Class 10th Surface Areas and Volumes Notes

13

Surface Areas and Volumes

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Chapter Roadmap

1. Basic Ideas & Units 2. Cuboid and Cube 3. Cylinder 4. Cone 5. Sphere and Hemisphere 6. Combination of Solids 7. Conversion of Solids 8. Frustum of a Cone 9. Board-Style Questions 10. 10 Detailed Solved Questions 11. Board Exam Strategy 12. Quick Revision

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Chapter Goal

This chapter teaches us how to find the surface area and volume of three-dimensional solids. The most important skills are choosing the correct formula, finding missing dimensions, handling combined solids, and using conservation of volume when a solid is melted and recast.

1. Basic Ideas, Terms and Units

1.1 What is Surface Area?

The surface area of a solid is the total area of all the surfaces that are exposed. It is measured in square units such as \(cm^2\), \(m^2\), etc.

1.2 What is Volume?

The volume of a solid is the amount of space occupied by it. It is measured in cubic units such as \(cm^3\), \(m^3\), etc.

CSA vs TSA

Curved/Lateral Surface Area: only the curved or side surface is counted.

Total Surface Area: all required exposed surfaces are counted.

1.3 Important Unit Conversions

\[ 1\,m=100\,cm \qquad 1\,m^2=10,000\,cm^2 \qquad 1\,m^3=1,000,000\,cm^3 \]

Board Tip: Always convert all dimensions into the same unit before applying a formula.

Solved Example 1 — Length Conversion

Question: Convert \(4.5\,m\) into centimetres.

1. Use \(1\,m=100\,cm\).
2. Therefore \(4.5\,m=4.5\times100\,cm\).
3. \(=450\,cm\).
4. Answer: \(\boxed{450\,cm}\).

Solved Example 2 — Area Conversion

Question: Convert \(2\,m^2\) into \(cm^2\).

1. Since \(1\,m^2=10,000\,cm^2\).
2. \(2\,m^2=2\times10,000\,cm^2\).
3. \(=20,000\,cm^2\).
4. Answer: \(\boxed{20,000\,cm^2}\).

Solved Example 3 — Volume Conversion

Question: Convert \(0.25\,m^3\) into \(cm^3\).

1. Use \(1\,m^3=1,000,000\,cm^3\).
2. \(0.25\times1,000,000=250,000\).
3. Answer: \(\boxed{250,000\,cm^3}\).

2. Cuboid and Cube

2.1 Cuboid

A cuboid has six rectangular faces. Let its length be \(l\), breadth be \(b\), and height be \(h\).

\[ \boxed{\text{LSA}=2h(l+b)} \] \[ \boxed{\text{TSA}=2(lb+bh+hl)} \] \[ \boxed{V=lbh} \]

2.2 Cube

A cube is a special cuboid in which all edges are equal. If each edge is \(a\):

\[ \boxed{\text{LSA}=4a^2} \qquad \boxed{\text{TSA}=6a^2} \qquad \boxed{V=a^3} \]

Solved Example 1 — Volume of Cuboid

Question: Find the volume of a cuboid with \(l=12\,cm,\ b=5\,cm,\ h=4\,cm\).

1. Formula: \(V=lbh\).
2. Substitute: \(V=12\times5\times4\).
3. \(V=240\,cm^3\).
4. Answer: \(\boxed{240\,cm^3}\).

Solved Example 2 — TSA of Cuboid

Question: Find the total surface area of a cuboid \(10\,cm\times6\,cm\times4\,cm\).

1. \(\text{TSA}=2(lb+bh+hl)\).
2. \(=2(10\times6+6\times4+4\times10)\).
3. \(=2(60+24+40)\).
4. \(=2(124)=248\,cm^2\).
5. Answer: \(\boxed{248\,cm^2}\).

Solved Example 3 — TSA of Cube

Question: Find the TSA of a cube of side \(7\,cm\).

