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A cylinder has a base with a circumference of 20pi meters
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Updated on: 07 May 2012, 04:06
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A cylinder has a base with a circumference of 20pi meters and an equilateral triangle inscribed on the interior side of the base. A marker is dropped into the tank with an equal probability of landing on any point on the base. If the probability of the marker landing inside the triangle is (Sqrt 3)/4 , what is the length of a side of the triangle? A. 3(sqrt 2pi) B. 3(sqrt 3pi) C. 10*sqrt (pi) D. 10(sqrt 3pi) E. 20 pi
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Originally posted by Joy111 on 07 May 2012, 02:12.
Last edited by Joy111 on 07 May 2012, 04:06, edited 1 time in total.




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Re: A cylinder has a base with a circumference of 20pi meters
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07 May 2012, 03:38
Joy111 wrote: A cylinder has a base with a circumference of 20pi meters and an equilateral triangle inscribed on the interior side of the base. A marker is dropped into the tank with an equal probability of landing on any point on the base. If the probability of the marker landing inside the triangle is (Sqrt 3)/4 , what is the length of a side of the triangle?
A. 3(sqrt 2pi) B. 3(sqrt 3pi) C. 10 pi D. 10(sqrt 3pi) E. 20 pi Since the probability of the marker landing on the portion of the base inside the triangle is \(\frac{\sqrt{3}}{4}\) then the portion of the base (circle) inside the triangle must be \(\frac{\sqrt{3}}{4}\) of the area of the base. Next: \(circumference=20\pi=2\pi{r}\) > \(r=10\) > \(area_{base}=\pi{r^2}=100\pi\); The area of the equilateral triangle is \(\frac{\sqrt{3}}{4}\) of the base: \(area_{equilateral}=\frac{\sqrt{3}}{4}*100\pi\) > also the ares of the equilateral triangle is \(area_{equilateral}=a^2*\frac{\sqrt{3}}{4}\), where \(a\) is the length of a side > \(area_{equilateral}=a^2*\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}*100\pi\) > \(a=10{\sqrt\pi}\). Answer: C. Similar question to practice: acylindricaltankhasabasewithacircumferenceof105453.htmlHope it helps.
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Re: Geometry, probablity
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07 May 2012, 02:39
Prob of marker landing in the triangle = area of triangle / area of base = sqrt(3)/4 Circumference of base = 20pi meters => Radius of base = 10 meters => Area of base = 100pi sq m Therefore area of the triangle = 100pi * sqrt(3)/4 As the area of an equilateral triangle is sqrt(3)*(side^2)/4, any side of the triangle = 10*sqrt(pi) This doesn't seem to be any of the answer choices.
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Re: Geometry, probablity
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07 May 2012, 04:05
GyanOne wrote: Prob of marker landing in the triangle = area of triangle / area of base = sqrt(3)/4
Circumference of base = 20pi meters => Radius of base = 10 meters => Area of base = 100pi sq m
Therefore area of the triangle = 100pi * sqrt(3)/4
As the area of an equilateral triangle is sqrt(3)*(side^2)/4, any side of the triangle = 10*sqrt(pi)
This doesn't seem to be any of the answer choices. sorry missed the sqrt in option C It has now been edited , My bad A. 3(sqrt 2pi) B. 3(sqrt 3pi) C. 10 sqrt (pi)D. 10(sqrt 3pi) E. 20 pi



