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# The figure shown above consists of three identical circles that are ta

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Re: The figure shown above consists of three identical circles that are ta [#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, 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 that are ta [#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 that are ta [#permalink]
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 that are ta [#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 that are ta [#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

Took me 5 minutes though
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Re: The figure shown above consists of three identical circles that are ta [#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 82209 times ]

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The figure shown above consists of three identical circles that are ta [#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

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 that are ta [#permalink]
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There is a very simple of way of solving this.

Basically the equation is nothing but the Area of the Triangle - The 3 Sectors = 64\sqrt{3}-32π

So area of the equilateral triangle is 64\sqrt{3}

Area of an equilateral triangle = side^2*\sqrt{3}/4

side^2=64*4
side=16

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Re: The figure shown above consists of three identical circles that are ta [#permalink]
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 $$64\sqrt{3}-32\pi$$, 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

Hello,

can you help me understand how did we arrive at pi*r^2/2? Usually the area of the sector formula is angle/360* pi*r^2
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Re: The figure shown above consists of three identical circles that are ta [#permalink]
Jaya6 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 $$64\sqrt{3}-32\pi$$, 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

Hello,

can you help me understand how did we arrive at pi*r^2/2? Usually the area of the sector formula is angle/360* pi*r^2

We have three 60-degree sectors. Together they make 180-dgeree sector, or half the circle, which is 360-degree. The area of half a circle is pi*r^2/2.
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The figure shown above consists of three identical circles that are ta [#permalink]
Area of equilateral triangle is given by $$\frac{s^2\sqrt{3}}{4}$$, where s is the length of a side.

In this case, the length of the side is 2r. S = 2r -->

$$\frac{(2r)^2\sqrt{3}}{4}$$ = $$r^2\sqrt{3}$$

Now we know that the irrational terms in the expression for the area is not likely to be affected by each other: $$64\sqrt{3} - 32Pi$$

So we can conclude that r^2 = 64.
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Re: The figure shown above consists of three identical circles that are ta [#permalink]
I have taken shortcut to get the answer, which takes 1-2 min to solve.

If you see 64√3 represents area of triangle , whereas 32 pi represents Area of 3 (1/6)th of a circle .
taking just 32 pi ( 3*1/6 =1/2) gives us half Area of circle , multiple by 2 gives 64 pi .
therefore , Area of circle 64 pi and radius is 8.
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The figure shown above consists of three identical circles that are ta [#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

Attachment:
Untitled.png

Why are y'all doing so much math?

The shaded area is the area of the dashed triangle minus three little pie slices. The part we're subtracting is 32(pi), so that's the pie slices. Just eyeballing it, we can see that the three pie slices add up to half of one of the circles (yeah, we know this for sure because each of the angles of the dashed triangle is 60 degrees, so each slice is 1/6th of a circle since 60 is 1/6th of 360...but we don't need to even know this to figure it out, since we can just eyeball it).

Anyway, half of the area of a circle = 32(pi), so the area of a circle is 64(pi). A=(pi)r^2. So r=8.

GetMathQuestionsRightWithoutDoingAllTheMath

ThatDudeKnowsBallparking

Originally posted by ThatDudeKnows on 13 May 2022, 12:42.
Last edited by ThatDudeKnows on 10 Jun 2022, 15:05, edited 1 time in total.
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Re: The figure shown above consists of three identical circles that are ta [#permalink]
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 the area 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

Hey Bunuel, what makes us spot 'pi. r^2/2 as 32pi, what is the reasoning behind it
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Re: The figure shown above consists of three identical circles that are ta [#permalink]
JeffTargetTestPrep 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

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

MartyTargetTestPrep

Silly question... why do you have to multiply both sides of the equation by 2? To confirm, whatever you to do one side of the equation, you have to do to the other side, correct? Happy Holidays
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Re: The figure shown above consists of three identical circles that are ta [#permalink]
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woohoo921 wrote:
MartyTargetTestPrep

Silly question... why do you have to multiply both sides of the equation by 2? To confirm, whatever you to do one side of the equation, you have to do to the other side, correct? Happy Holidays

Yes, you have to do the same thing to both sides.
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Re: The figure shown above consists of three identical circles that are ta [#permalink]
JeffTargetTestPrep

once we get to (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

can we just say r^2)√3 = 64√3

and therefore r = 8? or do we need to multiply both sides by 2 and follow the remaining steps

Thank you
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