NCERT Solutions for Class 10 Maths Chapter 12: Surface Areas and Volumes (NCERT 2026–27)

These Class 10 Maths Chapter 12 solutions cover Surface Areas and Volumes from the latest NCERT textbook (Reprint 2026–27). Every question in Exercise 12.1 and Exercise 12.2 is solved step by step, showing how to break a combination of solids into a cuboid, cone, cylinder, sphere or hemisphere, so you can find surface areas, volumes and capacities with confidence.

Class: 10 Subject: Mathematics Chapter: 12 Chapter Name: Surface Areas and Volumes Exercises: 12.1 (9 Q), 12.2 (8 Q) Session: 2026–27
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Chapter 12 Overview

Chapter 12, Surface Areas and Volumes, extends what you learnt in Class 9 about the cuboid, cone, cylinder, sphere and hemisphere to combinations of solids — objects such as a capsule, a tent, a toy, a bird-bath or a decorative block. You learn two key ideas: (1) the surface area of a combined solid is the sum of the curved/exposed surfaces only (the joined faces vanish), and (2) the volume of a combined solid is simply the sum of the volumes of its parts. The chapter has two exercises — Exercise 12.1 on surface area and Exercise 12.2 on volume and capacity — both fully solved below.

Key Concepts & Definitions

Combination of solids: a real object built by joining two or more basic solids (e.g. a cylinder with two hemispherical ends).

Total Surface Area (TSA): the area of all outer surfaces of a solid. When solids are joined, only the exposed surfaces are added — the surfaces that touch are removed.

Curved Surface Area (CSA): the area of the curved (non-flat) part only, e.g. the side of a cylinder or cone, or the dome of a hemisphere.

Volume: the space occupied by a solid. For a combined solid, volume = sum of the volumes of all the parts (nothing is lost on joining).

Capacity: the volume a hollow container can hold. A raised or hollowed portion (e.g. a hemispherical depression) reduces the capacity.

Slant height (cone): l = √(r2 + h2), the distance from the apex to the edge of the base.

Important Formulas (Chapter 12)

Cuboid: Volume = l × b × h  •  TSA = 2(lb + bh + hl).

Cube: Volume = a3  •  TSA = 6a2.

Cylinder: Volume = πr2h  •  CSA = 2πrh  •  TSA = 2πr(h + r).

Cone: Volume = ⅓πr2h  •  CSA = πrl  •  TSA = πr(l + r), where l = √(r2 + h2).

Sphere: Volume = ⅔πr3  •  Surface area = 4πr2.

Hemisphere: Volume = ⅔πr3  •  CSA = 2πr2  •  TSA = 3πr2.

Combined solid: TSA = sum of exposed CSAs  •  Volume = sum of part volumes.

Exercise 12.1 Solutions

Unless stated otherwise, take π = 22/7. Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified.

1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

SOLUTION Edge of each cube: a3 = 64 ⇒ a = 4 cm. Joining two cubes end to end gives a cuboid of length 4 + 4 = 8 cm, breadth 4 cm, height 4 cm. Surface area = 2(lb + bh + hl) = 2(8×4 + 4×4 + 4×8) = 2(32 + 16 + 32) = 2 × 80. ∴ Surface area = 160 cm2.

2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

SOLUTION Radius r = 14/2 = 7 cm. Height of cylinder h = total height − radius of hemisphere = 13 − 7 = 6 cm. Inner surface area = CSA of cylinder + CSA of hemisphere = 2πrh + 2πr2 = 2πr(h + r). = 2 × (22/7) × 7 × (6 + 7) = 2 × 22 × 13 = 572. ∴ Inner surface area = 572 cm2.

3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

SOLUTION Radius r = 3.5 cm. Height of cone h = 15.5 − 3.5 = 12 cm. Slant height l = √(r2 + h2) = √(3.52 + 122) = √(12.25 + 144) = √156.25 = 12.5 cm. TSA of toy = CSA of cone + CSA of hemisphere = πrl + 2πr2 = πr(l + 2r). = (22/7) × 3.5 × (12.5 + 7) = 11 × 19.5 = 214.5. ∴ Total surface area = 214.5 cm2.

4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

SOLUTION The greatest diameter of the hemisphere equals the side of the cube = 7 cm, so its radius r = 3.5 cm. Surface area = TSA of cube − base area of hemisphere + CSA of hemisphere = 6a2 − πr2 + 2πr2 = 6a2 + πr2. = 6 × 72 + (22/7) × 3.5 × 3.5 = 294 + 38.5 = 332.5. ∴ Surface area of the solid = 332.5 cm2.

5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

SOLUTION Let the edge of the cube = l, so the hemisphere has diameter l and radius r = l/2. Surface area of remaining solid = TSA of cube − area of circular top of depression + CSA of hemisphere = 6l2 − πr2 + 2πr2 = 6l2 + πr2. = 6l2 + π(l/2)2 = 6l2 + πl2/4 = (l2/4)(24 + π). ∴ Surface area = (l2/4)(24 + π) = ¼l2(24 + π) square units.

6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 12.10). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

SOLUTION Radius r = 5/2 = 2.5 mm. The two hemispheres at the ends have radius 2.5 mm each, so cylinder length h = 14 − 2(2.5) = 14 − 5 = 9 mm. Surface area = CSA of cylinder + 2 × CSA of hemisphere = 2πrh + 2(2πr2) = 2πr(h + 2r). = 2 × (22/7) × 2.5 × (9 + 5) = 2 × (22/7) × 2.5 × 14 = 220. ∴ Surface area = 220 mm2 (= 220/7 × 7, exactly 220 mm2).

