Question 62

A rectangle with the largest possible area is drawn inside a semicircle of radius 2 cm. Then, the ratio of the lengths of the largest to the smallest side of this rectangle is


Let us assume the length of the rectangle is 'l' and breadth of the rectangle is 'b'.

The radius, l/2 and b in the above diagram form a right-angled triangle.

=> $$\left(\frac{l}{2}\right)^2+b^2=2^2$$

We know that the area of the rectangle is l*b, which can be obtained by considering 2 times the geometric mean of $$\left(\frac{l}{2}\right)^2$$ and $$b^2$$.

Therefore, for the maximum area, the equality condition of AM-GM inequality should be satisfied

=> $$\left(\frac{l}{2}\right)^2=b^2$$ => l = 2b.
=> l/b = 2/1.

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