are equal)
When two sides and the angle made by them, in a triangle are equal then the third sides are also equal.
∴ QR = PS
Similarly if ∆ DSR and ∆ BQP are considered, PQ = RS is obtained.
Question 9.
P is the midpoint of the sides AB and DF in the figure.
(a) Prove that BD = AF
(b) Is EF parallel to BC? Why?
(c) If D is the midpoint of BC, A is the midpoint of EF and Q is the mid point of DE then can Q be the midpoint of AC? Why?
Solution:
(a) Consider ∆APF and ∆DBP
FP = DP and AP = PB
∠APF = ∠DPB. The sides and the angle made by them in the triangles are equal. So the third sides BD and AF are equal.
(b) FP = PD. So the angles opposite to them are also equal.
∠FAP and ∠DBP are equal.
∴ FA||BD andBC|| EF.
(c) Consider BD = AF, in Question (a)
∴ BC = EF, Consider ∆ AEQ and ∆ DCQ
AE = DC, QE = DQ
∠AEQ = ∠CDQ
Two sides of a triangle and the angle made by them are equal. So the third sides are also equal, ie AQ = QC.
∴ So Q is the midpoint of AC
Question 10.
Draw a parallelogram if one of its diagonal is 8 cm length and one side is 6 cm. and the angle formed by the side and the diagonal is 40.
Solution:
Draw a diagonal of length 8 cm. Draw a line of 6 cm with 40 angle at its one end. Draw the same in its opposite direction.
Question 11.
One angle of an isosceles triangle is 80. Find the other possible angles of the triangle.
Solution:
8o°, 8o°, 20°
8o°, 50°, 50°
Question 12.
O is the centre of the circle in the diagram. Radiaus is 3 cm and ∠AOB = 60°. Find the perimeter of ∆ AOB?
Solution:
∆ OAB is an isosceles triangle.
∴ ∠A =∠B
∠O = 60
∴ ∠OAB is an equilateral triangle so perimeter = 3 + 3 + 3 = 9 cm.
Question 13.
OM is perpendicular to AB in the diagram. Prove that M is the mid point of AB.
Solution:
OA = OB
∴ ∆ OAB is an isosceles triangle.
∴ ∠A = ∠B
When ∆ OMA and ∆ OMB are considered, OM is the common side
∠AMO = ∠BMO = 90 ∠AOM = ∠BOM
One side of the triangle and angles at the ends of sides are equal. So the other two sides are also equal.
∴ AM = MB
∴ M is the midpoint of AB.
Question 14.
In the figure ∠ABC = ∠ADC and AB = AD. Prove that A BCD is an isosceles triangle?
Solution:
AB = AD
∴ ∆ ABD is an isosceles triangle.
∴ ∠ABD = ∠ADB
It is given that ∠ABC = ∠ADC
∴ ∠CBD = ∠CDB
∴ CD = CB
∴ ∆ BCD is an isosceles triangle.
Question 15.
In ∆ ABC, AB = AC = 10 cm. M is the midpoint of BC. If BC = 12 cm, Find AM? Also find the area of ∆ ABC?
Solution:
Question 16.
Show that, the triangle obtained by joining the mid points of the sides of an isosceles triangle is also an isosceles triangle.
Can we get an equilateral triangle by joining the mid points of the sides of an equilateral triangle?
Solution:
∆ ABC is an isosceles triangle. P, Q, R are the mid points of the sides of the triangle. Consider ∆ PBR and ∆ QRC.
PB = QC
BR = CR
∠B = ∠C
Two sides and the angle made by them are equal. The third sides PR and QR also equal.
∴ ∆ PQR is an isosceles triangle. Similarly the triangle obtained by joining the mid points of the sides of an equilateral triangle is an equilateral triangle.
Question 17.
In the figure ∠B = ∠C = 90. If AB = CE and BE = CD, then find angles of ∆AED?
Solution:
If ∆ ABE and ∆ ECD are considered, AB = EC, BF = DC and ∠B = ∠C. Two sides of the triangle and the angle made by them are equal. The third sides AE and DE are also equal. ∆ADE is an isosceles triangle.
∆BAE = ∆DEC
∠BEA = ∠EDC (Angles opposite to the equal sides are also equal)
∠BAE + ∠BEA = 90
∴ ∠BEA + ∠+ BEA = 90
∴ ∠AED = 90°
∴ ∆AED is an isosceles triangle.
∴ ∠EAD = ∠EDA = 45°
Question 18.
Construct the following triangles by using only scale and compass.
(a) In ∆ ABC, AB = 6 cm, ∠A = 45°, ∠B = 75°
(b) In ∆ PQR, PQ = 7 cm,
∠P = 52\( \frac{1}{2}\)° , ∠ Q = 82\( \frac{1}{2}\)° 2
Solution:
(a) Draw AB = 6 cm
Draw AP making angle 45° with AB.
Draw BQ making angle 75° with AB.
Let AP and BQ intersect at C.
∴ ABC is the required triangle.
Students can Download Chapter 3 Recording of Transactions – I & Recording of Transactions – II Questions and Answers, Plus One Accountancy Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
Plus One Accountancy Chapter 3 Questions And Answers Question 1.
The source documents provide information about the nature of ……………. involved in it.
(a) Transactions
(b) Accounts
(c) Journal
Answer:
(a) Transactions.
Plus One Accountancy Chapter 3 Question 2.
Which of the following equation is correct ?
a. Assets = Equities
b. Assets = Capital + Liabilities
c. Assets + Expenses = Liabilities + Capital + Income
d. Capital = A – L
(i) a & b
(ii) a, b & c
(iii) a, b, c & d
(iv) d only
Answer:
(iii) All the above four
Plus One Accountancy Chapter 3 Notes Question 3.
Journal is the book of ……………
(a) Original entry
(b) Secondary entry
(c) Only cash transaction
Answer:
(a) Original entry.
Books Of Prime Entry Questions And Answers Pdf Question 4.
An ……………. is a formal record of all transactions relating to change in a particular item,
(a) Account
(b) Ledger
(c) Journal
Answer:
(a) Account.
