The height of a cone is 20 cm.


The height of a cone is 20 cm. A small cone is cut off from the top by a plane parallel to the base. If its volume be 1/125 of the volume of the original cone, determine at what height above the base the section is made.


We have the following situation as shown in the figure

Let VAB be a cone of height h1 VO1 =20cm. Then from the symmetric triangles VO 1A and VOA1, we have

$\frac{\mathrm{VO}_{1}}{\mathrm{VO}}=\frac{\mathrm{O}_{1} \mathrm{~A}}{\mathrm{OA}_{1}}$

$\Rightarrow \frac{20}{V O}=\frac{O_{1} A}{O A_{1}}$

It is given that, volume of the cone $\mathrm{VA}_{1} \mathrm{O}$ is $\frac{1}{125}$ times the volume of the cone $\mathrm{VAB}$. Hence, we have

$\frac{1}{3} \pi \mathrm{OA}_{1}{ }^{2} \times \mathrm{VO}=\frac{1}{125} \times \frac{1}{3} \pi \mathrm{O}_{1} \mathrm{~A}^{2} \times 20$

$\Rightarrow\left(\frac{\mathrm{OA}_{1}}{\mathrm{O}_{1} \mathrm{~A}}\right)^{2} \times \mathrm{VO}=\frac{4}{25}$

$\Rightarrow\left(\frac{\mathrm{VO}}{20}\right)^{2} \times \mathrm{VO}=\frac{4}{25}$

$\Rightarrow \quad \mathrm{VO}^{3}=\frac{400 \times 4}{25}$

$\Rightarrow \quad \mathrm{VO}^{3}=16 \times 4$

$\Rightarrow \quad \mathrm{VO}=4$

Hence, the height at which the section is made is 20 − 4 = 16 cm.

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