Question

In the adjoining figure, $\text{ABCD}$ is a trapezium in which parallel sides are $\text{AB}=78\text{cm},\text{DC = 52 cm}$ and the non parallel sides are $\text{BC = 30 cm}$ and $\text{AD = 28 cm}$. find the area of the trapezium.

Answer 1

Let us construct $=DE||CB$

$\therefore DC=EB=52\text{cm}$

$DE=CB=30\text{cm}$
$AE=AB-EB$
$=78-52=26\text{cm}$

For $\Delta ADE,s=\frac{28+30+26}{2}=42$

$\Delta DAE$ is a scalene triangle

$Area\left(\Delta DAE\right)=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{42(42-30)(42-28)(42-26)}$
$=\sqrt{42\times 12\times 14\times 16}$
$=\sqrt{\sqrt{6\times 7\times 6\times 2\times 7\times 2\times 4\times 4}}$
$=\sqrt{6\times 6\times 7\times 7\times 2\times 2\times 4\times 4}$
$=6\times 7\times 2\times 4$
$=336{\text{cm}}^2$

But we know $Area\left(\Delta DAE\right)=\frac{1}{2}\times AE\times Height$
$\Rightarrow 336=\frac{1}{2}\times 26\times h$

Area of trapezium $ABCD=\frac{1}{2}\times h\times \left(sumofparallelsides\right)$
$=\frac{1}{2}\times \frac{336}{13}\times 130=1680{\text{cm}}^2$