Question

4. If the volume of a right circular cone of height 9 cm is 48 r cm, find
the diameter of its base.​

Answer 1

Given -

• Volume of right circular cone = 48 cm
• Height (h) of right circular cone = 9 cm

To find -

• The diameter of the base

Solution -

We know that,

$Volume \: of \: cone = \frac{1}{3} \pi {r}^{2} h$

By putting the values we will get -

$48 = \frac{1}{3} \times \frac{22}{7} \times {r}^{2} \times 9$

$48 = \frac{66}{7} \times {r}^{2}$

${r}^{2} = 48 \div \frac{66}{7}$

${r}^{2} = 48 \times \frac{7}{66}$

${r}^{2} = \frac{56}{11}$

${r}^{2} = 5.1$

$r = \sqrt{5.1}$

$r = 2.3$

We know that,

Diameter =2×radius(r)

=2×2.3

=4.6

Therefore the diameter of the base is 4.6 cm

Related Formulae -

$1)Volume \: of \: cone = \frac{1}{3} \pi {r}^{2} h$

$2)C. S. A \: of \: cone = \pi \: rl$

$3)T. S. A \: of \: cone = \pi \: r(l + r)$

Answer 2

Answer:

Given -

Volume of right circular cone = 48 cm

Height (h) of right circular cone = 9 cm

To find -

The diameter of the base

Solution -

We know that,

Volume \: of \: cone = \frac{1}{3} \pi {r}^{2} hVolumeofcone=

31πr 2 h

By putting the values we will get -

48 = \frac{1}{3} \times \frac{22}{7} \times {r}^{2} \times 948=

31 × 722 ×r 2×9

48 = \frac{66}{7} \times {r}^{2}48=

766×r 2

{r}^{2} = 48 \div \frac{66}{7}r

2 =48÷ 766

{r}^{2} = 48 \times \frac{7}{66}r

2=48× 667

{r}^{2} = \frac{56}{11}r

2= 1156

{r}^{2} = 5.1r 2=5.1

r = \sqrt{5.1}r= 5.1

r = 2.3r=2.3

We know that,

Diameter =2×radius(r)

=2×2.3

=4.6

Therefore the diameter of the base is 4.6 cm

Related Formulae -

1)Volume \: of \: cone = \frac{1}{3} \pi {r}^{2} h1)Volumeofcone=

31 πr 2 h

2)C. S. A \: of \: cone = \pi \: rl2)C.S.Aofcone=πrl

3)T. S. A \: of \: cone = \pi \: r(l + r)3)T.S.Aofcone=πr(l+r)