Question

Find the area and the perimeter of quadrilateral $\text{ABCD}$, if $\text{AB= 8 cm}$, $\text{AD= 10 cm}$, $\text{BD = 12 cm}$, $\text{DC = 13 cm}$, and $\Delta \text{DBC}={90}^{\circ }$

Answer 1

Area of the quadrilateral $=$ area of the triangle ABD $+$ area of the triangle DBC.

From $\Delta DBC$, ${BC}^2=D{C}^2-B{D}^2$ [Using Pythagoras theorem]

$⟹B{C}^2={13}^2-{12}^2=25$

$⟹BC=\sqrt{25}=5\text{cm}$

Area of the $\Delta DBC$$=\frac{1}{2}\times b\times h$

$=\frac{1}{2}\times 5\times 12$

$=30{\text{cm}}^2$

Area of the $\Delta ABD$$=\sqrt{s(s-a)(s-b)(s-c)}$

Here, $s=\frac{10+12+8}{2}$

$=15\text{cm}$

$\therefore$ Area of the $\Delta ABD$$=\sqrt{15(15-10)(15-12)(15-8)}$

$=15\sqrt{7}{\text{cm}}^2$

$\therefore$ area of the quadrilateral$=15+15\sqrt{7}=15(1+\sqrt{7}){\text{cm}}^2$