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A cylindrical tank has a base with a circumference of meters and an i

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A cylindrical tank has a base with a circumference of meters and an i  [#permalink]

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New post 01 Dec 2018, 21:49
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A cylindrical tank has a base with a circumference of \(4\sqrt{3π}\) meters and an isosceles right 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, which of the following is the length of at least one side of the triangle?

A.3
B.12
C.\(\sqrt{2}\)
D.\(\sqrt{3}\)
E.\(\sqrt{6}\)

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New post 01 Dec 2018, 23:48
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\(2\pi r = 4\sqrt{3\pi}\)
square both sides ,
2(πr)^2 = 12π
πr^2 = 6

Let the area of isosceles right triangle be a.
\(\frac{a}{πr^2} = \frac{1}{4}\)
a = 6/4 = 3/2

Let the equal sides be x. Third side = \(x\sqrt{2}\)
Area of triangle = 1/2 * x * x = 3/2
\(x = \sqrt{3}\)
\(x\sqrt{2} = \sqrt{6}\)

E
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Re: A cylindrical tank has a base with a circumference of meters and an i  [#permalink]

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New post 02 Dec 2018, 03:02
AnkitOrYadav wrote:
A cylindrical tank has a base with a circumference of \(4\sqrt{3π}\) meters and an isosceles right 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, which of the following is the length of at least one side of the triangle?

A.3
B.12
C.\(\sqrt{2}\)
D.\(\sqrt{3}\)
E.\(\sqrt{6}\)



given :
2*pi*r= 4\sqrt{3pi}

r= 2\sqrt{3*pi}/pi

since its an isosceles triangle so side x:x:x\sqrt{2}

x\sqrt{2}: 2\sqrt{3*pi}/pi
x=\sqrt{6}/\sqrt{pi}

IMO option E is close option..

I m not able to figure out on how to use the given information on Probability of falling grain
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A cylindrical tank has a base with a circumference of meters and an i  [#permalink]

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New post 08 Apr 2019, 03:32
pandeyashwin wrote:
\(2\pi r = 4\sqrt{3\pi}\)
square both sides ,
2(πr)^2 = 12π
πr^2 = 6

Let the area of isosceles right triangle be a.
\(\frac{a}{πr^2} = \frac{1}{4}\)
a = 6/4 = 3/2

Let the equal sides be x. Hypotenuse = \(x\sqrt{2}\)
Area of triangle = 1/2 * x * x = 3/2
\(x = \sqrt{3}\)
\(x\sqrt{2} = \sqrt{6}\)

E


In the highlighted part, why did you exclude the 2 in LHS and 4 in RHS? Is that allowed?

Anyhow, here's the solution without exclusion.

\(2\pi r = 4\sqrt{3\pi}\)
square both sides ,
(2πr)^2 = 48π
πr^2 = 12

Let the area of isosceles right triangle be a.
\(\frac{a}{πr^2} = \frac{1}{4}\)
a = 12/4 = 3.

Let the equal sides be x. Third side = \(x\sqrt{2}\) (Sides of a right isosceles: 1:1:\(\sqrt{2}\))
Area of triangle = 1/2 * x * x = 3
\(x = \sqrt{6}\)

Answer E.
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Re: A cylindrical tank has a base with a circumference of meters and an i  [#permalink]

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New post 08 Apr 2019, 04:30
Archit3110 wrote:
AnkitOrYadav wrote:
A cylindrical tank has a base with a circumference of \(4\sqrt{3π}\) meters and an isosceles right 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, which of the following is the length of at least one side of the triangle?

A.3
B.12
C.\(\sqrt{2}\)
D.\(\sqrt{3}\)
E.\(\sqrt{6}\)



given :
2*pi*r= 4\sqrt{3pi}

r= 2\sqrt{3*pi}/pi

since its an isosceles triangle so side x:x:x\sqrt{2}

x\sqrt{2}: 2\sqrt{3*pi}/pi
x=\sqrt{6}/\sqrt{pi}

IMO option E is close option..

I m not able to figure out on how to use the given information on Probability of falling grain



\(4\sqrt{3π} = 2*\pi*r\) = Circumference of base
We need to find the area of the base so we need r^2. Let's square both sides to get
\(16*3*\pi = 4*\pi^2 * r^2\)
\(\pi*r^2 = 12\)

Now, here is how the given probability helps: the probability of the grain landing outside the triangle is 3/4. So the probability of the grain landing inside the triangle is 1/4.
Hence, area of base outside the triangle must be 3/4 of total while area of base inside the triangle is 1/4 of total.

The total area of base is 12, so inside triangle, the area is 3.

Area of an isosceles right triangle = (1/2)*s^2 = 3 (where s is the leg of the triangle)
\(s = \sqrt{6}\)
So the hypotenuse will be \(s*\sqrt{2} = \sqrt{6}*\sqrt{2} = 2*\sqrt{3}\)
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Re: A cylindrical tank has a base with a circumference of meters and an i   [#permalink] 08 Apr 2019, 04:30
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