Find the altitude of a trapezium whose area is 65 cm
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 $=\frac{130}{39}=\frac{10}{3} \mathrm{~cm}$

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