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The figure shown above consists of three identical circles t

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

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New post 14 Jul 2018, 09:39
1
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: 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, \(r^2=\frac{2(64sqrt{3}-32\pi)}{2\sqrt{3}-\pi}=\frac{64(2\sqrt{3}-\pi)}{(2\sqrt{3}-\pi)}=64\) --> \(r=8\)

Hope this clears your confusion.

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

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New post 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

Answer: B



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
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Joined: 09 Mar 2016
Posts: 1053
The figure shown above consists of three identical circles t  [#permalink]

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New post 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

Answer: B




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

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New post 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

Answer: B

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: \(64(2√3-π)\)
That's your numerator.

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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New post 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

Answer: B

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: \(64(2√3-π)\)
That's your numerator.

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 :lol: , mais je croyais fortement que l`equipe francaise remporterait la coupe du monde :-) D`ailleurs, 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 :lol:
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 :-)
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The figure shown above consists of three identical circles t &nbs [#permalink] 16 Jul 2018, 05:29

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