1. Formula: \(\text{TSA}=6a^2\).
2. Substitute \(a=7\): \(6(7)^2\).
3. \(=6\times49=294\,cm^2\).
4. Answer: \(\boxed{294\,cm^2}\).

3. Cylinder

A cylinder has two equal circular bases and one curved surface. Let \(r\) be the radius and \(h\) be the height.

\[ \boxed{\text{CSA}=2\pi rh} \] \[ \boxed{\text{TSA}=2\pi r(h+r)} \] \[ \boxed{V=\pi r^2h} \]

Solved Example 1 — Cylinder Volume

Question: Find the volume of a cylinder with \(r=7\,cm,\ h=10\,cm\), taking \(\pi=\frac{22}{7}\).

1. Use \(V=\pi r^2h\).
2. \(V=\frac{22}{7}\times7^2\times10\).
3. \(=\frac{22}{7}\times49\times10\).
4. \(=1540\,cm^3\).
5. Answer: \(\boxed{1540\,cm^3}\).

Solved Example 2 — Cylinder CSA

Question: Find the curved surface area of a cylinder with \(r=3.5\,cm,\ h=8\,cm\).

1. Use \(\text{CSA}=2\pi rh\).
2. \(=2\times\frac{22}{7}\times3.5\times8\).
3. \(=176\,cm^2\).
4. Answer: \(\boxed{176\,cm^2}\).

Solved Example 3 — Find Height

Question: A cylinder has volume \(1540\,cm^3\) and radius \(7\,cm\). Find its height.

1. Use \(V=\pi r^2h\).
2. \(1540=\frac{22}{7}\times49\times h\).
3. \(1540=154h\).
4. \(h=\frac{1540}{154}=10\,cm\).
5. Answer: \(\boxed{10\,cm}\).

4. Cone

A cone has one circular base and one curved surface. Its vertical height is \(h\), radius is \(r\), and slant height is \(l\).

\[ \boxed{l=\sqrt{r^2+h^2}} \] \[ \boxed{\text{CSA}=\pi rl} \] \[ \boxed{\text{TSA}=\pi r(l+r)} \] \[ \boxed{V=\frac13\pi r^2h} \]

Important: Do not confuse \(h\) (vertical height) with \(l\) (slant height).

Solved Example 1 — Slant Height

Question: A cone has radius \(6\,cm\) and height \(8\,cm\). Find its slant height.

1. Use \(l=\sqrt{r^2+h^2}\).
2. \(l=\sqrt{6^2+8^2}\).
3. \(=\sqrt{36+64}=\sqrt{100}\).
4. \(l=10\,cm\).
5. Answer: \(\boxed{10\,cm}\).

Solved Example 2 — Cone Volume

Question: Find the volume of a cone with \(r=7\,cm,\ h=12\,cm\).

1. \(V=\frac13\pi r^2h\).
2. \(=\frac13\times\frac{22}{7}\times49\times12\).
3. \(=616\,cm^3\).
4. Answer: \(\boxed{616\,cm^3}\).

Solved Example 3 — Cone CSA

Question: Find the CSA of a cone with \(r=5\,cm,\ l=13\,cm\).

1. \(\text{CSA}=\pi rl\).
2. \(=\frac{22}{7}\times5\times13\).
3. \(=\frac{1430}{7}\,cm^2\).
4. \(\approx204.29\,cm^2\).
5. Answer: \(\boxed{\frac{1430}{7}\,cm^2}\).

5. Sphere and Hemisphere

5.1 Sphere

A sphere is a perfectly round solid. Every point on its surface is at the same distance from its centre.

\[ \boxed{\text{Surface Area}=4\pi r^2} \] \[ \boxed{V=\frac43\pi r^3} \]

5.2 Hemisphere

A hemisphere is half of a sphere.

\[ \boxed{\text{CSA}=2\pi r^2} \] \[ \boxed{\text{TSA}=3\pi r^2} \] \[ \boxed{V=\frac23\pi r^3} \]

Solved Example 1 — Sphere Area

Question: Find the surface area of a sphere of radius \(7\,cm\).

