Exercise 10.1 — Cube and Cuboid

Surface area and volume of cube and cuboid.

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Surface Areas and Volumes — What Exercise 10.1 Covers

Exercise 10.1 of Class 9 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) introduces students to calculating the surface areas and volumes of common 3D solids — the cube, cuboid, prism, and pyramid. These concepts are essential for real-world problem solving: from estimating paint needed for a room to finding the water capacity of a swimming pool.

The exercise contains 8 problems covering lateral surface area, total surface area, volume, and an important conceptual question about how surface area changes when dimensions are scaled. All formulas tested here are directly examinable in board exams.

Cube & Cuboid Lateral Surface Area Total Surface Area Volume of Prism Volume of Pyramid Heron's Formula

Solid	Lateral Surface Area	Total Surface Area	Volume
Cube (side = l)	4l²	6l²	l³
Cuboid (l × b × h)	2h(l + b)	2(lh + bh + lb)	l × b × h
Prism (base area A, height H)	Perimeter of base × H	2A + Lateral SA	A × H
Pyramid (base area A, height H)	—	—	(1/3) × A × H

💡 Memory Tip: For a cuboid, Lateral SA covers only the four walls (no top or bottom). Total SA covers all six faces. This distinction is tested every year in Telangana and AP board exams.

Question 1 — Lateral and Total Surface Area of Right Prisms

Two shapes are given — a cube and a cuboid. We apply the respective formulas for each.

Part (i) — Cube

A cube with side 4 cm. Find its Lateral Surface Area and Total Surface Area.

Cube

All sides equal: l = 4 cm

Given: Side l = 4 cm

Lateral Surface Area = 4l² = 4 × (4)² = 4 × 16 = 64 cm² Total Surface Area = 6l² = 6 × (4)² = 6 × 16 = 96 cm²

✅ Lateral SA = 64 cm² | Total SA = 96 cm²

Part (ii) — Cuboid

A cuboid with length 8 cm, breadth 6 cm, height 5 cm. Find its Lateral and Total Surface Area.

Cuboid

l = 8 cm, b = 6 cm, h = 5 cm

Given: l = 8 cm, b = 6 cm, h = 5 cm

Lateral Surface Area = 2h(l + b) = 2 × 5 × (8 + 6) = 10 × 14 = 140 cm² Total Surface Area = 2(lh + bh + lb) = 2[(8×5) + (6×5) + (8×6)] = 2[40 + 30 + 48] = 2 × 118 = 236 cm²

✅ Lateral SA = 140 cm² | Total SA = 236 cm²

Question 2 — Find Volume of a Cube from its Total Surface Area

Question 2

The total surface area of a cube is 1350 m². Find its volume.

This is a reverse problem — we are given the Total Surface Area and must first find the side length, then compute the volume.

Step 1: Total Surface Area = 6l² = 1350 m² l² = 1350 ÷ 6 = 225 l = √225 √225 = 15 (since 15 × 15 = 225) l = 15 m Step 2: Volume = l³ = 15³ = 15 × 15 × 15 = 3375 m³

📌 Finding √225: Factor 225 = 3 × 3 × 5 × 5 × 1 = (3 × 5)² = 15². So √225 = 15. Always verify: 15² = 225 ✓

✅ Side = 15 m | Volume = 3375 m³

Question 3 — Area of Four Walls of a Room

Question 3

Find the area of four walls of a room with length 12 m, breadth 10 m, and height 7.5 m. (No doors or windows.)

A room is modelled as a cuboid. The area of its four walls equals the Lateral Surface Area of that cuboid — it excludes the floor and ceiling.

Area of four walls = Lateral SA of cuboid = 2h(l + b)

Given: l = 12 m, b = 10 m, h = 7.5 m Area of four walls = 2h(l + b) = 2 × 7.5 × (12 + 10) = 15 × 22 = 330 m²

💡 Real-life use: This calculation tells a painter exactly how many square metres of wall surface needs to be painted. For plastering, tiling, or wallpaper estimation, this is the standard formula used by engineers and contractors.

✅ Area of four walls = 330 m²

Question 4 — Find the Height of a Cuboid from its Volume

Question 4

Volume of a cuboid is 1200 cm³. Length = 15 cm, Breadth = 10 cm. Find its height.

We rearrange the volume formula V = l × b × h to solve for the unknown height.

