Solids Class 9 ICSE — Maths Chapter 21
Solids class 9 ICSE: Chapter 21 of the Class 9 ICSE Mathematics book explains the surface area and volume of 3D shapes. It defines a solid, its volume, and its surface area. The chapter works through cuboid and cube formulas, then applies them to cost problems, uniform cross-sections, and water flow through pipes. Two exercise sets close the chapter. You can verify the syllabus on the official CISCE website.
ICSE Class 9 Mathematics Chapter 21 Solids Surface Area and Volume of 3d Solids PDF
What is inside Chapter 21
11 pages · 6 sections · 12 worked examples · 26 exercise questions
| Section | Title | Page |
|---|---|---|
| 21.1 | INTRODUCTION | |
| 21.2 | CUBOID | |
| 21.3 | CUBE | |
| 21.4 | COST OF AN ARTICLE | |
| 21.5 | CROSS-SECTION | |
| 21.5 | FLOW OF WATER (OR ANY OTHER LIQUID) | p. 10 |
- EXERCISE 21(A): 16 questions
- EXERCISE 21(C): 10 questions
21.1 INTRODUCTION
A solid is anything that occupies space and has a definite shape. Its volume is the space it occupies. These two ideas open the chapter, and every problem in it builds on them.
The capacity of a container equals its internal volume. The volume of material in a hollow body equals its external volume minus its internal volume. You will use both rules when a problem involves walls or thickness.
The surface area of a solid is the sum of the areas of all its surfaces. The chapter calls this the total surface area. You compute it by adding every face.
Dimensions explain why these measures differ. A solid has length, breadth and height, so it is three-dimensional. An area has two dimensions, length and breadth. A perimeter has one dimension, its length.
Keep the hollow-body rule in mind. External volume minus internal volume gives the material in the walls. It reappears in the wood box problem in Section 21.4. For the full set of Class 9 maths chapters, browse the Class 9 Mathematics book hub.
21.2 CUBOID
A cuboid is a rectangular solid whose six faces are all rectangles. Most boxes and rooms are cuboids in shape.
| Quantity | Formula |
|---|---|
| Volume | l \times b \times h |
| Total surface area | 2(lb + bh + hl) |
| Lateral surface area | 2(l + b) \times h |
| Diagonal | \sqrt{l^2 + b^2 + h^2} |
Lateral surface area is the sum of the four vertical faces. It excludes the top and bottom. You use it when painting walls or papering a room.
The diagonal comes from applying Pythagoras twice. In the base triangle ABD, BD^2 = l^2 + b^2. Then in triangle DBH, BH^2 = BD^2 + h^2 = l^2 + b^2 + h^2. Substituting gives BH = \sqrt{l^2 + b^2 + h^2}.
A cuboid has four equal diagonals: AG, BH, CE and DF. The diagonal also gives the length of the longest rod that fits inside a box or room.
Take a cuboid with l = 8\ \text{cm}, b = 6\ \text{cm}, h = 10\ \text{cm}. Its volume is 8 \times 6 \times 10 = 480\ \text{cm}^3. Its total surface area is 2(48 + 60 + 80) = 376\ \text{cm}^2. Its diagonal is \sqrt{64 + 36 + 100} = \sqrt{200} = 10\sqrt{2}\ \text{cm}.
21.3 CUBE
A cube is a rectangular solid in which every face is a square. That means length = breadth = height = a, where a is the edge.
Set l = b = h = a in the cuboid formulas. Volume becomes a^3. Total surface area becomes 6a^2. Lateral surface area becomes 4a^2. The diagonal becomes \sqrt{a^2 + a^2 + a^2} = a\sqrt{3}.
| Quantity | Cuboid (l, b, h) | Cube (edge a) |
|---|---|---|
| Volume | l \times b \times h | a^3 |
| Total surface area | 2(lb + bh + hl) | 6a^2 |
| Lateral surface area | 2(l + b) \times h | 4a^2 |
| Diagonal | \sqrt{l^2 + b^2 + h^2} | a\sqrt{3} |
The cube is the equal-edge case of the cuboid. Every cuboid formula reduces to a cube formula when the three dimensions match.
Use the lateral surface area when the problem mentions four walls or a hollow open-top shape. Use the total when every face counts, like a closed box.
Take a cube with edge 7 cm. Volume = 7^3 = 343\ \text{cm}^3. Total surface area = 6 \times 49 = 294\ \text{cm}^2. Lateral surface area = 4 \times 49 = 196\ \text{cm}^2. Diagonal = 7\sqrt{3}\ \text{cm}.
21.4 COST OF AN ARTICLE
Cost = rate × quantity. The chapter applies this to three everyday buys. Petrol is bought by volume. Land is bought by area. Cloth is bought by length.
The first problem pattern is a closed wooden box. Find the external volume, then the internal volume, then subtract. The difference is the volume of wood.
