Question

$ABCD$ is a trapezium in which $AB\parallel CD$; BC and AD are non-parallel sides. It is given that $AB=75cm,BC=42cm,CD=30cm$ and $AD=39cm$. Find the area of the trapezium.

Answer 1

Draw $CE\parallel DA$ and $CF\perp AB$.

Therefore, $ADCE$ is a parallelogram having $AE\parallel CD$ and $CE\parallel DA$.

$\Rightarrow AD=CE=39cm,DC=AE=30cm$ and $BE=75-30=45cm$

In $△BCE$, by heron's formula we have

$a=39cm,b=45cm$ and $c=42cm$

$\Rightarrow s=\frac{a+b+c}{2}=\frac{39+42+45}{2}=63cm$

$\therefore$ Area of $\Delta BEC=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{63(63-39)(63-42)(63-45)}c{m}^2$

$=\sqrt{63\times 24\times 21\times 18}c{m}^2$

$=756c{m}^2$

Also, Area of $\Delta BEC=\frac{1}{2}\times base\times height$

$\Rightarrow \frac{1}{2}\times 45\times h=756c{m}^2\Rightarrow h=33.6cm$

So, Area of trapezium:-

$=\frac{1}{2}\times (AB+CD)\times height=\frac{1}{2}\times (75+30)\times 33.6=1764{cm}^2$