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In the figure shown, the radius of the bigger circle is 9

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In the figure shown, the radius of the bigger circle is 9  [#permalink]

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New post 02 Aug 2018, 10:27
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Question Stats:

79% (01:45) correct 21% (02:16) wrong based on 28 sessions

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In the figure shown, the radius of the bigger circle is 9 and the triangle is equilateral.
What is the radius of the smaller circle?

A. 1
B. 2
C. 3
D. 4
E. 6

Source: EXperts Global

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Re: In the figure shown, the radius of the bigger circle is 9  [#permalink]

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New post 02 Aug 2018, 11:48
pushpitkc wrote:
Attachment:
The attachment 010818.JPG is no longer available


In the figure shown, the radius of the bigger circle is 9 and the triangle is equilateral.
What is the radius of the smaller circle?

A. 1
B. 2
C. 3
D. 4
E. 6

Source: EXperts Global


OA: C
Attachment:
triangle.PNG
triangle.PNG [ 75.27 KiB | Viewed 390 times ]

\(O_s\) be center of small circle, \(O_b\) be center of big circle.
\(O_sD\) = Radius of small circle=\(r\) , \(O_bB\)=Radius of big circle(R)=\(9\)cm

\(\angle DAO_s= \angle BAO_b =\frac{60^{\circ}}{2}=30^{\circ}\)

that means \(\triangle DAO_s\) and \(\triangle BAO_b\) are \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle.
their sides will be ratio of \(1:\sqrt{3}:2\)

In \(\triangle BAO_b\)
\(AO_b=2O_bB=2*9=18\)
\(AO_b =O_bO_s +O_sA\)
\(O_sA =AO_b-O_bO_s=18-(9+r) =9-r\) [Using \(O_bO_s =R+r =9+r\)]

In \(\triangle DAO_s\)
\(O_sA=2O_sD\)
\(9-r = 2r\)
\(r=\frac{9}{3}=3\)
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Re: In the figure shown, the radius of the bigger circle is 9  [#permalink]

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New post 03 Aug 2018, 02:22
pushpitkc wrote:

In the figure shown, the radius of the bigger circle is 9 and the triangle is equilateral.
What is the radius of the smaller circle?

A. 1
B. 2
C. 3
D. 4
E. 6

Source: EXperts Global


In-radius of equilateral triangle, OG=OD=\(\frac{a}{2\sqrt{3}}\) (Where 'a' is the side of equilateral triangle)
So, GD=2*OG=\(\frac{a}{\sqrt{3}}\)

AD=height of equilateral triangle=\(\sqrt{3}a/2\)

We have AG=AD-GD=\(\sqrt{3}a/2\)-\(\frac{a}{\sqrt{3}}\)=\(\frac{a}{2\sqrt{3}}\)
Or, \(AG=\frac{1}{3}AD\)
Or, AG=1/3(AG+18)
Or, AG=9

Refer the enclosed figure, AEF is an equilateral triangle with center \(O^'\), So, \(O^'G=in-radius=\frac{1}{3}*AG=\frac{1}{3}*9=3\)

Ans. (C)
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Re: In the figure shown, the radius of the bigger circle is 9 &nbs [#permalink] 03 Aug 2018, 02:22
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In the figure shown, the radius of the bigger circle is 9

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