Melting a Sphere into Cones: Volume and Similar Solids

Grades 10-11

Interactive Melting a Sphere into Cones: Volume and Similar Solids graph — click to explore — Click the graph to explore it interactively →

Question: Give me the graph

To solve this problem, we first determine the dimensions of the spheres using the properties of similar solids, then apply the principle of conservation of volume to analyze the resulting cones.

(a) Volume of the Smaller Sphere

For any two similar solids, the ratio of their surface areas is equal to the square of the ratio of their linear dimensions (such as radii). Let

r

be the radius of the smaller sphere and

R

be the radius of the larger sphere.

\begin{align*} \frac{\text{Surface Area (smaller)}}{\text{Surface Area (larger)}} &= \left( \frac{r}{R} \right)^2 \\ \frac{4}{9} &= \left( \frac{r}{9} \right)^2 \\ \sqrt{\frac{4}{9}} &= \frac{r}{9} \\ \frac{2}{3} &= \frac{r}{9} \\ r &= 6 \text{ cm} \end{align*}

Now, we calculate the volume of the smaller sphere using the formula

V = \frac{4}{3}\pi r^3

\begin{align*} V_{\text{smaller}} &= \frac{4}{3} \pi (6)^3 \\ V_{\text{smaller}} &= \frac{4}{3} \pi (216) \\ V_{\text{smaller}} &= 288\pi \text{ cm}^3 \end{align*}

(b) Similarity of Cones A and B

First, we find the total volume of the metal available from both spheres:

\begin{align*} V_{\text{larger}} &= \frac{4}{3} \pi (9)^3 = 972\pi \text{ cm}^3 \\ V_{\text{total}} &= 288\pi + 972\pi = 1260\pi \text{ cm}^3 \end{align*}

Next, we calculate the volume of Cone

A

(

r_A = 6, h_A = 10

\begin{align*} V_A &= \frac{1}{3} \pi (r_A)^2 h_A \\ V_A &= \frac{1}{3} \pi (6)^2 (10) = 120\pi \text{ cm}^3 \end{align*}

Since the total metal is recast into Cones

A

and

B

, the volume of Cone

B

is:

\begin{align*} V_B &= V_{\text{total}} - V_A \\ V_B &= 1260\pi - 120\pi = 1140\pi \text{ cm}^3 \end{align*}

Given the base radius of Cone

B

12

cm (

r_B = 12

), we find its height

h_B

\begin{align*} 1140\pi &= \frac{1}{3} \pi (12)^2 h_B \\ 1140 &= 48 h_B \\ h_B &= 23.75 \text{ cm} \end{align*}

To check if the cones are similar, the ratio of their radii must equal the ratio of their heights:

\begin{align*} \frac{\text{Radius}_A}{\text{Radius}_B} &= \frac{6}{12} = 0.5 \\ \frac{\text{Height}_A}{\text{Height}_B} &= \frac{10}{23.75} \approx 0.421 \end{align*}

Since

0.5 \neq 0.421

, the ratios are not equal. Therefore, the claim is incorrect; Cones

A

and

B

are not similar.

Visual Guide

The diagram illustrates the "Before and After" state of the metal.

On the left, you can see the two spheres. The smaller sphere (teal) has a radius of $6$ cm, while the larger sphere (navy) has a radius of $9$ cm.
On the right, the recast cones are shown. Cone A (teal) maintains a radius of $6$ cm but has a height of $10$ cm. Cone B (maroon) has a larger radius of $12$ cm and a calculated height of $23.75$ cm.
Note that while Cone B is wider and taller, its proportions are different from Cone A, which is why they are not similar solids.