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

Solution

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.

Video Solution

video

Create a FREE account and get:

  • All Quant CAT complete Formulas and shortcuts PDF
  • 35+ CAT previous year papers with video solutions PDF
  • 5000+ Topic-wise Previous year CAT Solved Questions for Free

cracku

Boost your Prep!

Download App