1. \(A=4\pi r^2\).
2. \(=4\times\frac{22}{7}\times49\).
3. \(=616\,cm^2\).
4. Answer: \(\boxed{616\,cm^2}\).

Solved Example 2 — Sphere Volume

Question: Find the volume of a sphere of radius \(3\,cm\).

1. \(V=\frac43\pi r^3\).
2. \(=\frac43\pi(3)^3\).
3. \(=\frac43\pi(27)=36\pi\,cm^3\).
4. Answer: \(\boxed{36\pi\,cm^3}\).

Solved Example 3 — Hemisphere TSA

Question: Find the total surface area of a hemisphere of radius \(7\,cm\).

1. \(\text{TSA}=3\pi r^2\).
2. \(=3\times\frac{22}{7}\times49\).
3. \(=462\,cm^2\).
4. Answer: \(\boxed{462\,cm^2}\).

6. Combination of Solids

A combination of solids is a new solid formed by joining two or more basic solids such as a cylinder, cone, hemisphere, sphere, cuboid or cube.

Golden Rule for Surface Area: Do not count a surface that becomes internal after two solids are joined.

Golden Rule for Volume: Add the volumes of the parts when they do not overlap.

6.1 Cylinder + Hemisphere

For a cylinder and hemisphere with the same radius joined at the circular face, the joining circle is internal.

\[ V=\pi r^2h+\frac23\pi r^3 \] \[ \text{Exposed Area}=2\pi rh+2\pi r^2+\pi r^2 \] \[ \boxed{\text{Exposed Area}=2\pi rh+3\pi r^2} \]

Solved Example 1 — Combined Volume

Question: A solid consists of a cylinder of \(r=3\,cm,\ h=10\,cm\) and a hemisphere of the same radius. Find its volume.

1. Cylinder volume \(=\pi r^2h=90\pi\).
2. Hemisphere volume \(=\frac23\pi r^3=18\pi\).
3. Total volume \(=90\pi+18\pi=108\pi\,cm^3\).
4. Answer: \(\boxed{108\pi\,cm^3}\).

Solved Example 2 — Exposed Area

Question: For the same solid, find the exposed surface area.

1. Exposed area \(=\) cylinder CSA + hemisphere CSA + bottom circle.
2. \(=2\pi rh+2\pi r^2+\pi r^2\).
3. \(=2\pi(3)(10)+3\pi(9)\).
4. \(=60\pi+27\pi=87\pi\,cm^2\).
5. Answer: \(\boxed{87\pi\,cm^2}\).

Solved Example 3 — Cylinder + Cone Volume

Question: A solid is made of a cylinder and a cone with common radius \(3\,cm\). Their heights are \(8\,cm\) and \(4\,cm\), respectively. Find total volume.

1. Cylinder volume \(=\pi(3)^2(8)=72\pi\).
2. Cone volume \(=\frac13\pi(3)^2(4)=12\pi\).
3. Total \(=72\pi+12\pi=84\pi\,cm^3\).
4. Answer: \(\boxed{84\pi\,cm^3}\).

7. Conversion / Recasting of Solids

When a solid is melted and recast into another shape, the amount of material remains the same if there is no loss. Therefore, volume is conserved.

\[ \boxed{\text{Volume of original solid}=\text{Volume of new solid}} \]

7.1 Standard Steps

1. Write the volume formula for the first solid.
2. Write the volume formula for the second solid.
3. Equate the two volumes.
4. Substitute known values.
5. Solve for the unknown quantity.
6. Write the final answer with the correct unit.

Solved Example 1 — Cylinder Recast as Sphere

Question: A cylinder of radius \(3\,cm\) and height \(8\,cm\) is melted and recast into a sphere. Find the radius of the sphere.

