Let a rectangle of length l and breadth b be inscribed in the circle of radius a.

Then, the diagonal passes through the centre and is of length 2a cm.

Now, by Pythagoras theorem, we get,

(2a)² = l² + b²

⇒ b² = 4a² – l²

⇒ b =

Therefore, Area of rectangle, A =

.

d,

.

.

w, gives 4a² = 2l²

⇒ l = a

⇒ b =

w, when l =

Then, = -4 < 0

Then, by second derivative test, when l =, then the area of the rectangle is the maximum.

Since, l = b =,

Therefore, the rectangle is square.

Hence proved.