A cylindrical tank has a base with a circumference of

Question

A cylindrical tank has a base with a circumference of 4sqrt^(pi * sqrt^3) meters and an equilateral triangle painted on the interior side of the base. If a stone is dropped inside the tank and the probability of the stone hitting the portion of the base outside the triangle is 3/4, what is the length of a side of the triangle?

a.) Sqrt^(2*sqrt^6)

b.) Sqrt^(6*sqrt^6)/2

c.) Sqrt^(2*sqrt^3)

d.) Sqrt^3

e.) 2

A.  \sqrt{2{\sqrt{6}}}

B.  \frac{\sqrt{6{\sqrt{6}}}}{2}

C.  \sqrt{2{\sqrt{3}}}

D.  \sqrt{3}

E.  2

Bunuel · Accepted Answer

A cylindrical tank has a base with a circumference of 4sqrt^(pi * sqrt^3) meters and an equilateral triangle painted on the interior side of the base. If a stone is dropped inside the tank and the probability of the stone hitting the portion of the base outside the triangle is 3/4, what is the length of a side of the triangle?

a.) Sqrt^(2*sqrt^6)

b.) Sqrt^(6*sqrt^6)/2

c.) Sqrt^(2*sqrt^3)

d.) Sqrt^3

e.) 2

A.  \sqrt{2{\sqrt{6}}}

B.  \frac{\sqrt{6{\sqrt{6}}}}{2}

C.  \sqrt{2{\sqrt{3}}}

D.  \sqrt{3}

E.  2

Given:  circumference=4\sqrt{\pi{\sqrt{3}}}  and  P(out)=\frac{3}{4}

Now, as the probability of the grain of sand landing on the portion of the base  outside  the triangle is 3/4 then the the portion of the base (circle) outside the triangle must be 3/4 of the are of the base and the triangle itself 1/4 of the are of the base.

Square both sides:
  16\pi{\sqrt{3}}=4{\pi}^2{r}^2 ;

4{\sqrt{3}}={\pi}{r}^2 ;

area_{base}=\pi{r^2}=4{\sqrt{3}} .

The area of the equilateral triangle is 1/4 of the area of the base:

area_{equilateral}=\frac{1}{4}*4{\sqrt{3}}=\sqrt{3} ;

Also the area 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}=\sqrt{3} ;

a=2 .

Answer: E.

tinki · Answer

Hello Bunuel

ur explanation is perfect . i just cant understand one thing. i know that formula for the side of equilateral triangle inscribed in the circle 
should be   a= √3 * r   where r is radius and a is side of the triangle, but when using this formula i am not getting the right answer in the above exmpl. what could be the problem? is somth. wrong with formula ? thanks

fluke · Answer

The triangle is not necessarily inscribed because it is not mentioned in the question. It can be any equilateral triangle drawn within the base. The vertices of the triangle may not touch the circle.

shrouded1 · Answer

Let the radius of the base be r
Right so the circumference =  4 \sqrt{\pi * \sqrt{3}}=2 \pi r 
Simplifying we get  r = \frac{4\sqrt{3}}{\sqrt{\pi}} 
Area of base =  \pi r^2 = 4 \sqrt{3}

Let the side of the triangle be x, the area of the triangle is  \frac{\sqrt{3}x^2}{4}

Probably that the stone will fall inside the triangle = 1 - (3/4) = (1/4) = (Area of triangle)/(Area of base)

Substituting the areas calculated above :  \frac{\frac{\sqrt{3}x^2}{4}}{4\sqrt{3}} = \frac{1}{4} 
Which simplifies to give :  x^2=4

Hence  x=2

Answer = E

LalaB · Answer

could u please explain me why we cant use this formula here ? The radius of the circumscribed circle is R=a*\sqrt{3}/3 math-triangles-87197.html

Bunuel · Answer

Because we are told that equilateral triangle is painted so not necessarily inscribed on the interior side of the base.

WarriorGmat · Answer

Hi Bunuel,
can you explain
how can you consider P(out) as fraction of total base?

Posted from my mobile device

Bunuel · Answer

Bigger the area bigger the probability of a grain landing there. P(out)=3/4 simply means that the the portion of the base (circle) outside the triangle must be 3/4 of the are of the base.

ronr34 · Answer

Hi Bunuel,
The equation for the area of a triangle is only for an equilateral inscribed in a circle, is it not? is it for any triangle painted within a circle?

Bunuel · Answer

area_{equilateral}=side^2*\frac{\sqrt{3}}{4} is for ANY EQUILATERAL triangle. Check Triangles chapter of our Math Book: math-triangles-87197.html

Manman2012 · Answer

Hi, I was just looking at this exercise. I am a bit surprised we are expected to solve this in less than 2mn. Do a lot of people succeed to do that ? Just taking into account the time to : 
- read the exercise
- understand it
- maybe draw quickly on your paper
- formulate the approach
- solve
- maybe read again some parts of the problem in the middle to confirm you are on track and not missing important information, that would take more around 2:30-3:00 even for a rapid problem solver. What do you think ?

luffy_ueki · Answer

Circumference = 4\sqrt{\pi{\sqrt{3}} = 2*pi*r
pi*r*r = 4*sqrt(3)/pi

probability of lying outside the painted area = 3/4 
Thus lying inside= 1/4

probability of lying in the painted area = area of equilateral triangle / area of circle= (sqrt(3)*side*side/4) / pi*r*r
=>r =2

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A cylindrical tank has a base with a circumference of

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