BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Difficult triangles/circles problem

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 158
Joined: Mon Nov 02, 2009 5:49 pm
Thanked: 2 times
Followed by:3 members

Difficult triangles/circles problem

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
Attachments
3circles.png
Top
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 139
Joined: Fri Nov 26, 2010 4:45 pm
Location: Boston
Thanked: 20 times
Followed by:1 members
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
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
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 = 64√3 - 32�

Looking at the equation above, we can see that the 3 circle sectors = 32�.
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 = 32�, circle area = 64�.
Thus, �r² = 64�, and r=8.

The correct answer is B.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Master | Next Rank: 500 Posts
Posts: 158
Joined: Mon Nov 02, 2009 5:49 pm
Thanked: 2 times
Followed by:3 members

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 = 64√3 - 32�

Looking at the equation above, we can see that the 3 circle sectors = 32�.
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 = 32�, circle area = 64�.
Thus, �r² = 64�, and r=8.

The correct answer is B.
You really have a gift for explaining this very simply Thanks for your help!
Top
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.
Top
Post new topic Post Reply
• Page 1 of 1