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Re: A cylinder has a base with a circumference of 20pi meters
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21 May 2012, 20:45
Bunuel wrote: Joy111 wrote: A cylinder has a base with a circumference of 20pi meters and an equilateral triangle inscribed on the interior side of the base. A marker is dropped into the tank with an equal probability of landing on any point on the base. If the probability of the marker landing inside the triangle is (Sqrt 3)/4 , what is the length of a side of the triangle?
A. 3(sqrt 2pi) B. 3(sqrt 3pi) C. 10 pi D. 10(sqrt 3pi) E. 20 pi Since the probability of the marker landing on the portion of the base inside the triangle is \(\frac{\sqrt{3}}{4}\) then the portion of the base (circle) inside the triangle must be \(\frac{\sqrt{3}}{4}\) of the area of the base. Next: \(circumference=20\pi=2\pi{r}\) > \(r=10\) > \(area_{base}=\pi{r^2}=100\pi\); The area of the equilateral triangle is \(\frac{\sqrt{3}}{4}\) of the base: \(area_{equilateral}=\frac{\sqrt{3}}{4}*100\pi\) > also the ares of the equilateral triangle is \(area_{equilateral}=a^2*\frac{\sqrt{3}}{4}\), where \(a\) is the length of a side > \(area_{equilateral}=a^2*\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}*100\pi\) > \(a=10{\sqrt\pi}\). Answer: C. Similar question to practice: acylindricaltankhasabasewithacircumferenceof105453.htmlHope it helps. Hello All, See my solution below and please tell me where I am going wrong: Radius of base = 10 (as derived by you) Now, if we draw the equilateral triangle inscribed in a circle (as shown in my attachment which is not to scale), then: OA = radius = 10 and O is the centroid Centroid divides a median in the ratio 2:1. Hence, OD = OA / 2 = 5 Now, AD is the height of the triangle and in equilateral triangle, [math]height = a * sqrt (3) /2 where a = side of triangle Hence, 15 = a * sqrt (3) /2 Hence a = 10 * sqrt (3) Where am I going wrong?
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Re: A cylinder has a base with a circumference of 20pi meters
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18 Jul 2012, 04:44
In this question , Triangle is inscribed inside circle so if I am not wrong, centre of the circle will be the circumcentre. As per Maths book of GMAT club relationship between Equilateral triangle and Circcumradius is R = a/3^1/2 . We know R = 10 , so why putting the values in this formula getting the right answer ?



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Re: A cylinder has a base with a circumference of 20pi meters
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04 Sep 2012, 04:48
In this question , Triangle is inscribed inside circle so if I am not wrong, centre of the circle will be the circumcentre. As per Maths book of GMAT club relationship between Equilateral triangle and Circcumradius is R = a/3^1/2 . We know R = 10 , so why putting the values in this formula NOT getting the right answer ?



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Re: A cylinder has a base with a circumference of 20pi meters
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09 Sep 2012, 16:01
Bunuel wrote: Joy111 wrote: A cylinder has a base with a circumference of 20pi meters and an equilateral triangle inscribed on the interior side of the base. A marker is dropped into the tank with an equal probability of landing on any point on the base. If the probability of the marker landing inside the triangle is (Sqrt 3)/4 , what is the length of a side of the triangle?
A. 3(sqrt 2pi) B. 3(sqrt 3pi) C. 10 pi D. 10(sqrt 3pi) E. 20 pi Since the probability of the marker landing on the portion of the base inside the triangle is \(\frac{\sqrt{3}}{4}\) then the portion of the base (circle) inside the triangle must be \(\frac{\sqrt{3}}{4}\) of the area of the base. Next: \(circumference=20\pi=2\pi{r}\) > \(r=10\) > \(area_{base}=\pi{r^2}=100\pi\); The area of the equilateral triangle is \(\frac{\sqrt{3}}{4}\) of the base: \(area_{equilateral}=\frac{\sqrt{3}}{4}*100\pi\) > also the ares of the equilateral triangle is \(area_{equilateral}=a^2*\frac{\sqrt{3}}{4}\), where \(a\) is the length of a side > \(area_{equilateral}=a^2*\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}*100\pi\) > \(a=10{\sqrt\pi}\). Answer: C. Similar question to practice: acylindricaltankhasabasewithacircumferenceof105453.htmlHope it helps. Hey Bunuel quick question, If the radius of a circle that inscribed an equaliteral is \(r=S\sqrt{3}/3\), where r is the radius and S is the side of the equilateral, should int the answer be \(30/\sqrt{3}=S\)?



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Re: A cylinder has a base with a circumference of 20pi meters
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25 Mar 2013, 10:21
Could anyone pl clarify my doubt??
For an equilateral triangle circumscribed in a circle, side = R * root 3 where R is the circumradius.
In this problem, R = 10, So side of the triangle should be 10 * root 3..
Pl point where I am going wrong.



