Question

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

Answer 1

Let ABCD be a trapezium with,

$AB\parallel CD$

$AB=25m$

$CD=10m$

$BC=14m$

$AD=13m$

Draw $CE\parallel DA$. So, ADCE is a parallelogram with,

$CD=AE=10m$

$CE=AD=13m$

$BE=AB-AE=25-10=15m$

In $\Delta BCE$, the semi perimeter will be,

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

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

$s=21m$

Area of $\Delta BCE$,

$A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

$=\sqrt{21\left(21-14\right)\left(21-13\right)\left(21-15\right)}$

$=\sqrt{21\left(7\right)\left(8\right)\left(6\right)}$

$=\sqrt{7056}$

$=84m^2$

Also, area of $\Delta BCE$ is,

$A=\frac{1}{2}\times base\times height$

$84=\frac{1}{2}\times 15\times CL$

$\frac{84\times 2}{15}=CL$

$CL=\frac{56}{5}m$

Now, the area of trapezium is,

$A=\frac{1}{2}\left(sumofparallelsides\right)\left(height\right)$

$A=\frac{1}{2}\times \left(25+10\right)\left(\frac{56}{5}\right)$

$A=196m^2$

Therefore, the area of the trapezium is $196m^2$.