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# In the figure above, AB is a diameter of the circle with center O, and

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Joined: 02 Sep 2009
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In the figure above, AB is a diameter of the circle with center O, and [#permalink]

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14 Nov 2017, 01:01
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In the figure above, AB is a diameter of the circle with center O, and ABCD is a square. What is the area of the shaded region in terms of r?

(A) π(r^2 – 4)
(B) π(4 – π)
(C) r^2(π – 2)
(D) r^2(4π – 2)
(E) r^2(4 – π/2)

[Reveal] Spoiler:
Attachment:

2017-11-14_1154.png [ 12.62 KiB | Viewed 730 times ]
[Reveal] Spoiler: OA

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Re: In the figure above, AB is a diameter of the circle with center O, and [#permalink]

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14 Nov 2017, 02:29
Bunuel wrote:

In the figure above, AB is a diameter of the circle with center O, and ABCD is a square. What is the area of the shaded region in terms of r?

(A) π(r^2 – 4)
(B) π(4 – π)
(C) r^2(π – 2)
(D) r^2(4π – 2)
(E) r^2(4 – π/2)

[Reveal] Spoiler:
Attachment:
2017-11-14_1154.png

Area of shaded region = Area of Square - Area of semicircle with radius equal to half of side of square

If radius of circle = r
then, Side of square = 2r

Area of the square = (2r)^2 = 4r^2

Area of Semicircle = (1/2)*π*r^2

So area of the shaded region = 4r^2 - (1/2)*π*r^2 = r^2 (4-π/2)

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In the figure above, AB is a diameter of the circle with center O, and [#permalink]

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14 Nov 2017, 02:35

In the square ABCD, the side of the square is equal to the diameter of the circle.

We have been asked to the value of shaded region in terms of r.

The area of the square in terms of the diameter is $$(2r)^2 = 4r^2$$ since the diameter is twice the radius
Similarly, the area of the semicircle is $$\frac{1}{2}*π*r^2$$

The area of the shaded region is the difference between the area of the square and the area of semicircle,
which is $$4r^2 - \frac{1}{2}*π*r^2 = r^2(4 - π/2)$$ (Option E)
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In the figure above, AB is a diameter of the circle with center O, and   [#permalink] 14 Nov 2017, 02:35
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