Surface Areas & Volumes: Cuboid Formed by Joining Cubes End to End Class 10 Maths

In secondary solid geometry and packaging design, modular manufacturing creates large containers by bolting together identical cubic compartments end to end. When two identical solid cubes—each having six square faces—are glued together along a common face, a new solid rectangular prism (a cuboid) is formed.

A common intuition is that joining two identical cubes simply doubles the total surface area:

($2 \times 6a^2 = 12a^2$)

But in geometry, when two faces touch, they become internal boundaries sealed inside the solid, vanishing completely from the exposed exterior surface!

In CBSE Class 10 Mathematics, Chapter 12 (Surface Areas and Volumes), the Cuboid Formed by Joining Cubes End to End problem represents one of the most frequently asked questions in Section B (2 marks) and Section C (3 marks).

In this master guide, we break down geometric edge transformations, surface area conservation versus loss, and algebraic formulas for (n) combined cubes.

What You Will Learn

• How the dimensions of a solid transform when cubes are joined end to end
• Step-by-step solution to the classic NCERT problem: 2 cubes, each of volume ($64\text{cm}^3$), joined end to end
• Calculating the dimensions of the resulting cuboid:
  • Length ($l = 8\text{cm}$)
  • Breadth ($b = 4\text{cm}$)
  • Height ($h = 4\text{ cm}$)
• Calculating the Total Surface Area (TSA) of the resulting cuboid:
  • ($\mathbf{\text{TSA} = 160\text{ cm}^2}$)
• Understanding the Vanishing Face Rule:
  • ($\text{TSA}_{\text{cuboid}} = 2(6a^2) - 2a^2 = 10a^2$)
• Generalizing the formula for (n) identical cubes joined in a single row:
  • ($\mathbf{TSA = (4n + 2)a^2}$)
• Calculating the longest diagonal of the cuboid:
  • ($\mathbf{d = \sqrt{l^2 + b^2 + h^2}}$)

1. Dimensional Transformation: Joining Two Cubes

Let a cube have side length (edge) $a$:

• Volume of cube:

  ($V = a^3$)

• Surface area of one cube: ($A = 6a^2$)

What Happens to the Dimensions?

When two cubes of side $a$ are placed side-by-side, end to end:

1. Length increases: The lengths add together: $\mathbf{l= a + a = 2a}$

2. Breadth remains unchanged: $\mathbf{b = a}$

3. Height remains unchanged: $\mathbf{h = a}$

2. Master Problem: The Iconic NCERT Problem (Ex. 12.1, Q1)

Problem Statement:

2 cubes, each of volume $64\text{cm}^3$, are joined end to end. Find the surface area of the resulting cuboid.

Step-by-Step Solution:

Step 1: Find the Edge Length $a$ of Each Cube:

Given volume of one cube: $V = a^3 = 64\text{cm}^3$

Take the cube root of both sides: $a = \sqrt[3]{64} = \sqrt[3]{4 \times 4 \times 4} = \mathbf{4\text{cm}}$

• The edge length of each cube is $a = 4\text{cm}$.

Step 2: Determine Dimensions of the Resulting Cuboid:

When the two cubes are joined end to end:

• Length of cuboid: $\mathbf{l = 4 + 4 = 8\text{cm}}$
• Breadth of cuboid: $\mathbf{b = 4\text{cm}}$
• Height of cuboid: $\mathbf{h = 4\text{cm}}$

Step 3: Calculate Total Surface Area (TSA) of the Cuboid:

$\mathbf{\text{TSA} = 2(lb + bh + hl)}$

Substitute the dimensions: $\text{TSA} = 2[(8 \times 4) + (4 \times 4) + (4 \times 8)]$ $\text{TSA} = 2[32 + 16 + 32]$ $\text{TSA} = 2[80] = \mathbf{160\text{cm}^2}$

Final Answer:

The surface area of the resulting cuboid is $160\text{cm}^2$.

3. The Vanishing Face Rule: Why Surface Area is NOT $192\text{cm}^2$ ?

The Question (Conceptual Understanding): The surface area of one cube is $6a^2 = 6(4)^2 = 96\text{cm}^2$. Why is the surface area of two joined cubes $160\text{cm}^2$ instead of $2 \times 96 = 192\text{cm}^2$?

    Total Surface Area of 2 Independent Cubes:  96 cm²  +  96 cm²  =  192 cm²
    Area of TWO Touching Square Faces Sealed:    - [ 4²  +  4² ]    =  - 32 cm²
                                                                    ─────────
    SURFACE AREA OF RESULTING CUBOID:                                 160 cm²!

• When the two cubes are glued together along one face, two square faces (one from Cube 1 and one from Cube 2) are hidden inside the solid.
• They cease to be part of the outer exposed surface: $\mathbf{\text{TSA}_{\text{cuboid}} = 2(6a^2) - 2a^2 = 12a^2 - 2a^2 = 10a^2}\\$ $\text{TSA}_{\text{cuboid}} = 10 \times (4)^2 = 10 \times 16 = \mathbf{160\text{cm}^2}$
• The resulting cuboid has exactly 10 exposed square faces!

4. Generalization: Joining $n$ Identical Cubes in a Single Row

If $n$ identical cubes of edge length $a$ are joined end to end in a single line:

• Length: $l = na$
• Breadth: $b = a$
• Height: $h = a$
• Total Surface Area: $\text{TSA} = 2[(na \times a) + (a \times a) + (a \times na)] \\=2[na^2 + a^2 + na^2] = 2[(2n + 1)a^2]$

  $\mathbf{\text{TSA}\_{n\text{-cubes}} = (4n + 2)a^2}$

Verification for Different Counts:

• For $n = 1$ cube: $\text{TSA} = (4(1) + 2)a^2 = 6a^2$ (Single cube formula!)
• For $n = 2$ cubes: $\text{TSA} = (4(2) + 2)a^2 = 10a^2 = 10(16) = \mathbf{160\text{cm}^2}$ (Matches our problem!)
• For $n = 3$ cubes: $\text{TSA} = (4(3) + 2)a^2 = 14a^2 = 14(16) = \mathbf{224\text{cm}^2}$
• For $n = 5$ cubes: $\text{TSA} = (4(5) + 2)a^2 = 22a^2$

5. Finding the Longest Diagonal of the Resulting Cuboid

The Question: What is the maximum length of a rigid straight rod that can be placed inside this resulting cuboid?

The longest straight segment inside any cuboid is its space diagonal: $\mathbf{d = \sqrt{l^2 + b^2 + h^2}}$ Substitute $l = 8\text{cm}, b = 4\text{cm}, h = 4\text{cm}$: $d = \sqrt{8^2 + 4^2 + 4^2} = \sqrt{64 + 16 + 16} = \sqrt{96}$ Factor out $16$: $d = \sqrt{16 \times 6} = \mathbf{4\sqrt{6}\text{cm} \approx\approx 9.80\text{cm}}$

• The maximum length of a rod that can be placed in the cuboid is $4\sqrt{6}\text{cm}$.

6. Summary and Examination Tips

Exam Tip: In questions stating 2 cubes joined end to end, remember that ONLY the length changes: $l = 4+4=8\text{cm}$. The breadth and height remain $4\text{cm}$!

Common Mistake: Forgetting units. Surface area has units of $\text{cm}^2$, while volume has units of $\text{cm}^3$! Writing $160\text{cm}$ or $160\text{cm}^3$ will lose half a mark.