Question

Find the area of the trapezium $ABCD$ in which $AB\parallel DC,AB=18cm,\angle B=\angle C={90}^o,CD=12cm$ and $AD=10cm$.

Answer 1

$\begin{array}{c}MD=18-12=6cm\\ AD=10cm\\ A{M}^2=A{D}^2-M{D}^2={\left(10\right)}^2-{\left(6\right)}^2\\ AM=8cm\\ AM=BC=8cm\end{array}$

area of trapezium$=\frac{1}{2}\times \left(sumofparallelsides\right)\times Altitude$

$\begin{array}{c}=\frac{1}{2}\left(AB+CD\right)\times BC\\ =\frac{1}{2}\left(18+12\right)\times 8\\ =\frac{1}{2}\times 30\times 8=15\times 8=120c{m}^2\end{array}$