Understanding the Problem: Melting and Recasting Spheres

This problem involves the principle of conservation of volume. When a solid object, like a sphere, is melted and reshaped into other objects, the total volume of the material remains the same. Here, a large solid sphere is melted and recast into three smaller spheres, referred to as "shorts". We are given the dimensions (diameters) of the original sphere and two of the resulting shorts. We need to find the surface area of the third short.

Key Geometric Formulas for Spheres

To solve this problem, we need the formulas for the volume and surface area of a sphere:

• Volume of a sphere with radius \(r\): \(V = \frac{4}{3}\pi r^3\)
• Surface area of a sphere with radius \(r\): \(A = 4\pi r^2\)

Remember that the radius \(r\) is half of the diameter \(d\), so \(r = d/2\).

Calculating Volumes

First, let's calculate the radius and volume of the original large sphere and the two given smaller spheres.

Large Sphere (Original)

• Diameter \(D_{large} = 12\) cm
• Radius \(R_{large} = \frac{12}{2} = 6\) cm
• Volume \(V_{large} = \frac{4}{3}\pi (R_{large})^3 = \frac{4}{3}\pi (6)^3 = \frac{4}{3}\pi (216)\)
• \(V_{large} = 4\pi (72) = 288\pi\) cm\(^3\)

First Small Sphere (Short 1)

• Diameter \(d_1 = 6\) cm
• Radius \(r_1 = \frac{6}{2} = 3\) cm
• Volume \(V_1 = \frac{4}{3}\pi (r_1)^3 = \frac{4}{3}\pi (3)^3 = \frac{4}{3}\pi (27)\)
• \(V_1 = 4\pi (9) = 36\pi\) cm\(^3\)

Second Small Sphere (Short 2)

• Diameter \(d_2 = 10\) cm
• Radius \(r_2 = \frac{10}{2} = 5\) cm
• Volume \(V_2 = \frac{4}{3}\pi (r_2)^3 = \frac{4}{3}\pi (5)^3 = \frac{4}{3}\pi (125)\)
• \(V_2 = \frac{500}{3}\pi\) cm\(^3\)

Finding the Volume of the Third Sphere

Let the volume of the third short be \(V_3\). According to the principle of conservation of volume:

\[ V_{large} = V_1 + V_2 + V_3 \] \[ 288\pi = 36\pi + \frac{500}{3}\pi + V_3 \]

Now, we solve for \(V_3\):

\[ V_3 = 288\pi - 36\pi - \frac{500}{3}\pi \] \[ V_3 = (288 - 36)\pi - \frac{500}{3}\pi \] \[ V_3 = 252\pi - \frac{500}{3}\pi \] \[ V_3 = \left(252 - \frac{500}{3}\right)\pi \] To subtract the fractions, find a common denominator, which is 3:

\[ V_3 = \left(\frac{252 \times 3}{3} - \frac{500}{3}\right)\pi \] \[ V_3 = \left(\frac{756}{3} - \frac{500}{3}\right)\pi \] \[ V_3 = \frac{756 - 500}{3}\pi \] \[ V_3 = \frac{256}{3}\pi \) cm\(^3\)

Finding the Radius of the Third Sphere

Now that we have the volume of the third short, we can find its radius, \(r_3\), using the volume formula \(V_3 = \frac{4}{3}\pi r_3^3\).

\[ \frac{256}{3}\pi = \frac{4}{3}\pi r_3^3 \] Divide both sides by \(\frac{4}{3}\pi\):

\[ \frac{256}{3} \div \frac{4}{3} = r_3^3 \] \[ \frac{256}{3} \times \frac{3}{4} = r_3^3 \] \[ \frac{256}{4} = r_3^3 \] \[ 64 = r_3^3 \] To find \(r_3\), take the cube root of 64:

\[ r_3 = \sqrt[3]{64} \] \[ r_3 = 4 \) cm

Calculating the Surface Area of the Third Sphere

Finally, we calculate the surface area of the third short using its radius \(r_3 = 4\) cm and the surface area formula \(A_3 = 4\pi r_3^2\).

\[ A_3 = 4\pi (4)^2 \] \[ A_3 = 4\pi (16) \] \[ A_3 = 64\pi \) cm\(^2\)

Summary of Results

The surface area of the third short is \(64\pi\) cm\(^2\).

Revision Table: Sphere Formulas

Additional Information: Conservation of Volume

The concept of conservation of volume is fundamental in problems involving melting and recasting solids. It states that the total amount of substance (and therefore its volume, assuming density is constant) does not change during a phase transition like melting or when its shape is altered. This principle applies whether the solid is reshaped into a single new object or multiple objects, as in this problem where a sphere is melted and forms three smaller spheres. The sum of the volumes of the new shapes equals the volume of the original shape. This is a crucial idea in many geometry and mensuration problems.