Question

18. The height of a right circular cone is 20 cm. A small cone is cut off at the
top by a plane parallel to the base. If its volume be y1/8of the volume of
the given cone, at what height above the base is the section made?
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Answer 1

$\huge\bigstar\huge\mathcal{\underline{ \underline{QUESTION}}}\red\bigstar$

The height of a right circular cone is 20 cm. A small cone is cut off at the

top by a plane parallel to the base. If its volume be (1/8)of the volume of

the given cone, at what height above the base is the section made?

$\huge\red\bigstar\huge\mathcal{\underline{ \underline{SOLUTION}}}\bigstar$

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the height(H) of the large cone is =20 cm(given)

let's its base radius and slant height are R and l respectively.

and the base radius and height of the cone be

r and h

therefore volume of the large cone is

$V_{l.cone}=\frac{1}{3}\pi R{}^{2}H$

and the volume of the small cone is,

$V_{s.cone}=\frac{1}{3}\pi r{}^{2}h$

now from the similar triangle principles...

$\frac{r}{h}=\frac{R}{H}\\ \implies r=\frac{Rh}{H}$

now....

according to the problem

$V_{l.cone}=8(V_{s.cone})$

$\implies \frac{1}{3} \pi R{}^{2} H = 8( \frac{1}{3} \pi r {}^{2} h) \\ \implies R{}^{2} H = 8r {}^{2} h \\ \implies R{}^{2} H = 8( \frac{R{}^{2}h {}^{2} }{H {}^{2} }) h \\ \implies H {}^{3} = 8h {}^{3} \\ \implies H = 2h \\ \implies h = \frac{H}{2} \\ \implies h = \frac{\cancel{20}}{\cancel{2}} \: cm \: \: \: \: ( \because H = 20 \: cm) \\ \ \boxed{\implies h = 10 \: cm}$

$\large\mathfrak{...hope\: this \:helps\: you.....}$