Find the perimeter and area of a quadrilateral ABCD in which BC = 12 cm,

Solution:

We know that $\triangle A B D$ is a right-angled triangle.

$\therefore A B^{2}=\sqrt{A D^{2}-D B^{2}}=\sqrt{17^{2}-15^{2}}=\sqrt{289-225}=\sqrt{64}=8 \mathrm{~cm}$

Now,

Area of triangle $A B D=\frac{1}{2} \times$ Base $\times$ Height

$=\frac{1}{2} \times A B \times B D$

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

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

Let :

$a=9 \mathrm{~cm}, b=15 \mathrm{~cm}$ and $c=12 \mathrm{~cm}$

$s=\frac{a+b+c}{2}=\frac{9+15+12}{2}=18 \mathrm{~cm}$

By Heron's formula, we have:

Area of triangle $D B C=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{18(18-9)(18-15)(18-12)}$

$=\sqrt{18 \times 9 \times 3 \times 6}$

$=\sqrt{6 \times 3 \times 3 \times 3 \times 3 \times 6}$

$=6 \times 3 \times 3$

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

Now,

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

$=(60+54) \mathrm{cm}^{2}=114 \mathrm{~cm}^{2}$

And,

Perimeter of quadrilateral $A B C D=A B+B C+C D+A D=17+8+12+9=46 \mathrm{~cm}$