Surface Areas and Volumes – Class 10 Maths Chapter 12 Notes & NCERT PDF

Class 10MathematicsChapter 12NCERT book: Mathematics - Class 10

Notes and a simple summary of Chapter 12 of the NCERT Class 10 Mathematics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

🎧 Listen to the content

Chapter 12: Surface Areas and Volumes

Introduction

In Class IX you learnt the surface areas and volumes of the basic solids — cuboid, cone, cylinder and sphere. But real objects are rarely just one solid. An oil tanker on a truck is a cylinder with two hemispheres at its ends; a test tube is a cylinder with a hemisphere; a playing top (lattu) is a cone surmounted by a hemisphere. This chapter teaches you to handle such combinations of solids: first their surface areas (12.2), then their volumes (12.3).

The key idea, in the book's words: whenever we come across a new problem, we first try to see if we can break it down into smaller problems we have earlier solved.

Key Concepts

1. Surface area of a combination (Section 12.2)

When two solids are joined, the faces that are stuck together disappear from the outside. So you add only the surfaces you can actually see (or paint).

Tanker (cylinder + two hemispheres): TSA = CSA of hemisphere + CSA of cylinder + CSA of other hemisphere

Toy (cone on a hemisphere): TSA = CSA of hemisphere + CSA of cone

The total surface area of the combined solid is not the sum of the total surface areas of its parts.

Sometimes you must subtract a hidden face and add a new curved one. For a cube with a hemisphere on top: surface area = TSA of cube − base area of hemisphere + CSA of hemisphere = 6a² + πr².

2. Volume of a combination (Section 12.3)

Volume behaves differently: no volume disappears when solids are joined, so the volume of the new solid is simply the sum of the volumes of its parts (and a difference when something is scooped out or hollowed).

Methods Step by Step

Identify the basic solids and draw a rough sketch.

Write down every radius, height and slant height. Watch out for "diameter" in the question — halve it.

For a cone, find the slant height with l = √(r² + h²) if it is not given.

Surface area: list only the exposed faces, then add (and subtract hidden flat faces).

Volume: add the volumes of the parts; subtract any cavity or depression.

Use the value of π the question asks for (22/7 by default in this chapter, 3.14 when stated), and keep units consistent (convert m to cm, etc.).

Worked Examples (from the chapter)

Example 1 — Rasheed's top. Height 5 cm, diameter 3.5 cm, so r = 1.75 cm. Cone height = 5 − 1.75 = 3.25 cm, slant height l ≈ 3.7 cm.

TSA = 2πr² + πrl = (22/7) × 1.75 × (3.5 + 3.7) ≈ 39.6 cm².

Example 2 — Decorative block. Cube of edge 5 cm with a hemisphere of diameter 4.2 cm (r = 2.1 cm).

Area = 150 − πr² + 2πr² = 150 + πr² = 150 + 13.86 = 163.86 cm².

Example 3 — Toy rocket (π = 3.14). Cone: r = 2.5 cm, h = 6 cm, l = 6.5 cm; cylinder: r = 1.5 cm, h = 20 cm.

Orange part = πrl + πr² − π(1.5)² = 3.14 × 20.25 = 63.585 cm²; yellow part = π × 1.5 × (2 × 20 + 1.5) = 195.465 cm². A ring of the cone's base is painted because the cone's base is wider than the cylinder.

Example 4 — Bird-bath. Cylinder (h = 145 cm, r = 30 cm) with a hemispherical depression: TSA = 2πrh + 2πr² = 2πr(h + r) = 33000 cm² = 3.3 m².

Example 5 — Shanta's shed. Cuboid 15 m × 7 m × 8 m plus a half cylinder (diameter 7 m, length 15 m):

Volume = 840 + ½ × (22/7) × 3.5 × 3.5 × 15 = 1128.75 m³. Removing machinery (300 m³) and 20 workers (20 × 0.08 = 1.6 m³) leaves 827.15 m³ of air.