Plus One Accountancy Chapter Wise Questions And Answers Pdf Question 5.
If a firm borrows a sum of money, there will be
(a) Increase in Capital
(b) Decrease in Capital
(c) No effect on Capital
Answer:
(c) No effect on capital
Plus One Accountancy Textbook Questions And Answers Question 6.
The Purchase day book contains
(a) All Purchases
(b) Cash Purchases
(c) Credit Purchases
Answer:
(c) Credit Purchases
Transaction Textbook Answers Question 7.
Double column cash book records:-
(a) All transactions
(b) Cash and bank transaction
(c) Only cash transaction
(d) Only Credit transaction
Answer:
(b) Cash and bank transaction.
Plus One Accountancy Chapter 1 Question 8.
Goods purchased on cash are recorded in the
(a) Purchase book
(b) Sales book
(c) Cashbook
Answer:
(c) Cashbook
Books Of Original Entry Questions And Answers Pdf Question 9.
Cashbook does not record transactions of –
(a) Cash nature.
(b) Credit nature.
(c) Cash and credit nature.
Answer:
(b) Credit nature.
Plus One Accountancy Chapter Wise Questions And Answers Question 10.
Balancing of account means:
(a) Total of debit side
(b) Total of credit side
(c) Difference in total of debit and credit
(d) None of these
Answer:
(c) The difference in total debit and credit
Accounting Transactions Exercises With Answers Question 11.
In sales daybook, all ………… sales of goods are recorded.
Answer:
Credit Sales.
Plus One Accountancy Textbook Answers Question 12.
The petty cashier generally works on ……… system.
Answer:
Imprest system.
Plus One Accountancy Chapter 2 Question 13.
The cash book serves the dual purposes of …………..
Answer:
a journal and a ledger.
Journal Entries Questions And Answers Question 14.
When a transaction is recorded on both sides of the cash book but in different columns, such entry is called ……….
Answer:
Contra Entry.
Plus One Accountancy Important Questions Question 15.
The source document of purchase return is ………….
Answer:
Debit Note.
Journal And Ledger Class 11 Questions Question 16.
Cashbook is a …………. journal.
Answer:
Subsidiary.
Question 17.
Return of goods purchased on credit to the suppliers will be entered in …………. journal.
Answer:
Purchase Return/ Return outward book.
Question 18.
Assets are sold on credit are recorded in ……………….
Answer:
Journal Proper.
Question 19.
Credit balance shown by a bank column in cash book is ………..
Answer:
Overdraft.
Question 20.
A book maintained to record transactions, which do not find place in special journal, is known as …………..
Answer:
Journal proper/ Journal Residual.
Question 21.
………………… records the page number of the original book of entry on which relevant transaction is recorded.
Answer:
Journal Folio.
Question 22.
Purchased office stationery for Rs. 10,000. The account to be credited is ……….
Answer:
Cash.
Question 23.
The book in which all accounts are maintained is known as ……….
Answer:
Ledger.
Question 24.
Trade discount is allowed by …………….
Answer:
Manufacturer.
Question 25.
…………….. is process of transferring entries from book of original entry to the ledger.
Answer:
Posting.
Question 26
Cash discount is allowed by …………
Answer:
Creditor.
Complete the following on the basis of hints given:
Question 27.
Invoice – Source document of purchases
_____ – Source of sales return
Answer:
Credit Note
Question 28.
Bank Balance – Debit balance in cash book
__________ – Debit balance in pass book
Answer:
Bank overdraft
Question 29.
Pay-in-slip – Used for depositing cash in bank – account
________ – Used for withdrawing money from bank account
Answer:
Cheque
Find the Odd one and State reason.
Question 30.
Sales Return, Return inward, Credit Note, Debit Note
Answer:
Debit Note, it is related to purchase return but all others are related to sales return.
Question 31.
Journal, Ledger, Purchase Bill, Cash Book
Answer:
Purchase Bill.
Question 32.
…………. discount is allowed by the wholesalers to retailers for bulk purchase.
Answer:
Trade discount.
Question 33.
The process of balancing .of an account involves equalisation of both sides of the account. If the debit side of an account exceeds the credit side, the difference is put on the credit side. The said balance is
(a) A credit balance
(b) A debit balance
(c) None of the above
Answer:
(b) debit balance.
Question 34.
Find out the missing one on the basis of hint given Salary, Electricity Charges, Rent, Telephone Charges = ……..
Answer:
Indirect Expenses
Question 35.
What does the following entry mean?
Drawings A/c Dr 5,000
To Office furniture A/c 5,000
Answer:
The entry stands for withdrawn furniture for personal use of the owner.
Question 1.
Explain the meaning of Double Entry Book-keeping?
Answer:
The recording of the debit aspect and the credit aspect of a transaction in the books of accounts is called double-entry bookkeeping. In this system, every transaction affects at least two accounts or each and every transaction has at least two aspects – a receiving aspect and a giving aspect.
Question 2.
What is a source documents?
Answer:
A voucher or source document is a written document to be used in support of entry made in the accounts. They provide information about the transaction involved and help in verifying the correctness of books of accounts. For example the receipts, bills, cash memos, invoices, salaries bill, wage bills, counterfoils of cheques, registration deeds, etc.
Question 3.
Indicate whether the following accounts will have ‘debit’ or ‘credit’ balances.
Answer:
Question 4.
X Co. Ltd supplied furniture to Mr. Ravi (a trader) on credit. At the time of examining the qualities of the furniture by Mr. Ravi, it was not as per the specification and he returns them to the X Co. Ltd.
Answer:
Question 5.
Pass Journal entry in the books of Saleem Stores during the month of April 2007.
Answer:
[Being Commenced business with cash, furniture, and building]
Question 6.
A cash A/c will never show a credit balance. Why?
Answer:
As a person cannot spend more than his receipts, the receipt side or the debit side of the cash book will always be more than or equal to the credit side. But, it will never be less than the credit side. Hence a cash book always shows a debit balance or nil balance but never credit balance.
Question 7.
