The DAV Maths Book Class 7 Solutions Pdf and **DAV Class 7 Maths Chapter 8 Worksheet 2** Solutions of Triangle and Its Properties offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 8 WS 2 Solutions

Question 1.

In ΔABC, ∠A = 44°. If AB = AC, find ∠B and ∠C.

Answer:

AB = AC [given]

∠B = ∠C

∠A + ∠B + ∠C = 180° [Angles sum property]

44° + ∠B + ∠B = 180°

2∠B = 180° – 44°

2∠B = 136°

∠B = \(\frac{136^{\circ}}{2}\) = 68

∠B = ∠C = 68°.

Question 2.

ΔPQR is an isosceles triangle with PQ = PR. If ∠R = 42°, find the measure of ∠P.

Answer:

PQ = PR [given]

∠Q = ∠R [Angles opposite to equal sides]

But ∠R = 42°

Q = 42° [given]

∠P + ∠Q + ∠R = 180° [Angles sum property]

∠P + 42° + 42° = 180°

∠P + 84° = 180°

∠P = 180° – 84°

∠P = 96°

∠P = 96°.

Question 3.

ΔABC is an isosceles triangle with AB = AC. If ∠B of ∠A and ∠C?

Answer:

AB = AC [given]

∠B = ∠C [Angles opposite to equal sides]

But ∠B = 40° [given]

∠C = 40°

∠A + ∠B + ∠C = 180°

∠A + 40° + 40° = 180°

∠A + 80° = 180°

∠A = 180° – 80° = 100°

Hence ∠A = 100° and ∠C = 40°.

Question 4.

The vertical angle of an isosceles triangle is 100°. Find its base angles.

Answer:

∠A + ∠B + ∠C = 180° [Angle Sum Property]

⇒ 100° + ∠B + ∠C = 180°

⇒ ∠B + ∠C = 180° – 100°

⇒ ∠B + ∠C = 80°

But ∠B = ∠C (Base angles of an isosceles triangle)

∠B = ∠C = 40°

Question 5.

In given figure, ΔABC is isosceles with AB = AC. Find the values of x and y.

Answer:

In ΔABC,

AB = AC

∠B = ∠C

∠A + ∠B + ∠C = 180°

30° + ∠B + ∠B = 180°

2∠B = 180° – 30° = 150°

∠C = 75°

∠B + ∠x = 180° [Linear pairs]

75° + x = 180°

x = 180° – 75° = 105°

∠C + ∠y = 180° [Linear Pair]

75° + ∠y = 180°

∠y = 180° – 75° = 105°

Hence ∠x = ∠y = 105°.

Question 6.

In given figure, find ∠ACB and ∠A and write which sides of ΔABC are equal?

Answer:

∠ACD + ∠ACB = 180°

130° + ∠ACB = 180°

∠ACB = 180° – 130° = 50°

∠A + ∠B + ∠ACB = 180° [Angle Sum Property]

∠A + 80° + 50° = 180°

∠A = 180° – 80° – 50°

= 180° -130°

= 50°

∠A = 50°.

Question 7.

In given figure, AB = AC, and AP ⊥ BC and ∠B = 60°. Then find (i) ∠BAP (ii) ∠ACB.

Answer:

AB = AC [given]

∠B = ∠C (Angle opposite to equal sides of a triangle)

∠B = 60°

∠C = 60°

∠A + ∠B + ∠C = 180° (Angle Sum Property)

∠A + 60° + 60° = 180°

∠A + 120° = 180°

∠A = 180° – 120° = 60°

(i) ∠BAP = \(\frac{1}{2}\) ∠A = \(\frac{1}{2}\) × 60° = 30°

(ii) ∠ACB = 60° = ∠ABC

Question 8.

In given figure, find x and y. Is ΔABC isosceles? If so, name its equal sides.

Answer:

∠ABX = ∠BAC + ∠ACB [Exterior angle of a Δ is equal to the sum of its interior opposite angles]

∠ABX = 40° + y°

⇒ 110° = 40° + y°

⇒ y° = 110° – 40° = 70°

⇒ x° + 110° = 180° [Linear pairs] = 180°

⇒ x° = 180° – 110° = 70°

Hence, ΔABC is an isosceles triangle and AB = AC