GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 28 May 2020, 14:10

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# The figure shown above consists of three identical circles t

Author Message
TAGS:

### Hide Tags

Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3253
Location: India
GPA: 3.12
The figure shown above consists of three identical circles t  [#permalink]

### Show Tags

14 Jul 2018, 09:39
1
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 $$\frac{r^2}{2} = 32$$ -> $$r^2 = 32*2$$ -> $$r = \sqrt{64} = 8$$
_________________
You've got what it takes, but it will take everything you've got
VP
Joined: 09 Mar 2016
Posts: 1238
Re: The figure shown above consists of three identical circles t  [#permalink]

### Show Tags

15 Jul 2018, 03:17
JeffTargetTestPrep wrote:
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

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
VP
Joined: 09 Mar 2016
Posts: 1238
The figure shown above consists of three identical circles t  [#permalink]

### Show Tags

15 Jul 2018, 03:19
JeffTargetTestPrep wrote:
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

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
Senior SC Moderator
Joined: 22 May 2016
Posts: 3848
The figure shown above consists of three identical circles t  [#permalink]

### Show Tags

15 Jul 2018, 20:25
1
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!
_________________
Visit SC Butler, here! Get two SC questions to practice, whose links you can find by date.

Our lives begin to end the day we become silent about things that matter. -- Dr. Martin Luther King, Jr.
VP
Joined: 09 Mar 2016
Posts: 1238
The figure shown above consists of three identical circles t  [#permalink]

### Show Tags

16 Jul 2018, 05:29
1
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

belle journee
Intern
Joined: 26 Dec 2018
Posts: 7
Re: The figure shown above consists of three identical circles t  [#permalink]

### Show Tags

24 Jun 2019, 16:46
You guys all have great answers that rely on knowing some formulas and facts about geometry that I didn't know off of my head. So this is what I did. Area of shaded region is 64*sqrt(3) - 32pi. So 64*sqrt(3) I estimated at 110, minus 32pi, I estimated as 100. So the area of the shaded region is about 10, and seems to be about 1/5 of the total area of the triangle. So I guessed that the area of the triangle is about 50. Double that to get the area of a square that has a side equal to one side of the triangle. 100. Side length of 10, radius of circle 5. Chose between A and B but went with B.

I see now the importance of knowing even the formula for the area of an equilateral triangle would have my my life much easier.
Manager
Joined: 03 Mar 2017
Posts: 116
Location: India
GMAT 1: 650 Q45 V34
Re: The figure shown above consists of three identical circles t  [#permalink]

### Show Tags

24 Jun 2019, 21:18
we know that 3 circles are identical therefore the triangle will have 3 sides equal to 2r and thereby an equilateral triangle with all angles equal to 60degrees.
area of all sector=((pi*r^2)/6)*3=(pi*r^2)/2
Area of equilateral triangle=root(3)(4 r^2)/4=root(3) r^2
subtracting area of triangle from the area of 3 sectors and equating the given equation with it we get
r^2=64 => r=8
r=8
Intern
Joined: 19 Jul 2019
Posts: 16
Re: The figure shown above consists of three identical circles t  [#permalink]

### Show Tags

31 Jul 2019, 22:01
I am overall lost on this whole thing. I get that we assume that this is an equilateral triangle and that you have to find the area of the triangle and then subtract that from the area of the shaded triangle, right? Is this the first step?
Intern
Joined: 19 Jul 2019
Posts: 16
The figure shown above consists of three identical circles t  [#permalink]

### Show Tags

01 Aug 2019, 16:59
I don't know if this is right, but if you are still confused after looking at the above posts: here is what I did.

Area of triangle equals (1/2) times base times height, the height line goes from the middle of the triangle to the top of the triangle. Equilateral triangles all have 60 degree angles because all triangles have a total of 180 degrees. Since you are cutting the triangle in half, the triangle becomes a 30-60-90 degree triangle. The height of a 30-60-90 triangle is x times square root of 3, I don't know the symbol button on the keyboard for square root.

Then (1/2) times (2 times r--- this is the base) times height (r times the square root of 3--- look at the rules for 30-60-90 triangles). Then you are left with r squared times square root of 3 minus (64 times square root of 3) minus (32 times pie)

Since square root of 3 is the common base of the first two numbers, then you can factor and get (r squared minus 64). That equals (r minus 8)(r plus 8). r can't be negative 8, so it has to be positive 8, which is the radius.
The figure shown above consists of three identical circles t   [#permalink] 01 Aug 2019, 16:59

Go to page   Previous    1   2   [ 29 posts ]