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

Redeem

draw the figure and solve?

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
GMAT Instructor
Posts: 3650
Joined: Wed Jan 21, 2009 4:27 am
Location: India
Thanked: 267 times
Followed by:80 members
GMAT Score:760

draw the figure and solve?

by sanju09 » Fri Apr 29, 2011 2:54 am
Regular hexagon ABCDEF has a perimeter of 36. O is the center of the hexagon and of circle O. Circles A, B, C, D, E, and F have centers at A, B, C, D, E, and F respectively. If each circle is tangent to the two circles adjacent to it and to circle O, what is the area of the shaded region (inside the hexagon but outside the circles)?
(A) 118 - 18 π
(B) 54 √3 - 9 π
(C) 54 √3 - 18 π
(D) 108 - 27 π
(E) 54 √3 - 27 π


[spoiler]Is the stem sufficient to draw the figure and solve?[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Top
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pm
Location: Milpitas, CA
Thanked: 1854 times
Followed by:523 members
GMAT Score:770

by Anurag@Gurome » Fri Apr 29, 2011 4:52 am
sanju09 wrote:Regular hexagon ABCDEF has a perimeter of 36. O is the center of the hexagon and of circle O. Circles A, B, C, D, E, and F have centers at A, B, C, D, E, and F respectively. If each circle is tangent to the two circles adjacent to it and to circle O, what is the area of the shaded region (inside the hexagon but outside the circles)?
(A) 118 - 18 π
(B) 54 √3 - 9 π
(C) 54 √3 - 18 π
(D) 108 - 27 π
(E) 54 √3 - 27 π
[spoiler]Is the stem sufficient to draw the figure and solve?[/spoiler]
See the figure.

Image

Since ABCDEF is a regular hexagon with perimeter 36, so each side of the hexagon = 6
So, the radius of circles O, A, B, C, D, E, and F is 6/2 = 3
Hence, area of circle with center at O = (pi)(3)² = 9(pi)
Now, each internal angle of hexagon = (6 - 2) * 180º/6 = 120º {The Interior Angle (or Internal Angle) of a regular polygon with "n" sides can be calculated using: (n - 2) * 180° / n}
Next, Area of Sector = 1/2 * (θ * pi/180) * r² (when θ is in degrees)
So, Area of the circles that are inside the hexagon = 6 * {1/2 * (120 * pi/180) × 3²} = 18(pi)

A regular hexagon has 6 equilateral triangles. Now, we need to find the area of hexagon.
Look at the following figure.

Image

First we'll find the area of 1 equilateral triangle.
By Pythagoras Theorem, OD = √{s² - (s/2)²} = √(3s²/4) = s√3/2
So, area of 1 equilateral triangle = (1/2)* s * s√3/2 = s²√3/4
Therefore, area of hexagon ABCDEF = 6s²√3/4 = 6(36)√3/4 = 54√3

So, Area of shaded region = 54√3 - {(9(pi) + 18(pi)} = 54√3 - 27(pi)

The correct answer is E.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
User avatar
Legendary Member
Posts: 582
Joined: Tue Mar 08, 2011 12:48 am
Thanked: 61 times
Followed by:6 members
GMAT Score:740

by force5 » Fri Apr 29, 2011 4:55 am
this is the figure if i got it correct.

area of the hexagon is 54 Rt3

each circles on the sides have diameter equal to side of the hexagon which is 6 , hence radius =3
area theta/360 * Pi r^2 = 3pi.
total 6 circles= 6*3pi = 18 pi

area of central circle.
the length of diagonal FC is 12 since they form equilateral triangle
hence radius of central circle = 3

area= 9pi

total area of left over space

54 rt3- (18pi+ 9pi)= 54 rt3- 27pi
Image
Top
User avatar
GMAT Instructor
Posts: 3650
Joined: Wed Jan 21, 2009 4:27 am
Location: India
Thanked: 267 times
Followed by:80 members
GMAT Score:760

by sanju09 » Sat Apr 30, 2011 1:13 am
great answers

the figure is comprehended extremely well

Each of the 7 circles has a radius of 3. Each interior angle of the regular hexagon ABCDEF is 120º, hence one-third of each of the six circles with centers at A, B, C, D, E, and F plus one circle at O is inside the hexagon.

Area of the regular hexagon ABCDEF = 6 × √3/4 × (6)^2 = 54 √3

And the area shared by the 7 circles = 6 × 1/3 × π × (3)^2 + π × (3)^2 = 27 π

Hence, area of the shaded region (inside the hexagon but outside the circles) = [spoiler]54 √3 - 27 π

Hence E[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Top
Post new topic Post Reply
• Page 1 of 1