Question

A rectangle having the longer side as 9 cm makes an angle of ${30}^{\circ }$ with the diagonal. The length of the other side and the area of the rectangle will be equal to

• A. $6\sqrt{3}cm,30\sqrt{3}c{m}^2$
• B. $9cm,18\sqrt{3}c{m}^2$
• C. $3\sqrt{3}cm,27\sqrt{3}c{m}^2$
• D. $3\sqrt{3}cm,9\sqrt{3}c{m}^2$

Answer 1

The correct option is C

$3\sqrt{3}cm,27\sqrt{3}c{m}^2$

Given ABCD is rectanlge

$AB=CD,AD=CB,\angle B={90}^{\circ },\angle CAB={30}^{\circ }$

$\angle BCA=180-(90+{30}^{\circ })={60}^{\circ }$

In $\Delta ABC$, the angles are ${30}^{\circ },{60}^{\circ },{90}^{\circ }$

$\Rightarrow \sin \left(30\right):\sin \left(60\right):\sin \left(90\right)$

$\Rightarrow \frac{1}{2}:\frac{\sqrt{3}}{2}:1$

$\Rightarrow 1:\sqrt{3}:2$

So, the ratio of the sides CB,AB,AC will be $1:\sqrt{3}:2$, the corresponding sides can be calculated as

$\begin{array}{ccccc}{30}^{\circ }& & {60}^{\circ }& & {90}^{\circ }\\ 1& :& \sqrt{3}& :& 2\\ x& :& x\sqrt{3}& :& 2x\\ CB& & AB& & AC\\ \downarrow & & \downarrow & & \downarrow \\ 3\sqrt{3}cm& & 9cm& & 6\sqrt{3}cm\end{array}$

$(x\sqrt{3}=9\Rightarrow x=\frac{9}{\sqrt{3}}=3\sqrt{3}$)

Hence, the other side BC = $3\sqrt{3}$

Area $=AB\times BC=9\times 3\sqrt{3}=27\sqrt{3}c{m}^2$