Math, asked by Anonymous, 1 month ago

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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Answers

Answered by Anonymous
67

\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 :)

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Answered by roshni542
1

Answer:

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

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