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# The figure shown above consists of three identical circles

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The figure shown above consists of three identical circles [#permalink]

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26 Aug 2010, 23:25
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The figure shown above consists of three identical circles that are tangent to each other. If the area of the shaded region is $$64sqrt{3}-32\pi$$, what is the radius of each circle?

(A) 4
(B) 8
(C) 16
(D) 24
(E) 32

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-figure-shown-above-consists-of-three-identical-circles-t-168576.html
[Reveal] Spoiler: OA
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Re: Geometry problem from QR 2nd PS145 [#permalink]

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26 Aug 2010, 23:29
The triangle formed is equilateral with sides 2r. Therefore the area is $$1/2*2r*r\sqrt{3}$$ = $$r^2sqrt{3}$$. I'm not sure how to proceed after this.
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Re: Geometry problem from QR 2nd PS145 [#permalink]

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26 Aug 2010, 23:34
Area of shaded region = Area of equilateral triangle - 3 * Area of sector

Solve the above equation to find r. Thanks.
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Re: Geometry problem from QR 2nd PS145 [#permalink]

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27 Aug 2010, 06:50
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jpr200012 wrote:
The figure shown above consists of three identical circles that are tangent to each other. If the area of the shaded region is $$64sqrt{3}-32\pi$$, what is the radius of each circle?

(A) 4
(B) 8
(C) 16
(D) 24
(E) 32

Let the radius of the circle be $$r$$, then the side of equilateral triangle will be $$2r$$.

Area of the shaded region equals to area of the equilateral triangle minus area of three 60 degrees sectors.

Area of a 60 degree sector is 1/6 of the are of whole circle (as whole circle is 360 degrees and 60 is 1/6 of it), hence are of 3 such sectors will be 3/6=1/2 of the area of whole circle, so $$area_{sectors}=\frac{\pi{r^2}}{2}$$ (here if you could spot that $$\frac{\pi{r^2}}{2}$$ should correspond to $$32\pi$$ then you can write $$\frac{\pi{r^2}}{2}=32\pi$$ --> $$r=8$$);

Area of equilateral triangle equals to $$a^2\frac{\sqrt{3}}{4}$$, where $$a$$ is the length of a side. So in our case $$area_{equilateral}=(2r)^2*{\frac{\sqrt{3}}{4}}=r^2\sqrt{3}$$;

Area of the shaded region equals to $$64sqrt{3}-32\pi$$, so $$area_{equilateral}-area_{sectors}=r^2\sqrt{3}-\frac{\pi{r^2}}{2}=64sqrt{3}-32\pi$$ --> $$r^2=\frac{2(64sqrt{3}-32\pi)}{2\sqrt{3}-\pi}=\frac{64(2\sqrt{3}-\pi)}{(2\sqrt{3}-\pi)}=64$$ --> $$r=8$$.

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Re: Geometry problem from QR 2nd PS145 [#permalink]

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27 Aug 2010, 08:35
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I have a simpler solution to the above explanation your solution rocks but we wont have that much time in the GMAT exam to solve this question anyways
here it goes....
64(3)^1/2 - 32pie is the area of the shaded region hence the area of triangle will be equal to 64(3)^1/2 and area of three corresponding circle is 32pie
i will take the area of triangle you can take the area of circles segments

Therefore, 64(3)^1/2 = area of Equilateral Triangle = ((3)^1/2)/4 * Side^2

Hence (side)^2 = 64*4 = 2^8 ==> Side = 2^4 =16

Remember this is the side of the Triangle
Side of Triangle = 2(Radius of the Circle)
therefore radius of circle = 8
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Re: Geometry problem from QR 2nd PS145 [#permalink]

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27 Aug 2010, 15:33
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Always look at the numbers in the problem and at the answer choices. Do only as much math as needed.

shaded region = triangle - 3 sectors = $$64sqrt{3}-32\pi$$
so 3 sectors = $$32\pi$$
triangle is equilateral because each side = $$2r$$
so each angle = 60, and each sector = 60/360 = 1/6 each circle
so 3 sectors = 3*(1/6) = 1/2 each circle
so $$32\pi$$ = $$\frac{\pi{r^2}}{2}$$
$$r=8$$

The correct answer is B. No need to worry about the area of the triangle.
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Quant Rev. #145. Difficult triangles/circles problem [#permalink]

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14 Jan 2011, 11:58
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What is the quickest way to solve these types of problems. Please advise.

