In the given figure, ABCD is a rectangle in which diag.

Solution:

Given: In $\Delta A B C$,

$A C=17 \mathrm{~cm}$

$\sin \theta=\frac{8}{17}$

Since, $\sin \theta=\frac{P}{H}$

$\Rightarrow P=8$ and $H=17$

Using Pythagoras theorem,

$P^{2}+B^{2}=H^{2}$

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

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

$\Rightarrow B^{2}=225$

$\Rightarrow B=15$

Therefore,

$A B=8 \mathrm{~cm}$ and $B C=15 \mathrm{~cm}$

Therefore,

(i) Area of rectangle $A B C D=A B \times B C$

$=8 \times 15$

$=120 \mathrm{~cm}^{2}$

(ii) Perimeter of rectangle $A B C D=2(A B+B C)$

$=2(8+15)$

$=2(23)$

$=46 \mathrm{~cm}$