Kerala Syllabus Class 8 Mathematics - Unit 2 Equal Triangles - Questions and Answers | Teaching Manual 

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Kerala Syllabus STD 8 Maths: Unit 2 Equal Triangles - Textbook Solutions

♦ Textbook Activities (Textbook Page No: 18, 19)
Sides and angles

1) In each pair of triangles below, find the angles of the second triangle equal to the angles of the first triangle and write these pairs. 
Answer:
(i) ∠L = ∠Y (angles opposite to the 4 cm side)
    ∠M = ∠Z (angles opposite to the 2 cm side)
    ∠N = ∠X (angles opposite to the 3 cm side)
(ii) 
    ∠A = ∠R (angles opposite to the 3 cm side)
    ∠B = ∠P (angles opposite to the 2.4 cm side)
    ∠C = ∠Q (angles opposite to the 3.6 cm side)

(2) In the triangles shown below,
AB = QR, BC = RP, CA = PQ
Calculate ∠C of triangle ABC and all angles of triangle PQR
Answer:
    ∠A = 40°, ∠B = 60°
    ∴ ∠C = 180 − (40 + 60)
            = 180 − 100 = 80°
(In a triangle, the sum of the angles is 180°)
AB = QR
Since AB = QR, ∠C = ∠P. ∴ ∠P = 80°
Since BC = RP, ∠A = ∠Q. ∴ ∠Q = 40°
Since CA = PQ, ∠B = ∠R. ∴ ∠R = 60°

(3) In the triangles below,
 AB = QR, BC = PQ, CA = RP
Calculate the other two angles of each triangle.
Answer:
Since BC = PQ, ∴ ∠A = ∠R = 60°
Since CA = RP, ∠B = ∠Q. = 70°
In ∆ABC, ∠Q. = 180° − (60 + 70) = 180 − 130 = 50°
Since AB = QR, ∠C = ∠P = 50°

(4) The diagonal of a quadrilateral splits it into two triangles as shown below:
Do these triangles have the same angles? Why?
Answer:
In both ∆ABC and ∆ABD, AC = AD, BC = BD
The third side is AB in both triangles.
Since the sides of triangle ABC are equal to the sides of triangle ABD, the angles opposite to the equal sides are also equal. That is, the angles of triangle ABC and triangle ABD are equal.

(5) In the quadrilateral ABCD shown below,
AB = AD,  CB = CD
Calculate all the angles of the quadrilateral.
Answer:
Since AB = AD, the angles opposite to these sides are equal.
The angle opposite to AB = ∠ACB = 50°
The angle opposite to AD = ∠ACD = 50°
Since CB = CD, ∠BAC = ∠DAC = 30°
In ∆ABC, ∠B = 180° − (30° + 50°) = 180° − 80° = 100°
In ∆ADC, ∠D = 180° − (30° + 50°) = 180° − 80° = 100°
∴ Angles of a quadrilateral are ∠A = 60°,
∠B = 100°, ∠C = 100°, ∠D = 100°

♦ Textbook Activities (Textbook Page No: 24)
Two angles

1) In each pair of triangles below, find the sides of the triangle on the right equal to the sides of the triangle on the left. 
Answer:
One side and the angles at both ends of that side in ∆ABC are equal to one side and the angles at both ends of that side in ∆PQR.
AB = QR (Both sides are 3 cm each)
AC = PR (The sides opposite to the 70° angle)
BC = PQ (The sides opposite to the50° angle)
Answer:
In ∆LMN, ∠N = 180 − (30 + 80) = 70°
In ∆XYZ, ∠Z = 180 − (70 + 30) = 80°
LM = YZ (The sides opposite to the 70° angle)
LN = XY (The sides opposite to the 80° angle)
MN = XZ (The sides opposite to the 30° angle)

