# PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then ar (PAS) = 30 cm². Is the given statement true or false and justify your answer.

**Solution:**

It is given that

PS = 5 cm

Radius SQ = 13 cm

In ∆ SPQ,

Using the Pythagoras theorem

SQ² = PQ² + PS²

Substituting the values

13² = PQ² + 5²

PQ² = 169 - 25 = 144

PQ = 12 cm

We know that

Area of ∆ APS = 1/2 bh

Substituting the values

= 1/2 × PS × PQ

= 1/2 × 5 × 12

= 30 cm²

Therefore, the statement is true.

**✦ Try This: **PQRS is a rectangle inscribed in a quadrant of a circle of radius 12 cm. A is any point on PQ. If PS = 4 cm, then determine the ar (PAS).

**☛ Also Check: **NCERT Solutions for Class 9 Maths Chapter 9

**NCERT Exemplar Class 9 Maths Exercise 9.2 Problem 2**

## PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then ar (PAS) = 30 cm². Is the given statement true or false and justify your answer.

**Summary:**

The statement “PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then ar (PAS) = 30 cm²” is true

**☛ Related Questions:**

- PQRS is a parallelogram whose area is 180 cm² and A is any point on the diagonal QS. The area of ∆ A . . . .
- ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then ar (BDE) = 1/4 ar . . . .
- In Fig. 9.8, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = 1/2 . . . .

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