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Area of square = Area of Circle

=> (Diagnonal ^2 / 2) = Pi * Diameter^2/4
=> Diagonal/Diameter= SQRT (Pi/4)
=> Diagonal/Diameter = SQRT (1.57)
=> Diagonal/Diameter= ~1.26
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what am I doing wrong ?

area of sq = d^2/2
area of circle = pi r^2

equating we get d/r = under root(pi/2) , which is the answer

but we need d/2r so i am getting an additional 1/2 ???
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If the area of a square equals the area of a circle, which of the following is closest to the ratio of the diagonal of the square to the diameter of the circle?

A. 0.95
B. 1.26
C. 1.40
D. 1.57
E. 2.51

M19-21

If \(x\) is the side of the square, then the area of the square is \(x^2\) and the diagonal of the square is \(x\sqrt{2}\). If \(d\) is the diameter of the circle then the area of the circle is \(\pi(\frac{d}{2})^2\). Because \(x^2 = \pi(\frac{d}{2})^2\), \(d = \frac{2x}{\sqrt{\pi}}\). The required ratio \(=\frac{x\sqrt{2}}{\frac{2x}{\sqrt{\pi}}} = \frac{\sqrt{\pi}}{\sqrt{2}} = \sqrt{\frac{\pi}{2}}\) or approximately \(\sqrt{1.57}\). This is slightly smaller than \(\sqrt{1.69} = 1.3\). The best answer is therefore B.

Answer: B
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Bunuel - how did d become 2x/√pi? Please could you elaborate?
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Bunuel
bigfernhead
If the area of a square equals the area of a circle, which of the following is closest to the ratio of the diagonal of the square to the diameter of the circle?

A. 0.95
B. 1.26
C. 1.40
D. 1.57
E. 2.51

M19-21

If \(x\) is the side of the square, then the area of the square is \(x^2\) and the diagonal of the square is \(x\sqrt{2}\). If \(d\) is the diameter of the circle then the area of the circle is \(\pi(\frac{d}{2})^2\). Because \(x^2 = \pi(\frac{d}{2})^2\), \(d = \frac{2x}{\sqrt{\pi}}\). The required ratio \(=\frac{x\sqrt{2}}{\frac{2x}{\sqrt{\pi}}} = \frac{\sqrt{\pi}}{\sqrt{2}} = \sqrt{\frac{\pi}{2}}\) or approximately \(\sqrt{1.57}\). This is slightly smaller than \(\sqrt{1.69} = 1.3\). The best answer is therefore B.

Answer: B

\(x^2 = \pi(\frac{d}{2})^2\);

Take the square root to get \(x= \sqrt{\pi}*\frac{d}{2}\);

\(2x=\sqrt{\pi}*d\);

\(d = \frac{2x}{\sqrt{\pi}}\).
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