Surface Areas of Combinations of Solids for CBSE Class 10 Mathematics

In middle school, we learned how to compute the surface areas of standard geometric solids in isolation—cubes, cuboids, cylinders, cones, and spheres. However, the physical objects we encounter daily rarely exist as isolated primitive shapes. A circus tent is a cone mounted atop a cylinder; a medicine capsule is a cylinder capped by two hemispherical domes; a spinning top is a cone perched upon a hemisphere; and an architectural pillar is a column resting on a rectangular plinth.

In CBSE Class 10 Mathematics, Chapter 12 (Surface Areas and Volumes) focuses on combinations of solids. Calculating their surface area requires mastering a crucial conceptual shift: when two solids are glued together, their joint contact faces become hidden interior boundaries and are no longer exposed to the surface!

What You Will Learn

• The golden principle of composite surface areas: Exposed surfaces only!
• Solid Type 1: A conical top surmounted on a hemisphere of common radius
• Solid Type 2: A medicine capsule (cylinder with two hemispherical ends)
• Solid Type 3: A cube surmounted by a hemisphere (or with a hemispherical cavity scooped out)
• Solid Type 4: A solid wooden cylinder with hemispherical scoops at both ends
• Formula reference table for curved surface areas (CSA) and slant heights
• Step-by-step solved CBSE board examination problems and common traps

1. The Golden Rule of Composite Surface Areas

When students first encounter combination problems, their natural instinct is often to add the Total Surface Areas (TSA) of the two component solids. This is mathematically fatal!

    WRONG APPROACH:
    Total Surface Area ≠ TSA of Solid 1 + TSA of Solid 2   (INCORRECT!)

    CORRECT APPROACH:
    Total Surface Area = Sum of EXPOSED CURVED SURFACE AREAS of all parts!

The Golden Rule: <u>When two solids are joined together to form a composite solid, the contact surface between them disappears into the interior. The total surface area of the new solid is strictly the sum of the VISIBLE, EXPOSED SURFACES (usually the Curved Surface Areas, CSA) of the individual parts!</u>

2. Formula Quick-Reference Guide

3. High-Yield Solved Board Examination Problems

Solved Example 1: The Spinning Toy (Cone on Hemisphere)

Problem: A toy is in the form of a cone of radius $3.5\text{ cm}$ mounted on a hemisphere of the same radius. The total height of the toy is $15.5\text{ cm}$. Find the total surface area of the toy. (Use $\pi = 22/7$).

                                      /                                     /  \  <-- Cone (Height h = 12 cm)
                                    /    \      Radius r = 3.5 cm
                                   /                                        +--------+
                                 (          ) <-- Hemisphere (Radius r = 3.5 cm)
                                  \________/
                                  <-- 3.5 -->
                                  <------- Total Height = 15.5 cm ------->

Solution:

1. Analyze Component Dimensions:
   • Radius of both hemisphere and cone: $r = 3.5\text{ cm} = \frac{7}{2}\text{ cm}$.
   • The hemisphere extends downwards by a depth equal to its radius: $r = 3.5\text{ cm}$.
   • Vertical height of the conical part: $h = \text{Total Height} - \text{Radius of Hemisphere} = 15.5 - 3.5 = \mathbf{12\text{ cm}}$
2. Calculate Slant Height of the Cone ($l$): $l = \sqrt{r^2 + h^2} = \sqrt{(3.5)^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = \mathbf{12.5\text{ cm}}$
3. Formulate the Total Surface Area Equation: The base of the cone and the flat circular face of the hemisphere are glued together inside the toy. $\mathbf{\text{TSA of Toy} = \text{CSA of Cone} + \text{CSA of Hemisphere}}$ $\text{TSA} = \pi r l + 2\pi r^2 = \pi r (l + 2r)$
4. Substitute Numerical Values: $\text{TSA} = \frac{22}{7} \times \frac{7}{2} \times [12.5 + 2(3.5)]$ $\text{TSA} = 11 \times [12.5 + 7] = 11 \times 19.5 = \mathbf{214.5\text{ cm}^2}$
5. Therefore, <u>the total surface area of the toy is $214.5\text{ cm}^2$</u>.