Mr. Amal intends to start a computer software unit. He purchased a building for Rs. 3,00,000, for which a sum of Rs. 2,00,000 has been raised by availing a bank loan, and Rs. 75000 has been raised from his friend, Biju Mathew. Show this transaction by developing an accounting equation.
Answer:
Assets = Liabilities + Capital
Question 8.
List out the essential particulars to be contained in a sales bill.
Answer:
Question 9.
‘CH’ Ltd makes a purchases of Electrical goods from Arya electronics on credit. Being the goods despatched by Arya electronics is not as per specifications, CH Ltd returns them to the supplier.
Answer:
Question 10.
“Cash Book is both a journal and a ledger” Elucidate the statement.
Answer:
Cashbook is basically a journal because all entries relating to cash are first made in the cash book. But in the meantime, it serves the purpose of a ledger since it is drawn in the form of an account.
Question 11.
State any two transactions that increase and decrease capital.
Answer:
Question 12
State the three fundamental steps in the accounting process.
Answer:
The three fundamental steps in accounting are. ..
Question 13.
What is contra entry? Write contra entry transactions.
Answer:
When a transaction is recorded on both sides of the cash book but in different columns, such entry is called contra entry.
eg:
Question 1.
With the following details write up the capital account of Mr. X.
Answer:
Capital A/c of Mr.X
Question 2.
State the rules of debit and credit for all types of accounts.
Answer:
The term ‘debit’ is supposed to have derived from ‘debt’ and the term ‘credit’ from ‘creditable’. For convenience ‘Dr1 is used for debit and ‘Cr’ is used for credit. The effects of rules of debit and credit on various types of accounts are given as follows:
Question 3.
With the help of the given cash book identify the transactions of Sourya stores.
Answer:
Question 4.
Enter the following transactions in simple cash book for November 2014.
Answer:
Question5.
Give journal entries for the following transactions
Answer:
Question 6.
From the following, prepare the purchase return book of Madhamohan Readymade Garments.
Answer:
Question 7.
From the following prepare sales day book of Harihar Furniture Co.
Answer:
Question 8.
Enter the following transactions in Sales Return Day Book of Cochin furniture.
Answer:
Question 9.
List out the appropriate source documents for recording the following transactions.
Answer:
Question 1
Explain the terms:
Answer:
1. Journal:
Journal is the prime book in which transactions are entered first time from a source document. In other words, it is the book used for making primary record of day-to-day transactions chronologically. Recording of transactions in the journal is known as “Journalizing” and the recorded transactions are called journal entries. It is also called “Book of Original Entry or Book of Prime Entry”. A specimen form of a journal is given below.
2. Narration:
After entering each transaction in the journal, a brief explanation is provided below. This is called narration.
3. Account:
An account is a classified summary of transaction relating to a change in a particular item during a particular period.
4. Ledger:
A ledger is a collection of all accounts debited or credited in journals. The ledger is a book of second entry or final entry because transactions first entered in journal are finally incorporated in the ledger. It is also called “the Main or the Principal Book of Account of a business. The process of transferring journal entries into ledger is called “posting”.
Question 2.
What are the difference between books of original entry or journal and ledger?
Answer:
Journal | Ledger |
| 1. Book of primary or initial entry | 1. Book of secondary entry |
| 2. Transactions are entered in the order of their occurrence. | 2. Entries are recorded in an analytical order |
| 3. The process of entering transactions is called journalizing. | 3. The process of recording is called posting. |
| 4. Balancing is not done in the books of original entry. | 4. All ledger accounts are balanced |
Question 3.
What do you understand by petty cash book?
Answer:
The book maintained by the petty cashier to record small payments of repetitive nature (petty payments) ie stationery, postage, carriage, traveling, etc are made by him during a particular period is called the “petty cash book”.
The petty cashier works under the supervision and control of the main cashier who advances a certain amount to the petty cashier in the beginning of a specific period. The petty cashier is permitted to make payments only below a particular limit.
Question 4.
Give an example of a business transaction that causes one asset to increase and another asset to decrease, with no effect on either liabilities or capital.
Answer:
Purchased Machinery for cash Rs. 10,000 or any other similar transaction. Here Machinery increases and cash decreases. Therefore no change happens on the total value of assets.
Question 1.
Prove that the accounting equations is satisfied in all the following transaction of Chikku.
Answer:
Question 2.
Record the following transactions in the purchase Journal of Suraj Agencies.
Answer:
Question 3.
What is a special (subsidiary) Journal? Name the important special journals.
Answer:
The journals in which transactions of similar nature only are recorded may be termed as special journals or daybooks. The division of original entry into different Daybook is called the subdivision of the journal. The important special journals are:
Question 4.
What is journal proper? Write the transactions recorded in journal proper.
Answer:
A book maintained to record transactions which do not find place in special journal is known as journal proper or journal residual.
Transactions recorded in journal proper are as follows:
Question 5.
1. Remesh owes Kumar Rs. 5,000. He makes payments of the amount before the due date and Kumar allow him a discount of Rs. 500.
2. Narendran sold to Ravi goods of the catalog price of Rs. 50,000 at a trade discount of 10%. Pass Journal entries in the two cases.
Answer:
Question 6.
What do you understand by petty cash book? Write the advantages of petty cash book.
Answer:
The book maintained by the petty cashier to record small payments of repetitive nature (petty payments) ie stationery, postage, carriage, traveling, etc are made by him during a particular period is called the “petty cash book”.
The petty cashier works under the supervision and control of the main cashier who advances a certain amount to the petty cashier in the beginning of a specific period. The petty cashier is permitted to make payments only below a particular limit.
Advantages of maintaining petty cash book:
Question 1.
Prepare necessary subsidiary books from the following transactions
Answer:
Question 2.
Answer:
Question 3.
Record the following transactions in a Double column Cash Book.
Answer:
Question 1.
Journalise the following transactions:- 2008 April
Answer:
Question 2.
Prepare a cash book with cash and bank column from the following transactions.
Answer:
Students can Download Kerala SSLC Maths Model Question Paper 1 Malayalam Medium, Kerala SSLC Maths Model Question Papers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
You can Download अंदर के और बाहर के Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Hindi Solutions Unit 3 Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.
