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In the figure above, O is the center of the circle, and B is a point

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In the figure above, O is the center of the circle, and B is a point  [#permalink]

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14 Nov 2017, 01:17
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Difficulty:

15% (low)

Question Stats:

94% (01:24) correct 6% (01:22) wrong based on 43 sessions

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In the figure above, O is the center of the circle, and B is a point on the circle. In rectangle OABC, if OA = 4 and OC = 5, what is the area of the circle?

(A) 9π
(B) 16π
(C) 25π
(D) 41π
(E) 64π

Attachment:

2017-11-14_1157.png [ 6.09 KiB | Viewed 759 times ]

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In the figure above, O is the center of the circle, and B is a point  [#permalink]

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14 Nov 2017, 01:57
D 41.pi

OA=4 and OC=5 are the sides of a right triangle formed by the diagonal AC. We know that OA=OB=radius. Hence OB by pythogoras theorem is sqrt(41). Hence area=41.pi

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In the figure above, O is the center of the circle, and B is a point  [#permalink]

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14 Nov 2017, 09:46
Bunuel wrote:

In the figure above, O is the center of the circle, and B is a point on the circle. In rectangle OABC, if OA = 4 and OC = 5, what is the area of the circle?

(A) 9π
(B) 16π
(C) 25π
(D) 41π
(E) 64π

Attachment:
2017-11-14_1157.png

Rectangle OABC's diagonal = hypotenuse of right triangle ABO = radius of the circle.

If OC = 5, AB = 5 = one leg
OA = 4 = other leg
The sum of the squares of the legs of a right triangle equal the hypotenuse squared, where the hypotenuse = the radius.

$$4^2 + 5^2 = r^2$$
$$16 + 25 = r^2$$
$$r = \sqrt{41}$$

Circle area =
$$\pi*r^2 = \pi*(\sqrt{41})^2 = 41\pi$$

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In the figure above, O is the center of the circle, and B is a point &nbs [#permalink] 14 Nov 2017, 09:46
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