(2) In the picture, the top two sides and the bottom side of a pentagon, with equal angles and equal sides, are extended to form a triangle:
(i) Are the sides of the small triangle on the left equal to the sides of the
small triangle on the right? Why?
(ii) Are the left and right sides of the large triangle equal? Why?
Answer:
(i) Here, we take the small triangle on the left is BPC and the small triangle on the right is EQD
In the figure, the pentagon has equal angles and sides. ABCDE is a regular pentagon.
BC = DE, ∠ABC = ∠AED (Regular pentagon)
∠PBC = ∠QED (Linear pair)
∠BCD = ∠EDC (Regular pentagon)
∠BCP = ∠EDQ (Linear pair)
One side and the angles at both ends of that side in ∆BPC are equal to one side and the angles at both ends of that side in ∆EQD. So the third angle of these triangles are equal.
∴ ∠BPC = ∠EQD 
And the sides opposite to the equal angles are also equal.
PC = QD (The sides opposite to ∠PBC and ∠QED)
PB = QE (The sides opposite to ∠BCP and ∠EDQ)
The sides of ∆BPC and ∆EQD are equal in length.
That is the sides of the small triangle on the left equal to the sides of the small triangle on the right. 

(ii) AB = AE (Regular pentagon)
PB = QE (Sides of ∆BPC and ∆EQD are equal.)
Adding,
AB+ PB = AE + QE
∴ AP = AQ
The left and right sides of the large triangle APQ in the figure are equal in length.

3. The sides of a triangle are equal to the sides of another triangle.
(i) Is the height from each side of one triangle to the opposite vertex equal to the height from the equal side of the other triangle to its opposite vertex? Why?
(ii) Are the areas of the two triangles equal? Why? 
Answer:
(i) Let ∆ABC and ∆PQR are the two triangles, the sides of ∆ABC and ∆PQR are equal length.
That is, AB = QR
BC= PR
AC = PQ
Draw the heights DC from AB to C and SP from QR to P.
∠ADC 90°, ∠PSQ = 90°
Since the sides of ∆ABC are equal to the sides of ∆PQR, the angles opposite the equal sides are also equal.
Let ∠CAD = ∠PQS = x° (The angles opposite to the equal sides BC and PR)
∠ACD = 180° − (90° + x°)
∠QPS = 180° − (90° + x°)
∴ ∠ACD = ∠QPS
In ∆ADC, side AC and the angles at both ends of it are equal to side PQ in ∆QSP and the angles at both ends of it. Therefore, the sides opposite to the equal angles in both triangles are also equal. 
∴ CD = PS (The sides opposite to ∠CAD and ∠PQS)
In triangle ABC, the height CD from side AB to vertex C is equal to the height PS from side QR to vertex P in triangle PQR.

(ii) Area of ∆ABC = ½ x AB x CD
Area of ∆PQR = ½ x QR x PS 
= ½ x AB x CD (QR = AB, PS = CD)
∴ The area of both triangles are equal.

♦ Textbook Activities (Textbook Page No: 28, 29)
Two Sides
(1) The green lines in the picture below are parallel and of the same length; one is drawn from the end of the horizontal blue line, and the other is drawn from the midpoint of the blue line :
(i) Are the lengths of the red lines in the picture equal? Why?
(ii) Are the red lines parallel?
Answer:
(i) Let's consider the green lines in the figure as AC and BE, the blue line as AD and the red lines, BC and DE. Since B is the midpoint of the blue line AD, then AB = BD. When the line AD intersects the parallel lines AC and BE of equal length, corresponding angles, ∠CAB and ∠EBD formed are equal.
In ∆ABC and ∆BDE, AB = BD, AC BE and ∠A = ∠B. That is two sides, and the included angle of ∆ABC are equal to two sides and the included angle of ∆BDE. Therefore, the third side of these triangles is also equal.
∴ BC = DE, the red lines are equal in length.

(ii) Other angles of ∆ABC and ∆BDE are also equal. 
Therefore,
∠ABC = ∠BDE
∠ABC and ∠BDE are a pair of corresponding angles formed when the blue line AD intersects the red lines BC and DE. Since these angles are equal, the red lines BC and DE are parallel. So the red lines are parallel.