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Re: A cylinder has a base with a circumference of 20pi meters
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25 Mar 2013, 17:21
Quote: Hey Bunuel quick question,
If the radius of a circle that inscribed an equaliteral is \(r=S\sqrt{3}/3\), where r is the radius and S is the side of the equilateral, should int the answer be \(30/\sqrt{3}=S\)? I did the same thing as Alphabeta1234 and Friend29. Can someone explain why this is incorrect? \(S=\frac{30}{\sqrt{3}}\) \(S=10\sqrt{3}\). Thanks



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Re: A cylinder has a base with a circumference of 20pi meters
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13 Oct 2014, 17:15
C= 20pi =2pi*10 area of circle= pir^2 = pi10^2=100pi Now as question mentions that probability of dropping a marker anywhere on the base is same and in particular in the equilateral triangle is sqrt3/4. So, equilateral triangle's area is sqrt3/4 of total area of circle (100pi) = sqrt3/4 *100pi( for e.g we do 20% of 100). Now area of equilateral triangle can also be represented = as per its formula = a^2*sqrt3/4 So now when will be equating both the equations a^2*sqrt3/4 =sqrt3/4 *100pi a^2 = 100 pi a= sqrt 10pi



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Re: A cylinder has a base with a circumference of 20pi meters
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30 Oct 2014, 02:19
Bunuel wrote: Joy111 wrote: A cylinder has a base with a circumference of 20pi meters and an equilateral triangle inscribed on the interior side of the base. A marker is dropped into the tank with an equal probability of landing on any point on the base. If the probability of the marker landing inside the triangle is (Sqrt 3)/4 , what is the length of a side of the triangle?
A. 3(sqrt 2pi) B. 3(sqrt 3pi) C. 10 pi D. 10(sqrt 3pi) E. 20 pi Since the probability of the marker landing on the portion of the base inside the triangle is \(\frac{\sqrt{3}}{4}\) then the portion of the base (circle) inside the triangle must be \(\frac{\sqrt{3}}{4}\) of the area of the base.Next: \(circumference=20\pi=2\pi{r}\) > \(r=10\) > \(area_{base}=\pi{r^2}=100\pi\); The area of the equilateral triangle is \(\frac{\sqrt{3}}{4}\) of the base: \(area_{equilateral}=\frac{\sqrt{3}}{4}*100\pi\)> also the ares of the equilateral triangle is \(area_{equilateral}=a^2*\frac{\sqrt{3}}{4}\), where \(a\) is the length of a side > \(area_{equilateral}=a^2*\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}*100\pi\) > \(a=10{\sqrt\pi}\). Answer: C. Similar question to practice: acylindricaltankhasabasewithacircumferenceof105453.htmlHope it helps. I cannot seem to understand the highlighted part. How do we actually relate the given probability to the area of the base?



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Re: A cylinder has a base with a circumference of 20pi meters
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31 Oct 2014, 05:48
earnit wrote: Bunuel wrote: Joy111 wrote: A cylinder has a base with a circumference of 20pi meters and an equilateral triangle inscribed on the interior side of the base. A marker is dropped into the tank with an equal probability of landing on any point on the base. If the probability of the marker landing inside the triangle is (Sqrt 3)/4 , what is the length of a side of the triangle?
A. 3(sqrt 2pi) B. 3(sqrt 3pi) C. 10 pi D. 10(sqrt 3pi) E. 20 pi Since the probability of the marker landing on the portion of the base inside the triangle is \(\frac{\sqrt{3}}{4}\) then the portion of the base (circle) inside the triangle must be \(\frac{\sqrt{3}}{4}\) of the area of the base.Next: \(circumference=20\pi=2\pi{r}\) > \(r=10\) > \(area_{base}=\pi{r^2}=100\pi\); The area of the equilateral triangle is \(\frac{\sqrt{3}}{4}\) of the base: \(area_{equilateral}=\frac{\sqrt{3}}{4}*100\pi\)> also the ares of the equilateral triangle is \(area_{equilateral}=a^2*\frac{\sqrt{3}}{4}\), where \(a\) is the length of a side > \(area_{equilateral}=a^2*\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}*100\pi\) > \(a=10{\sqrt\pi}\). Answer: C. Similar question to practice: acylindricaltankhasabasewithacircumferenceof105453.htmlHope it helps. I cannot seem to understand the highlighted part. How do we actually relate the given probability to the area of the base? \(P = \frac{(favorable)}{(total)} =\frac{(area \ of \ triangle)}{(area \ of \ circle)} = \frac{\sqrt{3}}{4}\) > \((area \ of \ triangle)=(area \ of \ circle)*\frac{\sqrt{3}}{4}\).
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Re: A cylinder has a base with a circumference of 20pi meters
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21 Aug 2018, 15:50
Why do we need the extra information about probability if finding the radius is enough?




Re: A cylinder has a base with a circumference of 20pi meters &nbs
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