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

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Difficulty:   45% (medium)

Question Stats: 73% (02:34) correct 27% (02:40) wrong based on 1778 sessions

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

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We'll be glad if you participate in development of this project:
2. Please vote for the best solutions by pressing Kudos button;
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The figure shown above consists of three identical circles t  [#permalink]

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27 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 $$64\sqrt{3}-32\pi$$, so $$area_{equilateral}-area_{sectors}=r^2\sqrt{3}-\frac{\pi{r^2}}{2}=64\sqrt{3}-32\pi$$ --> $$r^2=\frac{2(64\sqrt{3}-32\pi)}{2\sqrt{3}-\pi}=\frac{64(2\sqrt{3}-\pi)}{(2\sqrt{3}-\pi)}=64$$ --> $$r=8$$.

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

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5
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

Time Taken - 2:00
Difficulty level - 700
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Re: The figure shown above consists of three identical circles t  [#permalink]

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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)

$$\sqrt{3}r^2 - \frac{\pi r^2}{2} = 64\sqrt{3} - 32\pi$$

Originally posted by PareshGmat on 09 Mar 2014, 17:55.
Last edited by PareshGmat on 16 Oct 2014, 18:17, 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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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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Option B.
The triangle formed by the circles will be an equilateral triangle.
Area of triangle=$$\sqrt{3}/4*(side)^2$$
Now area of triangle=area shaded+area of 3 sectors of the circles.
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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Attached is a visual that should help.
Attachments Screen Shot 2016-08-10 at 7.23.07 PM.png [ 89.13 KiB | Viewed 31483 times ]

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

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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

GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

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We'll be glad if you participate in development of this project:
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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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The figure shown above consists of three identical circles t  [#permalink]

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dave13 wrote:
Bunuel wrote: 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 $$area_{equilateral}-area_{sectors}=r^2\sqrt{3}-\frac{\pi{r^2}}{2}=64sqrt{3}-32\pi$$ --> $$r^2=\frac{2(64sqrt{3}-32\pi)}{2\sqrt{3}-\pi}=\frac{64(2\sqrt{3}-\pi)}{(2\sqrt{3}-\pi)}=64$$ --> $$r=8$$.

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 )

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

many thanks Hi dave13

If x = y, then 2x = 2y

So, the same principle will hold good for this case: $$r^2 * \sqrt{3} - \frac{\pi * r^2}{2}$$ = $$64*\sqrt{3} - 32\pi$$

The above expression will become $$2(r^2\sqrt{3} - \frac{\pi{r^2}}{2}) = 2(64sqrt{3} - 32\pi)$$

The LHS will be become $$r^2*2\sqrt{3} - 2*\frac{\pi{r^2}}{2}$$ or $$r^2*2\sqrt{3} - \pi * {r^2}$$. This will become $$r^2(2\sqrt{3} - \pi)$$ when further simplified.

Therefore, $$r^2=\frac{2(64sqrt{3}-32\pi)}{2\sqrt{3}-\pi}=\frac{64(2\sqrt{3}-\pi)}{(2\sqrt{3}-\pi)}=64$$ --> $$r=8$$

Alternate solution:
When you equate L.H.S to R.H.S in the original equation(the expression on both sides which contain $$\pi$$)
you will get $$\frac{r^2}{2} = 32$$ -> $$r^2 = 32*2$$ -> $$r = \sqrt{64} = 8$$
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The figure shown above consists of three identical circles t  [#permalink]

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Bunuel wrote:

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

dave13 wrote:
JeffTargetTestPrep wrote:

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

hi generis its been raining here... so after the rain i feel like solving algebra regarding this "$$r^2(2√3 - π) = 128√3 − 64π$$ divide both sides by (2√3 - π) "

$$r^2= \frac{128√3 − 64π}{(2√3 - π)}$$ after this step i am getting confused (when i dvide numerator by denominator)

so this is what i get in numerator first i smplify numerator 128√3 − 64π ( i subtract 128 - 64 = 64) --->

$$r^2=\frac{64√3-π}{2√3 - π}$$ now when i divide numerator by denominator i divide 64 by 2

$$r^2= 32$$ and get this why have a great sunny weekend La France va gagner la coupe du monde 3:1 Hi dave13 - et le score final était de 4-2! Es-tu heureux que la France a remporté la coupe? (Can't remember; do I need "de"??)