Solved Example 2: The Medicine Capsule (NCERT Classic)

Problem: A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is $14\text{ mm}$ and the diameter of the capsule is $5\text{ mm}$. Find its surface area.

                   Hemisphere             Cylinder             Hemisphere
                     (-----[=====================================]-----)
                     < 2.5 > <------------- 9 mm --------------> < 2.5 >
                     <---------------------- 14 mm -------------------->

Solution:

1. Analyze Dimensions:
   • Diameter $d = 5\text{ mm} \implies$ Radius of cylinder and hemispheres: $r = \frac{5}{2} = \mathbf{2.5\text{ mm}}$.
   • The two hemispherical ends each occupy a length equal to the radius ($2.5\text{ mm}$).
   • Length (height) of the central cylindrical part: $h = 14 - (2.5 + 2.5) = 14 - 5 = \mathbf{9\text{ mm}}$
2. Formulate Surface Area Strategy: The surface area of the capsule consists of the curved cylinder body plus the two curved hemispherical domes: $\mathbf{\text{Surface Area} = \text{CSA of Cylinder} + 2 \times (\text{CSA of Hemisphere})}$ Notice that two hemispheres of the same radius combine to form a full sphere: $\text{Surface Area} = 2\pi r h + 2(2\pi r^2) = 2\pi r h + 4\pi r^2 = \mathbf{2\pi r (h + 2r)}$
3. Substitute Values: $\text{Surface Area} = 2 \times \frac{22}{7} \times \frac{5}{2} \times [9 + 2(2.5)]$ $\text{Surface Area} = \frac{110}{7} \times [9 + 5] = \frac{110}{7} \times 14 = 110 \times 2 = \mathbf{220\text{ mm}^2}$
4. Therefore, <u>the surface area of the medicine capsule is $220\text{ mm}^2$</u>.

Solved Example 3: Cube Surmounted by a Hemisphere

Problem: A decorative block is made of two solids—a cube and a hemisphere. The base of the block is a cube with edge $5\text{ cm}$, and the hemisphere fixed on the top has a diameter of $4.2\text{ cm}$. Find the total surface area of the block. (Use $\pi = 22/7$).

Solution:

1. Analyze the Surfaces:
   • Edge of cube $a = 5\text{ cm}$.
   • Total surface area of 6 faces of the cube $= 6a^2$.
   • On the top face of the cube, the circular base of the hemisphere covers a portion of area $\pi r^2$.
   • Rising above the top face is the curved dome of the hemisphere with area $2\pi r^2$.
2. Formulate the Master Equation: $\mathbf{\text{TSA of Block} = \text{TSA of Cube} - \text{Base Area of Hemisphere} + \text{CSA of Hemisphere}}$ $\text{TSA of Block} = 6a^2 - \pi r^2 + 2\pi r^2 = \mathbf{6a^2 + \pi r^2}$
3. Substitute Values ($a = 5\text{ cm}, r = 2.1\text{ cm}$):
   • $\text{TSA of Cube} = 6 \times (5)^2 = 6 \times 25 = 150\text{ cm}^2$.
   • $\text{Added Net Area} = \pi r^2 = \frac{22}{7} \times 2.1 \times 2.1 = 22 \times 0.3 \times 2.1 = 13.86\text{ cm}^2$. $\text{TSA of Block} = 150 + 13.86 = \mathbf{163.86\text{ cm}^2}$
4. Therefore, <u>the total surface area of the block is $163.86\text{ cm}^2$</u>.

4. Summary and Examination Tips

Exam Tip: In problems where a hemisphere is scooped out of a cylinder or cube, students often think the surface area decreases because material was removed. This is false! Scooping creates an exposed interior curved cavity, which INCREASES the total surface area! The formula remains identical: add the $2\pi r^2$ of the cavity!

Common Mistake: Calculating height of a cone using the total height directly. Remember: the height of the cone $h$ is the total height MINUS the radius of the hemisphere ($h = H - r$).