Exercise 14.2 — Volumes

Volumes of cube and cuboid.

Lesson Notes PDF

1 / —

Loading PDF…

Exercise 14.2 – Volume of Cube and Cuboid

Exercise 14.2 from Chapter 14, Surface Area and Volume (Cube and Cuboid), of Class 8 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) focuses entirely on calculating the volume of solid shapes. Volume tells us how much space a three-dimensional object occupies, or how much liquid/material it can hold. This exercise builds on the concept of a unit cube and develops the formulas V = l × b × h for a cuboid and V = l³ for a cube, then applies them to ten real-life word problems.

By the end of this lesson, you will be able to calculate the volume of any box-shaped object, convert between units of volume (cm³, m³, litres), and solve practical problems involving tanks, walls, bricks, and wooden boxes — all of which are common 2-mark and 4-mark questions in board exams.

Unit Cube Volume of Cuboid = l × b × h Volume of Cube = l³ Unit Conversion (cm³, m³, litres)

💡 Golden Rule: Volume is always measured in cubic units (cu. units, cm³, m³). To find the volume of any cuboid-shaped object, multiply its length, breadth, and height — making sure all three are in the same unit before multiplying.

Understanding the Unit Cube and Volume of a Cuboid

A unit cube is a cube whose every side (length, breadth, and height) measures exactly 1 unit. Its volume is taken as 1 cubic unit, and it acts as the basic "building block" for measuring the volume of larger shapes. If you stack unit cubes together to build a bigger box, the total number of unit cubes used gives you the volume of that box.

Unit Cube (1 × 1 × 1)

Volume = 1 cu. unit

Row of 7 unit cubes (7 × 1 × 1)

Volume = 7 cu. units

Slab (7 × 4 × 1)

Volume = 28 cu. units

By extending this idea to a box with length l, breadth b, and height h (all measured in the same unit), the table below shows how volume grows as each dimension increases — this is exactly how the formula V = l × b × h is derived.

Length (l) units	Breadth (b) units	Height (h) units	Volume (V) cu. units
1	1	1	1 × 1 × 1 = 1
7	1	1	7 × 1 × 1 = 7
7	4	1	7 × 4 × 1 = 28
7	4	3	7 × 4 × 3 = 84

Volume of a Cuboid (V) = length (l) × breadth (b) × height (h)

Similarly, when length = breadth = height = l (i.e., a cube), the same pattern applies. A cube of side 5 units made up of unit cubes stacks up as 5 × 1 × 1 = 5 (a row), then 5 × 5 × 1 = 25 (a slab), and finally 5 × 5 × 5 = 125 (a full cube). This leads directly to the cube volume formula:

Volume of a Cube (V) = l × l × l = l³

✅ Quick Memory Trick: A cuboid has three different dimensions, so multiply all three (l × b × h). A cube has all equal sides, so it's simply l × l × l = l³ (the cube of the side length).

Units of Volume and Capacity – Conversion Table

Before solving the exercise problems, it's essential to know how units of volume relate to each other, especially when converting between millimetres, centimetres, decimetres, metres, and kilometres. Capacity (the volume of liquid a container can hold) is usually measured in millilitres (ml), litres (l), or kilolitres (kl).

Length Relation	Volume Relation
10 mm = 1 cm	1000 mm³ = 1 cm³
10 cm = 1 dm	1000 cm³ = 1 dm³
10 dm = 1 m	1000 dm³ = 1 m³
100 cm = 1 m	1,000,000 cm³ = 1 m³
1000 m = 1 km	1,000,000,000 m³ = 1 km³

📌 Capacity Conversion (used heavily in this exercise):
1 cm³ = 1 ml | 1000 cm³ = 1 litre | 1 m³ = 1,000,000 cm³ = 1000 litres = 1 kilolitre (kl)

Question 1 — Find the Volume of Each Cuboid

This question gives three sets of dimensions (length, breadth, height) and asks for the volume of each cuboid. Apply V = l × b × h directly, keeping all measurements in metres.

