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The figure shown above consists of three identical circles t

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The Official Guide For GMAT® Quantitative Review, 2ND Edition

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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 \(64\sqrt{3}-32\pi\), 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


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New post 09 Mar 2014, 15:15
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Image
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 \(area_{equilateral}-area_{sectors}=r^2\sqrt{3}-\frac{\pi{r^2}}{2}=64\sqrt{3}-32\pi\) --> \(r^2=\frac{2(64sqrt{3}-32\pi)}{2\sqrt{3}-\pi}=\frac{64(2\sqrt{3}-\pi)}{(2\sqrt{3}-\pi)}=64\) --> \(r=8\).

Answer: B.

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

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New post 10 Mar 2014, 22:15
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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
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Re: The figure shown above consists of three identical circles t  [#permalink]

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New post Updated on: 16 Oct 2014, 19:17
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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
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Originally posted by PareshGmat on 09 Mar 2014, 18:55.
Last edited by PareshGmat on 16 Oct 2014, 19:17, edited 2 times in total.
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Re: The figure shown above consists of three identical circles t  [#permalink]

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New post 10 Mar 2014, 07:45
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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
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Re: The figure shown above consists of three identical circles t  [#permalink]

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New post 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.
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Re: The figure shown above consists of three identical circles t  [#permalink]

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New post 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?
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Re: The figure shown above consists of three identical circles t  [#permalink]

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New post 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?!
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Re: The figure shown above consists of three identical circles t  [#permalink]

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New post 10 Jun 2015, 22:31
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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 expression
You 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 equation
Alternatively 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
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Re: The figure shown above consists of three identical circles t  [#permalink]

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New post 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)
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Re: The figure shown above consists of three identical circles t  [#permalink]

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New post 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 :-(
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Re: The figure shown above consists of three identical circles t  [#permalink]

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New post 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.
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Re: The figure shown above consists of three identical circles t  [#permalink]

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Attached is a visual that should help.
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Screen Shot 2016-08-10 at 7.23.07 PM.png [ 89.13 KiB | Viewed 12970 times ]


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

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New post 25 May 2017, 11:12
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

Problem Solving
Question: 145
Category: Geometry Circles; Triangles; Area
Page: 81
Difficulty: 600


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This problem is definitely not 600-level 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
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Re: The figure shown above consists of three identical circles t  [#permalink]

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New post 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
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Re: The figure shown above consists of three identical circles t  [#permalink]

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New post 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 60-degree 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
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Re: The figure shown above consists of three identical circles t  [#permalink]

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New post 31 Mar 2018, 13:46
Bunuel wrote:
Image
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: http://gmatclub.com/forum/new-geometry- ... 68426.html


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

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New post 31 Mar 2018, 13:57
dave13 wrote:
Bunuel wrote:
Image
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: http://gmatclub.com/forum/new-geometry- ... 68426.html


Bunuel - i clicked on the link provided but link doesnt exist :( any other options ? :)


GEOMETRY: Shaded Region Problems: https://gmatclub.com/forum/geometry-sha ... 75005.html
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The figure shown above consists of three identical circles t  [#permalink]

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New post 31 Mar 2018, 20:59
64sqrt{3}-32П, what is the radius of each circle?

S of shaded area = S of triangle - 1/6S of circle*3. SO in our formula only 32П looks like S of a circle.

32П = 1/6 ПR^2*3 (as we have 60 degree angle) ==> R=8 (B)

It took 40 sec for me.
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Re: The figure shown above consists of three identical circles t  [#permalink]

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New post 14 Jul 2018, 04:35
Bunuel wrote:
Image
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: https://gmatclub.com/forum/geometry-sha ... 75005.html


Hey pushpitkc :-)

can you please explain step by step how Bunuel from here \(r^2\sqrt{3}-\frac{\pi{r^2}}{2}=64sqrt{3}-32\pi\)[/m] -->
gets final result ? i got confused in in algebraic expression as always :) next to each your step (pls add a few explanatory words :) )

\(r^2=\frac{2(64sqrt{3}-32\pi)}{2\sqrt{3}-\pi}=\frac{64(2\sqrt{3}-\pi)}{(2\sqrt{3}-\pi)}=64\) --> [m]r=8

many thanks :-)
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Re: The figure shown above consists of three identical circles t &nbs [#permalink] 14 Jul 2018, 04:35

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