Question

Find the area of quadrilateral $ABCD$ given below ( in sq units) in nearest integer.

Answer 1

As ACD is an equilateral

triangle $\therefore ar\left(ACD\right)=\frac{\sqrt{3}}{4}(20{)}^2$

$\Rightarrow ar\left(ACD\right)=100\sqrt{3}sq.units$

$ar\left(ABC\right)=\frac{1}{2}\times 12\times x$

$[x^2+{12}^2={20}^{2}usingPythagorastherm]$

$\Rightarrow ar\left(ABC\right)=\frac{1}{2}\times 12\times \sqrt{{20}^2-{12}^2}$

$\Rightarrow ar\left(ABC\right)=96$

$\therefore ar\left(ABCD\right)=ar\left(ABC\right)+ar\left(ACD\right)$

$=96+100\sqrt{3}$

$ar\left(ABCD\right)=269.2sq.units$