Part (i)

l = 8.2 m, b = 5.3 m, h = 2.6 m

V = 8.2 × 5.3 × 2.6

112.996 m³

Part (ii)

l = 5.0 m, b = 4.0 m, h = 3.5 m

V = 5.0 × 4.0 × 3.5

70 m³

Part (iii)

l = 4.5 m, b = 2.0 m, h = 2.5 m

V = 4.5 × 2.0 × 2.5

22.5 m³

💡 Tip: When multiplying decimals, count the total number of decimal places in all the numbers being multiplied — that's how many decimal places your final answer should have. For example, 8.2 (1 decimal) × 5.3 (1 decimal) × 2.6 (1 decimal) gives an answer with 3 decimal places: 112.996.

Question 2 — Capacity of Tanks (Volume in m³ and Litres)

This question tests both the volume formula and unit conversion. The dimensions are given in metres and centimetres mixed together, so the first step is always to convert everything into metres. Once the volume in m³ is found, multiply by 1000 to convert it into litres (since 1 m³ = 1000 litres).

Part (i)

Length = 3 m 20 cm, Breadth = 2 m 90 cm, Depth = 1 m 50 cm

Convert: 3 m 20 cm = 3.20 m, 2 m 90 cm = 2.90 m, 1 m 50 cm = 1.50 m V = 3.20 × 2.90 × 1.50 = 13.92 m³ Capacity in litres = 13.92 × 1000 = 13,920 litres

Part (ii)

Length = 2 m 50 cm, Breadth = 1 m 60 cm, Depth = 1 m 30 cm

Convert: 2.50 m, 1.60 m, 1.30 m V = 2.50 × 1.60 × 1.30 = 5.2 m³ Capacity in litres = 5.2 × 1000 = 5,200 litres

Part (iii)

Length = 7 m 30 cm, Breadth = 3 m 60 cm, Depth = 1 m 40 cm

Convert: 7.30 m, 3.60 m, 1.40 m V = 7.30 × 3.60 × 1.40 = 36.792 m³ Capacity in litres = 36.792 × 1000 = 36,792 litres

📌 Why multiply by 1000? Since 1 m³ = 1000 litres, simply multiply the volume in cubic metres by 1000 to get the capacity in litres. This conversion is one of the most frequently tested concepts in board exams for this chapter.

Question 3 — What Happens When the Edge of a Cube is Halved?

This is a conceptual reasoning question. Let the original edge of the cube be l units, so its volume is V₁ = l³. If the edge is reduced to half, the new edge becomes l/2, and we need to find the new volume V₂ and compare it with V₁.

Solution

Effect of halving the edge of a cube

Original volume: V₁ = l³ New edge = l/2 New volume: V₂ = (l/2)³ = l³/8 So, V₂ = V₁/8 Reduced volume = V₁ − V₂ = V₁ − V₁/8 = (8V₁ − V₁)/8 = 7V₁/8

✅ Conclusion: Yes, the volume does reduce. The new cube's volume becomes only 1/8th of the original volume, which means the volume decreases by 7/8th (87.5%) of its original value — a much bigger drop than students often expect!

⚠️ Common misconception: Many students think halving the side halves the volume too. But because volume depends on the cube of the side (l³), halving the side actually makes the volume 8 times smaller, not 2 times smaller.

Question 4 — Find the Volume of Cubes with Given Side Lengths

Here, you are given the side length of a cube and asked to find its volume using V = l³. The challenge is mainly in multiplying decimal numbers accurately three times.

Part (i)

Side l = 6.4 cm

V = 6.4 × 6.4 × 6.4

262.144 cm³

Part (ii)

Side l = 1.3 cm

V = 1.3 × 1.3 × 1.3

2.197 cm³

💡 Tip: When cubing a decimal with 1 decimal place (like 6.4), the final answer will always have 3 decimal places (1 + 1 + 1), since you are multiplying the number by itself three times.

Question 5 — How Many Bricks Are Needed to Build a Wall?

This is a real-life application problem. A wall and a brick are both treated as cuboids. To find how many bricks fit into the wall, divide the volume of the wall by the volume of one brick. The key challenge is converting all measurements to the same unit (centimetres) before calculating.

