Finding Lateral Surface Area of Cubes from a Melted Cuboid

This problem involves understanding the properties of cuboids and cubes, specifically how their volumes relate when one is melted and recast into the other. We are given the areas of three adjacent faces of a cuboidal solid block and need to find the lateral surface area of a certain number of smaller cubes formed from it.

Step 1: Calculate the Volume of the Cuboid

Let the dimensions of the cuboid be length (\(l\)), width (\(w\)), and height (\(h\)). The areas of three adjacent faces are given as \(lw\), \(wh\), and \(hl\). We are given:

• Area 1 (\(lw\)) = 216 cm\(^2\)
• Area 2 (\(wh\)) = 96 cm\(^2\)
• Area 3 (\(hl\)) = 144 cm\(^2\)

The volume (\(V\)) of the cuboid is given by \(V = lwh\). To find the volume, we can multiply the three given areas:

\((lw) \times (wh) \times (hl) = 216 \times 96 \times 144\)

\(l^2 w^2 h^2 = 216 \times 96 \times 144\)

\((lwh)^2 = 216 \times 96 \times 144\)

\(V^2 = 216 \times 96 \times 144\)

Now, we calculate the value of \(V^2\):

\(216 = 6^3 = (2 \times 3)^3 = 2^3 \times 3^3\)

\(96 = 32 \times 3 = 2^5 \times 3\)

\(144 = 12^2 = (2^2 \times 3)^2 = 2^4 \times 3^2\)

\(V^2 = (2^3 \times 3^3) \times (2^5 \times 3^1) \times (2^4 \times 3^2)\)

\(V^2 = 2^{(3+5+4)} \times 3^{(3+1+2)}\)

\(V^2 = 2^{12} \times 3^6\)

To find \(V\), we take the square root:

\(V = \sqrt{2^{12} \times 3^6} = 2^{12/2} \times 3^{6/2} = 2^6 \times 3^3\)

\(V = 64 \times 27\)

\(V = 1728\) cm\(^3\).

The volume of the cuboidal block is 1728 cm\(^3\).

Step 2: Calculate the Volume and Side Length of Each Cube

The cuboidal block is melted and formed into 8 cubes of the same size. This means the total volume is conserved. The volume of the 8 cubes is equal to the volume of the cuboid.

Let the side length of each cube be \(s\). The volume of one cube is \(s^3\). The total volume of 8 cubes is \(8s^3\).

\(8s^3 = V\)

\(8s^3 = 1728\)

\(s^3 = \frac{1728}{8}\)

\(s^3 = 216\)

To find \(s\), we take the cube root of 216:

\(s = \sqrt[3]{216}\)

\(s = 6\) cm.

The side length of each cube is 6 cm.

Step 3: Calculate the Lateral Surface Area of One Cube

The lateral surface area (LSA) of a cube is the sum of the areas of its four vertical faces. Each face of a cube is a square with side length \(s\). The area of one face is \(s^2\).

LSA of one cube = Area of 4 faces = \(4 \times s^2\)

LSA of one cube = \(4 \times (6 \text{ cm})^2\)

LSA of one cube = \(4 \times 36 \text{ cm}^2\)

LSA of one cube = 144 cm\(^2\).

Step 4: Calculate the Lateral Surface Area of 3 Cubes

We need to find the lateral surface area of 3 such cubes. Since all 8 cubes are of the same size, their lateral surface areas are equal.

LSA of 3 cubes = \(3 \times\) LSA of one cube

LSA of 3 cubes = \(3 \times 144 \text{ cm}^2\)

LSA of 3 cubes = 432 cm\(^2\).

The lateral surface area of 3 such cubes is 432 cm\(^2\).

Revision Table: Cuboids and Cubes Formulas

Additional Information: Volume Conservation and Recasting

When a solid material is melted and recast into a different shape or shapes, the total volume of the material remains constant. This principle of volume conservation is fundamental in many mensuration problems involving transformation of solids.

In this specific problem:

• The volume of the original cuboid is equal to the total volume of the 8 cubes formed.
• The surface area, however, changes during transformation. The total surface area of the 8 small cubes is generally different from the surface area of the original cuboid. Melting and recasting typically increase the total surface area, especially when one large solid is broken into many smaller ones, as new surfaces are created.
• The problem requires finding the lateral surface area, which is a specific part of the total surface area, for a set number of the new shapes (cubes).

Understanding the difference between volume (the amount of space a substance occupies) and surface area (the total area of the exposed surface of a solid) is crucial for solving such geometry problems.