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Difficulty:
25%
(medium)
Question Stats:
74% (01:31) correct
26%
(01:41)
wrong
based on 176
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Attachment:
2014-06-24_1857.png [ 9.07 KiB | Viewed 4115 times ]
(1) PC = 3
(2) x = 60
24 Jun 2014, 08:05
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In the figure, O is the center of the circle of radius 3, and ABCD is a square. If angle AOP = x and the side BC of the square is tangent to the circle, then what is the area of the square ABCD ?
You can solve this question without actually computing anything there.
(1) PC = 3 --> the length of OC, which is the hypotenuse in triangle BCO, equals to OP + PC = radius + PC = 3 + 3 = 6. Thus we know the lengths of the two sides (OB and OC) in that triangles which enables us to find the third one, BC. Since BC is the side of the square then its area is BC^2. Sufficient.
(2) x = 60. Triangle BOC is 30-60-90 right triangle. We know one of the sides there (BO), hence we can find BC. Since BC is the side of the square then its area is BC^2. Sufficient.
Answer: D.
24 Jun 2014, 08:09
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smartyman
Attachment:
2014-06-24_1857.png
In the figure, O is the center of the circle of radius 3, and ABCD is a square. If angle AOP = x and the side BC of the square is tangent to the circle, then what is the area of the square ABCD ?(1) PC = 3
(2) x = 60
pc=3 then by pythagores theorem you can calculate the one side of square and there by area.
x=60 by tan formula= tan(angle)= p/b you can calculate one side of square and there by area.
so both independently give you answer.
