SCERT AP 7th Class Maths Solutions Pdf Chapter 5 Triangles Ex 5.3 Textbook Exercise Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangles Exercise 5.3

Question 1.

Write the exterior angles of ∆XYZ.

Answer:

Exterior angles of ∆XYZ are ∠PXY, ∠ZYQ and ∠XZR.

Exterior angle of X is ∠PXY. Exterior angle of Y is ∠QYZ.

Exterior angle of Z is ∠XZR.

Question 2.

Find the exterior angles in each of the following triangles :

Answer:

From the figure

∠A = 60°, ∠B = 73°

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

Exterior angle at

C = ∠A + ∠B

= 600 + 730

= 133°

Answer:

From the figure,

∠D = 90°, ∠E = 30°

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

Exterior angle at

F = ∠D + ∠E

= 90° +.30°

= 120°

Question 3.

Find the value of ‘x’ in the following figures.

Answer:

From the figure, ∠A = 35°, ∠B = x° and exterior angle at C = 70°

Exterior angle of a triangle is equal to the sum of its opposite interior angles. Exterior anjpNtt

C = ∠A + ∠B = 70°

⇒ 35° + x= 70°

⇒ 35° + x – 35° = 70° – 35°

∴ x = 35°

Answer:

From the figure, ∠P = 4x, ∠Q = 3x and exterior angle at R = 119°

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

Exterior angle at R = ∠P + ∠Q = 119°

⇒ 4x + 3x = 119°

⇒ 7x = 119°

⇒ \(\frac{7 x}{7}=\frac{119^{\circ}}{7}\)

∴ x = 17°

Question 4.

If the exterior angle of a triangle is 110° and it’s interior opposite angles are x° and (x + 10)°, then find the value of ‘x’.

Answer:

Given interior opposite angles are x° and (x + 10)°.

Exterior angle =110° .

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

x + x + 10° = 110°

⇒ 2x + 10°- 10° = 110°- 10°

⇒ 2x = 100°

⇒ \(\frac{2 x}{2}=\frac{100^{\circ}}{2}\)

∴ x = 50°

Question 5.

Find the values of V and ‘y’ in each of the following figures.

Answer:

From the figure, ∠A = 40°, ∠B = 60°, ∠D = 45°

Exterior angle at C = x°

Exterior angle at E = y°

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

Exterior angle at C = ∠A + ∠B

x = 40° + 60°

∴ x = 100°

Exterior angle at E = ∠C + ∠D

y = x + 45°

y = 100° + 45°

∴ y = 145°

Answer:

From the figure, ∠M = y, ∠N = 70°

Exterior angle at M = x°

Exterior angle at L = 120°

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

Exterior angle at L = ∠M + ∠N = 120°

⇒ y + 70° = 120°

⇒ y + 70 – 70 = 120 – 70

∴ y = 50°

∠QLN + ∠MLN = 180° (linear pair of angles)

120° + ∠MLN = 180°

120° + ∠MLN – 120° = 180°- 120°

∠MLN = 60°

∴ ∠L = 60°

Exterior angle at M = ∠L + ∠N

x = 60° + 70° = 130°

∴ x = 130°

∴ x = 130° and y = 50°