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A circle is inscribed in a square. If the square's diagnol

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A circle is inscribed in a square. If the square's diagnol [#permalink]

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17 Feb 2007, 22:35
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A circle is inscribed in a square. If the square's diagnol is 4cm long, what is the area of the square that is not occupied by the circle (approx)?

1.7 cm^2
2.7 cm^2
12 cm^2
24 cm^2
25 cm^2

The answer mentions that the area of the circle is 2(pi). How so?
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17 Feb 2007, 22:43
ggarr wrote:
A circle is inscribed in a square. If the square's diagnol is 4cm long, what is the area of the square that is not occupied by the circle (approx)?

1.7 cm^2
2.7 cm^2
12 cm^2
24 cm^2
25 cm^2

The answer mentions that the area of the circle is 2(pi). How so?

sqrt(2)*s = 4

so s=2*sqrt(2) which means radius = sqrt(2) So area of circle is 2*PI

Here side of square is 2 times radius of circle
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17 Feb 2007, 23:06
side of square = diagonal/sqrt(2) = 2*sqrt(2)
area of square = 8
area of circle = pi * (2*sqrt(2)/2)^2 = 2pi
remaining area = 8 - 6.28 = 1.7
A
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Joined: 12 Jun 2006
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18 Feb 2007, 00:24
Quote:
so s=2*sqrt(2) which means radius = sqrt(2)
Here's my reasoning:
The radius = sqrt(2) b/c radius = 2*(sqrt)2/2 (radius is 1/2 the diameter and the diameter of the circle equals the length of a side (2*(sqrt)2) b/c the circle is inscribed within the square) ***breathes*** ... correct?
18 Feb 2007, 00:24
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