Chapter 14 - Math of Space: Surface Area and Volume (Ganita Manjari)

Study Chapter 14: Math of Space: Surface Area and Volume from NCERT Ganita Manjari, Class 9 Maths, Part II, with Teachoo. Understand how to calculate the surface areas and volumes of cubes, cuboids, right circular cylinders, cones, pyramids, spheres and hemispheres, and apply these ideas to practical problems.

How much material is needed to cover a container? How much water can it hold? Can two objects have the same volume but different surface areas? How can you estimate the number of balls that fit inside a room?

Teachoo helps you connect each formula with the shape and the question, so that you know what to calculate and why. Learn to distinguish curved surface area from total surface area, use the correct dimensions, convert units and explain your calculations clearly.

Whether you are practising Exercise Sets 14.1–14.4, revising formulas or solving the End-of-Chapter Exercises, make Teachoo your study companion for understanding three-dimensional geometry with confidence.

What will you learn in Chapter 14?

Cuboids and cubes

Understand surface area by examining faces and nets, and understand volume through unit cubes and layers. Explore packing, joined cubes, painted cubes and changes in surface area when a solid is cut into smaller pieces.

Right circular cylinders

Learn how unrolling a cylinder’s curved surface produces a rectangle. Calculate curved surface area, volume and the surface area of cylinders open or closed at their ends.

Right circular cones

Distinguish the vertical height from the slant height. Understand the cone’s curved surface through its net and use the relationship between radius, height and slant height.

Pyramidal shapes

Explore pyramids with different polygonal bases. Calculate volume using the base area and perpendicular height, and find surface area by adding the areas of the faces.

Spheres and hemispheres

Learn the formulas for spherical surface area and volume. Distinguish the curved surface area of a hemisphere from its total surface area, which includes the circular base.

Areas and volumes around us

Apply the chapter to containers, packaging, water displacement, melting and recasting, and everyday objects.

Guesstimates

Make reasonable assumptions, choose suitable geometric models and calculate estimates. Understand why packing gaps and incomplete information can affect the answer.

Teachoo connects these topics so that you can move confidently from identifying a solid to interpreting your final result.

Surface area and volume formulas

Solid	Surface area	Volume
Cube, side a	Total surface area = 6a²	a³
Cuboid, length l, width w, height h	Total surface area = 2(lw + lh + wh)	lwh
Right circular cylinder, radius r, height h	Curved surface area = 2πrh; total surface area with both ends closed = 2πrh + 2πr²	πr²h
Right circular cone, radius r, vertical height h, slant height s	Curved surface area = πrs; total surface area = πrs + πr²	⅓πr²h
Pyramid, base area B, perpendicular height h	Total surface area = base area + areas of all triangular faces	⅓Bh
Sphere, radius r	Surface area = 4πr²	⁴⁄₃πr³
Hemisphere, radius r	Curved surface area = 2πr²; total surface area = 3πr²	⅔πr³

For a right circular cone:

s² = h² + r²

For a cylinder closed at one end and open at the other:

Surface area = 2πrh + πr²

With Teachoo, understand which surfaces are included before selecting a formula.

Understand surface area and volume with a simple example

Consider a cube of side 4 cm.

Total surface area = 6a²
= 6 × 4²
= 96 cm²

Volume = a³
= 4³
= 64 cm³

The surface area measures the area of the cube’s six outer faces. The volume measures the three-dimensional space it occupies.

This distinction matters: surface area is measured in square units, while volume is measured in cubic units. Teachoo helps you connect the calculation with what the question actually asks.

Exercise Sets 14.1–14.4 and End-of-Chapter Exercises

Ganita Manjari Class 9 Chapter 14 contains four exercise sets—14.1, 14.2, 14.3 and 14.4—followed by 22 numbered End-of-Chapter Exercises. Some questions have subparts, and several extension questions are starred.

The chapter includes problems involving:

• Packing smaller cubes or boxes into larger containers.

• Joining and cutting solids.

• Counting painted faces of small cubes.

• Comparing surface areas and volumes.

• Changing the dimensions of cylinders.

• Melting and recasting solids.

• Calculating areas and volumes of cones, spheres and hemispheres.

• Finding the quantity of milk or water required.

• Using water displacement to calculate volume.

• Estimating quantities using measurements and assumptions.

• Comparing the volumes of spherical objects.

• Investigating how changes in dimensions affect area and volume.

The attached material also provides nets of solids, helping you visualise their faces and surfaces.

Use Teachoo to identify the shape, choose the formula, substitute the measurements and check the units before interpreting the answer.

Why study Math of Space with Teachoo?

Knowing a formula is useful only when you know how to apply it. An open container needs a different surface-area calculation from a closed one. A cone’s slant height is used for curved surface area, while its perpendicular height is used for volume.

Teachoo teaches these distinctions clearly and helps you recognise common mistakes before they affect your answer. You learn to draw or interpret the solid, identify the relevant dimensions and show the calculation in a logical sequence.

Start with cubes and cuboids, then build towards curved solids and estimation problems. Choose Teachoo as your go-to learning platform for Ganita Manjari Class 9 Maths, and study surface area and volume with clarity and confidence.

