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29 Nov 2017, 23:07
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
79% (02:15) correct
21%
(02:20)
wrong
based on 102
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PQ, QR and PR are diameters of the three circles shown above. If QR = 4, and PQ = 2QR, what is the area of the shaded region?
(A) 6π
(B) 8π
(C) 10π
(D) 12π
(E) 16π
Attachment:
2017-11-30_1001_002.png [ 8.33 KiB | Viewed 4976 times ]
29 Nov 2017, 23:23
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D.
From question we can calculate PR=12 PQ=8.
Now shaded area= 0.5*(area of circle of dia PR-area of circle of dia PQ)+0.5*(area of circle of dia QR)
On solving this gives pi/8*(12^2-8^2+4^2)
= pi/8*(96)
=12pi
Sent from my iPhone using GMAT Club Forum
From question we can calculate PR=12 PQ=8.
Now shaded area= 0.5*(area of circle of dia PR-area of circle of dia PQ)+0.5*(area of circle of dia QR)
On solving this gives pi/8*(12^2-8^2+4^2)
= pi/8*(96)
=12pi
Sent from my iPhone using GMAT Club Forum
30 Nov 2017, 08:20
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Bunuel
Attachment:
eeeeee.png [ 13.05 KiB | Viewed 4302 times ]
To find shaded area: Use just the bottom half + \(\frac{1}{2}\) Circle QR
Circle QR diameter = 4, r = 2
Circle QR area = \(4\pi\)
\(\frac{1}{2} * 4\pi = 2\pi\) = \(\frac{1}{2}\) Area of Circle QR
Circle PQ diameter: (2*4) = 8, r = 4
Circle PQ area = \(16\pi\)
\(\frac{1}{2} * 16\pi = 8\pi\) = \(\frac{1}{2}\) Area of Circle PQ
Circle PR diameter: (4 + 8 = 12), r = 6
Circle PR area = \(36\pi\)
\(\frac{1}{2} * 36\pi = 18\pi\) = \(\frac{1}{2}\) Area of Circle PR
Shaded area = Use just the bottom half (\(\frac{1}{2}\) PR + \(\frac{1}{2}\) QR) - \(\frac{1}{2}\)PQ
\(\frac{1}{2}\) QR is the little shaded area in the top half
Shaded area = Areas of
(\(\frac{1}{2}\) of shaded large Circle PR + \(\frac{1}{2}\) shaded little Circle QR)* - \(\frac{1}{2}\) unshaded medium Circle PQ
\((18\pi + 2\pi) - 8\pi\)
\(20\pi - 8\pi = 12\pi\)
Answer D
No need to account for the \(\frac{1}{2}\) shaded little Circle QR in the bottom half; it is included in \(\frac{1}{2}\) big Circle QR in the bottom half
