Question

ABCD is a trapezium in which $AB\parallel DC$ $AB=78cm,CD=52cm,AD=28cm$ and $BC=30cm$ find the area of the trapezium.

Answer 1

In trapezium $ABCD$ construct a line parallel to $AD=28cm$ meeting $AB$ at point $E$ . Than $EB=AB-AE=78-52=26cm$.

In $△CEB$

$EC=28cm,CB=30cm,EB=26cm$

$S=\frac{(28+30+26)}{2}=42cm$

$Areaof△CEB=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{42(42-28)(42-30)(42-26)}$

$=\sqrt{42\times 14\times 12\times 16}$

$=336cm$

$Areaof△CBE=\frac{bh}{2}$

$\frac{bh}{2}=336$

$bh=\left(336\right)2$

$26\left(h\right)=\left(336\right)2$ ----- ($EB=b=26$)

$h=\frac{\left(336\right)2}{26}$

$h=\frac{\left(168\right)2}{13}cm$

Area of trapezium $=\frac{(sumof\parallel sides)h}{2}$

$=\frac{\left(\right(52+78)\times \frac{168\times 2}{13}}{2}$

$=\frac{(130\times 168\times 2}{13\times 2}$

$=1680sq.cm$