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02 Feb 2016, 01:59
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C
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Difficulty:
25%
(medium)
Question Stats:
80% (02:31) correct
20%
(02:31)
wrong
based on 75
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Square ABCD is inscribed in circle O, which is inscribed in larger square EFGH, as shown in the diagram above. If the radius of circle O is 1, then the shaded area is approximately what percent of the area of square EFGH?
A. 3%
B. 4%
C. 25%
D. 50%
E. 75%
Attachment:
2016-01-31_1806.png [ 5.79 KiB | Viewed 3940 times ]
02 Feb 2016, 09:21
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Shaded area = Area of the circle - Area of the square ABCD.
If r=1, Circle Area = \(\pi\) \(r^{2}\) = \(\pi\)
Square ABCD diagonal is equal 2 therefore it's sides are equal \(\sqrt[]{2}\), Square ABCD Area = 2
Square EFGH side = circle diameter = 2. Square EFGH Area = 4
% of shaded region = \(\frac{Shaded Area}{Square EFGH Area}\) = \(\frac{3,14 - 2}{4}\) = approx. \(\frac{1}{4}\) = 25%
If r=1, Circle Area = \(\pi\) \(r^{2}\) = \(\pi\)
Square ABCD diagonal is equal 2 therefore it's sides are equal \(\sqrt[]{2}\), Square ABCD Area = 2
Square EFGH side = circle diameter = 2. Square EFGH Area = 4
% of shaded region = \(\frac{Shaded Area}{Square EFGH Area}\) = \(\frac{3,14 - 2}{4}\) = approx. \(\frac{1}{4}\) = 25%
02 Feb 2016, 13:57
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Bunuel
Area of EFGH:
1)Let L lay on the half of HG. OL=OC=1 (each is radius of circle O).
2) FG=2OL=2, S EFGH= FG*FG=4
Area of shaded:
1)\(AB=\sqrt{AC^2/2}\)
\(AB=\sqrt{2}\), S ABCD=2
2) S shaded=\(πR^2-2=1,14\)
1,14 / 4=0,28
C.
