StudyDaddy Geometry

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QUESTION

Feb 03, 2017

At what distance from the base of a right circular cone must a plane be passed parallel to the base in order that the volume of the frustum formed shall be three-fifths of the volume of the given cone? (Ans: 0.26319h)

The plane should cut the cone at a distance of $0.26319h$ from the base.

This kind of problem can appear to be deceptively easy. The best way to solve it is to tell it "Resistance is futile. You will be solved!" :-)

Well, the first thing you should do is draw a diagram as follows:

We know that the volume of a cone with radius $r$ and height $h$ is given by $V=(\pi r^2 h)/3$ .

The volume of the frustum is given by:

$V_f=(\pit^2(k+x))/3 - (\pis^2k)/3$

We need this to be equal to $3/5$ of the total volume:

$(\pit^2(k+x))/3 - (\pis^2k)/3=(\pit^2(k+x))/5$ [A]

The RHS equals $(\pit^2(k+x))/5$ because $3/5$ of the total volume

$V_f=(\pit^2(k+x))/3$ is given by $3/5 * (\pit^2(k+x))/3 = (\pit^2(k+x))/5$ .

Continuing with [A], we have:

$t^2(k+x) - s^2k=(3t^2(k+x))/5$

Divide throughout by $s^2$ :

$(t^2(k+x))/s^2 - k=(3t^2(k+x))/(5s^2)$ [B]

From similar triangles, we know that $k/s = (k+x)/t$

$implies t/s = (k+x)/k implies t^2/s^2 =(k+x)^2/k^2$

Replacing $t^2/s^2$ in [B] with $(k+x)^2/k^2$ gives

$(k+x)^3/k^2 - k=(3(k+x)^3)/(5k^2)$

Multiply throughout by $k^2$ to get:

$(k+x)^3 - k^3=(3(k+x)^3)/5$

Simplifying:

$(k+x)^3/( k^3)=5/2 implies (x+k)/k = (5/2)^(1/3) approx 1.3572088$

From this we see that $x+k approx 1.3572088k$ and $x approx 0.3572088k$ .

We need the ratio $x/(x+k)$ , and so

$x/(x+k) approx (0.3572088k)/(1.3572088k ) approx 0.26319$ as expected.