# Show that of all

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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