अंदर के और बाहर के शब्दार्थ
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You can Download बात उस मंगलवार की Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Hindi Solutions Unit 3 Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.
बात उस मंगलवार की पाठ्यपुस्तक के प्रश्न और उत्तर
Bath Us Mangalvar Ki Kerala Syllabus 8th प्रश्ना 1.
मेहनत की कमाई का भोजन स्वदिष्ठ क्यों हो जाता है?
उत्तर:
श्रम के कारण भूख लगती है। भूख मिटाने के लिए जब खाना खाते हैं, वह अधिक स्वादिष्ठ होता है। बेकार में बैठकर खानेवाले को इसी प्रकार की अनुभूति नहीं होती।
बात उस मंगलवार की Notes Kerala Syllabus 8th प्रश्ना 2.
मरीज़ गैर ज़रूरी इंजेक्शन चाहते हैं। क्यों?
उत्तर:
बीमारी और इलाज़ के संबंध में बहुत गलतफहमियाँ हैं। आम जनता चाहती है कि इंजेक्शन से बीमारी जल्दी से दूर होती है। यह विचार चिकित्सा के संबंध में उनकी अज्ञता के कारण है।
Bath Use Mangalwar Ki Kerala Syllabus 8th प्रश्ना 3.
“यहीं जंगल के बीच ये सभी मेरे क्लीनिक हैं।” -इससे आपने क्या समझा?
उत्तर:
यहाँ डॉक्टर का मनोभाव प्रकट होता है। यह डॉक्टर चिकित्सा को व्यापार नहीं . मानती है। वे इसे सेवाकार्य मानती है। इसलिए शहर में उनकी अपनी क्लीनिक नहीं है। जंगल के अनपढ़, अशिक्षित, गरीब ही उनके मरीज़ हैं। वे उनके लिए काम – करती हैं।
Kerala Syllabus 8th Standard Notes Hindi प्रश्ना 1.
डॉ रमणी अटकुरी की चरित्रगत विशेषताएँ लिखें।
उत्तर:
रमणी अटकुरी एक ईमानदार डॉक्टर हैं। वे डॉक्टरी को व्यापार मानती नहीं। उनके अनुसार डॉक्टर को समाज की सेवा करनी चाहिए। इस आदर्श को वे अपने जीवन में निभाती है। वह एक आदर्श डॉक्टर हैं।
Hsslive Guru 8th Class Hindi Kerala Syllabus प्रश्ना 2.
आपके दृष्टिकोण में एक डॉक्टर के गुण क्या-क्या हैं? टिप्पणी लिखें।
उत्तर:
मेरी राय में डॉक्टर को ईमानदार होना चाहिए। अपने पेशे को धन कमाने का उपाय न मानना चाहिए। मरीज़ों के पास जाकर उनकी सेवा-सुश्रूषा करनी चाहिए। मरीज़ों के साथ सहानुभूति पूर्ण व्यवहार अपनाना चाहिए। अपने उत्तरदायित्व को निभाते हुए समाज के लोगों को स्वस्थ रखने के कामों में भाग लेना चाहिए।
Bath Us Mangalvar Ki Question Answer Kerala Syllabus 8th प्रश्ना 3.
संवाद लिखें।
उत्तर:
Hss Live Guru 8 Hindi Kerala Syllabus प्रश्ना 4.
संवाद लिखें।
उत्तर:
बात उस मंगलवार की शब्दार्थ Word meanings
You can Download Polygons Questions and Answers, Activity, Notes, Kerala Syllabus 8th Standard Maths Solutions Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.
Textbook Page No. 49
Class 8 Maths Polygon Kerala Syllabus Question 1.
Find the sum of angles of a polygon with 52 sides?
Solution:
Sum of angles = (52 – 2) 180
= 50 × 180
= 9000
Hsslive Guru 8th Class Maths Kerala Syllabus Question 2.
The sum of angles of a polygon is 8100°. Find the number of its sides?
Solution:
Let sides = n
∴ Sum of angles = 8100
(n – 2) 180 = 8100
n – 2 = \(\frac{8100}{180}\)
= 45
n = 47, Number sides of the polygon = 47
Hss Live Maths 8th Standard Kerala Syllabus Question 3.
Can 1600 be the sum of angles of a polygon. Can 900° be the sum?
Solution:
1600 is not the multiple of 180°. So it cannot be the sum of the angles. 900° is the multiple of 180. So 900° can be the sum of angles of a polygon.
Polygons Chapter For Class 8 Kerala Syllabus Question 4.
All the angles of a polygon with 20 sides are equal. Find the measure of each angle?
Solution:
Sum of angles of a polygon with 20 sides. = (20 – 2) × 180
= 18 × 180
Each angle = \(\frac{18 \times 180}{20}\) = 162°
Hss Live Guru 8th Maths Kerala Syllabus Question 5.
The sum of angles in a polygon is 1980. Find the sum of angles of the polygon with one side more. Find the sum with one side less?
Solution:
When one side of a polygon increases, the sum of angle increases by 180°.
∴ Sum of angles = 1980 + 180 = 2160°
When the number of sides decreases by 1, the sum of angles also decreases by 180°.
∴ Sum of angles = 1980 – 180 = 1800°
Textbook Page No. 51
Polygon Chapter Class 8 Kerala Syllabus Question 1.
Two angles of a triangle are 40°, 60° each. Find the measure of the exterior angles?
Solution:
Third angle = 180 – (40 + 60) = 80°
Exterior angles = 180 – 40, 180 – 60,
180 – 80, 140,
ie, 140°, 120° and 100°
Class 8 Polygon Chapter Kerala Syllabus Question 2.
Find all the angles in the figure.
Solution:
∠ABC = 180 – 105 = 75°
∠C = 180 – (35 + 75)
= 180° – 110° = 70°
Exterior angle at A = 180° – 35° = 145°
Exterior angle at B = 180° – 75° = 105°
Exterior angle at C = 180° – 70°
= 110°
Hsslive Maths Class 8 Kerala Syllabus Question 3.
