Question

1. The volume of cone is 18480 cubic cm if the height of the cone is 40 cm find the radius of the base​

Answer 1

• The radius of the base of a cone = 21 cm.

Given :

• The volume of the cone = 18480 cm³.
• The height of the cone = 40 cm.

To Find :

• The radius of the base of a cone.

Solution :

Given,

The volume of the cone is 18480 cm³.

We know that,

$\huge{ \pink\star} \: \: \large \boxed{\bf v = \dfrac{1}{3} \times \pi \times {r}^{2} \times h}$

Where,

• V = volume = 18480 cm³.
• π = 22 / 7.
• r = radius = ?
• h = height = 40 cm.

$\bf \implies 18480 \: {cm}^{3} = \dfrac{1}{3} \times \dfrac{22}{7} \times {r}^{2} \times 40 \: cm$

$\bf \implies 18480 \: {cm}^{3} = \dfrac{1}{3} \times \dfrac{22}{7} \times {r}^{2} \times 40 \: cm \\ \\ \\ \bf \implies \dfrac{18480 \: {cm}^{3} }{40 \: cm} = \dfrac{1}{3} \times \dfrac{22}{7} \times {r}^{2} \\ \\ \\ \bf \implies462 \: {cm}^{2} = \dfrac{1}{3} \times \dfrac{22}{7} \times {r}^{2} \\ \\ \\ \bf \implies \dfrac{462 \: {cm}^{2} }{( \dfrac{1}{3}) } = \dfrac{22}{7} \times {r}^{2} \\ \\ \\ \bf \implies 462 \: {cm}^{2} \times \dfrac{3}{1} = \dfrac{22}{7} \times {r}^{2}$

$\bf \implies 1386 \: {cm}^{2} = \dfrac{22}{7} \times {r}^{2} \\ \\ \\ \bf \implies \dfrac{1386 \: {cm}^{2} }{( \dfrac{22}{7})} = {r}^{2} \\ \\ \\ \bf \implies 1386 \: {cm}^{2} \times \dfrac{7}{22} = {r}^{2} \\ \\ \\ \bf \implies 693 \: {cm}^{2} \times \frac{7}{11} = {r}^{2} \\ \\ \\ \bf \implies 63 \: {cm}^{2} \times 7 = {r}^{2} \\ \\ \\ \bf \implies 441 \: {cm}^{2} = {r}^{2} \\ \\ \\ \bf \implies \sqrt{441 \: {cm}^{2} } = r \\ \\ \\ \bf \implies 21 \: cm = r \\ \\ \\ \: \: \therefore \bf\: \: r = 21 \: cm$

Hence,

The radius of the base of a cone is 21 cm.

Verification :

Given,

The volume of the cone is 18480 cm³.

We know that,

$\huge{ \pink\star} \: \: \large \boxed{\bf v = \dfrac{1}{3} \times \pi \times {r}^{2} \times h}$

Where,

• V = volume = 18480 cm³.
• π = 22 / 7.
• r = radius = 21 cm.
• h = height = 40 cm.

$\bf \implies 18480 \: {cm}^{3} = \dfrac{1}{3} \times \dfrac{22}{7} \times {(21 \: cm)}^{2} \times 40 \: cm \\ \\ \\ \bf \implies 18480 \: {cm}^{3} = \dfrac{1}{ \not3} \times \dfrac{22}{7} \times 21 \: cm \times \not21 \: cm \times 40 \: cm \\ \\ \\ \bf \implies 18480 \: {cm}^{3} = 1 \times \dfrac{22}{ \not7} \times 21 \times 7 \times 40 \: {cm}^{3} \\ \\ \\ \bf \implies 18480 \: {cm}^{3} = 22 \times 21 \times 40 \: {cm}^{3} \\ \\ \\ \bf \implies 18480 \: {cm}^{3} = 462 \times 40 \: {cm}^{3} \\ \\ \\ \bf \implies 18480 \: {cm}^{3} = 18480 \: {cm}^{3}$

Hence Verified !

Answer 2

$\huge\bold{\underline{Question}}$

The volume of cone is 18480 cubic cm if the height of the cone is 40 cm find the radius of the base.

$\huge\bold{\underline{Answer}}$

$\sf\large\underline\red{Given:}$

• The volume of cone is 18480 cubic cm

• The height of the cone is 40 cm.

$\sf\large\underline\red{To\:find:}$

Find the radius of the base.

$\sf\large\underline\red{Solution:}$

Let the radius of the cone be ' r ' cm

Height ( h ) = 40 cm

Volume ( v ) = 18480 cm³

We know that,

$\boxed{\bf{\blue{Volume\:of\:a\:right\:circular\:cone\:=\dfrac{1}{3}πr²h}}}$

Where,

v ⭆ Volume

r ⭆ The radius of Circular Base

h ⭆ height of cone

π ⭆22/7

$\sf\green{★\:According\:to\:the\:question}$

$\sf{\implies \dfrac{1}{3} × π × r² × 40 = 18480}$

$\sf{\implies \dfrac{1}{3} × \dfrac{22}{7} × r² × 40 = 18480}$

$\sf{\implies r² × \dfrac{880}{21} = 18480}$

$\sf{\implies r² = 18480 × \dfrac{21}{880}}$

$\sf{\implies r² = 21×21}$

$\sf{\implies r² = 441}$

$\sf{\implies r = 21}$

$\huge\bold{\boxed{\rm{\pink{r\:=21\:cm}}}}$

Therefore,

The radius of the base of the cone is 21 cm .

__________________________________