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# Circle O is inscribed within square ABCD. Square EFGH is

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Manager
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Circle O is inscribed within square ABCD. Square EFGH is [#permalink]

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09 May 2006, 20:09
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Circle O is inscribed within square ABCD. Square EFGH is inscribed within circle O. What is ratio of the area ABCD to the area of EFGH?
Again i am dumb in math so please explain me the answer.I will appreciate your help guys. Thanks-------------
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09 May 2006, 20:26
Can anyone explain to me pls if you have time.
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09 May 2006, 20:30
MA466 wrote:
Circle O is inscribed within square ABCD. Square EFGH is inscribed within circle O. What is ratio of the area ABCD to the area of EFGH?

Again i am dumb in math so please explain me the answer.I will appreciate your help guys. Thanks-------------

length of a side of square ABCD = x (suppose 10) = hypoteneous of square EFGH
side of the square EFGH = s
hypoteneous of square EFGH = sqrt (2s^2)
x^2 = sqrt (2s^2)
10^2 = sqrt (2s^2)
2s^2 = 100
s^2 = 50 = 5 sqrt(2)

area of square ABCD = 10x10= 100
area of square EFGH = (5sqrt(2))^2 = 50

area of square ABCD: area of square EFGH = 100:50 or 2:1.
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09 May 2006, 22:12
Assume sides of square ABCD = x
Then area of ABCD = x^2

Assume sides of sqaure EFGH = y
Then area of ABCD = y^2

We also know the diameter of the circle = hypotenuse of the square EFGH

so y^2 + y^2 = x^2
2y^2 = x^2

So ABCD:EDFG = x^2/y^2 = 2y^2/y^2 = 2/1
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10 May 2006, 07:39
Good Explanation guys. OA is 2:1.Thanks.
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10 May 2006, 10:40
If the diagonal are perpandicular then the area = (diagonal1Xdiagonal2)/2.

Since the side of the cube ABCD is also the diagonal of cube EFGH. we can find the ratio.

let side L=6 of ABCD, then Area of ABCD= 36.
diagonal of EFGH=6, therefore Area of EFGH = 18.
2:1

Please correct me if i am wrong.
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11 May 2006, 00:18
Hi,

The diameter of a circle serves both as a side of ABCD, and as a diagonal of EFGH.
Suppose, the side of abcd = a, the diameter=d, the side of efgh=b.
We have to find a^2/^b2.
a=d,
sqrt2*b=d, b = d/sqrt2
11 May 2006, 00:18
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