Find all the exterior angles of the quadrilateral in the figure.
Solution:
4th angle = 360 – (130 + 70 + 60) = 100°
Exterior angles = 120°, 110°, 50°, 8o°
8th Class Maths Notes Kerala Syllabus Question 4.
Find all the angles of the given diagrams
Solution:
(i)
Angles of the triangle are 70°, 35°, 75°
Exterior angles are 35°, 110°, 105°
(ii)
Angles of the quadrilateral are
∠C = 85°, ∠B = 85°, ∠A = 100°, ∠D = 90°
Exterior angles are 95°, 80°, 90°, 95°
All angles at D are 90°
Angles at C are 85°, 95°, 85°, 95°
(iii)
Angles of the quadrilateral are :
∠ A = 65°, ∠B = 75s°, ∠C = 100°, ∠D = 120°
Exterior angles are 115°, 105°, 80°, 60°
Angles at C are 100°, 80°, 100°, 80°
Hss Live Guru 8 Maths Kerala Syllabus Question 5.
Prove that in a triangle, the exterior angle at a vertex is the sum of interior angles at the other two vertices
Solution:
∠DBC + ∠ABC = 180° (linear pair)
∠A + ∠C + ∠ABC = 180° (Angles of a triangle)
∴ ∠A + ∠C = ∠DBC
So in a triangle the exterior angle at a vertex is the sum of interior angles at the other two vertices.
Textbook Page No. 54
Hsslive Guru 8th Maths Kerala Syllabus Question 1.
All the angles of an eighteen sided polygon are equal. Find each exterior and interior angles?
Solution:
The angles of the polygon are equal. So the exterior angles are also equal. Sum of the exterior angles = 360°
Measurement of an interior angle = \(\frac{360}{18}\) = 20°
Hsslive Guru 8 Maths Kerala Syllabus Question 2.
The sides PQ and RS of the quadrilateral PQRS are parallel. Find all the exterior and interior angles of the quadrilateral.
Solution:
PQ and RS are parallel to each other ∠P = 500, The exterior angle at S = 500
∴ ∠S = 130°
∠Q = 360 – (50 + 130 + 110)
= 360 – 290 = 70°
The angles of the quadilateral are ∠P = 50°, ∠Q = 70°, ∠R = 110°, and ∠S = 130°
The exterior angles are 130°, 110°, 70°, 50°
Hss Live Guru Class 8 Maths Kerala Syllabus Question 3.
Draw a quadrilateral and mark the exterior angles at two vertices. Is there any relationship between the sum of these angles and the sum of the interior angles at the other two vertices?
Solution:
Sum of the angles of the quadrilateral = 360°
Sum of the exterior angles = 360°
Let the sum of two exterior angles = x
The sum of two interior angles at the same vertex = 360 – x. So the sum of the other two interior angles = x. So sum of two exterior angles at two vertices is equal to the sum of two interior angles at the other two vertices.
Kerala Syllabus 8th Standard Maths Notes Question 4.
An exterior angle in a polygon with all angles are equal is twice of an interior angle.
(i) Find the measure of each angle in it?
(ii) Find the number of sides?
Solution:
The exterior angles are equal as all the angles are equal.
Let the interior angle is x, then the exterior angle is 2x.
x + 2x = 3x = 180
∴ x = 60
(i) Interior angles are 6o° each and exterior angles are 120° each.
(ii) The polygon has 3 sides. It is a triangle.
Hsslive Guru Maths 8th Standard Kerala Syllabus Question 5.
The sum of exterior angles in a polygon is twice the sum of the interior angles.
(i) Find how many sides the polygon has?
(ii) Find the number of sides, if the sum of the exterior angles is half of the sum of the interior angles.
(iii) Find the number of sides if the sum of the exterior angles is equal to the sum of the interior angles?
Solution:
The sum of exterior angles in a polygon is 360°.
(i) The sum of the interior angles in a triangle is 180°. It is a triangle and has 3 sides.
(ii) If the sum of the interior angles is 720°, the polygon has six sides.
(iii) If the sum of the interior angles is 360°, the polygon is a quadrilateral it has 4 sides.
Textbook Page No. 58
Hss Live Guru Maths 8th Kerala Syllabus Question 1.
Draw a hexagon with all sides equal but angles different?
Solution:
The sum of angles of a hexagon is 720. Make 720, the sum of 6 different angles. Draw a line with a definite length and make an angle at its end. Then draw the next line and angle. Draw 6 lines continuously and you get a hexagon.
Std 8 Maths Kerala Syllabus Kerala Syllabus Question 2.
Draw a hexagon with all the angles are equal and sides are different.
Solution:
Draw a line. Draw another line by taking an angle 1200 at its one end. Then draw different lines by taking angles 1200. By joining we get a hexagon.
Hsslive 8th Class Maths Kerala Syllabus Question 3.
Find the measurements of each angle in an equal polygon with 15 sides. Find each exterior angles?
Solution:
The sum of angles of a polygon with
15 sides = (15 – 2) × 180
= 13 × 180 = 2340
One angle = \(\frac{2340}{15}\) = 156°
Exterior angle = 180 – 156 = 24°
Kerala Syllabus 8th Standard Maths Notes Pdf Question 4.
One angle of a regular polygon is 1680. Find the number of its sides?
Solution:
Exterior angle = 180 – 168 = 12°
Sum of exterior angles = 360°
∴ Number of sides = \(\frac{360}{12}\) = 30
Hsslive Class 8 Maths Kerala Syllabus Question 5.
Can you draw a regular polygon with each exterior angle 6°. Can you draw it if the exterior angle is 7? .
Solution:
Sum of exterior angles = 360°
One of the exterior angle = 6°
Number of sides = \(\frac{360}{6}\) = 60, yes we can draw; If one exterior angle is 7°
Number of sides = \(\frac{360}{7}\) = 51.42
Not a whole number. The polygon cannot be drawn.
Question 6.
A regular pentagon and regular hexagon are jointly drawn in the figure. Find PQR?
