# Find the area of the trapezium PQRS with height PQ given in Fig. 12.3

**Solution:**

Given, PQRS is a trapezium with height PQ.

We have to find the area of trapezium PQRS.

Draw RT perpendicular to PS

Now, ST = PS - PT

= 12 - 7

ST = 5 m

Also, PT = RQ

Area of trapezium PQRS = area of triangle STR + area of rectangle PQRT

Considering triangle STR,

STR is a right angle with T at right angle.

By using Pythagorean theorem,

SR² = ST² + TR²

(13)² = (5)² + TR²

169 = 25 + TR²

TR² = 169 - 25

TR² = 144

Taking square root,

TR = 12 cm

Area of triangle = 1/2 × base × height

Area of triangle STR = 1/2 × TR × ST

= 1/2 × 12 × 5

= 6 × 5

Area of triangle STR = 30 cm²

Area of rectangle = length × width

Area of rectangle PQRT = PQ × RQ

= 12 × 7

Area of rectangle PQRS = 84 cm²

Now, area of trapezium PORS = 30 + 84

= 114 cm²

Therefore, the area of trapezium is 114 cm²

**✦ Try This:** The sides of a triangle are 56 cm, 60 cm and 52 cm long. Find the area of the triangle.

**☛ Also Check: **NCERT Solutions for Class 9 Maths Chapter 12

**NCERT Exemplar Class 9 Maths Exercise 12.3 Problem 10**

## Find the area of the trapezium PQRS with height PQ given in Fig. 12.3

**Summary:**

The area of the trapezium PQRS with height PQ given in Fig. 12.3 is 114 cm²

**☛ Related Questions:**

- If each side of a triangle is doubled, then find the ratio of area of the new triangle thus formed a . . . .
- How much paper of each shade is needed to make a kite given in Fig. 12.4, in which ABCD is a square . . . .
- The perimeter of a triangle is 50 cm. One side of a triangle is 4 cm longer than the smaller side an . . . .

visual curriculum