So close! One mistake.
Quote:
first i smplify numerator
128√3 − 64π
(i subtract 128 - 64 = 64) --->

$$r^2=\frac{64√3-π}{2√3 - π}$$

This numerator is not correct:
$$64√3-π$$

$$(128√3 − 64π)=$$
$$64(2√3 -π)$$

Put the terms $$128√3$$ and $$64π$$ in factors to make dividing 64 out easier ("factorize" the terms)

In the numerator, we have
$$(128*√3)-(64*π)=$$
$$(64*2*√3)-(64*π)$$

Factor out $$64$$, thus $$64(2√3)-64(π)$$

Remove the common factor again. UNlike terms go together in parentheses with the minus sign: $$64(2√3-π)$$

So now we have $$r^2=\frac{64(2√3-π)}{(2√3-π)}$$

You should see it now. Hope that helps! _________________
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The figure shown above consists of three identical circles t  [#permalink]

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generis wrote:
Bunuel wrote:

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

dave13 wrote:
JeffTargetTestPrep wrote:

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

hi generis its been raining here... so after the rain i feel like solving algebra regarding this "$$r^2(2√3 - π) = 128√3 − 64π$$ divide both sides by (2√3 - π) "

$$r^2= \frac{128√3 − 64π}{(2√3 - π)}$$ after this step i am getting confused (when i dvide numerator by denominator)

so this is what i get in numerator first i smplify numerator 128√3 − 64π ( i subtract 128 - 64 = 64) --->

$$r^2=\frac{64√3-π}{2√3 - π}$$ now when i divide numerator by denominator i divide 64 by 2

$$r^2= 32$$ and get this why have a great sunny weekend La France va gagner la coupe du monde 3:1 Hi dave13 - et le score final était de 4-2! Es-tu heureux que la France a remporté la coupe? (Can't remember; do I need "de"??)

So close! One mistake.
Quote:
first i smplify numerator
128√3 − 64π
(i subtract 128 - 64 = 64) --->

$$r^2=\frac{64√3-π}{2√3 - π}$$

This numerator is not correct:
$$64√3-π$$

$$(128√3 − 64π)=$$
$$64(2√3 -π)$$

Put the terms $$128√3$$ and $$64π$$ in factors to make dividing 64 out easier ("factorize" the terms)

In the numerator, we have
$$(128*√3)-(64*π)=$$
$$(64*2*√3)-(64*π)$$

Factor out $$64$$, thus $$64(2√3)-64(π)$$

Remove the common factor again. UNlike terms go together in parentheses with the minus sign: $$64(2√3-π)$$

So now we have $$r^2=\frac{64(2√3-π)}{(2√3-π)}$$

You should see it now. Hope that helps! generis
Salut! non, je dirais que je suis ni heureux ni malheureux , mais je croyais fortement que lequipe francaise remporterait la coupe du monde Dailleurs, le score final 3-1 etait ma prognostique je l`ai ecrit avant le debut du match je suis heureux de voir ta solution fantastique merci beaucoup What if i would not factor it, how should we solve this ? i like to choose the hard way , the most sinuous ways i tried to solve it myself but something went wrong $$r^2= \frac{128√3 − 64π}{(2√3 - π)}$$ = $$r^2= \frac{128√3 − 64π}{(2√3 - π)} * \frac{(2√3 + π)}{(2√3 + π)}$$ = $$\frac{256√3 +128n√3 -128n√3+64n^2}{4√9+2n√3-2n√3-n^2}$$ = so here i will cancel out +/- 128n√3 in numerator and +/-2n√3 in denominator = i get $$\frac{256*3+64n^2}{27-n}$$ so from here i get wrong answer

i followed the rule / pattern in the last example here https://www.cliffsnotes.com/study-guide ... xpressions

P.S.
And no need to use "de' in your sentence
The French Partitive Articles
When you are talking about a portion of an item (food), or something that cannot be quantified (e.g. qualities, like patience), use a partitive article:
du (+ masculine word)
de la (+ feminine word),
de l’ (followed by a vowel),
des (+ plural word).
FYI: https://www.frenchtoday.com/blog/french ... e-articles

Also you cant say " Es-tu heureux que la France a remporté la coupe? " you need to use here Passe du subjonctive since action took place in the past "Es-tu heureux que la France ait remporté la coupe" is correct version After verbs expressing feelings such as etre heureux, etre contente que, regretter que, avoir envie de, avoir hate de, etre sur, etc you need to use either present du Subjonctif or passe du Subjonctif. if action takes place in the past use "passe du Subjonctif" if action takes place in the present use present du Subjonctif

Example: Je suis content (présent de l’indicatif) en ce moment qu'elle soit (présent du subjonctif) ici aujourd'hui.
i am glad now that she is today here.

Example :
Je suis triste (présent de l’indicatif) maintenant qu’elle soit partie (passé du subjonctif) hier.
i am sad now that she left yesterday

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

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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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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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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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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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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

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

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

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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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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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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

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

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Bunuel wrote: 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 $$area_{equilateral}-area_{sectors}=r^2\sqrt{3}-\frac{\pi{r^2}}{2}=64sqrt{3}-32\pi$$ --> $$r^2=\frac{2(64sqrt{3}-32\pi)}{2\sqrt{3}-\pi}=\frac{64(2\sqrt{3}-\pi)}{(2\sqrt{3}-\pi)}=64$$ --> $$r=8$$.

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 ?  Re: The figure shown above consists of three identical circles t   [#permalink] 31 Mar 2018, 12:46

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# The figure shown above consists of three identical circles t  