Question

A square $S_1$ encloses another square $S_2$ in such a manner that each corner of $S_2$ is at the midpoint of the side of $S_1$ if $A_1$ is the area of $S_1$ and $A_2$ is the area of $S_2$ then

• A. $A_1=A_2$
• B. $A_2=2A_1$
• C. $A_1=2A_2$
• D. $A_1=4A_2$

Answer 1

The correct option is C $A_1=2A_2$

Let each side of the larger square $S_1$ be $a$ units.

$\therefore A_1=a^2$ ....(i)

In $△LMD$,

By Pythagoras theorem,

$L{M}^2=L{D}^2+D{M}^2$

$={\left(\frac{a}{2}\right)}^2+{\left(\frac{a}{2}\right)}^2$

$=\frac{2a^2}{4}=\frac{a^2}{2}$

$\Rightarrow LM=\frac{a}{\sqrt{2}}$

$\therefore A_2={\left(\frac{a}{\sqrt{2}}\right)}^2=\frac{a^2}{2}=\frac{A_1}{2}$ ...from (i)

$\Rightarrow A_1=2A_2$