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Annie
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Post subject: Probability Triangle  Posted: Fri Mar 28, 2008 2:54 pm |
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A cylindrical tank has a base with a circumference of 4(sqrt(pi sqrt(3)) meters and an equilateral triangle painted on the interior side of the base. A grain of sand is dropped into the tank, and has an equal probability of landing on any particular point on the base. If the probability of the grain of sand landing on the portion of the base outside the triangle is 3/4, what is the length of a side of the triangle?
a. root (2 (root 6))
b. (root 6 (root 6))/2
c. root (2 root 3)
d. root 3
e. 2
Answer is e.
I understand the whole explanation except for "here the triangle has an area of root 3"
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StaceyKoprince
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Post subject: Posted: Wed Apr 02, 2008 9:49 pm |
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Posts: 6077
Location: San Francisco
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Can you double check the problem? You've typed above "has a base with a circumference of meters..." How many meters? (Note: if it included symbols, and you just copied and pasted, it won't come through properly - you have to type it in yourself.)
I can guess, though that it might have something to do with comparing the area of the triangle to the area of the circular base. If the sand has a 3/4 probability of landing outside of the triangle, then it also has a 1/4 probability of landing inside the triangle. If we say the total area of the circle = area of triangle + area of circle OUTSIDE of triangle, then the 1/4 probability corresponds to the triangle's area and the 3/4 probability corresponds to the area outside the triangle. If I can calculate the area of hte circle, I can calculate the area of the triangle (that is, 1/4 the area of the circle).
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Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT
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matt.mcmahon
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Post subject: RE: Posted: Tue Apr 29, 2008 10:23 pm |
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It is 4(root(pi(root 3))) meters
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RonPurewal
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Post subject: Re: RE: Posted: Wed Apr 30, 2008 5:09 am |
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matt.mcmahon@gsmsigma.com wrote: It is 4(root(pi(root 3))) meters
if that's the circumference, then the radius is this quantity divided by 2p. (here 'p' stands for pi)
which is
(4√(P√3)) / 2P
= 2√(P√3)) / P **
= 2√√3 / √P *** - if you don't understand this step, i'll also show the work starting from (**).
starting from (***):
circle area = P(r^2)
= P * 4√3/P
= 4√3
so triangle area = 1/4 of this = √3
starting from (**):
circle area = P(r^2)
= P * 4P√3 / P^2
= 4√3
so triangle area = 1/4 of this = √3
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blaad
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Post subject: Posted: Wed Jun 18, 2008 2:36 am |
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How will you find out the length of a side of an equilateral triangle with area root 3?
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shoboy
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Post subject: area of equilateral traingle with area = root3 Posted: Fri Jun 20, 2008 9:11 am |
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area of a triangle = 1/2*base*height. If one side of equilateral triangle = a, then to get the height, bissect the triangle and use pythatgoras theory to solve for the height, i.e height = root(a squared - a squared/4) = root(3a squared/4) =a(root3)/2. We can then solve for "a" :- 1/2*base*height = root3, where base =a, and height =a(root3)/4, so 1/2*a*a(root3)/2 = root3. If you solve the equation, you get a = 2.
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rfernandez
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Post subject: Posted: Fri Jun 27, 2008 4:05 am |
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Posts: 386
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shoboy correctly derives the formula for the area of an equilateral triangle in terms of its side.
Alternatively, I have found it helpful to memorize this formula: A = s^2 * root(3) / 4, where s is the length of the triangle's side.
If A is given as root(3), then:
root(3) = s^2 * root(3) / 4
1 = s^2 / 4
4 = s^2
2 = s
I know it's yet another formula to memorize, but it comes in handy.
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vrajesh.dave
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Post subject: Re: Probability Triangle Posted: Sun Apr 04, 2010 11:03 pm |
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Here is the explaination from the MGMAT Cat  I understand that the area of triangle = 1/2 * base * height hence Area of equilateral triangle can be express as 1/2 * S * (S *root(3))/2 But how do we get area of triangle as root(3).
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Ben Ku
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Post subject: Re: Probability Triangle Posted: Wed Apr 28, 2010 11:16 pm |
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Posts: 823
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vrajesh.dave wrote: Here is the explaination from the MGMAT Cat I understand that the area of triangle = 1/2 * base * height hence Area of equilateral triangle can be express as 1/2 * S * (S *root(3))/2 But how do we get area of triangle as root(3). In the explanation, the area of the circular base was found to be 4 sqrt(3). Because the equilateral triangle is 1/4 of the area of the base, therefore it has an area of sqrt(3). Hope that helps.
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Ben Ku
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ManhattanGMAT
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vjsharma25
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Post subject: Re: Probability Triangle Posted: Thu Apr 07, 2011 4:05 am |
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I approached it differently and was getting somewhat different result.
You don't need the probability statement at all for solving this question.
You can calculate the radius as the circumference of the base is given
radius = (4√(P√3)) / 2P
= 2√(P√3)) / P
= 2√√3 / √P
Now we know that an equilateral triangle has circumradius = S/√3,where S is the side of the triangle.
So
S/√3 = 2√√3 / √P ----------(1)
There is a unique relation between height and side of an equilateral triangle
Height(H) = (√3/2)S ----------(2)
We can solve the two equations 1 and 2 for the value of H.
But the final expression which I got after solving this is
H^2 = 9√3/P and if you solve it for H,it gives a value(>2) different from 2.
Can someone explain where I am messing up.
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vjsharma25
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Post subject: Re: Probability Triangle Posted: Thu Apr 07, 2011 4:08 am |
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OK,I think i got my answer. The question statement just states that equilateral triangle is drawn inside it,it may or may not be touching the circumference of the circle.
Hmmm,thats why we need to calculate the area of the triangle from the probability statement.
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jnelson0612
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Post subject: Re: Probability Triangle Posted: Wed Apr 13, 2011 3:08 pm |
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Correct, vjsharma. Good work in figuring out your error.
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Jamie Nelson
ManhattanGMAT Instructor
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