Example 6 — Juice glass (π = 3.14). Apparent capacity = πr²h = 196.25 cm³; the raised hemisphere takes ⅔πr³ ≈ 32.71 cm³, so actual capacity ≈ 163.54 cm³.

Example 7 — Toy in a cylinder. Hemisphere + cone with r = 2 cm, cone height 2 cm: volume = ⅔π(8) + ⅓π(4)(2) = 25.12 cm³. The circumscribing cylinder (r = 2, h = 4) holds 50.24 cm³, so the difference is 25.12 cm³.

Board Exam Focus

Most questions give you a picture of a real object — capsule, tent, pen stand, gulab jamun, vessel. Name the parts before you calculate.

Tent questions: the base is not covered with canvas, so leave it out.

"Hollowed out" or "scooped out" means subtract volume but add the curved inner surface to the area.

Two cubes of volume 64 cm³ joined end to end give a cuboid 8 cm × 4 cm × 4 cm with surface area 160 cm².

Formula Sheet

Cuboid: TSA = 2(lb + bh + hl); V = lbh. Cube: TSA = 6a²; V = a³

Cylinder: CSA = 2πrh; TSA = 2πr(r + h); V = πr²h

Cone: l = √(r² + h²); CSA = πrl; TSA = πr(l + r); V = ⅓πr²h

Sphere: SA = 4πr²; V = (4/3)πr³

Hemisphere: CSA = 2πr²; TSA = 3πr²; V = (2/3)πr³

Common Mistakes

Adding full TSAs of the parts — the joined faces must be left out.

Using the cone's height instead of its slant height in πrl.

Forgetting to subtract the hemisphere's radius from the total height to get the cone's or cylinder's height (as in the top and the capsule).

Using the diameter as the radius.

Mixing units (metres and centimetres) or forgetting that area is in square units and volume in cubic units.

💡 Key Learning Points

✓Break a complex object into basic solids (cuboid, cube, cylinder, cone, sphere, hemisphere) before calculating.
✓Surface area of a combined solid counts only exposed faces — joined faces are left out, hidden flat faces are subtracted.
✓Volume of a combined solid is the sum of the volumes of its parts, minus any cavity.
✓Find the slant height of a cone with l = √(r² + h²) and use it in the curved surface area πrl.
✓Apply the correct π (22/7 or 3.14) and keep units consistent in real-life problems like tents, capsules and tanks.

👨‍🏫 Teaching Tips

→Bring real objects — a capsule, a test tube, a lattu, a ice-cream cone — and ask students to name the basic solids in each.
→Use clay or paper models: join a cone to a hemisphere and show physically which faces disappear.
→Make students paint (or shade) the exposed faces on a sketch before writing any formula.
→Build a class 'formula wall' for all basic solids and refer to it while solving.
→Contrast Example 1 (surface area) with Example 7 (volume) to show why areas are not simply added but volumes are.
→Give a 'units check' routine: underline every measurement and convert to one unit first.
→Pair students: one identifies the parts, the other writes the formula, then they swap.

📋 Assessment Questions

Can the student list which faces are visible when two solids are joined?
Can the student find the surface area of a toy made of a cone on a hemisphere?
Can the student calculate the volume of air in a shed shaped like a cuboid with a half cylinder on top?
Can the student handle 'hollowed out' or 'depression' problems correctly for both area and volume?
Does the student use l = √(r² + h²) when the slant height is not given?

Ask the AI tutor about Class 10 Mathematics Chapter 12

Get question answers, important questions, a quick MCQ quiz or a simple explanation of “Surface Areas and Volumes”. Free — sign in with Google.

How the AI tutor works →

👩‍🏫 Class 10 Mathematics classes near you — online or at home

Prefer a real teacher? TalentJR teachers take live online classes anywhere in India and visit homes in selected cities. Tell us what you need and our team will call you with timings and fees.

Live online classesSmall groups or 1:1 with a teacher, from home, any city.
Teacher at homeA TalentJR teacher visits your home (selected cities).
At a centreClasses at a TalentJR partner centre near you, where available.