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

Question

In the figure, $A BC D$ is a rectangle. Calculate the area of $A BC D$ .

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In the figure,

A BC D

is a rectangle. Calculate the area of

A BC D

8 mins ago

Discuss this question LIVE

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Text solutionVerified

The diagonal of a rectangle divides it into two equal right triangles,Triangle

A BC

and Triangle

A D C

In triangle

A D C

sin 3 0^{0} = \frac{C D}{2}

\Rightarrow \frac{1}{2} = \frac{C D}{2}

C D = 1

Applying Pythagorean theorem, we get

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

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

\Rightarrow A D = 3

So, height of triangle

A C D

C D = 1

Length of the base of

△ A C D,

A D = 3

Area of triangle

A C D

= \frac{1}{2} \times

base

\times

height

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

= \frac{3}{2}

Similarly,the area of right triangle

A BC

= \frac{3}{2}

So, the area of rectangle

=

area of right triangle

A BC

+

Area of triangle

A C D =

\frac{3}{2}

+ \frac{3}{2} = 3

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Question Text	In the figure, $A BC D$ is a rectangle. Calculate the area of $A BC D$ .
Answer Type	Text solution:1
Upvotes	150