Bunuel wrote:

The volume of a cone is S × H/3, where S is the area of the base and H is the height. According to the figure above, if AB = BC, what is the ratio of the volume of the right section of the cone (with AB as height) to the volume of the left section of the cone (with BC as height)? (Note: Figure not drawn to scale.)

A. 1 : 3

B. 1 : 4

C. 1 : 6

D. 1 : 7

E. 1 : 8

Kudos for a correct solution.Let AB = BC = h. Then AC = 2h

Let the line extended from B meet at D and from C meet at E as shown in fig below

Attachment:

cone-1.gif [ 6.52 KiB | Viewed 1661 times ]
Triangle ABD and Triangle ACE are similar triangles ( ∠BAD = ∠ CAE , ∠B = ∠C =90)

Thus \(\frac{AB}{AC}\) = \(\frac{1}{2}\)

And \(\frac{BD}{CE}\) = \(\frac{1}{2}\) ==> \(\frac{x}{R}\) = \(\frac{1}{2}\)

R = 2x

volume of the right section of the cone (with AB as height) = π\(r^2\) * \(\frac{AB}{3}\)

= π\(x^2\) * \(\frac{h}{3}\)

volume of the left section of the cone (with BC as height) = volume of the left section of the cone (with AC as height) - volume of the right section of the cone (with AB as height)

volume of the left section of the cone (with AC as height) = π\(r^2\) * \(\frac{AC}{3}\)

=π\(R^2\) * \(\frac{2h}{3}\)

=π\(2x^2\) * \(\frac{2h}{3}\)

=π\(x^2\) * \(\frac{8h}{3}\)

Thus volume of the left section of the cone (with BC as height) = π\(x^2\) * \(\frac{8h}{3}\) - π\(x^2\) * \(\frac{h}{3}\)

= π\(x^2\) * \(\frac{7h}{3}\)

Thus Ratio = 1:7

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