1. Cylinder volume \(=\pi(3)^2(8)=72\pi\).
2. Let sphere radius be \(R\).
3. Sphere volume \(=\frac43\pi R^3\).
4. Equate volumes: \(\frac43\pi R^3=72\pi\).
5. Cancel \(\pi\): \(R^3=54\).
6. \(R=\sqrt[3]{54}\,cm\).
7. Answer: \(\boxed{\sqrt[3]{54}\,cm}\).

Solved Example 2 — Cuboid into Cubes

Question: A cuboid \(12\,cm\times8\,cm\times6\,cm\) is divided into cubes of side \(4\,cm\). Find the number of cubes.

1. Cuboid volume \(=12\times8\times6=576\,cm^3\).
2. One cube volume \(=4^3=64\,cm^3\).
3. Number of cubes \(=\frac{576}{64}\).
4. \(=9\).
5. Answer: \(\boxed{9}\) cubes.

Solved Example 3 — Cone Recast into Cylinder

Question: A cone of radius \(6\,cm\) and height \(12\,cm\) is recast into a cylinder of the same radius. Find the cylinder height.

1. Cone volume \(=\frac13\pi(6)^2(12)=144\pi\).
2. Cylinder volume \(=\pi(6)^2H=36\pi H\).
3. Equate: \(36\pi H=144\pi\).
4. Cancel \(36\pi\): \(H=4\,cm\).
5. Answer: \(\boxed{4\,cm}\).

8. Frustum of a Cone

A frustum of a cone is obtained when the top part of a cone is cut off by a plane parallel to its base.

Let the larger radius be \(R\), smaller radius be \(r\), height be \(h\), and slant height be \(l\).

\[ \boxed{l=\sqrt{h^2+(R-r)^2}} \] \[ \boxed{\text{CSA}=\pi(R+r)l} \] \[ \boxed{\text{TSA}=\pi(R+r)l+\pi R^2+\pi r^2} \] \[ \boxed{V=\frac13\pi h(R^2+r^2+Rr)} \]

Solved Example 1 — Frustum Slant Height

Question: Find the slant height when \(R=5\,cm,\ r=2\,cm,\ h=4\,cm\).

1. \(l=\sqrt{h^2+(R-r)^2}\).
2. \(=\sqrt{4^2+(5-2)^2}\).
3. \(=\sqrt{16+9}=\sqrt{25}\).
4. \(l=5\,cm\).
5. Answer: \(\boxed{5\,cm}\).

Solved Example 2 — Frustum Volume

Question: Find the volume when \(R=4\,cm,\ r=2\,cm,\ h=3\,cm\).

1. \(V=\frac13\pi h(R^2+r^2+Rr)\).
2. \(=\frac13\pi(3)(16+4+8)\).
3. \(=28\pi\,cm^3\).
4. Answer: \(\boxed{28\pi\,cm^3}\).

Solved Example 3 — Frustum CSA

Question: Find the CSA when \(R=5\,cm,\ r=3\,cm,\ l=4\,cm\).

1. \(\text{CSA}=\pi(R+r)l\).
2. \(=\pi(5+3)(4)\).
3. \(=32\pi\,cm^2\).
4. Answer: \(\boxed{32\pi\,cm^2}\).

9. Important Board-Exam Question Patterns

Pattern 1 — Direct Formula Questions

Find CSA, TSA or volume of a cuboid, cube, cylinder, cone, sphere or hemisphere.

Pattern 2 — Missing Dimension

Volume, surface area or another dimension is given and you have to find radius, height, side or slant height.

Pattern 3 — Combination of Solids

A toy, vessel or solid is made by joining two shapes. Add volumes and carefully count only exposed surfaces.

Pattern 4 — Recasting

A solid is melted and changed into another shape. Use conservation of volume.

