In figure, compute the area of quadrilateral ABCD.

Solution:

Given:

DC = 17 cm, AD = 9 cm and BC = 8 cm

In ΔBCD we have

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

$\Rightarrow 17^{2}=B D^{2}+8^{2}$

$\Rightarrow B D^{2}=289-64$

$\Rightarrow B D=15$

In ΔABD we have

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

$\Rightarrow 15^{2}=\mathrm{AB}^{2}+9^{2}$

$\Rightarrow \mathrm{AB}^{2}=225-81=144$

⇒ AB = 12

ar(quad ABCD) = ar(ΔABD) + ar(ΔBCD)

$\operatorname{ar}(q u a d ~ A B C D)=1 / 2(12 \times 9)+1 / 2(8 \times 17)=54+68=122 \mathrm{~cm}^{2}$

$\operatorname{ar}(q u a d \mathrm{ABCD})=1 / 2(12 \times 9)+1 / 2(8 \times 15)=54+60=114 \mathrm{~cm}^{2}$