Question

A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. Find the radius of the sphere.

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Answer 1

Solution:-

Given:-

• Height of the cone is 8.4 cm.
• Base of the cone is 2.1 cm.
• The cone is melted and recast into a sphere.

To Find:-

The radius of the sphere.

Formula:-

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

( Where r is radius and h is height of the cone.)

$Volume \: of \: a \: sphere = \frac{4}{3} \pi \: {r}^{3}$

( Where r is the radius of the sphere.)

Concept zone:-

The cone is melted and recast into a sphere . So, naturally their volumes will be same .

By the Problem:-

$Volume \: of \: a \: cone = \frac{1}{3} \pi \: {r}^{2} h \\ \\ \implies \: \frac{1}{3} \times \frac{22}{7} \times {(2.1)}^{2} \times 8.4$

So, Volume of the sphere is also

$\implies \: \frac{1}{3} \times \frac{22}{7} \times {(2.1)}^{2} \times 8.4$

$Volum e \: of \: the \: Sphere= \frac{1}{3} \times \frac{22}{7} \times {(2.1)}^{2} \times 8.4 \\ \\ \implies \: \frac{4}{3} \pi \: {r}^{3} = \frac{1}{3} \times \frac{22}{7} \times {(2.1)}^{2} \times 8.4 \: \\ \\ \implies \: {r}^{3} = \frac{1}{3} \times \frac{22}{7} \times {(2.1)}^{2} \times 8.4 \div \frac{4}{3} \div \frac{22}{7} \\ \\ \implies \: {r}^{3} = \frac{1}{\cancel 3} \times \cancel\frac{22}{7} \times {(2.1)}^{2} \times \cancel{8.4 } \times \frac{3}{ \cancel 4} \times \cancel\frac{7}{22} \\ \\ \implies \: {r}^{3} = {(2.1)}^{2} \times 0.7 \times 3 \: \\ \\ \implies \: {r}^{3} = {(2.1)}^{2} \times 2.1 \\ \\ \implies{r}^{3} = {(2.1)}^{3} \\ \\ \implies \: r = 2.1 \: cm \: (cancel \: cubes \: both \: sides)$

Answer:-

$\\ \implies \boxed{ Radius = 2.1 \: cm\:}$

Answer 2

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