Solution:
Sum of angles of a regular pentagon = (5 – 2) 180°
= 3 × 180 = 540°
One angle of a regular pentagon
= \(\frac{540}{5}\) = 108°
Sum of angles of a regular hexagon = (6 – 2) 180°
= 4 × 180 = 720°
One of its angle = \(\frac{720}{6}\) = 120°
∴ ∠PQR = 360 – (108 + 120)
= 360 – 228 = 132°
Question 7.
A square, a regular pentagon and a regular hexagon are jointly drawn in the figure Find ∠BAC
Solution:
One angle of a regular pentagon = 108°
One angle of a regular hexagon = 120°
One angle of a square = 90°
∴ ∠BAC = 360 – (108 + 20 + 90)
360 – 318 = 42°.
Question 8.
In the figure below, ABCDEF is a regular hexagon. Prove that ∆ BDF is an equilateral triangle?
Solution:
Sum of angles of a regular hexagon = (6 – 2) 180 = 720°
= 4 × 180 = 720°
One angle = \(\frac{720}{6}\) = 120°
consider ∆ EFO
∠E = 120°
∠EFD = ∠EDF = 30°(Angles opposite to equal sides of an isosceles triangle are equal)
Similarly ∠AFB = 30°
∴∠DFB = 120 – (30 + 30) = 60°
∴ ∠FBD = 60°,
∠FDB = 60°
∴ ∆ FDB is an equilateral triangle.
Question 9.
In the figure below ABCDEF is a regular hexagon. Prove that ACDF is rectangle.
Solution:
One angle of a regular hexagon = 120°
∠EFD = ∠EDF = 30°; ∠F = 120°
∴ ∠AFD = 90°
Similarly all the angles of ACDF is 90°
∴ ACDF is a rectangle.
Question 1.
How many triangles can be for made if all the diagonals from a vertex of a 10 sided polygon are drawn?
Solution:
7 diagonals 8 triangles.
Question 2.
The sum of angles of a polygon is x°. Find the sum of angles of the polygon with one side is more. Find the number of sides if one side is less?
Solution:
(x + 180°), (x – 180°)
Question 3.
Can it possible that the external angle of a polygon be 13?
Solution:
\(\frac{360}{13}\) is a fraction. So it is not possible.
Question 4.
The external angles of a triangle from the three vertices are (2x + 30°), (3x – 10°), 100°. Find the value of x?
Solution:
2x + 30 + 3x – 10 + 100 = 360
5x + 120 = 360
5x = 360 – 120 = 240
∴ x = \(\frac{240}{5}\) = 48
Question 5.
Draw a circle and construct a regular pentagon with all the vertices in it?
Solution:
Draw circle and construct the pentagon.
Question 6.
Find the sum of angles of a heptagon? Find an angle if all the angles are equal?
Solution:
Sum of angles of a heptagon = (7 – 2) × 180°
= 5 × 180° = 900°
All the angles are equal. So one angle
= \(\frac{900}{7}\) = 128.571………..
Question 7.
Find the sum of the angles of the polygon with the given sides?
(a) 12
(b) 15
(c) 20
(d) 24
Solution:
(a) Sum of angles = (12 – 2 ) × 180
1800°
(b) Sum of angles = (15 – 2) × 180
= 2340°
(c) Sum of angles = (20 – 2) × 180
= 3240°
(d) Sum of angles = (24 – 2) × 180
= 3960°
Question 8.
The angles of a hexagon are (x – 10)°, x°, (x + 10)°, (x + 20)°, (x + 30)°, and (x + 40)°. Find the value of x?
Solution:
Sum of the angles of a hexagon = (6 – 2) × 180° = 720°
ie, (x – 10) + x + (x + 10) + (x + 20) + (x + 30) + (x + 40) = 720
6x + 90 = 720
ie, 6x = 630; x = 630 ÷ 6
∴ x = 105°
Question 9.
The sum of the angles of a polygon with 12 sides is 1800°. Find the sum of the angles of a polygon with 13 sides?
Solution:
When one side is increased, the angle is increased by 180°.
∴ Sum of the angles of a 13 sides Polygon = 1800° + 180 = 1980°
Kerala Syllabus 10th Standard Social Science Textbook Question 1.
‘The revenue policy of the British was the major cause for the decline of agricultural sector in India’. Examine this statement analysing the features of the Permanent Settlement.
Answer:
The Permanent Settlement was introduced by Lord Cornwallis, the Governor General of British India, in Bengal, Bihar and Orissa. Its features were:
Farmers were the immediate victims of British rule. It was the land revenue system implemented by the British that destroyed the backbone of farmers. The aim of their tax policy was to maximize income.
When the farmers were unable to pay tax in cash before ‘ the deadline, they had to take loan from money lenders at a high rate of interest. The loans were obtained by mortgaging the agricultural land. The agricultural land of the farmers, who could not pay back the loan and interest, was seized by money lenders.
Question 2.
Match Column A with Column B.
A | B |
| 1. Santhal rebellion | 1. Malabar |
| 2. Mappila rebellion | 2. Dadabhai Naoroji |
| 3. Kurichiya rebellion | 3. Rajmahal Hills |
| 4. Drain Theory | 4. Wayanad |
Answer:
| A | B |
| 1. Santhal rebellion | 3. Rajmahal Hills |
| 2. Mappila rebellion | 1. Malabar |
| 3. Kurichiya rebellion | 4. Wayanad |
| 4. Drain Theory | 2. Dadabhai Naoroji |
Question 3.
What were the circumstances that led to the commercialisation of agriculture during the British period?
Answer:
Question 4.
Analyse the causes of the Indigo Revolt.
Answer:
The demand for indigo increased after the progress in textile industry. The British industrialists gave the farmers a good amount as advance for the cultivation of indigo. The farmers yielded to the temptation of the British and widely planted indigo as they were in trouble with no other means to pay the heavy land tax. Each farmer who accepted the advance amount from the British was liable to plant indigo in a fixed portion of his land.
The farmers were also compelled to cultivate it at the most fertile part of the agricultural land. Due to the interference of the British agents in the harvesting season, the farmers received only a lower price for indigo. When artificial colours were invented, indigo became obsolete. This exploitation of the British forced the indigo farmers for a revolt.
