UPSC 2014 CDS I Math Paper

Let ABCD be the inscribed square whose side is x units.Length of diagonal of square
=AC=2a=√2×‌side‌
⇒x=

2a

√2

=√2a
Therefore, area of inner square
⇒x2=(√2a)2=2a2
Side of the outer square
=PQ=a+a=2a
Therefore, area of outer square
=(2a)2=4a2
Difference between the areas of the outer square and inner square
=4a2−2a2=2a2