Question

The sides of a quadrilateral $ABCD$ $6cm$, $8cm$,$12cm$,$14cm$, respectively. The angle between the first two sides is a right angle, Find area? (taken in order)

Answer 1

$\therefore AC=\sqrt{6^2+8^2}=\sqrt{100}=10cm$

In $△ACD$,

Semi-perimeter \left(s\right)=\dfrac{14+12+10}{12}$

$=18$

$\therefore$ Area of quadrilateral $ABCD=$Area of $△ABC$+ Area of $△ACD$

$=\frac{1}{2}\times 66\times 8+\sqrt{18(18-14)(18-12)(18-10)}$

$=24+\sqrt{18\times 4\times 6\times 8}$

$=24+\sqrt{9\times 8\times 8\times 6}$

$=24+24\sqrt{6}$

$=24(\sqrt{6}+1)sq.units$ Ans.