MAT Exam 2022 Paper

Concept:The least-sized square inscribed in a square has its vertices at the midpoints of the sides of the outer square.Explanation:Let the outer square $ABCD$ have side $20$ m.Place it as $A(0,0)$, $B(20,0)$, $C(20,20)$, $D(0,20)$.Let the inner square have vertices $P(x,0)$ on $AB$, $Q(20,x)$ on $BC$, $R(20-x,20)$ on $CD$, and $S(0,20-x)$ on $DA$.For every $x$, $PQRS$ is a square.Now, using the distance formula,$$PQ^2 = (20-x)^2 + x^2$$$$PQ^2 = 400 - 40x + 2x^2$$Rewrite as:$$PQ^2 = 2(x-10)^2 + 200$$This is minimum when $x = 10$.At $x=10$, the vertices become the midpoints of the sides of $ABCD$.Then,$$PQ^2 = 10^2 + 10^2 = 200$$Thus, side of the inscribed square is $10\sqrt{2}$ m.Required area $= PQ^2 = 200\ \text{m}^2$.Answer:C. $200\ \mathrm{m}^{2}$