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23 Dec 2020, 06:00
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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:
55%
(hard)
Question Stats:
67% (02:45) correct
33%
(02:40)
wrong
based on 48
sessions
History
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Time
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Not Attempted Yet
In the diagram below, O1 and O2 are concentric circles, and AB is tangent to O1 at C. If the radius of O1 is r and the radius of O2 is twice as long, what is the area of the shaded region?
(A) \(\frac{πr^2}{2}\)
(B) \(πr^2\)
(C) \(1.5πr^2\)
(D) \(2πr^2\)
(E) \(3πr^2\)
Attachment:
2020-12-21_16-35-40.png [ 31.98 KiB | Viewed 2315 times ]
23 Dec 2020, 07:04
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Join OC, Using Pythagoras Theorem,
\(OC^2 + AC^2=AO^2\)
Given \(AO =2r; OC=r\)
\(AC^2=4r^2-r^2=3r^2\)
\(AC=\sqrt{3}r\)
\(\angle OAC=\frac{OC}{AC}=\frac{1}{\sqrt{3}}=tan 30^{\circ}\); As, \(tan 30^{\circ}=\frac{1}{\sqrt{3}}\)
\(\angle AOC = 60^{\circ}\)
\(\angle AOB = 120^{\circ}\)
Area of the shaded region \(= \frac{(360-120)}{360}* π ((2r)^2-r^2) = \frac{240}{360}*π*3r^2=2πr^2\)
Ans D
\(OC^2 + AC^2=AO^2\)
Given \(AO =2r; OC=r\)
\(AC^2=4r^2-r^2=3r^2\)
\(AC=\sqrt{3}r\)
\(\angle OAC=\frac{OC}{AC}=\frac{1}{\sqrt{3}}=tan 30^{\circ}\); As, \(tan 30^{\circ}=\frac{1}{\sqrt{3}}\)
\(\angle AOC = 60^{\circ}\)
\(\angle AOB = 120^{\circ}\)
Area of the shaded region \(= \frac{(360-120)}{360}* π ((2r)^2-r^2) = \frac{240}{360}*π*3r^2=2πr^2\)
Ans D
25 Dec 2020, 09:23
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Hi Meanup,
How come subtracting 360 - 120 /360 of (Area of bigger circle - Area of smaller circle) will give area of shaded region ? i didn't got this thing ?
How come subtracting 360 - 120 /360 of (Area of bigger circle - Area of smaller circle) will give area of shaded region ? i didn't got this thing ?
