Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 44320

The figure shown above consists of three identical circles t [#permalink]
Show Tags
09 Mar 2014, 15:14
Question Stats:
76% (02:09) correct 24% (02:29) wrong based on 903 sessions
HideShow timer Statistics
The Official Guide For GMAT® Quantitative Review, 2ND EditionAttachment:
Untitled.png [ 10.71 KiB  Viewed 18839 times ]
The figure shown above consists of three identical circles that are tangent to each other. If the area of the shaded region is \(64\sqrt{3}32\pi\), what is the radius of each circle? (A) 4 (B) 8 (C) 16 (D) 24 (E) 32 Problem Solving Question: 145 Category: Geometry Circles; Triangles; Area Page: 81 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you!
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Math Expert
Joined: 02 Sep 2009
Posts: 44320

The figure shown above consists of three identical circles t [#permalink]
Show Tags
09 Mar 2014, 15:15
The figure shown above consists of three identical circles that are tangent to each other. If the area of the shaded region is \(64\sqrt{3}32\pi\), 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 \(area_{equilateral}area_{sectors}=r^2\sqrt{3}\frac{\pi{r^2}}{2}=64sqrt{3}32\pi\) > \(r^2=\frac{2(64sqrt{3}32\pi)}{2\sqrt{3}\pi}=\frac{64(2\sqrt{3}\pi)}{(2\sqrt{3}\pi)}=64\) > \(r=8\). Answer: B. GEOMETRY: Shaded Region Problems: newgeometryshadedregionproblems168426.html
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1838
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: The figure shown above consists of three identical circles t [#permalink]
Show Tags
09 Mar 2014, 18:55
7
This post received KUDOS
2
This post was BOOKMARKED
Let the radius of each circle = r The triangle formed is a equilateral triangle with each side = 2r So Area of the Triangle = \(\frac{\sqrt{3}}{4} * (2r)^2 = \sqrt{3} r^2\) ....................... (1) This triangle consists of 3 sectors each of 60 Degrees i.e total 180 Deg Area of these sectors = \(\frac{\pi r^2}{2}\) .................. (2) Area of shaded region = (1)  (2) \(\sqrt{3}r^2  \frac{\pi r^2}{2} = 64\sqrt{3}  32\pi\) Solving the above, radius r = 8 = Answer = B
_________________
Kindly press "+1 Kudos" to appreciate
Last edited by PareshGmat on 16 Oct 2014, 19:17, edited 2 times in total.



Senior Manager
Joined: 20 Dec 2013
Posts: 266
Location: India

Re: The figure shown above consists of three identical circles t [#permalink]
Show Tags
10 Mar 2014, 07:45
2
This post received KUDOS
Option B. The triangle formed by the circles will be an equilateral triangle. Area of triangle=\(\sqrt{3}/4*(side)^2\) And side =2*radius Now area of triangle=area shaded+area of 3 sectors of the circles. Area shaded=given Area of 3 sectors=\(3*[60/360*pi*r^2]\) We'll equate the 2 sides[Area triangle=Area shaded+area of 3 sectors] and simplify to get r=8



Current Student
Joined: 25 Sep 2012
Posts: 281
Location: India
Concentration: Strategy, Marketing
GMAT 1: 660 Q49 V31 GMAT 2: 680 Q48 V34

Re: The figure shown above consists of three identical circles t [#permalink]
Show Tags
10 Mar 2014, 22:15
10
This post received KUDOS
1
This post was BOOKMARKED
Find the are of equilateral triangle with side as 2r Find the are of the three sectors (eq. trianlge has \(\angle\) = \(60^{\circ}\) ) Subtract both to get the area of shaded region. You will get the equation as
\(\sqrt{3}r^{2}  \frac{ \pi r^{2}}{2} = 64\sqrt{3}  32 \pi\)
Don't even need to solve the equation you can see \(r^{2} = 64\) OR \(\frac{r^{2}}{2} = 32\) Hence, r = 8
Answer  B Time Taken  2:00 Difficulty level  700



Intern
Joined: 07 Jul 2013
Posts: 14
Concentration: Operations, Marketing

Re: The figure shown above consists of three identical circles t [#permalink]
Show Tags
03 Oct 2014, 13:06
area of the shaded region = area of the triangle formed  area of the three sectors. therefore, 64\sqrt{3} = area of the triangle.
the triangle is an equilateral triangle since the sides are equal to 2 * radius.
area of an equilateral triangle = \sqrt{3} / 4 * (side)^2
from above we can calculate radius = 8.