2. Is the quadrilateral below a parallelogram? Why?
Answer:
Consider ∆ABC and ∆ABD.
AC = BD = 4 cm
AB = AB (Common side) 
∠CAB = ∠DBA = 35°
Two sides and the included angle of ∆ABC are equal to two sides and the included angle of ∆ABD. Therefore the third sides of these triangles will also be equal. 
∴ BC = AD.
Now in quadrilateral ACBD, we have AC = BD and BC = AD. Since opposite sides are equal, this is a parallelogram.

3. In the figure below, the lines AB and CD are parallel. M is the midpoint of AB.
(i) Are the quadrilaterals AMCD and MBCD parallelograms? Why?
(ii) Calculate all the angles of the triangles AMD, MBC and DCM.
Answer:
(i) Since the alternate angles ∠ADM and ∠CMD formed when the lines AD and MC are cut by a third line DM are equal, the lines AD and MC are parallel. Since both pairs of opposite sides are parallel, quadrilateral AMCD is a parallelogram.
We can prove that the sides MD and BC of the quadrilateral MBCD are parallel. Then, quadrilateral MBCD is also a parallelogram.

(ii) In Qn. (i), we see that quadrilateral AMCD and MBCD are parallelograms.
In parallelogram AMCD,  AM = CD
∴ ∠AMD = ∠CDM, ∠ADM, ∠CMD
∴ ∠A = ∠DCM = 40°
∠CMB = ∠DCM = 40°, ∠CBM = 60°
   ∠MCB = 180° − (600° + 40°)
             = 180° − 100° = 80°
∠CDM = ∠CBM =60°, ∠AMD = 60°
   ∠MCB = ∠GMD = ∠ADM = 80° 
The angles of the triangles AMD, MBC and DCM are 40°, 60°, 80 each.

4. In the picture, 0 is the centre of the circle and A, B, C, D are points on the circle.
Are the lines AB and CD equal? Why?
Answer:
In ∆AOB and ∆COD
OA = OC (Radius of the circle) 
OB = OD (Radius of the circle) 
∠AOB = ∠COD = 80° (Given)
Two sides and the included angles of ∆AOB are equal to two sides and the included angles of ∆COD. Therefore, the third side of these triangles will also be equal. 
∴ AB = CD
That is the lines AB and CD are equal in length. 

5. In the picture, O is the centre of the circle and A, B, C are points on the circle. ∠AOB and ∠AOC are equal.
Are the lines AB and AC equal? Why?
Answer:
In ∆AOB and ∆AOC,
OA= OB, OA= OC (Radius of the circle)
∠AOB = ∠AOC (Given)
The two sides and included angles of ∆AOB are equal to two sides and the included angles of ∆AOC. 
Therefore, the third side of these triangles will also be equal.
∴ AB = AC
That is, the lines AB and AC are equal in length.

♦ Textbook Activities (Textbook Page No: 32)
Isosceles triangles 
(1) Find the other angles in each of the isosceles triangles below? 
Answer:
(A) In the figure, since AC = BC, ∠B = ∠A = 30°
(Angles opposite equal sides are equal)
∠C = 180° − (30° + 30°)
      = 180° − 60° = 120°

(B) In the figure, since DF = EF, ∠D = ∠E 
(Angles opposite equal sides are equal)
∠D + ∠E = 180° − 40° = 140°
∠D = ∠E = 140°/2 = 70°

(C) In the figure, since PQ = QR, ∠R = ∠P 
(Angles opposite equal sides are equal)
∠P + ∠R = 180° − 100° = 80°
∠P = ∠R = 80°/2 = 40°

(D) In the figure, since XY = YZ, ∠Z = ∠X
(Angles opposite equal sides are equal)
∠X + ∠Z = 180° − 90° = 90°
∠X = ∠Z = 90°/2 = 45°

♦ Textbook Activities (Textbook Page No: 35, 36)
(1) One angle of an isosceles triangle is 120°. What are the other two 
angles? 
Answer:
One angle 120°
Sum of the other two angles = 180° − 120° = 60°
Each of the other angles = 60°/2 = 30°

