Question

Find the area of a quadrilateral $ABCD$ is which $AB=3cm$, $BC=4cm$, $CD=4cm$, $DA=5cm$ and $AC=5cm$.

Answer 1

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In $△ABC$

semi-perimeter, $S=\frac{3+4+5}{2}=\frac{12}{2}=6$

$ar\left(△ABC\right)=\sqrt{6(6-3)(6-4)(6-5)}=\sqrt{6\times 3\times 2\times 1}$

$=\sqrt{36}=6c{m}^2$

In $△ACD$,

semi-perimeter, $S=\frac{5+5+4}{2}=\frac{14}{2}=7$

Applying Heron's formula : $\sqrt{S(S-a)(S-b)(S-c)}$

$ar\left(△ACD\right)=\sqrt{7(7-5)(7-5)(7-4)}=\sqrt{7\times 2\times 2\times 3}$

$=\sqrt{84}=2\sqrt{21}c{m}^2$

$\therefore$ Area of quadrilateral $ABCD=ar\left(△ABC\right)+ar\left(△ACD\right)=(6+2\sqrt{21})c{m}^2$