10. 10 Detailed Solved Questions

Question 1 — Cuboid TSA

Find the total surface area of a cuboid with \(l=15\,cm,\ b=10\,cm,\ h=8\,cm\).

1. \(\text{TSA}=2(lb+bh+hl)\).
2. \(=2(15\times10+10\times8+8\times15)\).
3. \(=2(150+80+120)\).
4. \(=2(350)=700\,cm^2\).
5. Answer: \(\boxed{700\,cm^2}\).

Question 2 — Cube Volume

The edge of a cube is \(9\,cm\). Find its volume and total surface area.

1. Volume \(=a^3=9^3=729\,cm^3\).
2. TSA \(=6a^2=6(81)=486\,cm^2\).
3. Answer: Volume \(=\boxed{729\,cm^3}\), TSA \(=\boxed{486\,cm^2}\).

Question 3 — Cylinder CSA and Volume

Find the CSA and volume of a cylinder with \(r=7\,cm,\ h=15\,cm\), taking \(\pi=\frac{22}{7}\).

1. \(\text{CSA}=2\pi rh=2\times\frac{22}{7}\times7\times15=660\,cm^2\).
2. \(V=\pi r^2h=\frac{22}{7}\times49\times15=2310\,cm^3\).
3. Answer: CSA \(=\boxed{660\,cm^2}\), Volume \(=\boxed{2310\,cm^3}\).

Question 4 — Cone Height

A cone has radius \(5\,cm\) and slant height \(13\,cm\). Find its height.

1. \(l^2=r^2+h^2\).
2. \(13^2=5^2+h^2\).
3. \(169=25+h^2\).
4. \(h^2=144\).
5. \(h=12\,cm\).
6. Answer: \(\boxed{12\,cm}\).

Question 5 — Sphere Surface Area

Find the surface area of a sphere of diameter \(28\,cm\).

1. Radius \(r=\frac{28}{2}=14\,cm\).
2. Surface area \(=4\pi r^2\).
3. \(=4\times\frac{22}{7}\times14^2\).
4. \(=2464\,cm^2\).
5. Answer: \(\boxed{2464\,cm^2}\).

Question 6 — Hemisphere Volume

Find the volume of a hemisphere of radius \(6\,cm\).

1. \(V=\frac23\pi r^3\).
2. \(=\frac23\pi(6)^3\).
3. \(=\frac23\pi(216)\).
4. \(=144\pi\,cm^3\).
5. Answer: \(\boxed{144\pi\,cm^3}\).

Question 7 — Cylinder + Hemisphere

A solid consists of a cylinder of radius \(3\,cm\) and height \(8\,cm\) with a hemisphere of the same radius on top. Find its total volume.

1. Cylinder volume \(=\pi(3)^2(8)=72\pi\).
2. Hemisphere volume \(=\frac23\pi(3)^3=18\pi\).
3. Total \(=72\pi+18\pi=90\pi\,cm^3\).
4. Answer: \(\boxed{90\pi\,cm^3}\).

Question 8 — Exposed Area

For Question 7, find the exposed surface area of the solid.

1. Exposed area = cylinder CSA + hemisphere CSA + bottom circle.
2. \(=2\pi rh+2\pi r^2+\pi r^2\).
3. \(=2\pi(3)(8)+3\pi(9)\).
4. \(=48\pi+27\pi=75\pi\,cm^2\).
5. Answer: \(\boxed{75\pi\,cm^2}\).

Question 9 — Recasting

A metal cube of side \(6\,cm\) is melted to form spheres of radius \(2\,cm\). How many complete spheres can be formed?

1. Cube volume \(=6^3=216\,cm^3\).
2. One sphere volume \(=\frac43\pi(2)^3=\frac{32\pi}{3}\,cm^3\).
3. Number \(=\frac{216}{32\pi/3}=\frac{648}{32\pi}\approx6.45\).
4. Only complete spheres can be counted.
5. Therefore, \(6\) complete spheres can be formed.
6. Answer: \(\boxed{6}\) complete spheres.