Manager
Joined: 26 Feb 2015
Posts: 123

Re: The figure shown above consists of three identical circles t [#permalink]
Show Tags
09 Apr 2015, 05:51
Is this a valid way of approaching the question, or just a lucky coincidence:
Area of equilateral triangle is \((S^2\sqrt{3})/4\)
So \((S^2\sqrt{3})/4\) = \(64\sqrt{3}\)
Simplify this to\(S = 16\sqrt{3}/\sqrt{3}\) > = 16. One side is 16, divide by 2 to get 8.
I simply ignored the "32"  is this a valid way of doing it?



Current Student
Joined: 03 Aug 2011
Posts: 294
Concentration: Strategy, Finance
GMAT 1: 640 Q44 V34 GMAT 2: 700 Q42 V44 GMAT 3: 680 Q44 V39 GMAT 4: 740 Q49 V41
GPA: 3.7
WE: Project Management (Energy and Utilities)

Re: The figure shown above consists of three identical circles t [#permalink]
Show Tags
10 Jun 2015, 11:58
I like the problem and managed to solve it, but my question here is whether staying below 2 mins is doable in this case. Average time for a correct answer shown by the system also confirms, taht most people need about 3 mins?!
_________________
Thank you very much for reading this post till the end! Kudos?



eGMAT Representative
Joined: 04 Jan 2015
Posts: 865

Re: The figure shown above consists of three identical circles t [#permalink]
Show Tags
10 Jun 2015, 22:31
3
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
erikvm wrote: Is this a valid way of approaching the question, or just a lucky coincidence:
Area of equilateral triangle is \((S^2\sqrt{3})/4\)
So \((S^2\sqrt{3})/4\) = \(64\sqrt{3}\)
Simplify this to\(S = 16\sqrt{3}/\sqrt{3}\) > = 16. One side is 16, divide by 2 to get 8.
I simply ignored the "32"  is this a valid way of doing it? Hi erikvm, The best approach would be to write down the equations for the area of shaded region and then equate both sides. Your method worked because the final expression was written in the form where you could equate two expressions from LHS and RHS. This may not be the case always and hence is not the best practice to follow. Also to answer bgpower's question, the final equation can be written as \(\sqrt{3}r^2  \frac{πr^2}{2} = 64\sqrt{3}  32π\) There are two easy ways to solve the equation for r. Equating the expressionYou can see the similarity in expression on both sides of the equation and can equate \(\frac{πr^2}{2} = 32π\) or \(\sqrt{3}r^2 = 64\sqrt{3}\) . This would give you the value of \(r = 8\). Solving the equationAlternatively the equation can also be solved very quickly. You just need to be aware of the possibilities of the terms canceling out. The equation can be written as \(r^2 =\) (\(64\sqrt{3}  32π)/(\sqrt{3} \frac{π}{2})\) If we take \(64\) common from the numerator we will have the expression as \(r^2 =\) \(64 *( \sqrt{3}  \frac{π}{2})/(\sqrt{3} \frac{π}{2})\) thus resulting in \(r^2 = 64\) and \(r = 8\) So with the above approach, it is possible to solve the question in less than 2 minutes. Hope this helps Regards Harsh
_________________
 '4 out of Top 5' Instructors on gmatclub  70 point improvement guarantee  www.egmat.com



Intern
Joined: 24 Aug 2015
Posts: 8

Re: The figure shown above consists of three identical circles t [#permalink]
Show Tags
03 Oct 2015, 10:24
I understand the explanation until I was stuck in the formula. It took me exactly 4 minutes to figure out how to do the calculation. 64\(\sqrt{3}\)  32phi = r2\(\sqrt{3}\)  1/2 phi. r2 then you want to multiple both sides with 2 to eliminate 1/2 2(64\sqrt{3}  32phi) = r2 (2\sqrt{3}  phi)



Manager
Joined: 05 Jul 2015
Posts: 106
Concentration: Real Estate, International Business
GPA: 3.3

Re: The figure shown above consists of three identical circles t [#permalink]
Show Tags
22 Nov 2015, 17:52
I used guesstimation which works for people who suck at math like me. All you need to know is √3 = 1.7 pi=3.14 64*1.7  32*3.14 Is more or less 8 The shaded region is about (1/4) of the triangle so a^2(√3/4)=4*8 a^2 = 32*(4/√3) a^2 = 128/1.7 is less than 260 but more than 192 16*16=256 so r = 16/2 = 8 Took me 5 minutes though



Current Student
Joined: 02 Jun 2015
Posts: 91
Location: Brazil
Concentration: Entrepreneurship, General Management
GPA: 3.3

Re: The figure shown above consists of three identical circles t [#permalink]
Show Tags
23 Nov 2015, 10:37
you have to see the resolution of the problem. It shows that is the area of the triangle  total area of the circle / 2
so, we just have to use the formula to find the area of the triangle
2R*R√3 / 2 > Rˆ2√3
Rˆ2√3 = 64√3 Rˆ2=64 R = 8
However it took me about 5 minutes to figure the whole thing out and come out with this fastest way.