The figure shown (please see attached) consists of three identical circles that are tangent to each other. If the area of the shaded region (center space where the three circles do not touch) is $$64\sqrt{3} - 32\pi$$, what is the radius of each circle?

a. 4
b. 8
c. 16
d. 24
e. 32
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Re: Quant Rev. #145. Difficult triangles/circles problem [#permalink]

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14 Jan 2011, 15:13
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Merging similar topics.
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Re: The figure shown above consists of three identical circles [#permalink]

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06 Mar 2014, 02:36
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Bumping for review and further discussion.

GEOMETRY: Shaded Region Problems!
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Re: Geometry problem from QR 2nd PS145 [#permalink]

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25 May 2014, 14:03
Bunuel wrote:
jpr200012 wrote:
The figure shown above consists of three identical circles that are tangent to each other. If the area of the shaded region is $$64sqrt{3}-32\pi$$, what is the radius of each circle?

(A) 4
(B) 8
(C) 16
(D) 24
(E) 32

Let the radius of the circle be $$r$$, then the side of equilateral triangle will be $$2r$$.

Area of the shaded region equals to area of the equilateral triangle minus area of three 60 degrees sectors.

Area of a 60 degree sector is 1/6 of the are of whole circle (as whole circle is 360 degrees and 60 is 1/6 of it), hence are of 3 such sectors will be 3/6=1/2 of the area of whole circle, so $$area_{sectors}=\frac{\pi{r^2}}{2}$$ (here if you could spot that $$\frac{\pi{r^2}}{2}$$ should correspond to $$32\pi$$ then you can write $$\frac{\pi{r^2}}{2}=32\pi$$ --> $$r=8$$);

Area of equilateral triangle equals to $$a^2\frac{\sqrt{3}}{4}$$, where $$a$$ is the length of a side. So in our case $$area_{equilateral}=(2r)^2*{\frac{\sqrt{3}}{4}}=r^2\sqrt{3}$$;

Area of the shaded region equals to $$64sqrt{3}-32\pi$$, so $$area_{equilateral}-area_{sectors}=r^2\sqrt{3}-\frac{\pi{r^2}}{2}=64sqrt{3}-32\pi$$ --> $$r^2=\frac{2(64sqrt{3}-32\pi)}{2\sqrt{3}-\pi}=\frac{64(2\sqrt{3}-\pi)}{(2\sqrt{3}-\pi)}=64$$ --> $$r=8$$.

Hi Bunuel,

How can you tell that the triangle is an equilateral triangle?

Thanks!
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Re: Geometry problem from QR 2nd PS145 [#permalink]

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25 May 2014, 14:06
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russ9 wrote:
Bunuel wrote:
jpr200012 wrote:
The figure shown above consists of three identical circles that are tangent to each other. If the area of the shaded region is $$64sqrt{3}-32\pi$$, what is the radius of each circle?

(A) 4
(B) 8
(C) 16
(D) 24
(E) 32

Let the radius of the circle be $$r$$, then the side of equilateral triangle will be $$2r$$.

Area of the shaded region equals to area of the equilateral triangle minus area of three 60 degrees sectors.

Area of a 60 degree sector is 1/6 of the are of whole circle (as whole circle is 360 degrees and 60 is 1/6 of it), hence are of 3 such sectors will be 3/6=1/2 of the area of whole circle, so $$area_{sectors}=\frac{\pi{r^2}}{2}$$ (here if you could spot that $$\frac{\pi{r^2}}{2}$$ should correspond to $$32\pi$$ then you can write $$\frac{\pi{r^2}}{2}=32\pi$$ --> $$r=8$$);

Area of equilateral triangle equals to $$a^2\frac{\sqrt{3}}{4}$$, where $$a$$ is the length of a side. So in our case $$area_{equilateral}=(2r)^2*{\frac{\sqrt{3}}{4}}=r^2\sqrt{3}$$;

Area of the shaded region equals to $$64sqrt{3}-32\pi$$, so $$area_{equilateral}-area_{sectors}=r^2\sqrt{3}-\frac{\pi{r^2}}{2}=64sqrt{3}-32\pi$$ --> $$r^2=\frac{2(64sqrt{3}-32\pi)}{2\sqrt{3}-\pi}=\frac{64(2\sqrt{3}-\pi)}{(2\sqrt{3}-\pi)}=64$$ --> $$r=8$$.

Hi Bunuel,

How can you tell that the triangle is an equilateral triangle?

Thanks!

The triangle is equilateral because each of its sides is equal to two radii of identical circles.

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-figure-shown-above-consists-of-three-identical-circles-t-168576.html
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Re: Geometry problem from QR 2nd PS145   [#permalink] 25 May 2014, 14:06
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