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Class 10 · Maths · Chapter 12 · NCERT Class 10 Mathematics

Surface Areas and Volumes Class 10 Notes

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Chapter mind map

The whole chapter at a glance: the big idea, then each branch and what sits under it.

Surface Areas and Volumes of Combined Solids

Analyzing complex objects by breaking them down into basic solids like cuboids, cones, cylinders, and spheres to calculate total area and space occupied.

  1. Surface Area Principles

    Total surface area (TSA) depends on visible exterior surfaces; interfaces where solids meet are hidden and excluded from the sum.

    • Exposed Area Rule — Add every exposed curved and flat surface. Shared faces inside the object are not painted or counted.
    • Overlapping Bases — When a smaller solid sits on a larger face, net area increases by (CSA of smaller solid - Base area of smaller solid).

  2. Additive Volume Logic

    Volume is purely additive because it represents total space occupied; no 'hidden' volume exists at the interfaces.

    • Sum of Constituents — Total Volume = V1 + V2 + ... regardless of orientation, provided the shapes do not overlap in space.
    • Industrial Shed Example — Volume of a structure like a cuboid surmounted by a half-cylinder is the sum of both individual volumes.

  3. Subtractive Volume

    Used when a solid has a portion removed or an internal feature that reduces its capacity.

    • Apparent vs Actual Capacity — Apparent capacity is based on outer dimensions; actual capacity subtracts internal obstructions like a raised bottom.
    • Hollowed Volume Formula — Actual Volume = Outer Volume - Inner Volume (e.g., Cylinder volume minus internal Hemisphere volume).

  4. Practical Case Studies

    Applying geometric formulas to real-world objects by identifying their constituent basic solids.

    • The Playing Top (Lattu) — A cone surmounted by a hemisphere. Height of cone = Total height - Hemisphere radius.
    • Medicine Capsule — Composed of a central cylindrical part with a hemisphere attached to each of its two ends.

  5. Formula & Unit Mastery

    Essential mathematical constants and procedural checks for accurate calculation.

    • Standard Volume Formulas — Cylinder: πr²h; Cone: (1/3)πr²h; Sphere: (4/3)πr³; Hemisphere: (2/3)πr³.
    • Calculation Rigor — Use consistent units; use π = 22/7 unless 3.14 is specified; distinguish clearly between diameter and radius.

Chapter notes

An exploration of calculating the surface area and volume of complex objects formed by combining basic geometric solids like cuboids, cones, cylinders, and spheres.

01 · Explore

Introduction to Combined Solids

In our daily lives, we rarely encounter objects that are perfectly simple geometric shapes. Instead, most items are combinations of basic solids.

From previous studies, we are familiar with basic three-dimensional shapes such as the cuboid, cone, cylinder, and sphere. However, a truck's oil tanker is not just a cylinder; it is often a cylinder with two hemispheres attached to its ends. Similarly, a test tube used in a laboratory is a combination of a cylinder and a hemisphere.

To analyze these complex shapes, we must learn how to break them down into their constituent parts. By understanding the properties of the individual basic solids, we can calculate the total surface area and volume of the combined object. This chapter focuses on the mathematical methods required to solve such practical problems.

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Question

Identify the basic solids that make up a typical medicine capsule.

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NCERT reference: chapter PDF pages 1, 2.

02 · Explore

Surface Area of a Combination of Solids

Calculating the surface area of a combined solid requires identifying which surfaces are visible and which are hidden by the joining process.

When two solids are joined together, the surfaces that meet at the interface disappear from the total surface area of the new object. For example, if we join a hemisphere to the base of a cone to make a toy, the flat circular bases of both shapes are hidden inside the toy. Therefore, the total surface area (TSA) of the toy is the sum of the curved surface areas (CSA) of the individual parts.

The general rule is to add every exposed curved and flat surface. Alternatively, add the total surface areas of the original solids and subtract both copies of each shared face. Curved surface areas alone are enough only when no flat faces remain exposed.

