Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm,

Solution:

In the right-angled ∆">ΔACB:

$A B^{2}=B C^{2}+A C^{2}$

$\Rightarrow 17^{2}=B C^{2}+15^{2}$

$\Rightarrow 17^{2}-15^{2}=B C^{2}$

$\Rightarrow 64=B C^{2}$

 

$\Rightarrow B C=8 \mathrm{~cm}$

Perimeter $=A B+B C+C D+A D$

$=17+8+12+9$

= 46 cm

Area of $\Delta A B C=\frac{1}{2}(b \times h)$

$=\frac{1}{2}(8 \times 15)$

$=60 \mathrm{~cm}^{2}$

In $\Delta A D C:$

$A C^{2}=A D^{2}+C D^{2}$

So, $\Delta A D C$ is a right - angled triangle at D.

Area of $\Delta A D C=\frac{1}{2} \times b \times h$

$=\frac{1}{2} \times 9 \times 12$

$=54 \mathrm{~cm}^{2}$

$\therefore$ Area of the quadrilateral = Area of $\triangle A B C+$ Area of $\triangle A D C$

$=60+54$

$=114 \mathrm{~cm}^{2}$