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

Question

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.\(\sqrt{6}\)

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

pandeyashwin · Answer

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

Archit3110 · Answer

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

OhsostudiousMJ · Answer

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.

KarishmaB · Answer

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}

CEO2021 · Answer

Hi  pandeyashwin

Your approach is absolutely correct but some calculation error is there (highlighted):)

bumpbot · Answer

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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

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