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Re: A is the center of the top face of the right circular cylinder in the [#permalink]
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SOLUTION:

Let us assume ∠ACB=x, therefore ∠BAC=4x,
∠ABC=x (△ABC is Isosceles with AC=AB (radii))

In triangle ABC,x+4x+x=180° , x= 30° ,
Thus, ∠BAC=4*30=120°

Volume is proportional to Cross sectional area(Circle for the entire cylinder and a sector for the shaded part).
This cross sectional area is proportional to angle which is 360 for the complete cylinder and
120 for the sector that represents the cross section area of the triangle.


Hence, Volume proportion of shaded region to entire Cylinder
= Angle of the sector/Angle of the full circle

=120/360 =1/3 (E)

Hope this helps :thumbsup:
Devmitra Sen(Math)
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Re: A is the center of the top face of the right circular cylinder in the [#permalink]
why did we consider volume of cylinder as 360 degree and volume of shaded part as 120 degree ? Please give detail explanation
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Re: A is the center of the top face of the right circular cylinder in the [#permalink]
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Bunuel wrote:

A is the center of the top face of the right circular cylinder in the figure above. If the degree measure of \(∠BAC\) is four times that of \(∠ACB\), and the height of the cylinder is equal to the diameter of its base, then the volume of the shaded region is what fraction of the volume of the entire cylinder?


A. 1/7
B. 1/6
C. 1/5
D. 1/4
E. 1/3








Solution:

Triangle ABC is an isosceles triangle because AB and AC are each a radius of the circle. Since ∠BAC is four times that of ∠ACB, we see that ∠BAC is 120 degrees (and ∠ACB and ∠ABC are each 30 degrees). Since ∠BAC is 120 degrees, the sector ABC is ⅓ of the area of the top face of the cylinder, and therefore, the the volume of the shaded region will also be ⅓ of the volume of the cylinder.

Answer: E
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Re: A is the center of the top face of the right circular cylinder in the [#permalink]
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Re: A is the center of the top face of the right circular cylinder in the [#permalink]
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