Solution

Wall: 8 m long, 6 m high, 22.5 cm thick. Brick: 25 cm × 11.25 cm × 6 cm

Convert wall dimensions to cm: Length = 8 m = 800 cm, Height = 6 m = 600 cm, Breadth = 22.5 cm Volume of wall = 800 × 22.5 × 600 cm³ Volume of one brick = 25 × 11.25 × 6 cm³ Number of bricks = Volume of wall ÷ Volume of brick = (800 × 22.5 × 600) ÷ (25 × 11.25 × 6) = 6400 bricks

📌 Real-world connection: This type of problem reflects how civil engineers and contractors estimate building materials. Always convert all dimensions (the wall and the brick) into the same unit before dividing the volumes.

Question 6 — Difference Between Volume of a Cuboid and a Cube

This question asks you to calculate the volumes of two separate solids — a cuboid and a cube — and then find the difference between them.

Solution

Cuboid: 25 cm × 15 cm × 8 cm. Cube: edge 16 cm

Volume of cuboid = l × b × h = 25 × 15 × 8 Volume of cuboid = 3000 cm³ Volume of cube = a³ = 16³ = 16 × 16 × 16 Volume of cube = 4096 cm³ Difference = Volume of cube − Volume of cuboid = 4096 − 3000 = 1096 cm³

✅ Approach reminder: Always calculate each solid's volume separately first, then subtract the smaller volume from the larger one to find the difference. Don't try to combine the formulas before calculating.

Question 7 — Volume of Wood Used in a Closed Box

This is one of the trickiest problems in the exercise. A closed wooden box has an outer volume and an inner (hollow) volume. Since the wood itself has thickness, the inner dimensions are smaller than the outer dimensions on both sides of each measurement.

Closed wooden box (1 cm thick walls)

Solution

Outer dimensions: 5 cm × 4 cm × 7 cm, wood thickness = 1 cm

Volume of box (outer) = 5 × 4 × 7 = 140 cm³ Inner length = 5 − 1 − 1 = 3 cm Inner breadth = 4 − 1 − 1 = 2 cm Inner height = 7 − 1 − 1 = 5 cm Volume of box (inner) = 3 × 2 × 5 = 30 cm³ Volume of wood = Outer volume − Inner volume = 140 − 30 = 110 cm³

💡 Why subtract 1 twice? The wood thickness applies on both sides of every dimension — for example, a 5 cm outer length includes 1 cm of wood on the left and 1 cm of wood on the right, leaving only 5 − 1 − 1 = 3 cm of hollow space inside.

Questions 8 & 9 — Cutting Smaller Cubes/Cuboids from a Bigger Cuboid

Both these questions follow the same logic: divide the volume of the larger solid by the volume of the smaller solid to find how many smaller pieces can be cut from the larger one.

Question 8

How many cubes of edge 4 cm can be cut from a cuboid of 20 cm × 18 cm × 16 cm?

Volume of cuboid = 20 × 18 × 16 cm³ Volume of one cube = 4³ = 4 × 4 × 4 cm³ Number of cubes = Volume of cuboid ÷ Volume of cube = (20 × 18 × 16) ÷ (4 × 4 × 4) = 90 cubes

Question 9

How many cuboids of 4 cm × 3 cm × 2 cm can be made from a cuboid of 12 cm × 9 cm × 6 cm?

Volume of big cuboid = 12 × 9 × 6 cm³ Volume of small cuboid = 4 × 3 × 2 cm³ Number of small cuboids = Volume of big cuboid ÷ Volume of small cuboid = (12 × 9 × 6) ÷ (4 × 3 × 2) = 27 small cuboids

📌 Pattern: "How many smaller pieces fit into a bigger solid?" questions are always solved using: Number of pieces = Volume of bigger solid ÷ Volume of one smaller piece.

Question 10 — Finding the Height of a Vessel from Its Capacity

This question reverses the usual process. Instead of finding the volume from given dimensions, you are given the capacity (in litres) and two of the three dimensions, and asked to find the missing height. The key step is converting litres to cm³ using 1 litre = 1000 cm³.

