(i) Calculate the area of quad. ABCD, given in Fig.

(i) In △">△△BCD, 

$\mathrm{DB}^{2}+\mathrm{BC}^{2}=\mathrm{DC}^{2}$

$\Rightarrow \mathrm{DB}^{2}=17^{2}-8^{2}=225$

$\Rightarrow \mathrm{DB}=15 \mathrm{~cm}$

$\operatorname{Ar}(\triangle B C D)=\frac{1}{2} \times b \times h=\frac{1}{2} \times 8 \times 15=60 \mathrm{~cm}^{2}$

In $\triangle B A D$

$\mathrm{DA}^{2}+\mathrm{AB}^{2}=\mathrm{DB}^{2}$

$\Rightarrow \mathrm{AB}^{2}=15^{2}-9^{2}=144$

$\Rightarrow \mathrm{AB}=12 \mathrm{~cm}$

$\operatorname{Ar}(\triangle \mathrm{DAB})=\frac{1}{2} \times b \times h=\frac{1}{2} \times 9 \times 12=54 \mathrm{~cm}^{2}$

Area of quad. $A B C D=\operatorname{Ar}(\triangle D A B)+\operatorname{Ar}(\triangle B C D)=54+60=114 \mathrm{~cm}^{2}$

(ii) Area of trap(PQRS) $=\frac{1}{2}(8+16) \times 8=96 \mathrm{~cm}^{2}$