Question

ABCD is a trapezium with $AB\parallel CD$. $AD=BC=5cm,AB=12cm$ and $CD=7cm$, find the area of trapezium ABCD.

Answer 1

In trapezium $ABCD,$ $AB||CD$, $AD=BC$

Let $DE$ and $CF$ be the perpendiculars on $AB$ from $D$ and $C.$

Then as $DC=EF,EF=7$ cm, therefore $AE=FB=\frac{12-7}{2}=\frac{5}{2}=2.5cm$

Then in triangle $ADE$ by pythagoras theorem,

$A{D}^2=A{E}^2+D{E}^2$

$5^2=(2.5{)}^2+D{E}^2$

$D{E}^2=18.75$

$DE=4.34cm$

Therefore height $=4.34$ cm

Hence area of trapezium,

$=\frac{1}{2}\times Sumofparallelsides\times H$

$=\frac{1}{2}\times (7+12)\times 4.34$

$=41.23c{m}^2$