Question

In the figure, $ABCD$ is a rectangle. Calculate the area of $ABCD$.

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $\sqrt{6}$
• D. $2\sqrt{2}$
• E. $2\sqrt{3}$

Answer 1

Answer: (B)

$\sqrt{3}$

The diagonal of a rectangle divides it into two equal right triangles,Triangle $ABC$ and Triangle $ADC$.

In triangle $ADC$,

$\sin {30}^0=\frac{CD}{2}$

$\Rightarrow \frac{1}{2}=\frac{CD}{2}$

$CD=1$

Applying Pythagorean theorem, we get

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

$\Rightarrow A{D}^2=2^2-1^2$

$\Rightarrow AD=\sqrt{3}$

So, height of triangle $ACD$, $CD=1$

Length of the base of $△ACD,$ $AD=\sqrt{3}$

Area of triangle $ACD$,

$=\frac{1}{2}\times$ base $\times$ height

$=\frac{1}{2}\times \sqrt{3}\times 1$

$=\frac{\sqrt{3}}{2}$

Similarly,the area of right triangle $ABC$ $=\frac{\sqrt{3}}{2}$

So, the area of rectangle$=$ area of right triangle $ABC$ $+$ Area of triangle $ACD=$ $\frac{\sqrt{3}}{2}$$+\frac{\sqrt{3}}{2}=\sqrt{3}$.