Question

find the altitude of trapezium whose area is 45 square cm and the sum of the lengths of the parallel bases is 18

Answer 1

Hello!

A = ½ x ( a + b ) x h

h = $\dfrac{\sf 2A}{\sf (a+b)}$

h = $\dfrac{\sf 2 \:x\:45}{\sf 18} = \boxed{\sf 5\:cm}$

Hence,
Altitude of the trapezium is $\bf{5\:cm}$

Cheers!