Question

The height of a cone is 32 cm. A small cone is cut off at the top by a plane
parallel to its base . If its volume is ⅟64 of the volume of the given cone, at
what height above the base , is the cone cut? (use π = 22/7 )

Answer 1

Let the height, radius and voleume of larger cone be H, R and V
and the height, radius and volume of smaller cone is h, r and v.

Given that H = 32 cm, v/V =1/64, 
We need to find (H-h).

Note that $\frac{h}{H} = \frac{r}{R}$

$\frac{v}{V} = \frac{(1/3) \pi r^2h}{(1/3) \pi R^2H} \\ \frac{1}{64} = \frac{r^2h}{R^2H} \\ \frac{1}{64} = ( \frac{r}{R} )^{2} \frac{h}{H} \\ \frac{1}{64} = ( \frac{h}{H} )^{2} \frac{h}{H} = (\frac{h}{H})^3 \\ \sqrt[3]{\frac{1}{64}} = \frac {h}{H}\\\frac {h}{H}= \frac{1}{4} \\ \\h= \frac{H}{4}= \frac{32}{4}= 8\ cm\\ \\ thus\ (H-h) =32-8 = \boxed {24\ cm}$

Answer 2

Answer:

Let the height, radius and voleume of larger cone be H, R and V

and the height, radius and volume of smaller cone is h, r and v.

Given that H = 32 cm, v/V =1/64,

We need to find (H-h).

Note that \frac{h}{H} = \frac{r}{R}

H

h

=

R

r

\begin{lgathered}\frac{v}{V} = \frac{(1/3) \pi r^2h}{(1/3) \pi R^2H} \\ \frac{1}{64} = \frac{r^2h}{R^2H} \\ \frac{1}{64} = ( \frac{r}{R} )^{2} \frac{h}{H} \\ \frac{1}{64} = ( \frac{h}{H} )^{2} \frac{h}{H} = (\frac{h}{H})^3 \\ \sqrt[3]{\frac{1}{64}} = \frac {h}{H}\\\frac {h}{H}= \frac{1}{4} \\ \\h= \frac{H}{4}= \frac{32}{4}= 8\ cm\\ \\ thus\ (H-h) =32-8 = \boxed {24\ cm}\end{lgathered}

V

v

=

(1/3)πR

2

H

(1/3)πr

2

h

64

1

=

R

2

H

r

2

h

64

1

=(

R

r

)

2

H

h

64

1

=(

H

h

)

2

H

h

=(

H

h

)

3

3

64

1

=

H

h

H

h

=

4

1

h=

4

H

=

4

32

=8 cm

thus (H−h)=32−8=

24 cm