# Quadrilaterals Chapter 8 Class 9 Maths NCERT Textbook With Solutions PDF

NCERT Solutions for Class 9 Maths Chapter 8‘ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 9 Maths Chapter 8 Exercise Solution’ using the download button.

#### 8.1 Introduction

You have studied many properties of a triangle in Chapters 6 and 7 and you know that on joining three non-collinear points in pairs, the figure so obtained is a triangle.

Now, let us mark four points and see what we obtain by joining them in pairs in some order.

#### 8.2 Angle Sum Property of a Quadrilateral

Let us now recall the angle sum property of a quadrilateral. The sum of the angles of a quadrilateral is 360º.

This can be verified by drawing a diagonal and dividing the quadrilateral into two triangles.

#### 8.4 Properties of a Parallelogram

Let us perform an activity. Cut out a parallelogram from a sheet of paper and cut it along a diagonal (see Fig. 8.7). You obtain two triangles.

What can you say about these triangles? Place one triangle over the other. Turn one around, if necessary. What do you observe? Observe that the two triangles are congruent to each other.

## NCERT Solutions Class 9 Maths Chapter 8 Quadrilaterals

1. The angles of a quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the quadrilateral.

Solution:

Let the common ratio between the angles be = x.

We know that the sum of the interior angles of the quadrilateral = 360°

Now,

3x+5x+9x+13x = 360°

⇒ 30x = 360°

⇒ x = 12°

, Angles of the quadrilateral are:

3x = 3×12° = 36°

5x = 5×12° = 60°

9x = 9×12° = 108°

13x = 13×12° = 156°

2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Solution:

Given that,

AC = BD

To show that, ABCD is a rectangle if the diagonals of a parallelogram are equal

To show ABCD is a rectangle we have to prove that one of its interior angles is right-angled.

Proof,

AB = BA (Common)

BC = AD (Opposite sides of a parallelogram are equal)

AC = BD (Given)

Therefore, ΔABC ≅ ΔBAD [SSS congruency]

∠A = ∠B [Corresponding parts of Congruent Triangles]

also,

∠A+∠B = 180° (Sum of the angles on the same side of the transversal)

⇒ 2∠A = 180°

⇒ ∠A = 90° = ∠B

Therefore, ABCD is a rectangle.

Hence Proved.