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# In rectangle ABCD shown above, sides AB and CD pass through the center

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Math Expert
Joined: 02 Sep 2009
Posts: 44599
In rectangle ABCD shown above, sides AB and CD pass through the center [#permalink]

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16 Nov 2017, 22:51
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100% (00:42) correct 0% (00:00) wrong based on 38 sessions

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In rectangle ABCD shown above, sides AB and CD pass through the centers of the two circles. If AB = 12 and AD = 16, what is the area of the shaded region?

(A) 120
(B) 156
(C) 192
(D) 192 – 36π
(E) 192 – 72π

[Reveal] Spoiler:
Attachment:

2017-11-17_0947.png [ 6.66 KiB | Viewed 786 times ]
[Reveal] Spoiler: OA

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Joined: 05 Dec 2016
Posts: 260
Concentration: Strategy, Finance
GMAT 1: 620 Q46 V29
Re: In rectangle ABCD shown above, sides AB and CD pass through the center [#permalink]

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17 Nov 2017, 00:25
Area of ABCD - Area of the circle (since rectangle includes two halfes of one circle) =
12*16-P*6^2=192-36P

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Joined: 22 May 2016
Posts: 1548
In rectangle ABCD shown above, sides AB and CD pass through the center [#permalink]

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17 Nov 2017, 17:09
Bunuel wrote:

In rectangle ABCD shown above, sides AB and CD pass through the centers of the two circles. If AB = 12 and AD = 16, what is the area of the shaded region?

(A) 120
(B) 156
(C) 192
(D) 192 – 36π
(E) 192 – 72π

[Reveal] Spoiler:
Attachment:
2017-11-17_0947.png

Area of rectangle = (12*16) = 192

Area of two semi-circles = Area of one circle
Radius = $$\frac{1}{2}(12)=6$$
Area = $$πr^2 = 36π$$

(Rectangle area) - (one circle area) = shaded area

$$192 - 36π$$

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Re: In rectangle ABCD shown above, sides AB and CD pass through the center [#permalink]

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18 Nov 2017, 01:46
The Area of the Rectangle is 12 x 16 = 192. The two circular halves. So it is 192-36Phi
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Re: In rectangle ABCD shown above, sides AB and CD pass through the center [#permalink]

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20 Nov 2017, 12:29
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KUDOS
Expert's post
Bunuel wrote:

In rectangle ABCD shown above, sides AB and CD pass through the centers of the two circles. If AB = 12 and AD = 16, what is the area of the shaded region?

(A) 120
(B) 156
(C) 192
(D) 192 – 36π
(E) 192 – 72π

[Reveal] Spoiler:
Attachment:
2017-11-17_0947.png

We see that the area of the shaded region is the area of rectangle ABCD minus twice the area of a semicircle. In other words, the area of the shaded region is the area of rectangle ABCD minus the area of one circle.

Area of shaded region = area of ABCD - area of one circle

Area of shaded region = 12 x 16 - 6^2 x π

Area of shaded region = 192 - 36π

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Re: In rectangle ABCD shown above, sides AB and CD pass through the center   [#permalink] 20 Nov 2017, 12:29
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# In rectangle ABCD shown above, sides AB and CD pass through the center

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