Director
Status: Professional GMAT Tutor
Affiliations: AB, cum laude, Harvard University (Class of '02)
Joined: 10 Jul 2015
Posts: 533
Location: United States (CA)
Age: 38
GMAT 1: 770 Q47 V48 GMAT 2: 730 Q44 V47 GMAT 3: 750 Q50 V42
GRE 1: 337 Q168 V169
WE: Education (Education)

Re: The figure shown above consists of three identical circles t [#permalink]
Show Tags
10 Aug 2016, 19:37
1
This post was BOOKMARKED
Attached is a visual that should help.
Attachments
Screen Shot 20160810 at 7.23.07 PM.png [ 89.13 KiB  Viewed 8332 times ]
_________________
Harvard grad and 770 GMAT scorer, offering highquality private GMAT tutoring, both inperson and online via Skype, since 2002.
You can download my official testtaker score report directly from the Pearson Vue website: https://tinyurl.com/y8zh6qby Date of Birth: 09 December 1979.
GMAT Action Plan  McElroy Tutoring



Manager
Joined: 23 Dec 2013
Posts: 228
Location: United States (CA)
GMAT 1: 710 Q45 V41 GMAT 2: 760 Q49 V44
GPA: 3.76

Re: The figure shown above consists of three identical circles t [#permalink]
Show Tags
25 May 2017, 11:12
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionAttachment: Untitled.png The figure shown above consists of three identical circles that are tangent to each other. If the area of the shaded region is \(64\sqrt{3}32\pi\), what is the radius of each circle? (A) 4 (B) 8 (C) 16 (D) 24 (E) 32 Problem Solving Question: 145 Category: Geometry Circles; Triangles; Area Page: 81 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you! This problem is definitely not 600level difficulty. Not even close. Anyway, that said, the solution can be worked out as follows. The side of the equilateral triangle has length 2r. So that means the area of the equilateral triangle is (2r)^2*sqrt(3)/4 = r^2*sqrt(3) We know that the area of the shaded region is the area of the triangle minus the area of three sectors of 60 degrees each. The next logical step is to realize that three sectors of equal circles of 60 degrees will sweep an area equal to half the circle's area (60*3 = 180 degrees). So if the area of the circle is pi*r^2, then we know that the area of half the circle is pi*r^2/2. This combined sector area is equal to 32pi. 32pi = pi r^2/2 64pi = r^2*pi 64 = r^2 8 = r. It's probably best to stop here, but you can also solve for 8 via the triangle area method: 64sqrt(3) = r^2 sqrt(3) r = 8



Manager
Joined: 21 Jun 2017
Posts: 78

Re: The figure shown above consists of three identical circles t [#permalink]
Show Tags
09 Oct 2017, 11:45
Area of shaded = area of triangle  the sum of the sectors area of shaded = 64\sqrt{3}  32pi area of triangle = 1/2(2r)(r\sqrt{3}) = r^2 \sqrt{3} sum of the sectors = (3(60/360 x pi r^2))
64\sqrt{3}  32pi = r^2\sqrt{3}  (3(1/6 x pi r^2)) 1/2 x 3r^2 = r^2  64 + 32 3r^2 = 2r^2  128 + 64 r^2 = 64
square each side r=8 thus, the answer is (B) 8
hope this helps



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2116

Re: The figure shown above consists of three identical circles t [#permalink]
Show Tags
12 Oct 2017, 18:13
Bunuel wrote: Attachment: Untitled.png The figure shown above consists of three identical circles that are tangent to each other. If the area of the shaded region is \(64\sqrt{3}32\pi\), what is the radius of each circle? (A) 4 (B) 8 (C) 16 (D) 24 (E) 32 We can let each radius = r, and so the side of each triangle = 2r. Notice that the area of the equilateral triangle consists of the central shaded region and three identical circular sectors, each of which is a 60degree sector from its circle. Using this information, we can create the following equation: (Area of equilateral triangle)  (3 x area of 1/6 of each circle) = area of shaded region (2r)^2√3/4  3(1/6 x π r^2) = 64√3 − 32π [(4r^2)√3]/4  (πr^2)/2 = 64√3 − 32π (r^2)√3  (πr^2)/2 = 64√3 − 32π Multiplying both sides by 2, we have: 2(r^2)√3  πr^2 = 128√3 − 64π r^2(2√3  π) = 128√3 − 64π Dividing both sides by (2√3  π), we have: r^2 = 64 r = 8 Answer: B
_________________
Jeffery Miller
Head of GMAT Instruction
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions




Re: The figure shown above consists of three identical circles t
[#permalink]
12 Oct 2017, 18:13






