Question

the height and the radius of the base of a right circular cone is how is half of the corresponding height and radius of another big account find the ratio of their Volume II the ratio of their at the surface area

Answer 1

height of big account = h
radius " ". ". =r
heigh of another cone =h 1
radius " ". ". =r 1
so ratio of volume =

Answer 2

$\longrightarrow \: \: \: \: \: \: \: \: { \purple{{ \underline{ \mathbb{ \: \: \: \: Q}{ \rm{uestion↓ \: \: \: \: }}}}}}$

The height and the radius of the base of a right circular cone is half the corresponding height and radius of another bigger one. Find:

(i) the ratio of their volumes.

(ii) the ratio of their lateral surface areas.

$\longrightarrow \: \: \: \: \: \: \: \: { \purple{{ \underline{ \mathbb{ \: \: \: \:A }{ \rm{nswer↓ \: \: \: \: }}}}}}$

i) Let the radius and height of a bigger cone be r and h.

A/Q, the height and radius of the smaller cone is half the corresponding height and radius of bigger one.

∴ Radius of smaller cone = 1/2 r

Height of smaller cone = 1/2 h.

$\rm \: ∴ \: Ratio \: of \: their \: volumes = \frac{volume \: of \: smaller \: cone}{volume \: of \: bigger \: cone}$

$\rm \: ➺ \: \frac{ \frac{1}{3}\pi {r}^{2}h }{ \frac{1}{3} \pi {r}^{2}h }$

$\rm \: ➺ \: \frac{ \frac{1}{3} \: .\cancel{\pi} \: . { (\frac{1}{2} r)}^{2}. \: \frac{1}{2} h }{ \frac{1}{3} \: . \cancel{\pi }. \: {r}^{2} .\: h }$

$\rm \: ➺ \: \frac{ \frac{1}{24} \: \cancel{ {r}^{2} h }}{ \frac{1}{3} \: \cancel{ {r}^{2} h} }$

$\rm \: ➺ \: \frac{1 \times \cancel{3}}{ \cancel{24} {} \: ^{8} \times 1} \: \: = \: \: \frac{1}{8} \: \: \: \: \: \: { \pink{ \boxed{ \implies{1 : 8}}}}$

________________________

ii) Let the slant height of smaller cone be l and bigger cone be L.

$\rm \: ∴ \: l = \sqrt{ {r}^{2} + {h}^{2} } \\ \\ \rm \: \: \: \: \: \: \: \: \: \: = \sqrt{ \frac{ {r}^{2} }{2} + \frac{ {h}^{2} }{2} }$

$\rm \: ∴ L = \sqrt{ {r}^{2} + {h}^{2} }$

We can also write that,

$\rm \: l = \dfrac{ L}{2} \: \: \: \: \: \: \: \: \: \: \: \: { \tt { \gray{ (\sqrt{ {r}^{2} + {h}^{2} } = L)}}}$

$\rm \: ∴ \: Ratio \: of \: lateral \: surface \: area = \frac{lateral \: surface \: area \: of \: smaller \: cone}{lateral \: surface \: area \: of \: bigger \: cone}$

$\rm \:➺ \: \frac{\pi r l}{\pi rL}$

$\rm \:➺ \: \frac{ \cancel{\pi}. \: \frac{1}{2} \cancel{ r}. \: \frac{ \cancel{L}}{2} }{ \cancel{\pi}. \: \cancel{r}. \: \cancel{ L}} \: = \: \frac{ \frac{1}{4} }{1} = \frac{1}{4} \: \: \: \: \: { \boxed{ \pink{= 1: 4}}}$