Question

What will be the ratio of the areas of squares, when the diagonal of one square is double the other?

• A. $1:3$
• B. $1:2$
• C. $1:5$
• D. $1:4$

Answer 1

The correct option is D

$1:4$

Let the length of one diagonal be $x$ units.
The length of the other diagonal is $2x$ units

Applying Pythagoras theorem in $△BCD$:
$B{C}^2+C{D}^2=x^2$

$\Rightarrow B{C}^2+B{C}^2=x^2$ [$\because BC=CD$]

$\Rightarrow 2B{C}^2=x^2$

$\Rightarrow \text{Area of square}ABCD=B{C}^2=\frac{x^2}{2}$ sq. units ... (i)

Applying Pythagoras Theorem in $△FGH$:
$F{G}^2+G{H}^2=4x^2$

$\Rightarrow F{G}^2+F{G}^2=4x^2$ [$\because FG=GH$]

$\Rightarrow 2F{G}^2=4x^2$

$\Rightarrow \text{Area of square}EFGH=F{G}^2=2x^2$ sq. units ... (ii)

Now, ratio of the areas of the squares $=\frac{\left(i\right)}{\left(ii\right)}=\frac{\frac{x^2}{2}}{2x^2}=\frac{1}{4}$
$=1:4$