Question:
Find the altitude of a trapezium whose area is 65 cm2 and whose bases are 13 cm and 26 cm.
Solution:
Given:
Area of the trapezium $=65 \mathrm{~cm}^{2}$
The lengths of the opposite parallel sides are $13 \mathrm{~cm}$ and $26 \mathrm{~cm}$.
Area of trapezium $=\frac{1}{2} \times($ Sum of parallel bases $) \times($ Altitude $)$
On putting $g$ the values:
$65=\frac{1}{2} \times(13+26) \times($ Altitude $)$
$65 \times 2=39 \times$ Altitude
Altitude $=\frac{130}{39}=\frac{10}{3} \mathrm{~cm}$