Question

In the given figure, calculate the area of quadrilateral ABCD.

Answer 1

(i) ​
ar(quadrilateral ABCD) = ar(∆ ABD) + ar(∆ BDC)
∆ ABD and ∆ BDC are right-angled triangles. So,

BD = $\sqrt{\left({17}^2-8^2\right)}=\sqrt{225}=15\mathrm{cm}$
and AB = $\sqrt{\left({15}^2-9^2\right)}=\sqrt{144}=12\mathrm{cm}$
Now,
ar(∆ ABD) = $\frac{1}{2}$⨯AD ⨯ AB
= $\frac{1}{2}$⨯9 ⨯ 12
= 54 cm²

^​ar(∆ BDC) = $\frac{1}{2}$⨯BD ⨯ BC
= $\frac{1}{2}$ ⨯15 ⨯ 8
= 60 cm²

Hence, area of quadrilateral ABCD = 54 + 60 = 114 cm².