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
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