(2) The picture shows a triangle drawn by joining the centre of a circle and
two points on the circle:
Calculate the other two angles of the triangle.
Answer:
In the figure, 0 is the centre of the circle, A and B are points on the circle. Since 0A and 0B are radii of the same circle, 0A = 0B.
In triangle AB0, two sides are equal. So the angles opposite these sides are also equal.
∴ ∠A = ∠B, ∠A + ∠B = 180° − 60° = 120°
So ∠A = ∠B = 120°/2 = 60°

(3) The picture shows the triangle drawn by joining three points on a circle.
Two of the angles formed by joining these points to the centre of the circle are also given:
(i) Calculate the third angle at the centre.
Answer: 
In the figure 0 is the centre of the circle A, B and C are points on the circle.
∠A0B = 110°, ∠A0C = 120°
∠A0B + ∠A0C + ∠B0C = 360°
110°+ 120° + ∠B0C = 360°
230° + ∠B0C = 360°
∠B0C = 360° − 230° = 130° 
∴ The third angle at the centre of the circle = 130° 

(ii) Calculate all the angles of the large (green) triangle in the circle.
Answer:
In ∆A0B, 0A = 0B (Radius of the circle)
∠0AB + ∠0BA = 180° − 110° = 70°
So, ∠0AB = ∠0BA = 70°/2 = 35°

In ∆A0C, 0A = 0C (Radius of the circle)
∠0AC + ∠0CA = 180° − 120° = 60°
So, ∠0AC = ∠0CA = 60°/2 = 30°

In ∆B0C, 0B = 0C (Radius of the circle)
∠0BC + ∠0CB = 180° − 130° = 50°
So, ∠0BC = ∠0CB = 50°/2 = 25°

∴ The angles of the large green triangle in the circle:
∠A = ∠0AB + ∠0AC = 35° + 30° = 65°
∠B = ∠0BA + ∠0BC = 35° + 25° = 60°
∠C = ∠0CA + ∠0CB = 30° + 25° = 55°

(4) Calculate the areas of each of the triangles below : 
(a) In the figure, ∆ABC is an isosceles triangle.
AB = AC = 5 cm
BC = 8 cm
Draw a perpendicular AD from A to BC.
∠ADB = ∠ADC = 90° 
BD = CD = 4 cm
In right triangle ADB,
AD = √5² - 4² = √25 − 16 = √9 = 3cm
The height of the triangle = 3 cm
The area of the triangle = ½ x BC x AD = ½ x 8 x 3 = 12 sq. cm

(b) In the figure, ∆ABC is an isosceles triangle.
AB = AC = 5 cm
BC = 6 cm
Draw a perpendicular AD from A to BC.
∠ADB = ∠ADC = 90° 
BD = CD = 3 cm
In right triangle ADB,
AD = √5² - 3² = √25 − 9 = √16 = 4cm
The height of the triangle = 4 cm
The area of the triangle = ½ x BC x AD = ½ x 6 x 4 = 12 sq. cm

(5) Given that one angle of an isosceles triangle is 70°. What can we say about the other angles?
Answer:
(i) One angle = 90° 
Sum of the other two angles = 180° − 70° = 110° 
Two angles in an isosceles triangle are equal,
The measure of each angle = 110°/2 = 55° 

(ii) In an isosceles triangle, two sides and two angles are equal.
Since one angle is 70°, one of the other two angles must be 70°
Third angle = 180 − (70 + 70) = 40°
That is, one angle of an isosceles triangle is 70°, and the other two angles must be 55° each, or it must be 70° and 40°

(6) How many non - equal isosceles triangles can be drawn with one angle 70° and one side 8 centimetres? 
Answer:
When one angle is 70°, the other two angles will be 55° each. When one angle is 40°, the other two angles will be 70° each. Here, one angle should be 70° and one side should be 8 cm. Let's draw four non-equal isosceles triangles.