Question

In the given figure, $ABCD$ is a quadrilateral in which $AB||CD$ and $CD\perp DA$. If $AB=16cm,DC=BC=10cm$, find the area of the quadrilateral $ABCD$.

Answer 1

R.E.F. Image.

$\to AM=CD=10cm$

$BM=AB-AM$

$=16-10$

$=6cm$

$\to$ In $△BMC$ using Pythagoras

theorem

$\therefore B{C}^2=C{M}^2+B{M}^2$

$\therefore (10{)}^2=C{M}^2+(6{)}^2$

$\therefore C{M}^2=100-36$

$\therefore C{M}^2=64$

$\therefore CM=8$

$\to CM=AD=8cm$

$\to$ Area of ABCD $=\frac{1}{2}(AB+CD)\times AD$

$=\frac{1}{2}(16+10)\times 8$

$=26\times 4$

$=104c{m}^2$