This question has been taken from Book:- ML aggarwal, Avichal publication, class10th, quadratic equation in one variable, chapter 5, exercise 5.5

This is an important ques

The lengths of the parallel sides of a trapezium are (x + 9) cm and (2x – 3) cm and the distance between them is (x + 4) cm. If its area is 540 cm², find x.

Question no.17 , ML Aggarwal, chapter 5, exercise 5.5, quadratic equation in one variable, ICSE board,

Solution:We know that,

Area of a trapezium = ½ × (sum of parallel sides) × (height)

Given, the length of parallel sides are (x + 9) and (2x – 3)

And height = (x + 4)

Now, according the conditions in the problem

½ × (x + 9 + 2x – 3) × (x + 4) = 540

(3x + 6) (x + 4) = 540 × 2

3x

^{2}+ 12x + 6x + 24 = 10803x

^{2}+ 18x – 1056 = 0x

^{2}+ 6x – 352 = 0 [Dividing by 3]By factorization method, we have

x

^{2}+ 22x – 16x – 352 = 0x(x + 22) – 16(x + 22) = 0

(x – 16) (x + 22) = 0

So,

x – 16 = 0 or x + 22 = 0

x = 16 or x = -22

As measurements cannot be negative x = -22 is not possible

Therefore, x = 16