AP State Syllabus AP Board 7th Class Maths Solutions Chapter 12 Quadrilaterals Ex 1 Textbook Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 12th Lesson Quadrilaterals Exercise 1

Question 1.

In quadrilateral PQRS

(i) Name the sides, angles, vertices and diagonals.

(ii) Also name all the pairs of adjacent sides, adjacent angles, opposite sides and opposite angles. .

Solution:

ï) In quadrilateral PQRS,

sides = PQ, QR, RS and SP

angles = ∠P, ∠Q, ∠R, and ∠S

vertices = P, Q, R and S

diagonals = PR and QS

ii) Pairs of adjacent sides = (PQ, QR), (QR. RS), (RS, SP), (SP, PQ)

Pairs of adjacent angles (∠P, ∠Q). (∠Q, ∠R), (∠R, ∠S), (∠S, ∠P)

Pairs of opposite sides = (PQ, RS) and (QR, PS)

Pairs of opposite angles = (∠P,∠R) and (∠Q, ∠S)

Question 2.

The three angles of a quadrilateral are 60°, 80° and 1200. Find the fourth angle?

Solution:

Given three angles = 60°,80°,120°

Sum of the given three angles = 60°+ 80° + 120° = 260°

Sum of the four Interior angles of a quadrilateral = 360°

∴ The fourth angle = 360° — 260° = 100°

Question 3.

The angles of a quadrilateral are in the ratio 2 : 3 : 4: 6. Find the measure of each of the four angles.

Solution:

Given that ratio of the four angles = 2 : 3 : 4 : 6

Sum of the terms in the ratio = 2 + 3 + 4 + 6 = 15

Sum of the four angles = 360°

∴ 1st angle = \(\frac { 2 }{ 15 }\) x 360° = 48°

2nd angle = \(\frac { 3 }{ 15 }\) x 360° = 72°

3rd angle = \(\frac { 4 }{ 15 }\) x 360° = 96°

4th angle = \(\frac { 6 }{ 15 }\) x 360° = 144°

Question 4.

The four angles of a quadrilateral are equal. Draw this quadrilateral in your notebook.

Find each of them.

Solution:

Sum of the four angles in a quadrilateral = 360°

Each of the angle = \(\frac{360^{\circ}}{4}\) = 90°

∠A = ∠B = ∠C = ∠D = 90°

Question 5.

In a quadrilateral, the angles arex°, (x + l0)°, (i+ 20)°, (x + 30)°. Find the angles.

Solution:

Given angles are x°, (x + 10)°, (x + 20)°, (x + 30)°

Sum of the four angles = x + x + 10° + x + 20° + x + 30° = 4x + 60°

But the sum of the four angles = 360°

4 x 60° = 60° = 360°

4x = 360° – 60° = 300°

x = \(\frac{300^{\circ}}{4}\) = 75°

∴ The angles are x = 750

x + 10°- 75°+ 10° = 85°

x + 20° = 75° +20° = 95°

x + 30° = 75° + 30° = 105°

Question 6.

The angles of a quadrilateral cannot be in the ratio 1: 2 : 3 : 6. Why? Give reasons.

(Hint: Try to draw a rough diagram of this quadrilateral)

Solution:

Given that the angles of a quadrilateral cant be in the ratio 1: 2 : 3 : 6

If the ratio is 1 : 2 : 3 : 6 then

the sum of terms of ratio = I + 2 + 3 + 6 = 12

Sum of the angles = 360°

∴ 1st angle = \(\frac{1}{12}\) x 360° = 30°

2nd angle = \(\frac{2}{12}\)x 360° = 60°

3rd angle = \(\frac{3}{12}\) x 360° = 90°

4th angle = \(\frac{3}{12}\) x 360° = I80

i.e., 4th angle is 180°, a straight angle.

A quadrIlateral cant he formed with these angles.