Question 5.
‘Kurichiya rebellion was a resistance by the tribes against the British exploitation ’. What were the circumstances that led to the Kurichiya rebellion?
Or
Evaluate the circumstances that led to the Kurichiyas and Kurumbas of Wayanad to turn against the British.
Answer:
Kurichiya revolt was a tribal insurgency against the British. It was organised by the Kurichiya and the Kurumba tribes of Wayanad in 1812. Its reasons were:
Question 6.
What were the causes of the decline of the Indian textile industry?
Answer:
Question 7.
Do you think that the famines in India were the creation of the British? Why?
Answer:
The main reasons for the famine and poverty in India were the economic policies introduced by the British in India and their exploitation. Most of the people lived on the brink of famine all throughout the British rule. The economic exploitation of the British, decline of traditional industries, high tax, drain of wealth and resources, decline of agricultural sector and the exploitation by the landlords and money lenders all pushed the people to poverty.
When poverty became acute, famines broke out in different parts of the country. About 2 crore people died in the 24 famines that occurred in the second half of the 19th century.
Question 8.
Evaluate the role of ‘Drain Theory’ by Dadabhai Naoroji in stimulating national feeling among the Indian masses.
Answer:
Through his studies, Dadabhai Naoroji publicized the facts on the deterioration of Indian economy under the British rule. His studies were based on empirical data. He established the fact that a huge amount of money was flowing to Britain every year. He proved that the drain of wealth was the root cause of poverty and starvation in India.
This was known as ‘Drain Theory’. Indian wealth flew to Britain by the export of Indian raw materials, salary and pension to the British officers in India, profit gained through the sale of the British products in India and tax from India. Dadabhai Naoroji could make the people aware that the economic policy of the British impoverished India.
The common people realized that the poverty and exploitation they faced had been the creation of the British. It reinforced their anti-. British attitude which finally led to the growth of nationalistic feeling among the people.
Question 9.
Analyse the causes of the Revolt of 1857.
Answer:
Historians termed the Revolt of 1857 as India’s First War of Independence. This was an organized agitation launched by peasants, handicrafts men, kings and soldiers who were dissatisfied with the harmful policies of the British. There were many causes that led to the revolt.
Dissatisfaction among Indian soldiers:
Poor salary and abuse by the British officers were the major reasons for the resentment of sepoys. The rumour that the cartridge in the newly supplied Enfield rifles were greased with the fat of cows and pigs provoked them. It wounded the religious sentiments of the Hindu and Muslim soldiers. The soldiers who were unwilling to use the new cartridges were punished by the officers.
Discontentment among native kings:
The British rule had adversely affected the kings too. In addition to the Doctrine of Lapse, the princely states were convicted of inefficient rule and were annexed by the British. This made the kings to lead the rebellion.
Miseries of farmers and craftsmen:
The tax policy of the British pushed the farmers to poverty. Cottage and handicrafts industry declined due to the British policies. In the second half of the 19th century, different sections of oppressed people, mobilized against the British and launched organized agitations.
Question 10.
What were the sources of economic drain from India to Britain?
Or
What are the methods by which the wealth of India drained to Britain according to the Drain Theory of Dadabhai Naoroji?
Answer:
Question 11.
Do you think that the Swadeshi Movement was a mass movement? Why?
Answer:
The Swadeshi Movement was organised as a part of anti-partition movement of Bengal in 1905. The method of Swadeshi resistance was the boycott of foreign goods and consumption of indigenous products. As part of the agitation, foreign goods were collected and burnt publicly. The extensive use of indigenous products by discarding foreign items rejuvenated Indian industry.
Massive participation of women, labourers and students were the remarkable feature of this movement. Washermen took a vow that they would not wash foreign clothes. The priests swore that they would not perform rituals and prayers using foreign items. Women boycotted foreign bangles and utensils. Students quit school to take part in the movement.
Question 12.
Prepare the diagram that depicts the features of the Permanent Settlement, the Ryotwari and the Mahalwari land revenue systems.
Answer:
Permanent Settlement
Ryotwari System
Mahalwari System
Question 13.
How did the British land revenue policy make the peasantry fall easy prey to the exploitation of moneylenders? Explain.
Or
Evaluate how the revenue system implemental by the British adversely affected the agricultural sector.
Answer:
The land revenue system implemented by the British destroyed the backbone of the farmers. The land revenue policies implemented by the British adversely affected the agricultural sector. When the farmers were unable to pay tax in the form of money before the deadline, they had to take loan from moneylenders at a high rate of interest. The loans were obtained by mortgaging agricultural land. The agricultural land of the farmers who could not pay back the loan and interest, was seized by moneylenders.
10 Th Std History Notes Question 14.
Conduct a discussion on ‘British policies and commercialization of agriculture ’.
Answer:
During the British rule, the farmers were compelled to cultivate crops according to the market needs. As a result, commercial crops were cultivated instead of food crops. This transformation is known as commercialisation of agriculture.
The tax policies of Permanent Settlement and the Ryotwari system had great impact on agricultural sector. The farmers had to pay high rate of tax in the form of cash before deadline. To meet this, they cultivated the crops that had higher market price. The products that had demand in the European markets were given higher price.
Thus the Indian lands became the cultivating fields of indigo, cotton, jute for Europe. The commercialisation of agriculture led to the emergence of interference of British agents. Due to their interference, the farmers were forced to sell their products at low price.
Question 15.
Imagine yourself as a journalist. Prepare a news report on the plight of indigo farmers of the 19th century.
Answer:
Indigo farmers in misery Bengal :
The farmers of Bengal became t producers of indigo. The British forced the farmers to cultivate indigo for the factories set up by the industrialists in Bengal. They gave farmers a good amount as advance for the cultivation of indigo.
The farmers were in trouble with no other means to pay the heavy land tax. So they yielded to the temptation of the British and widely planted indigo. Due to the interference of British agents during harvesting season, the farmers received only a lower price for indigo. Now artificial dyes are used instead of indigo. This made the plight of the farmers more miserable for they had used much of their land for indigo cultivation.
