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by kanwar86 » Sat Nov 19, 2011 11:10 pm
Hint - 1) Side of square ABCD is diameter of circle O.
2) Diameter of circle O is diagonal of square EFGH.

Try to come up with an answer, which is (2:1)
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by shankar.ashwin » Sat Nov 19, 2011 11:21 pm
Consider side of square ABCD = 4 (Area = 16)

Now side of square = Diameter of circle O

Diameter of circle O = Diagonal of square EFGH.

2*Side^2 (EFGH) = 16 (Phy Theorem)

Area(EFGH) = Side^2 = 16/2 = 8.

Now Area ABCD/ Area EFGH = 16/8 = 2:1
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by Anurag@Gurome » Sun Nov 20, 2011 8:06 pm
MBA.Aspirant wrote:how to approach this one?

Solution:
Let the side of the square ABCD be 2s.
Hence, the diagonal of the square EFGH is 2s.
This means that the side of the square is 2s/(sqrt2) = (sqrt2)s.
This means that the ratio of the sides of the square is 2s:(sqrt2)s = sqrt2:1.
Hence, the ratio of the areas of the square is (sqrt2)^2:1^2 = 2:1
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