What will be the ratio of the areas of squares, when the diagonal of one squar

Question

What will be the ratio of the areas of squares, when the diagonal of one square is twice of the other.

Options:

A . 1:3

B . 3:1

C . 5:1

D . 4:1

Answer: Option D
:
D
Let the length of one diagonal be

x

units.
The length of the other diagonal is

2 x

units
Applying Pythagoras theorem,

2 \times {(s i d e o f s q u a r e)}^{2} = {(l e n g t h o f d i a g o n a l)}^{2}

\Rightarrow {(s i d e o f s q u a r e)}^{2}

\frac{1}{2} \times {(l e n g t h o f d i a g o n a l)}^{2}

Sides of the squares will be

\sqrt{\frac{{4 x}^{2}}{2}} = \frac{2 x}{\sqrt{2}}

and

\sqrt{\frac{x^{2}}{2}} = \frac{x}{\sqrt{2}}

Area of square =

{(s i d e o f s q u a r e)}^{2}

Hence the ratio of the area of squares is

(\frac{2 x}{\sqrt{2}})^{2}

and

(\frac{x}{\sqrt{2}})^{2}

\frac{1}{2} \times (2 x)^{2} : \frac{1}{2} \times x^{2}

= 4: 1

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