As they had cultivated indigo in the most fertile part of their land, they were unable to cultivate food crops. Most of the farmers are on the brink of eviction due to high lease rate. The indigo farmers of Bengal prepare for an agitation against their unending misery and exploitation.
Question 16.
Analyse the circumstances that led to the Indigo Revolt.
Answer:
The British gave the farmers a good amount as advance for the cultivation of indigo. The farmers yielded to the temptation of the British and widely planted indigo as they were in trouble with no other means to pay the heavy land tax. The British gave only low price than the market price to the farmers. They had no freedom to cultivate the crops that give high profit. With the invention of artificial dyes, indigo became obsolete. This made the plight of the farmers miserable.
Question 17.
Analyse the reasons for the decline of Indian textile industry and complete the diagram below.
Answer:
Question 18.
Prepare an article on the problems faced by different sections of people due to the British policies in India.
Answer:
Weavers gave up their work massively due to the exploitation and torture of the British officers. They forced the weavers to work at meagre wages and to exchange the products to them at cheaper rate. So they searched for other jobs. The village industries like pottery, tanning and carpentry declined. This pushed the villages to famine and poverty.
In the 19th century, the British industrialists started modem industries like textile, jute, steal and paper. The labourers in these industries were also exploited. Prolonged working hours, meagre wages and unhealthy accommodation were the problems that they faced. The workers agitated when they suffered extreme exploitation.
Question 19.
Discuss the causes of the Revolt of 1857 based on the hints given below.
i) Miseries of farmers.
ii) Poverty of the craftsmen.
iii) Dissatisfaction of kings.
iv) Miseries of the sepoys.
Answer:
The historians termed the Revolt of 1857 as India’s First War of Independence. All sections of people were dissatisfied with the British rule.
i) Miseries of farmers:
The land revenue policies of the British in India namely the Permanent Settlement, Ryotwari System and Mahalwari System impoverished the peasants. When the farmers were unable to pay tax in cash before the deadline, they had to take loan from moneylenders at a high rate of interest by mortgaging their land.
The moneylenders seized the agricultural land of farmers who could not pay back loan and interest. The farmers who cultivated commercial crops as a result of the commercialization of agriculture also had to suffer losses.
ii) Poverty of the craftsmen:
The British policies completely ruined not only the agricultural sector, but also the handicrafts in India. Large scale import of machine-made British textiles was the major reason for the ruin of Indian texti le industry.
Weavers gave up their work massively due to the exploitation and torture of the British officers. They searched for other jobs. The village industries like pottery, tanning and carpentry also declined. The ruins of agricultural sector and handicrafts industry led India to famine and deaths due to starvation.
iii) Dissatisfaction of Kings:
The British rule had adversely affected the kings also. In addition to the Doctrine of Lapse, the princely states were convicted of inefficient rule and were annexed by the British. Lord Dalhousie, the British Governor General, annexed the princely states of Satara, Jhansi, Nagpur and Sambalpur using the policy of Doctrine of Lapse. This made them to lead the revolt.
iv) Miseries of the Sepoys:
Poor salary and abuse by the British officers were the major reasons for the Sepoy’s resentment. The rumour that the cartridge in the newly supplied Enfield rifles were greased with the fat of cows and pigs provoked them. It wounded the religious sentiments of Hindus and Muslim soldiers. The soldiers who were unwilling to use the new cartridges were punished by the officers.
Thus the Revolt of 1857 was the first organised rebellion in India against the wrong British policies.
Social Science Short Notes For Class 10 Kerala Syllabus Question 20.
Prepare a note on Drain Theory.
Answer:
Question 21.
Conduct a Seminar on ‘How the economic exploitation of the British earned the emergence of nationalism ’.
Answer:
Seminar
Topic: The Economic Exploitation of the British and the emergence of Indian Nationalism
Introduction:
The economic exploitation of the British created an anti-British feeling among different sections of people. This attitude was a major factor that led to the emergence of Indian nationalism in the 19th century.
Emergence of Indian nationalism :
Nationalism is the sense of unity among the people of a country irrespective of caste, creed, religion and region. Indian National Congress was an example for such an organised form of nationalism.
The land revenue policy of the British pushed the farmers to poverty. The high land tax made most of the peasants landless. Commercialisation of agriculture was another reason that led to the decline of agricultural sector. Traditional industries also suffered a set back due to the economic exploitation of the British. Anti-British feelings became intense, resulting in a state of unity among the people. This unity can be termed as nationalism.
Indian National Congress:
The Indian National Congress led the anti-British struggle from 1885 till India attained independence in 1947. The earlier leaders of the Congress were much conscious of the economic exploitation of the British and the resultant poverty faced by the Indians. They unveiled the British attempts to convert India as a market for selling British products, and a mere centre for collecting raw materials for British industries.
Dadabhai Naoroji and Drain Theory:
Dadabhai Naoroji through his studies publicized the facts on the deterioration of Indian economy under the British rule. He established the fact that a huge amount of money was flowing to Britain every year. He proved that the drain of wealth was the root cause of poverty and starvation in India. His finding known as ‘Drain Theory’ is included in his work ‘Poverty and Unbritish Rule in India’.
Ideas of early Congress leaders:
The earlier leaders like Dadabhai Naoroji, Ramesh Chandra Dutt and Gopal Krishna Gokhale had a pivotal role in making the common people aware of the economic policy of the British that impoverished India.
The common people realized that the poverty and exploitation they faced had been the creation of the British. It reinforced their anti-British attitude. The nationalism grown out of such awareness is termed as ‘Economic Nationalism’.
Swadeshi Movement and Nationalism:
To check economic drain, the early national leaders pleaded with the people to boycott foreign goods and strengthen Indian industry by consuming Indian products. The major strategy adopted for the anti-partition movement in Bengal in 1905 was the boycott of foreign goods and consumption of indigenous products.
Indian nationalism attained further strength from Swadeshi Movement. The protests against the British policy of looting India’s wealth and impoverishing the Indians transformed as Indian nationalism.
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