Question

The height of a cone is 30 cm. A small cone is cut-off at the top by a plane parallel to the base. If its volume be $\small\tt\dfrac{1}{27}$ of the volume of the given cone, then the height above the base, where the section is made, is ?

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Answer 1

$\huge\sf\underline\blue{Answer:-}$

Given,

height of the cone = 30cm

Let,

the radius of big cone = R

and, the radius of small cone = r

We know that, $\tt{\frac{1}{3} \pi \: r ^{2} h}$

Volume of the big cone

$\implies$ $\tt{v^ {1} = \frac{1}{3}\pi r R \times 30}$

$\implies$ $\tt\bold{10cm^3}$

Volume of the small cone

$\implies$ $\tt{v^{2}= \frac{1}{3}\pi r^{2}h}$

Solution,

As given, $\tt{v ^{2} = \frac{1}{27} \times v ^{1}}$

$\implies$ $\tt{\frac{{v}^{2} }{ {v}^{1} } = \frac{1}{27}}$

$\implies$ $\tt{\frac{ \frac{1}{3}\pi rh }{10 \times \pi rh} = \frac{1}{27}}$

$\implies$ $\tt{({\frac{r}{R}})^{2} \times \frac{h}{30} = \frac{1}{27}-----> (1)}$

From the figure,

△ACD∼△AOB (AA similarity criterion)

$\implies$ $\tt{({\frac{r}{R}}) \times \frac{h}{30} = \frac{1}{27}}$

Hence $\tt\small{{eq}^{n} (1)}$, will become

$\implies$ $\tt{(\frac{h}{30} )^{2} \times \frac{h}{30} = \frac{1}{27}}$

$\implies$ $\tt{( \frac{h}{30})^{2} = ( \frac{1}{3}) ^{2}}$

$\implies$ $\tt{\frac{h}{30} = \frac{1}{3}}$

∴ h = 10 cm.

~ 30 - 10 = 20cm.

Thus the cut is at a height 20 cm above the base of small cone.

[For the figure you may refer to the attachment]

Hope it helps :)

Answer 2

Answer:

$\huge \tt{refer \: the \: attachment}$