Apply the properties of isosceles and equilateral triangles to find the unknown angles in the given figures :

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asked Jun 15 in Class VII Maths by muskan15 (-2,549 points)

Apply the properties of isosceles and equilateral triangles to find the unknown angles in the given figures :

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answered Jun 15 by navnit40 (-3,424 points)

a = 70° (Angles opposite to equal sides)

But a + 70° + x = 180° (Angles of a triangle)

⇒ 70° + 70° + x = 180°

⇒ 140° + x = 180°

⇒ x = 180° - 140° = 40°

y = b (Angles opposite to equal sides)

But a = y + b

(Exterior angle of a triangle is equal to sum of its interior opposite angles)

⇒ 70° = y + y ⇒ 2y = 70°

⇒ y = 70°/2 = 35°

Hence x = 40°, y = 35°

(ii) In Fig. (ii),

In an equilateral triangle,

Each angle = 60°

In isosceles triangle.,

Let each base angle = a

∴ a + a + 100° = 180°

⇒ 2a + 100° = 180°

⇒ 2a = 180° – 100° = 80°

∴ a = 80°/2 = 40°  ∴ x = 60° + 40° = 100°

And y = 60° + 40° = 100°

(iii) In Fig. (iii), 130° = x + p

(Exterior angle of a triangle is equal to the sum of its interior opposite angles)

∵ Lines are parallel (Given)

∴ p = 60° (Alternate angle)

And  y = a

But a + 130° = 180° (Linear pair)

⇒ a = 180° – 130° = 50v  ∴ y = 50°

And x + p = 130°

⇒ x + 60° = 130° ⇒ x = 130° – 60° = 70°

Hence, x = 70°, y = 50° and p = 60°

(iv) In Fig. (iv),

x = a + b

but b = y (Angles opposite to equal sides)

Similarly a = c

But a + c + 30° = 180°

⇒ a + a + 30° = 180° ⇒ 2a + 30° = 180°

⇒ 2a = 180° – 30° = 150°

⇒ a = 150°/2 = 75° and b + y = 90°

⇒ y + y = 90° ⇒ 2y = 90°

⇒ y = 90°/2 = 45° ⇒ b = 45°

Hence, x = a + b = 75° + 45° = 120°

And  y = 45°

(v) In Fig.(v),

a + b + 40° = 180° (Angles of a triangle)

⇒ a + b = 180° – 40° = 140°

But a = b (Angles opposite to equal sides)

∴ a = b = 140°/2 = 70°

∴ x = b + 40° = 70° + 40° = 110°

(Exterior angle of a triangle is equal to the sum of its interior opposite angles)

Similarly y = a + 40°

= 70° + 40° = 110°

Hence, x = 110°, y = 110°

(vi)  In the Fig. (vi)

a = b (Angles opposite to equal sides)

∴ y = 120°

But a + 120° = 180° (Linear pair)

⇒ a = 180° – 120° = 60°

∴ b = 60°

But x + a + b = 180° (Angles of a triangle)

⇒ x + 60° + 60° = 180°

⇒ x + 120° = 180°

∴ x = 180° – 120° = 60°

B = z + 25

(Exterior angle of a triangle is equal to the sum of its interior opposite angles)

⇒ 60° = z + 25°

⇒ z = 60° – 25° = 35°

Hence x = 60°, y = 120° and z = 35°

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