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

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2 \sqrt{3}

\sqrt{3}

4 \sqrt{3}

answer is B.

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Detailed Solution

Given figure is,
Question Image

The diagonal of a rectangle divides it into two equal right triangles.
In the given figure, we have 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

C

= \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}

.
Hence, the correct option is (2).

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