Q. 194.0( 72 Votes )
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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.
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