Question

31. A cone of base radius 4 cm is divided into two parts by drawing a plane through
the mid-points of its height and parallel to its base. Compare the volume of the
two parts.​

Answer 1

$\underline{\huge{\underline{Given:-}}}$

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Radius of cone → 4 cm

Let the height of original cone → h cm

It is divided into two parts through the midpoint of it's axis and parallel to it's base .

So, A cone ADE at the top and a frustum DGCE is formed .

So, the height of small cone → $\frac{h}{2}cm$

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$\underline{\huge{\underline{Sol:-}}}$

In ∆ AFE and ∆ ABC

Angle AFE → Angle ABC ( Each 90° )

Angle FAE → Angle BAC ( Same angle )

So, ∆ AFE ≈ ∆ ABC ( By AA similarity )

So, the ratio of their corresponding sides will be equal.

$\frac{ \frac{h}{2} }{h} = \frac{ FE }{ BC } \\ \\ \frac{1}{2} = \frac{FE}{4} \\ \\ FE = 2 \: cm$

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Volume of cone = $\frac{1}{3} \pi\: {r}^{2} \:h$

Volume of cone AGC = $\frac{1}{3} \pi\: {4}^{2} \:h$

= $\frac{16}{3} \pi\: h \: {cm}^{3}$

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Volume of cone ADE = $\frac{1}{3} \pi\: {2}^{2} \:h$

= $\frac{4}{3} \pi\:\frac{h}{2}\: {cm}^{3}$

= $\frac{2}{3} \pi\: h\: {cm}^{3}$

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$Volume of frustum = \frac{16}{3} \pi\: h - \frac{2}{3} \pi\: h$

$= \frac{14}{3} \pi \: h$

Ratio of two parts =

=$\frac{2}{3} \pi\: h : \frac{14}{3} \pi \: h$

= 1 : 7

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