Given: V = 1200 cm³, l = 15 cm, b = 10 cm l × b × h = 1200 15 × 10 × h = 1200 150 × h = 1200 h = 1200 ÷ 150 h = 8 cm

✅ Height of the cuboid = 8 cm

📌 Tip: When any one of l, b, or h is unknown, simply divide the volume by the product of the two known dimensions. This technique applies equally to tanks, boxes, and containers.

Question 5 — How Does Surface Area Change When Dimensions Are Scaled?

This conceptual question explores what happens to the total surface area of a cuboid when all its dimensions are multiplied by the same factor. It reveals a beautiful mathematical pattern that holds true for all shapes — area scales as the square of the scale factor.

Original TSA: S = 2(lh + bh + lb)

Part (i) — Each dimension doubled

New dimensions: 2l, 2b, 2h

New TSA = 2[(2l)(2h) + (2b)(2h) + (2l)(2b)] = 2[4lh + 4bh + 4lb] = 4 × 2(lh + bh + lb) = 4S ← Surface area becomes 4 times

Part (ii) — Each dimension tripled

New dimensions: 3l, 3b, 3h

New TSA = 2[(3l)(3h) + (3b)(3h) + (3l)(3b)] = 2[9lh + 9bh + 9lb] = 9 × 2(lh + bh + lb) = 9S ← Surface area becomes 9 times

Part (iii) — Each dimension raised to n times

New dimensions: nl, nb, nh

New TSA = 2[(nl)(nh) + (nb)(nh) + (nl)(nb)] = 2[n²lh + n²bh + n²lb] = n² × 2(lh + bh + lb) = n²S ← Surface area becomes n² times

Scale Factor (n)	New Dimensions	New Surface Area	Multiplied by
n = 2 (doubled)	2l, 2b, 2h	4S	4 = 2²
n = 3 (tripled)	3l, 3b, 3h	9S	9 = 3²
n = 4	4l, 4b, 4h	16S	16 = 4²
n = 5	5l, 5b, 5h	25S	25 = 5²
n (general)	nl, nb, nh	n²S	n²

🔑 Golden Rule: When all dimensions of a solid are scaled by a factor of n, the surface area scales by n² and the volume scales by n³. This is a fundamental principle of geometry used in engineering, architecture, and biology.

Question 6 — Volume of a Triangular Prism Using Heron's Formula

Question 6

Base of a prism is a triangle with sides 3 cm, 4 cm, and 5 cm. Height of prism = 10 cm. Find the volume.

The base is a triangle, so we must first find its area using Heron's Formula, then multiply by the height of the prism.

Triangular Prism

Triangular base: 3-4-5 right triangle

Heron's Formula: A = √[s(s−a)(s−b)(s−c)] where s = (a+b+c)/2

Step 1: Semi-perimeter s = (3 + 4 + 5) / 2 = 12 / 2 = 6 cm Step 2: Area A = √[s(s−a)(s−b)(s−c)] = √[6 × (6−3) × (6−4) × (6−5)] = √[6 × 3 × 2 × 1] = √36 = 6 cm² Step 3: Volume of prism = Base Area × Height = 6 × 10 = 60 cm³

💡 Shortcut observed: The triangle with sides 3, 4, 5 is a right-angled triangle (since 3² + 4² = 5²). Its area can also be calculated directly as ½ × base × height = ½ × 3 × 4 = 6 cm². Heron's formula gives the same answer, confirming both methods.

✅ Volume of triangular prism = 60 cm³

Question 7 — Volume of a Square Pyramid

Question 7

A square pyramid has height 3 m and the perimeter of its square base is 16 m. Find its volume.

Square Pyramid

Square base, side = 4 m, height = 3 m

Volume of Pyramid = (1/3) × Base Area × Height

Step 1: Perimeter of square base = 4l = 16 m l = 16 ÷ 4 = 4 m Step 2: Area of square base = l² = 4² = 16 m² Step 3: Volume = (1/3) × base area × height = (1/3) × 16 × 3 = 16 × 1 ← the 3 and 1/3 cancel = 16 m³

✅ Volume of the square pyramid = 16 m³

📌 Compare with prism: A prism with the same base and height would have volume = 16 × 3 = 48 m³. A pyramid holds exactly one-third of what a prism of equal base and height holds. That 1/3 factor is the key difference.

