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The parallel sides of a trapezium are 20cm and 10cm. Its nonparallel sides are both equal, each being 13cm. Find the area of the trapezium.
Given:
The parallel sides of a trapezium are 20cm and 10cm. Its nonparallel sides are both equal, each being 13cm.
To do:
We have to find the area of the trapezium.
Solution:
From the figure,
$\triangle APC$ is a right angled triangle. Therefore, by using Pythagoras theorem,
$AC^2=AP^2+CP^2$ $13^2=AP^2+5^2$
$AP^2=169-25$
$AP=\sqrt{144}$
$AP=12\ cm$
ABCD is a rectangle of length $10\ cm$ and breadth $12\ cm$.
Area of trapezium ABCD$=$ Area of rectangle PQBA$+$ Area of triangle ACP$+$Area of triangle BQD
$=10\times12+\frac{1}{2}\times12\times5+\frac{1}{2}\times12\times5$ ($AP=BQ=12\ cm$)
$=120+30+30\ cm^2$
$=180\ cm^2$
The area of the trapezium is $180\ cm^2$.
Updated on: 10-Oct-2022
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