Question:
In the given figure, $A B C D$ is a rectangle in which diag. $A C=17 \mathrm{~cm}, \angle B C A=\theta$ and $\sin \theta=\frac{8}{17}$.
Find
(i) the area of rect. ABCD,
(ii) the perimeter of rect. ABCD.
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 $A B C D=2(A B+B C)$
$=2(8+15)$
$=2(23)$
$=46 \mathrm{~cm}$