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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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02 Aug 2018, 10:27
10
00:00

Difficulty:

45% (medium)

Question Stats:

66% (02:37) correct 34% (02:31) wrong based on 41 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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02 Aug 2018, 11:48
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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 [ 75.27 KiB | Viewed 840 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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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  [#permalink]

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21 Oct 2019, 21:29
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Re: In the figure shown, the radius of the bigger circle is 9   [#permalink] 21 Oct 2019, 21:29
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