Solution

A cuboidal vessel is 30 cm long and 25 cm wide. Find its height to hold 4.5 litres of water.

Capacity = 4.5 litres = 4.5 × 1000 = 4500 cm³ Capacity = l × b × h 4500 = 30 × 25 × h 4500 = 750 × h h = 4500 ÷ 750 = 6 cm

✅ Key takeaway: Whenever a problem gives capacity in litres but dimensions in cm, always convert litres to cm³ first (multiply by 1000), then use V = l × b × h to find the unknown dimension.

Common Mistakes to Avoid

Mixing units: Always convert all dimensions to the same unit (cm, m) before multiplying. Mixing metres and centimetres directly leads to wildly wrong answers.
Confusing cube and cuboid formulas: Use l³ only when all three dimensions are equal (a cube). Otherwise, always use l × b × h for a cuboid.
Forgetting "both sides" for thickness problems: In hollow box problems (like Q7), subtract the wall thickness twice from each outer dimension — once for each side.
Incorrect litre conversion: Remember 1 m³ = 1000 litres, but 1 cm³ = 1 ml. Don't confuse the conversion factor for cubic metres with cubic centimetres.
Decimal placement errors: When multiplying decimals (as in Q1, Q2, Q4), count decimal places carefully across all three numbers to place the decimal point correctly in the final answer.

⛔ Most common board exam error: In Question 3, students often say "the volume becomes half" when the edge is halved. The correct answer is that the volume becomes 1/8th of the original — a frequently tested conceptual trap.

Quick Reference — All Answers at a Glance

Question	Problem Type	Answer
Q1(i)	Volume of cuboid (8.2 × 5.3 × 2.6)	112.996 m³
Q1(ii)	Volume of cuboid (5.0 × 4.0 × 3.5)	70 m³
Q1(iii)	Volume of cuboid (4.5 × 2.0 × 2.5)	22.5 m³
Q2(i)	Tank capacity (3.20 × 2.90 × 1.50)	13.92 m³ = 13,920 l
Q2(ii)	Tank capacity (2.50 × 1.60 × 1.30)	5.2 m³ = 5,200 l
Q2(iii)	Tank capacity (7.30 × 3.60 × 1.40)	36.792 m³ = 36,792 l
Q3	Effect of halving cube's edge	Volume reduces to 1/8th; reduction = 7V/8
Q4(i)	Volume of cube (side 6.4 cm)	262.144 cm³
Q4(ii)	Volume of cube (side 1.3 cm)	2.197 cm³
Q5	Bricks needed for a wall	6400 bricks
Q6	Difference: cube vs cuboid volume	1096 cm³
Q7	Volume of wood in closed box	110 cm³
Q8	Cubes (edge 4 cm) from cuboid	90 cubes
Q9	Small cuboids from big cuboid	27 cuboids
Q10	Height of vessel from capacity	6 cm

What This Exercise Prepares You For

Mastering the volume of cubes and cuboids in Exercise 14.2 builds the foundation for the next part of Chapter 14, which covers the surface area of cubes and cuboids — where instead of finding the space inside a solid, you calculate the total area of all its outer faces. Both concepts together are essential for understanding three-dimensional geometry.

In Class 9 and Class 10, these ideas extend further to other solids such as cylinders, cones, and spheres, where the same logic of "filling with unit cubes" is replaced by formulas involving radius and height. For Telangana and Andhra Pradesh board exams, volume and capacity word problems (like Q2, Q5, and Q10 in this exercise) are extremely common as 2-mark and 4-mark questions.

To strengthen your algebra skills alongside geometry, you may also revisit Exponents and Powers, since calculating l³ and a³ relies on the same exponent rules. For square and cube relationships, the chapter on Square Roots and Cube Roots is also closely connected — especially useful when working backward from a known volume to find the side of a cube.

📐 Board Exam Tip (Telangana & AP): For word problems like bricks-in-a-wall (Q5) or cutting smaller cuboids (Q8, Q9), always write the "Number of pieces = Volume of big solid ÷ Volume of small solid" formula explicitly before substituting values — this earns method marks even if your final arithmetic has a small error.