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Circle O is inscribed with square ABCD. Square EFGH is inscribed within circle O. What is the ratio of the area of ABCD to the area of EFGH?

A. 6:5
B. 5:4
C. 4:3
D. 3:2
E. 2:1
Bunuel, getting the ratio 8:1.
Help, please!

ShahadatHJ

A characteristic of a circle inscribed in a square is that that the squares diagonal and the circle's diameter are equal (this is square ABCD).

A characteristic of square inscribed in a circle is that the circle's diameter and the squares side are equal (this is square EFGH).

To make things easier assign values:

Let the circle have a diameter of \(2\).

As square ABCD will have sides equal to the diameter of the circle, the area of ABCD is: \(2*2 = 4\)

Now as the the EFGH's diagonal is equal to the diameter of the circle (2), and we know that the diagonal of a square makes two 45-45-90 special right-angle triangles with sides \(x, x\)&\(x\sqrt{2}\), respectively, we know then that \(x\sqrt{2}=2\) and thus \(x = \frac{2}{\sqrt{2}}\).

The area for the EFGH will then be: \(\frac{2}{\sqrt{2}} * \frac{2}{\sqrt{2}}\)

\(=\frac{4}{2}\)

\(=2\).

So the ratio of the area of ABCD to the area of EFGH is: \(4:2\) which simplified is \(2:1\)

Answer E
Thank you!
I was calculating the diagonal using the radius (instead of the diameter) for the inscribed square ... damn me. Did not notice properly!
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