Question

If the volume of a right circular cone of height 9cm is 48<br />$\pi$<br />cm3. find the diameter of the base <br /><br />​

Answer 1

Answer:

$\large\boxed{\sf{8\:cm}}$

Explanation:

Let the diameter of base of right circular cone is 'd' cm.

Therefore, radius will be $\bold{\dfrac{d}{2}\:cm}$.

It's given that the volume is $\bold{48 \pi \:{cm}^{3}}$

And, height is 9 cm

But, Volume of cone is given by ,

$\large \boxed{ \sf{v = \frac{1}{3} \pi {r}^{2} h}}$

Therefore, we have the relation,

$\sf{= > \frac{1}{3} \times \pi \times {( \frac{d}{2}) }^{2} \times 9 = 48\pi }\\ \\ \sf{ = > \frac{3\pi {d}^{2} }{4} = 48\pi }\\ \\ \sf{ = > {d}^{2} = \frac{4 \times 48\pi}{3\pi}} \\ \\ \sf{= > {d}^{2} = 4 \times 16 = 64}\\ \\ \sf{ = > {d}^{2} = {8}^{2} }\\ \\ \sf{ = > d = 8 \: cm}$

Hence, diameter of the base is 8 cm

Answer 2

$\huge\underline\mathrm{SOLUTION:-}$

AnswEr:

• Diameter of the base = 8 Cm.

Given That:

• Height of the cone = 9 Cm
• Volume of cone = 48π Cm³

Need To Find:

• Diameter of the base = ?

ExPlanation:

Let, Radius of the cone = r Cm

Formula used here:

• Volume of cone = 1/3πr²h

Putting the values according to the given formula:

➠ $\mathsf {\frac{1}{3}\pi r^2 h = 48\pi}$

➠ $\mathsf {\frac{1}{3}\pi r^2 \times (9) = 48\pi}$

➠ $\mathsf {\pi r^23 = 48\pi}$

➠ $\mathsf {r^2 = \frac{48\pi}{3\pi} }$

➠ $\mathsf {r^2 = 16}$

➠ $\mathsf {r = \sqrt{16} }$

➠ $\mathsf {r =\sqrt{(4)^2} }$

➠ $\mathsf {r = 4\: Cm}$

Now, again formula used here:

• Diameter = 2 × Radius

Putting the values according to the given formula:

➠ Daimeter = 2 × 4

➠ Diameter = 8 Cm

ThereFore:

• Diameter of the base = 8 Cm.

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What is Radius?

• Radius is a line segment joining the centre of a circle to any point on the circle.

• $\mathsf {Radius = \frac{1}{2}\: Diameter}$

What is Diameter?

• Diameter is a line segment whose in points lie on the circle and which passes through the centre of the circle.

• Diameter = 2 × Radius

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