Question 10 — Frustum

Find the volume of a frustum with \(R=6\,cm,\ r=3\,cm,\ h=4\,cm\).

1. \(V=\frac13\pi h(R^2+r^2+Rr)\).
2. \(=\frac13\pi(4)(36+9+18)\).
3. \(=\frac43\pi(63)\).
4. \(=84\pi\,cm^3\).
5. Answer: \(\boxed{84\pi\,cm^3}\).

11. Board Exam Strategy

High-Scoring Method

1. Write Given: clearly write all dimensions.
2. Identify the solid: cuboid, cube, cylinder, cone, sphere or hemisphere.
3. Choose the formula: CSA, TSA or volume.
4. Substitute: put values carefully.
5. Calculate: show important intermediate steps.
6. Write unit: area in square units and volume in cubic units.

Most Important Checks

• Is the given value a radius or diameter?
• Did you use \(r=\frac d2\) when diameter was given?
• Did you use slant height \(l\) only where required for cone surface area?
• Did you remove internal joining surfaces in a combination?
• Did you equate volumes in a recasting problem?
• Did you use the value of \(\pi\) specified by the question?

12. Quick Revision Sheet

Solid / Concept	Formula / Key Fact
Cuboid LSA	\(2h(l+b)\)
Cuboid TSA	\(2(lb+bh+hl)\)
Cuboid Volume	\(lbh\)
Cube LSA	\(4a^2\)
Cube TSA	\(6a^2\)
Cube Volume	\(a^3\)
Cylinder CSA	\(2\pi rh\)
Cylinder TSA	\(2\pi r(h+r)\)
Cylinder Volume	\(\pi r^2h\)
Cone Slant Height	\(l=\sqrt{r^2+h^2}\)
Cone CSA	\(\pi rl\)
Cone TSA	\(\pi r(l+r)\)
Cone Volume	\(\frac13\pi r^2h\)
Sphere Area	\(4\pi r^2\)
Sphere Volume	\(\frac43\pi r^3\)
Hemisphere CSA	\(2\pi r^2\)
Hemisphere TSA	\(3\pi r^2\)
Hemisphere Volume	\(\frac23\pi r^3\)
Frustum Slant Height	\(l=\sqrt{h^2+(R-r)^2}\)
Frustum CSA	\(\pi(R+r)l\)
Frustum Volume	\(\frac13\pi h(R^2+r^2+Rr)\)
Recasting	\(\text{Volume before}=\text{Volume after}\)

30-Second Final Revision

• Cuboid: \(V=lbh\).
• Cube: \(V=a^3\).
• Cylinder: \(V=\pi r^2h\).
• Cone: \(V=\frac13\pi r^2h\).
• Sphere: \(V=\frac43\pi r^3\).
• Hemisphere: \(V=\frac23\pi r^3\).
• Combination: count only exposed surfaces.
• Recasting: volume remains the same.
• Units: area → square units; volume → cubic units.

Common Mistakes to Avoid

• Using diameter in place of radius.
• Forgetting the factor \(\frac13\) in cone volume.
• Using \(2\pi r^2\) as the TSA of a hemisphere instead of \(3\pi r^2\).
• Counting an internal joining face in the surface area of a combined solid.
• Forgetting to conserve volume during recasting.
• Writing \(cm^2\) for volume or \(cm^3\) for surface area.
• Using \(h\) instead of \(l\) in cone CSA/TSA.
• Mixing metres and centimetres in the same calculation.

Last-Minute Trick: Remember the order Given → Formula → Substitution → Calculation → Unit → Final Answer. Show these steps in board exams for neat, easy-to-check solutions.

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Class 10 Mathematics • Chapter 13: Surface Areas and Volumes

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Practise direct formula, combined-solid and recasting questions for maximum accuracy.