Question 8 — Water Capacity of an Olympic Swimming Pool

Question 8

An Olympic swimming pool is 50 m long, 25 m wide, and 3 m deep throughout. How many litres of water does it hold?

Swimming Pool (Cuboid)

l = 50 m, b = 25 m, h = 3 m

The swimming pool is in the shape of a cuboid. The volume of water it can hold equals the volume of the cuboid. We then convert cubic metres to litres.

1 cubic metre (m³) = 1000 litres

Given: l = 50 m, b = 25 m, h = 3 m Volume = l × b × h = 50 × 25 × 3 = 3750 m³ Converting to litres: 3750 m³ × 1000 litres/m³ = 37,50,000 litres (37.5 lakh litres)

✅ The Olympic swimming pool holds 37,50,000 litres of water.

💡 Did you know? A standard Olympic pool holds about 2.5 million litres, but this problem uses different dimensions. Always pay attention to given measurements — never assume "standard" sizes in board exam problems.

Common Mistakes to Avoid

Confusing Lateral SA and Total SA: Lateral SA excludes the top and bottom faces. Total SA includes all faces. In "four walls of a room" problems, use Lateral SA, not Total SA.
Forgetting units: Surface area is always in square units (cm², m²); volume is always in cubic units (cm³, m³). Never write volume in cm² or area in cm³.
Using diameter instead of radius: This is more relevant for cylinders and spheres (Chapter 10 later sections), but the habit of misidentifying given values also affects cuboid problems when mixed measurements are given.
Not converting m³ to litres: In Q8, students often stop at 3750 m³ and forget the conversion. Always convert when the question asks for "litres."
Scaling mistake in Q5: When dimensions double, surface area becomes 4× (not 2×). Area scales as the square of the scale factor, not the factor itself.
Missing ½ in Heron's formula: The semi-perimeter s = (a + b + c) ÷ 2. Writing s = (a + b + c) without dividing by 2 gives a completely wrong answer.

⛔ Most common board exam error: In the four-walls problem (Q3), students use Total SA (6 faces) instead of Lateral SA (4 walls). A room's floor and ceiling are not "walls" — only the four vertical sides count. This costs 2 marks per question.

Quick Reference — All Answers at a Glance

Q	Problem	Key Formula	Answer
1(i)	Cube, l = 4 cm	LSA = 4l², TSA = 6l²	LSA = 64 cm², TSA = 96 cm²
1(ii)	Cuboid 8×6×5 cm	LSA = 2h(l+b), TSA = 2(lh+bh+lb)	LSA = 140 cm², TSA = 236 cm²
2	Cube TSA = 1350 m²	6l² = 1350 → l³	Volume = 3375 m³
3	Room 12×10×7.5 m	4 walls = 2h(l+b)	330 m²
4	Cuboid V=1200, l=15, b=10	h = V ÷ (l×b)	h = 8 cm
5(i)	Dimensions doubled	TSA = n²S, n=2	4S (4 times)
5(ii)	Dimensions tripled	TSA = n²S, n=3	9S (9 times)
5(iii)	Dimensions × n	TSA = n²S	n²S
6	Triangular prism, 3-4-5 base, h=10	V = Area × H	60 cm³
7	Square pyramid, perimeter=16, h=3	V = (1/3) × l² × h	16 m³
8	Pool 50×25×3 m	V = l×b×h → litres	37,50,000 litres

What This Exercise Prepares You For

Exercise 10.1 lays the foundation for the rest of Chapter 10, which moves on to cylinders, cones, and spheres (Exercises 10.2 and 10.3). The habits built here — identifying the correct formula, substituting carefully, and converting units — are identical in those sections, just with curved surfaces instead of flat faces.

In Class 10, the entire chapter on Surface Areas and Volumes builds directly on these concepts, adding combinations of solids and conversion problems. For Telangana SSC and AP Board exams, surface area and volume questions account for 5–8 marks in every paper. Students who master Exercise 10.1 thoroughly find the remaining exercises in this chapter significantly easier.

📐 Board Exam Tips (Telangana & AP):

Always write the formula first before substituting values — this earns 1 mark even if arithmetic goes wrong.
Write units in every step, not just the final answer.
For Q5-type conceptual problems, show algebraic working clearly — a numerical example alone is not accepted as proof.