the question is basically saying that a triangle is inscribed in a circle
circumference of the circle = 2*pi*r = 4 sqrt [(pi*sqrt3)]
pi * r = 2 sqrt [(pi*sqrt3)]
(pi*r)/2 = sqrt [(pi*sqrt3)]
[(pi*r)/2]^2 = (pi*sqrt3)
pi^2 * r^2/ 4 = pi * sqrt 3
pi * r^2/4 = sqrt 3
pi * r^2 = 4 * sqrt3 ---------I
probability of sand falling outside the triangle = 3/4
this means
area of the circle excluding the triangle / total area of the circle = 3/4
Therefore,
area of the triangle/ total area of the circle = 1-3/4 = 1/4------II
area of the circle = pi * r^2 = 4 * sqrt 3 [refer I]
area of the quilateral triangle = side^2 sqrt3/4
side^2 sqrt3/4 / 4 * sqrt 3 = 1/4 ---refer II
side^2 / 16 = 1/4
side^2 = 16/4
side^2 = 4
side = 2
Hence E.
This was an awesome question, took me about 3 mins to solve, but nonetheless brilliant.
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Source: Beat The GMAT — Problem Solving |
the above post pretty much covers all the bases, but the lack of square root symbols makes it pretty hard to read.
HINT: google "square root symbol", copy one off one of the pages that appears (or even from the search results - you don't even have to click the links if the √ symbol appears in the first couple of lines of text), and then paste to your heart's delight.
you'll even save yourself work this way, because instead of typing "sqrt" (4 keystrokes), you can just hit ctrl-V (1 keystroke, or 2, depending on how you look at it) to paste the symbol.
i would've done the same for pi, but the pi character shows up as something very much resembling an 'n' on these forums.
--
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
HINT: google "square root symbol", copy one off one of the pages that appears (or even from the search results - you don't even have to click the links if the √ symbol appears in the first couple of lines of text), and then paste to your heart's delight.
you'll even save yourself work this way, because instead of typing "sqrt" (4 keystrokes), you can just hit ctrl-V (1 keystroke, or 2, depending on how you look at it) to paste the symbol.
i would've done the same for pi, but the pi character shows up as something very much resembling an 'n' on these forums.
--
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
Ron has been teaching various standardized tests for 20 years.
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Pueden hacerle preguntas a Ron en castellano
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Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
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Learn more about ron
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Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
--
Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
--
Learn more about ron
