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

Question

010818.JPG 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

Princ · Accepted Answer

OA: C triangle.PNG 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 riangle DAO_s and riangle BAO_b are 30^{\circ}-60^{\circ}-90^{\circ} triangle. their sides will be ratio of 1:\sqrt{3}:2 In riangle 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 O_bO_s =R+r =9+r ] In riangle DAO_s O_sA=2O_sD 9-r = 2r r=\frac{9}{3}=3

PKN · Answer

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)

Fdambro294 · Answer

Rule 1: whenever you have 2 similar triangles, each corresponding dimension or length of the two similar triangles will always be in a set ratio

(1st) draw a line parallel to the bottom of the equilateral triangle and tangent to both the large circle and small circle, such that this tangent line intersects right at the point of tangency of the 2 circles. Call the points it hits on the triangle A and B.

Call the top vertex of the entire equilateral triangle C. Thus, with this tangent line, we’ve created triangle ABC on top of the larger circle with radius 9, and inside the entire larger equilateral triangle.

Since this tangent line is parallel to the bottom base of the entire triangle, we can use the Rule for corresponding angles when 2 parallel lines are cut by a transversal and say that triangle ABC is similar to the entire overall triangle. To find the proportional ratio, we can use the Height Dimension of each triangle:

(2nd) the radius of a circle inscribed inside an equilateral triangle (in the case of the larger circle, this inradius = 9) can be found by taken (1/3)rd or the Height of the Equilateral Triangle (H)

H = (s) * sqrt(3) * (1/2)

—-where s = the side of the equilateral triangle

Thus, the in-radius of 9 is equal to:

9 = (1/3) (s * sqrt(3) * 1/2)

Re-arranging to find just the Height = H = (s) * sqrt(3) * 1/2 ———> 27

Thus, the height of the entire equilateral triangle = H = 27

The height of the smaller triangle ABC we created will be the distance left over after we remove the Diameter of the larger inscribed circle ———> Diameter = 18

h = height of smaller triangle ABC = 27 - 18 = 9

This means that any dimension or length of the two similar equilateral triangles (ABC and the entire equilateral triangle) will be in the set ratio of:

27/9 = 3/1

this proportionality ratio extends to measures such as the Inradius, which is the radius of the circle inscribed inside the equilateral triangle. The inradius of the larger circle (9) will be proportional to equilateral triangle ABC’s inradius (r) in the ratio of: 3/1

3/1 = 9/r

r = 3

C

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GBeniwal · Answer

This can be quickly solved (well within 2 minutes) by following property. The thing about GMAT is that it always (almost) has a simpler solution.

In an equilateral triangle, the radius of the Incircle is half the radius of Circumcircle.

Radius of incircle is given = 9

So, Radius of Circumcircle = 18

Hence, we can find the length from top vertex of the triangle to point where both the triangle meets.

Now, use the property again and you will have the answer as 3.

TBT · Answer

https://youtu.be/iX7SEp1v9OI 
Experts Global official explanation

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

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