For a cylinder of radius r and height h with a hemisphere on top, the exposed area is 2πrh + 2πr² + πr² = 2πrh + 3πr². The first two terms are the curved surfaces, and the last is the bottom. The common circular face is inside and is not painted.

rHemisphereCylinderhShared face: inside the solid
A hemisphere sits on a cylinder with the same radius. The shared circular face is hidden. Add the curved surfaces and the exposed bottom for total surface area; add both volumes for volume. Learning sketch; use the labels and stated dimensions, not measurements from the picture.

Process of Combining Solids for Surface Area

  1. 1

    Identify Parts

    Break the complex object into basic solids like cones, cylinders, or hemispheres.

  2. 2

    Determine Interfaces

    Identify which faces are joined together and thus hidden from the outside.

  3. 3

    Sum Exposed Areas

    Calculate and add only the curved surface areas or exposed flat faces.

The total surface area of a combined solid is the sum of its visible exterior surfaces.

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If a cylinder is surmounted by a cone, which surfaces contribute to the total surface area?

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NCERT reference: chapter PDF pages 2, 3.

03 · Explore

Case Study: The Playing Top (Lattu)

Applying surface area formulas to a real-world object like a spinning top.

Consider a playing top shaped like a cone surmounted by a hemisphere. To find the area to be colored, we need the TSA of the combined shape. The total height of the top is 5 cm and the diameter is 3.5 cm.

First, we find the radius (r) which is 3.5 / 2 = 1.75 cm. The height of the hemispherical part is equal to its radius (1.75 cm). Therefore, the height of the conical part (h) is 5 - 1.75 = 3.25 cm. We then calculate the slant height (l) of the cone using the Pythagorean theorem: l = √(r² + h²).

Calculating Area of a Top

TSA = 2πr² + πrl

Radius r = 1.75 cm. Height of cone h = 3.25 cm. Slant height l = √((1.75)² + (3.25)²) ≈ 3.7 cm. CSA of hemisphere = 2 * (22/7) * 1.75 * 1.75 = 19.25 cm². CSA of cone = (22/7) * 1.75 * 3.7 = 20.35 cm². Total Area = 19.25 + 20.35 = 39.6 cm² (approx).

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Why is the height of the cone 3.25 cm and not 5 cm?

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NCERT reference: chapter PDF pages 3, 4.

04 · Explore

Solids with Overlapping Bases

Sometimes a solid is placed on a larger surface, leaving part of that surface exposed.

Consider a decorative block made of a cube with a hemisphere fixed on one face. If the side of the cube is 'a' and the radius of the hemisphere is 'r', the hemisphere covers a circular area on the top face of the cube.

To find the total surface area of this block, we take the total surface area of the cube (6a²), subtract the area of the base of the hemisphere (πr²) because it is covered, and then add the curved surface area of the hemisphere (2πr²) because it is now an exposed outer surface.

Hemisphere on a Cube

TSA = 6a² + πr²

For a cube of side 5 cm and a hemisphere of diameter 4.2 cm (r = 2.1 cm): Area of 6 faces = 6 * 5² = 150 cm². Area covered by hemisphere base = π * 2.1². Exposed CSA of hemisphere = 2π * 2.1². Net change = +πr². Total Area = 150 + (22/7) * 2.1 * 2.1 = 150 + 13.86 = 163.86 cm².

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If a hemisphere is scooped out of a cube face instead of being placed on top, does the surface area formula change?

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NCERT reference: chapter PDF pages 4, 6.

The rest of this chapter

Keep reading Surface Areas and Volumes, free

  1. Locked: 1. Volume of a Combination of Solids
  2. Locked: 2. Industrial Application: Volume of a Shed
  3. Locked: 3. Subtractive Volume: The Juice Glass
  4. Locked: 4. Summary of Key Concepts

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