A field is in the shape of a trapezium whose parallel sides are $ 25 \mathrm{~m} $ and $ 10 \mathrm{~m} $. The non-parallel sides are $ 14 \mathrm{~m} $ and $ 13 \mathrm{~m} $. Find the area of the field.

Given:

A field is in the shape of a trapezium whose parallel sides are \( 25 \mathrm{~m} \) and \( 10 \mathrm{~m} \). The non-parallel sides are \( 14 \mathrm{~m} \) and \( 13 \mathrm{~m} \).

To do:

We have to find the area of the field.

Solution:

The area of the trapezium $=$ Area of the rectangle $+$ Area of triangle

Let $h$ be the height of the trapezium.

This implies, $h$ is the height of rectangle as well as triangle.

In triangle,

$s=\frac{a+b+c}{2}$

$s=\frac{14+13+15}{2}$

$s=\frac{42}{2}$

$s=21$

Area of triangle $=\sqrt{21 \times (21-14) \times (21-13) \times (21-15)}$

$=\sqrt{7\times3 \times 7 \times 8\times 6}$

$=\sqrt{7\times3\times7\times4\times2\times2\times3}$

$=7\times3\times2\times2$

$=84\ m^2$

Area of triangle with base $15\ m$ and height $h=\frac{1}{2}\times15\times h$

$84=\frac{15}{2}h$

$h=\frac{84\times2}{15}$

$h=\frac{56}{5}\ m$ Area of Rectangle $=\frac{56}{5} \times 10=112\ m^2$

Therefore,

Area of trapezium $=84+112=196\ m^2$

The area of the field is 196 sq.m.

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Updated on: 10-Oct-2022

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