Question

Calculate the area of the quadrilateral $ABCD$ in which $AB=32cm$, $AD=24cm$, $\angle A={90}^{\circ }$ and $BC=CD=52cm$.

Answer 1

$\Rightarrow$ Join $DB$

$\Rightarrow$ $△ADB$ is a right angled triangle.

$(DB{)}^2=(AD{)}^2+(AB{)}^2$ [ By Pythagoras theorem ]

$(DB{)}^2=(24{)}^2+(32{)}^2$

$(DB{)}^2=576+1024$

$(DB{)}^2=1600$

$\therefore$ $DB=40cm$

$CX\perp DB$

$DX=\frac{40}{2}=20cm$

$(DC{)}^2=(DX{)}^2+(XC{)}^2=$

$(52{)}^2=(20{)}^2-(XC{)}^2$

$2704=400-(XC{)}^2$

$(XC{)}^2=2304$

$\therefore$ $XC=48cm$

$\Rightarrow$ Area of quadrilateral $ABCD=Ar\left(△ADB\right)+Ar\left(△DBC\right)$

$\Rightarrow$ Area of quadrilateral $ABCD=\frac{1}{2}\times AB\times AD+\frac{1}{2}XC\times BD$

$\Rightarrow$ Area of quadrilateral $ABCD=\frac{1}{2}\times 32\times 24+\frac{1}{2}48\times 40$

$\Rightarrow$ Area of quadrilateral $ABCD=384+960=1344c{m}^2$