What is the area of the given quadrilateral?

The area of the quadrilateral can be obtained by using Heron’s formula which states that

Area of triangle = $\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}$ where, $\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}$.

Diagonal AC has to be drawn first.

It can be seen that ΔADC is a right-angled triangle.

∴ AC² = AD² + DC² (Pythagoras theorem)

⇒ AC² = (5 cm)² + (12 cm)² = (25 + 144) cm² = 169 cm²

⇒ AC = $\sqrt{169}$ = 13 cm

Area of ΔADC = $\frac{1}{2}\times \mathrm{AD}\times \mathrm{DC}=\frac{1}{2}\times 5\times 12=30 {\mathrm{cm}}^2$

In ΔABC, a = 7 cm, b = 8 cm, and c = 13 cm.

$s=\frac{a+b+c}{2}=\frac{7+8+13}{2}=14 cm$

⇒ Area of ΔABC = $\sqrt{14(14-7)(14-8)(14-13)}$

$$\begin{gathered} =\sqrt{14\times 7\times 6\times 1} \\ =14\sqrt{3} \end{gathered}$$

Area of quadrilateral ABCD = Area of ΔADC + Area of ΔABC

= $30+14\sqrt{3}=2\left(15+7\sqrt{3}\right)$

Thus, the area of the quadrilateral is $2\left(15+7\sqrt{3}\right)$ cm².