Question

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

• A. 1:3
• B. 3:1
• C. 5:1
• D. 4:1

Answer 1

The correct option is D

4:1

Let the length of one diagonal be $x$ units.
The length of the other diagonal is $2x$ units
Applying Pythagoras theorem,
$2\times {\left(sideofsquare\right)}^2={\left(lengthofdiagonal\right)}^2$
$\Rightarrow {\left(sideofsquare\right)}^2$
= $\frac{1}{2}\times {\left(lengthofdiagonal\right)}^2$
Sides of the squares will be $\sqrt{\frac{{4x}^2}{2}}=\frac{2x}{\sqrt{2}}$
and $\sqrt{\frac{x^2}{2}}=\frac{x}{\sqrt{2}}$
Area of square = ${\left(sideofsquare\right)}^2$

Hence the ratio of the area of squares is $(\frac{2x}{\sqrt{2}}{)}^2$ and $(\frac{x}{\sqrt{2}}{)}^2$

= $\frac{1}{2}\times (2x{)}^2:\frac{1}{2}\times x^2$
= 4: 1