Difficult triangles/circles problem

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tonebeeze: Master | Next Rank: 500 Posts; Posts: 158; Joined: Mon Nov 02, 2009 5:49 pmFollowed by: 3 members; Thanked: 2 times

by tonebeeze » Fri Jan 14, 2011 11:14 am

What is the quickest way to solve these types of problems. Please advise.

Quant Review PS #145

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)(square root 3) - 32 pi, what is the radius of each circle?

a. 4
b. 8
c. 16
d. 24
e. 32

OA: B

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Source: Beat The GMAT — Problem Solving | Fri Jan 14, 2011 11:14 am

stormier: Master | Next Rank: 500 Posts; Posts: 139; Joined: Fri Nov 26, 2010 4:45 pmFollowed by: 1 member; Location: Boston; Thanked: 20 times; GMAT Score:720

by stormier » Fri Jan 14, 2011 11:39 am

tonebeeze wrote:What is the quickest way to solve these types of problems. Please advise.

Quant Review PS #145

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)(square root 3) - 32 pi, what is the radius of each circle?

a. 4
b. 8
c. 16
d. 24
e. 32

OA: B

Assume that the radius of circle is r. The side of the equilateral triangle formed in the figure is 2r.

You may either remember (or quickly calculate) that area of an equilateral triangle of side a = a^2. sqrt(3)/4

The area of the equilateral triangle = sqrt(3). (side)^2/4 = sqrt(3).(2r)^2/4 = r^2.sqrt(3)

Now lets calculate the area of each of the three sectors inside the triangle. (i.e. the thtree symmetrical unshaded parts of the triangle)

Each sector is 60/360 = 1/6th of the area of the circle.

Thus area of three sectors = 3/6 = 1/2 area of circle = 1/2 pi . r^2

Area of shaded region = r^2. sqrt(3) -r^2.pi/2 = 64*sqrt(3) - 32.pi = 64*sqrt(3) - 64. pi/2

Thus r^2 = 64 => r = 8

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Fri Jan 14, 2011 2:25 pm

tonebeeze wrote:What is the quickest way to solve these types of problems. Please advise.

Quant Review PS #145

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)(square root 3) - 32 pi, what is the radius of each circle?

a. 4
b. 8
c. 16
d. 24
e. 32

OA: B

Shaded region = triangle - 3 circle sectors = 64âˆš3 - 32âˆ�

Looking at the equation above, we can see that the 3 circle sectors = 32âˆ�.
Since the triangle is equilateral, each of its angles is 60 degrees. Since 60/360 = 1/6, each circle sector is 1/6 the area of each circle.
Thus, the 3 sectors = 3*1/6 = 1/2 each circle area.
Since the 3 sectors = 32âˆ�, circle area = 64âˆ�.
Thus, âˆ�rÂ² = 64âˆ�, and r=8.

The correct answer is B.

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tonebeeze: Master | Next Rank: 500 Posts; Posts: 158; Joined: Mon Nov 02, 2009 5:49 pmFollowed by: 3 members; Thanked: 2 times

by tonebeeze » Fri Jan 14, 2011 3:49 pm

GMATGuruNY wrote:
tonebeeze wrote:What is the quickest way to solve these types of problems. Please advise.

Quant Review PS #145

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)(square root 3) - 32 pi, what is the radius of each circle?

a. 4
b. 8
c. 16
d. 24
e. 32

OA: B
Shaded region = triangle - 3 circle sectors = 64âˆš3 - 32âˆ�

Looking at the equation above, we can see that the 3 circle sectors = 32âˆ�.
Since the triangle is equilateral, each of its angles is 60 degrees. Since 60/360 = 1/6, each circle sector is 1/6 the area of each circle.
Thus, the 3 sectors = 3*1/6 = 1/2 each circle area.
Since the 3 sectors = 32âˆ�, circle area = 64âˆ�.
Thus, âˆ�rÂ² = 64âˆ�, and r=8.

The correct answer is B.

You really have a gift for explaining this very simply Thanks for your help!

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nehatandon: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Fri Jan 07, 2011 9:17 pm

by nehatandon » Sat Jan 15, 2011 2:31 pm

Nice question. Good explanation stormier.

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