Frequently asked questions

1. What is Chapter 14 in Class 9 Ganita Manjari?

Chapter 14 in Ganita Manjari, Class 9 Maths, Part II, is “Math of Space: Surface Area and Volume.” It covers cubes, cuboids, cylinders, cones, pyramids, spheres, hemispheres and practical estimation problems. Teachoo helps you understand the formulas and apply them to textbook questions.

2. What is the difference between surface area and volume?

Surface area measures the area of an object’s surfaces and is expressed in square units, such as cm² or m².

Volume measures the three-dimensional space an object occupies and is expressed in cubic units, such as cm³ or m³.

For a container, its internal volume determines its capacity.

3. What is the difference between curved surface area and total surface area?

Curved surface area includes only the curved part of a solid. Total surface area includes all its boundary surfaces.

For example, a closed cylinder has a curved surface and two circular ends. Its total surface area is 2πrh + 2πr².

4. What are the formulas for a cube and a cuboid?

For a cube of side a:

Total surface area = 6a²
Volume = a³

For a cuboid of length l, width w and height h:

Total surface area = 2(lw + lh + wh)
Volume = lwh

5. How do you calculate the surface area of an open cylinder?

For an ideal cylinder open at one end and closed at the other, include the curved surface and one circular base:

Surface area = 2πrh + πr²

If both ends are open, only the curved surface is included. Teachoo helps you identify which surfaces the question requires.

6. What is the difference between the height and slant height of a cone?

The height is the perpendicular distance from the vertex to the base. The slant height runs along the curved surface from the vertex to the rim of the base.

For a right circular cone:

Slant height² = height² + radius²

Use slant height for curved surface area and perpendicular height for volume.

7. What are the surface area and volume formulas for a cone?

For a right circular cone of radius r, perpendicular height h and slant height s:

Curved surface area = πrs
Total surface area = πrs + πr²
Volume = ⅓πr²h

8. How do you calculate the volume of a pyramid?

Use:

Volume = ⅓ × base area × perpendicular height

For a square pyramid with base side a and perpendicular height h:

Volume = ⅓a²h

The perpendicular height is the distance from the apex to the plane of the base.

9. What are the formulas for a sphere and a hemisphere?

For a sphere of radius r:

Surface area = 4πr²
Volume = ⁴⁄₃πr³

For a hemisphere of radius r:

Curved surface area = 2πr²
Total surface area = 3πr²
Volume = ⅔πr³

10. Can two solids have the same volume but different surface areas?

Yes. A cube and a cuboid can have equal volumes but different surface areas.

This matters when comparing storage capacity and the amount of material needed for packaging. Teachoo helps you calculate and interpret both quantities separately.

11. What happens when a cube is cut into smaller cubes?

The total volume remains unchanged if no material is lost. However, the combined surface area of the smaller cubes increases because cutting exposes new surfaces.

Do not confuse the original cube’s outer surface area with the sum of the surface areas of all the separate pieces.

12. How do you solve melting and recasting problems?

When the same material is melted and recast with no loss or relevant change in density, equate the original and final volumes.

For identical new objects:

Number of objects = original volume ÷ volume of one new object

If the question asks for complete objects, use the whole number that can actually be formed and account for any leftover material.

13. How do you convert cubic centimetres into litres?

Use:

1 litre = 1,000 cm³

Divide a volume in cm³ by 1,000 to express it in litres. Also remember:

1 m³ = 1,000 litres

Convert all measurements to compatible units before substituting them into a formula.

14. Why can’t you find the number of balls in a room by simply dividing their volumes?

Spheres leave gaps when packed together. Dividing the room’s volume by one ball’s volume gives an upper bound, not a realistic packing count.

A better estimate considers the room’s dimensions, the balls’ diameters and how they are arranged. This is one of the important modelling ideas explored in Chapter 14.

15. What is a guesstimate problem?

A guesstimate problem asks you to estimate a quantity using reasonable assumptions, approximate measurements and a mathematical model.

For example, estimating the number of bricks in a classroom wall requires assumptions about wall dimensions, openings and brick size. Different estimates can be reasonable when their assumptions and calculations are explained clearly.

16. Which exercises are included in Chapter 14?

The chapter includes Exercise Sets 14.1, 14.2, 14.3 and 14.4, followed by 22 numbered End-of-Chapter Exercises. It also includes practical activities, estimation questions and a project on finding the approximate volume of a house.

17. How should I write a surface area or volume solution?

Identify the solid and list the measurements. State the required formula, convert units where necessary, substitute the values and calculate.

Finish with the correct square or cubic units and explain what the result represents. For estimation problems, also state your assumptions.

18. Why choose Teachoo for Class 9 Surface Area and Volume?

Choose Teachoo to understand which formula applies, which dimensions to use and how to interpret the answer. Teachoo’s focus on clear reasoning makes it a strong choice for studying Ganita Manjari Class 9 Chapter 14, practising mensuration questions and revising confidently.

Learn Math of Space with Teachoo—visualise the solid, understand the formula and solve with confidence.