Solving the Remoulding Geometry Problem

This problem involves understanding how the volume of a solid shape remains constant when it is remoulded into another shape. We are given the dimensions of a cone and asked to find the radius of a sphere formed by remoulding that cone. The key principle here is conservation of volume.

Step 1: Calculate the Radius of the Cone's Base

The base of a cone is a circle. The area of the base is given as \(64\pi\) cm\(^2\). The formula for the area of a circle is \(A = \pi r^2\), where \(r\) is the radius.

Given base area = \(64\pi\).

So, \(\pi r_{cone}^2 = 64\pi\).

Dividing both sides by \(\pi\), we get:

\(r_{cone}^2 = 64\)

Taking the square root of both sides:

\(r_{cone} = \sqrt{64} = 8\) cm.

The radius of the cone's base is 8 cm.

Step 2: Calculate the Height of the Cone

We are given the slant height (\(l_{cone}\)) of the cone as 17 cm and we just found the base radius (\(r_{cone}\)) is 8 cm. The height (\(h_{cone}\)), radius, and slant height of a cone form a right-angled triangle, with the slant height as the hypotenuse. We can use the Pythagorean theorem:

\(l_{cone}^2 = r_{cone}^2 + h_{cone}^2\)

Substitute the known values:

\(17^2 = 8^2 + h_{cone}^2\)

\(289 = 64 + h_{cone}^2\)

Subtract 64 from both sides:

\(h_{cone}^2 = 289 - 64\)

\(h_{cone}^2 = 225\)

Taking the square root of both sides:

\(h_{cone} = \sqrt{225} = 15\) cm.

The height of the cone is 15 cm.

Step 3: Calculate the Volume of the Cone

The formula for the volume of a cone is \(V_{cone} = \frac{1}{3}\pi r_{cone}^2 h_{cone}\).

Substitute the values for \(r_{cone}\) and \(h_{cone}\):

\(V_{cone} = \frac{1}{3}\pi (8^2)(15)\)

\(V_{cone} = \frac{1}{3}\pi (64)(15)\)

Simplify the calculation:

\(V_{cone} = \pi (64)\left(\frac{15}{3}\right)\)

\(V_{cone} = \pi (64)(5)\)

\(V_{cone} = 320\pi\) cm\(^3\).

The volume of the cone is \(320\pi\) cm\(^3\).

Step 4: Equate Cone Volume and Sphere Volume

When the cone is remoulded into a solid sphere, the volume of the material remains the same. Therefore, the volume of the sphere is equal to the volume of the cone.

\(V_{sphere} = V_{cone}\)

\(V_{sphere} = 320\pi\) cm\(^3\).

Step 5: Calculate the Radius of the Sphere

The formula for the volume of a sphere is \(V_{sphere} = \frac{4}{3}\pi r_{sphere}^3\), where \(r_{sphere}\) is the radius of the sphere.

We have \(V_{sphere} = 320\pi\). So, we set up the equation:

\(\frac{4}{3}\pi r_{sphere}^3 = 320\pi\)

Divide both sides by \(\pi\):

\(\frac{4}{3} r_{sphere}^3 = 320\)

Multiply both sides by \(\frac{3}{4}\) to isolate \(r_{sphere}^3\):

\(r_{sphere}^3 = 320 \times \frac{3}{4}\)

\(r_{sphere}^3 = (320 \div 4) \times 3\)

\(r_{sphere}^3 = 80 \times 3\)

\(r_{sphere}^3 = 240\)

To find \(r_{sphere}\), take the cube root of 240:

\(r_{sphere} = \sqrt[3]{240}\)

We need to simplify \(\sqrt[3]{240}\). Find the cube factors of 240. We can see that \(240 = 8 \times 30\), and \(8\) is a perfect cube (\(2^3\)).

\(r_{sphere} = \sqrt[3]{8 \times 30}\)

Using the property \(\sqrt[3]{ab} = \sqrt[3]{a} \times \sqrt[3]{b}\):

\(r_{sphere} = \sqrt[3]{8} \times \sqrt[3]{30}\)

\(r_{sphere} = 2 \times \sqrt[3]{30}\)

\(r_{sphere} = 2\sqrt[3]{30}\) cm.

The radius of the sphere is \(2\sqrt[3]{30}\) cm.

Summary of Calculations

The calculated radius of the sphere is \(2\sqrt[3]{30}\) cm.

Revision Table: Key Mensuration Formulas

Additional Information: Conservation of Volume

The concept of conservation of volume is fundamental in problems where a solid object is melted, remoulded, or reshaped from one form to another. Assuming no material is lost or added during the process, the total amount of material, and thus the volume, remains constant. This principle allows us to equate the volume of the initial shape(s) to the volume of the final shape(s) and solve for unknown dimensions. This is applicable across various shapes like cubes, cuboids, cylinders, cones, spheres, etc.