Answer :

The figure is given below:


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 =


.





.


.





Gives 4a2 = 2l2


l = a


b =


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.

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