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18 Aug 2006, 13:24
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If ABDC and GHJI are squares and EF is a diameter of the circle, what is the ratio of the area of the shaded region to that of the circle?
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19 Aug 2006, 12:01
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Is the answer 7/(11*sqrt10) ?
I got it by the following method:
GJ is the diameter of the circle. It is also the diagonal of the square GHJI.
Now, let the side of this square be 'a' units long.
Then a^2 + a^2 = (2r)^2
=> a = sqrt(2)*r
Since ABCD is a square, mid point of CD should be the same as centre of circle (remember that CD is a part of the circle's diameter which definitely passes through the centre).
Let the centre of the circle be O.
Also assume OD = x. Then BD = 2x. Also OB = r
Then x^2 + (2x)^2 = r^2
=> 2x = 2r/sqrt(5)
Length of shaded area = CD = 2x
Breadth of shaded area = 0.5 * IG = r/sqrt(2)
Hence area of shaded region = [2r/sqrt(5)] * r/sqrt(2)
=> sqrt(2/5) * r^2
Area of circle = pi * r^2
Hence ratio = [sqrt(2/5) * r^2] : [pi * r^2]
=> 7/(11*sqrt10)
What is the OA pls ?
I got it by the following method:
GJ is the diameter of the circle. It is also the diagonal of the square GHJI.
Now, let the side of this square be 'a' units long.
Then a^2 + a^2 = (2r)^2
=> a = sqrt(2)*r
Since ABCD is a square, mid point of CD should be the same as centre of circle (remember that CD is a part of the circle's diameter which definitely passes through the centre).
Let the centre of the circle be O.
Also assume OD = x. Then BD = 2x. Also OB = r
Then x^2 + (2x)^2 = r^2
=> 2x = 2r/sqrt(5)
Length of shaded area = CD = 2x
Breadth of shaded area = 0.5 * IG = r/sqrt(2)
Hence area of shaded region = [2r/sqrt(5)] * r/sqrt(2)
=> sqrt(2/5) * r^2
Area of circle = pi * r^2
Hence ratio = [sqrt(2/5) * r^2] : [pi * r^2]
=> 7/(11*sqrt10)
What is the OA pls ?