A closed wooden box has external dimensions 30 cm × 24 cm × 18 cm. The wood is 2 cm thick. Each internal dimension loses 2 \times 2 = 4 cm, because thickness is subtracted on both sides. Internal dimensions are 26 × 20 × 14 cm.
External volume = 30 \times 24 \times 18 = 12960\ \text{cm}^3. Internal volume = 26 \times 20 \times 14 = 7280\ \text{cm}^3. Wood volume = 12960 - 7280 = 5680\ \text{cm}^3. At ₹3 per cm³, the cost is 5680 \times 3 = ₹17040.
The double subtraction is the common error. Subtract the thickness once and the internal dimensions come out too large. Always subtract twice on each side.
The section also solves three more patterns. Melting and recasting conserves volume. Joining cubes end to end hides their touching faces, so the cuboid surface area is less than the sum of the cube areas. When only l + b + h and the diagonal are known, use (l + b + h)^2 = l^2 + b^2 + h^2 + 2(lb + bh + hl) to recover the surface area.
21.5 CROSS-SECTION
A cross-section is the cut made through a solid perpendicular to its length or height. A solid has a uniform cross-section when every such cut gives the same shape and size.
A cuboid and a right circular cylinder have uniform cross-sections. A right circular cone does not. Its cuts are the same shape but not the same size.
For any solid with a uniform cross-section, volume = area of cross-section × length. Also, surface area excluding the cross-section = perimeter of cross-section × length.
A metal bar is 120 cm long. Its cross-section is a trapezium with parallel sides 10 cm and 6 cm and height 4 cm. Area = \frac{1}{2}(10 + 6) \times 4 = 32\ \text{cm}^2. Volume = 32 \times 120 = 3840\ \text{cm}^3.
Match the units before multiplying. The chapter converts 1 m to 100 cm when the cross-section area is in cm². A length in metres times an area in cm² gives a wrong volume.
The second formula handles the side surfaces of such solids. It appears in the shed and tunnel problems later in the chapter.
21.5 FLOW OF WATER (OR ANY OTHER LIQUID)
When water flows through a pipe or canal of uniform cross-section, the volume flowing per unit time equals the area of cross-section times the speed of flow.
A pipe has cross-section area 4 cm² and carries water at 25 cm/s. Each second, it delivers 4 \times 25 = 100\ \text{cm}^3. In one minute, it delivers 100 \times 60 = 6000\ \text{cm}^3. Since 1 litre = 1000 cm³, that is 6 litres per minute.
Check the units before converting. Cross-section in cm² and speed in cm/s give cm³ per second. Multiply by 60 for per minute. Divide by 1000 for litres.
The section also links flow to tank levels. Volume of water drawn off = length × width × fall in level. Divide the drawn volume by the base area to find how far the level falls.
The final solved example is a factory shed. Its volume content uses cross-section area × length for the long solid. The roofing and painting costs then use perimeter and area rates.
What to practise from this chapter
This chapter has two exercise sets. Work through both to cover the formula applications.
- EXERCISE 21(A) — 16 questions. Covers ratio dimensions, cube volume to surface area, air per person in a hall, melting and recasting cubes, wall bricks with doors and mortar, and the open tank iron sheet cost.
- EXERCISE 21(C) — 10 questions. Covers the cube from its face perimeter, auditorium capacity, the longest rod in a box, cubes packed in a box, rise in field level after digging, melting into cubes, the submerged cube, the petrol tank distance, and box dimensions from paper cost.
There is no Exercise 21(B) in this chapter. The two sets above are the full practice load. You can also explore all Class 9 books across subjects on the Class 9 books page.
Frequently asked questions
Why are the internal dimensions of a closed wooden box smaller than the external ones by twice the thickness?
The wood is present on both the inside and the outside of each face. So each internal dimension loses the thickness twice, once per side. A box with 2 cm wood has internal dimensions smaller by 4 cm on each side.
When a cube is melted and recast into a cuboid, which quantity stays the same?
Volume stays the same. Melting changes only the shape, not the amount of material. Set the cube volume equal to the cuboid volume, then solve for the unknown dimension.
How do I convert the volume of water flowing through a pipe into litres per minute?
Multiply the cross-section area by the speed to get cm³ per second. Multiply by 60 for per minute, then divide by 1000 to get litres. The chapter gives 1 litre = 1000 cm³.
What is a uniform cross-section and why does it matter for volume?
A uniform cross-section means every perpendicular cut has the same shape and size. A cuboid and a cylinder have one. A cone does not. When a solid has one, volume = area of cross-section × length.
How do I find the surface area of a cuboid when only l + b + h and the diagonal are given?
Use the identity (l + b + h)^2 = l^2 + b^2 + h^2 + 2(lb + bh + hl). The surface area is 2(lb + bh + hl). Subtract the squared diagonal from the squared sum to get it.
Reference: ICSE Class 9 Mathematics textbook material, chapter 21: Solids Surface Area and Volume of 3d Solids — CISCE-aligned.