7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ 500 per m2. (Note that the base of the tent will not be covered with canvas.)

SOLUTION Radius r = 4/2 = 2 m, cylinder height h = 2.1 m, cone slant height l = 2.8 m. Canvas area = CSA of cylinder + CSA of cone = 2πrh + πrl = πr(2h + l). = (22/7) × 2 × (2×2.1 + 2.8) = (22/7) × 2 × (4.2 + 2.8) = (22/7) × 2 × 7 = 44. Area of canvas = 44 m2. Cost = 44 × ₹500 = ₹22000.

8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.

SOLUTION Radius r = 1.4/2 = 0.7 cm, height h = 2.4 cm (same for cylinder and cone). Slant height of cone l = √(r2 + h2) = √(0.72 + 2.42) = √(0.49 + 5.76) = √6.25 = 2.5 cm. The remaining solid has: CSA of cylinder (outside) + area of one circular base + CSA of the conical cavity (inner surface). TSA = 2πrh + πr2 + πrl = πr(2h + r + l). = (22/7) × 0.7 × (4.8 + 0.7 + 2.5) = 2.2 × 8 = 17.6 cm2. ∴ TSA ≈ 18 cm2 (to the nearest cm2).

9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 12.11. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

SOLUTION Radius r = 3.5 cm, cylinder height h = 10 cm. A hemisphere of radius 3.5 cm is scooped from each end. TSA = CSA of cylinder + 2 × CSA of hemisphere (the two flat circular ends are replaced by the curved hollows) = 2πrh + 2(2πr2) = 2πr(h + 2r). = 2 × (22/7) × 3.5 × (10 + 7) = 22 × 17 = 374. ∴ Total surface area of the article = 374 cm2.

Exercise 12.2 Solutions

Unless stated otherwise, take π = 22/7. Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified.

1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

SOLUTION Radius r = 1 cm; height of cone h = r = 1 cm. Volume = volume of cone + volume of hemisphere = ⅓πr2h + ⅔πr3. = ⅓π(1)2(1) + ⅔π(1)3 = ⅓π + ⅔π = π. ∴ Volume of the solid = π cm3.

2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

SOLUTION Radius r = 3/2 = 1.5 cm. Each cone has height 2 cm, so the two cones together use 2 × 2 = 4 cm of the length; cylinder height h = 12 − 4 = 8 cm. Volume = volume of cylinder + 2 × volume of cone = πr2h + 2(⅓πr2hcone). = π(1.5)2(8) + 2 × ⅓π(1.5)2(2) = π(2.25)(8) + (2/3)π(2.25)(2) = 18π + 3π = 21π. = 21 × (22/7) = 66. ∴ Volume of air = 66 cm3.

3. A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see Fig. 12.15).

SOLUTION Radius r = 2.8/2 = 1.4 cm. The two hemispherical ends together make one sphere of radius 1.4 cm. Cylinder length h = 5 − 2(1.4) = 5 − 2.8 = 2.2 cm. Volume of one gulab jamun = πr2h + ⅔πr3 × … (two hemispheres = one sphere) = πr2h + (4/3)πr3. = (22/7)(1.96)(2.2) + (4/3)(22/7)(2.744) = 13.552 + 11.499 = 25.05 cm3 (approx.). Volume of 45 gulab jamuns = 45 × 25.05 = 1127.28 cm3. Syrup = 30% of 1127.28 = 0.30 × 1127.28 ≈ 338.18 cm3. ∴ Syrup ≈ 338 cm3.

4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see Fig. 12.16).

SOLUTION Volume of cuboid = 15 × 10 × 3.5 = 525 cm3. Volume of one conical depression = ⅓πr2h = ⅓ × (22/7) × (0.5)2 × 1.4 = ⅓ × (22/7) × 0.25 × 1.4 = ⅓ × 1.1 = 0.3667 cm3. Volume of 4 depressions = 4 × 0.3667 = 1.4667 cm3. Volume of wood = 525 − 1.4667 = 523.53 cm3. ∴ Volume of wood in the stand ≈ 523.53 cm3.

5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

SOLUTION Volume of cone (full water) = ⅓πr2h = ⅓π(5)2(8) = (200/3)π cm3. Water that flows out = ¼ of this = ¼ × (200/3)π = (50/3)π cm3. This equals the total volume of the lead shots. Volume of one lead shot = ⅔π(0.5)3 = ⅔π(0.125) = (1/6)π cm3. Number of shots = [(50/3)π] ÷ [(1/6)π] = (50/3) × 6 = 100. ∴ Number of lead shots = 100.

6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8g mass. (Use π = 3.14)

SOLUTION Lower cylinder: radius 24/2 = 12 cm, height 220 cm ⇒ V1 = π(12)2(220) = 31680π cm3. Upper cylinder: radius 8 cm, height 60 cm ⇒ V2 = π(8)2(60) = 3840π cm3. Total volume = (31680 + 3840)π = 35520π = 35520 × 3.14 = 111532.8 cm3. Mass = 111532.8 × 8 = 892262.4 g. ∴ Mass of the pole = 892262.4 g ≈ 892.26 kg.

7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

SOLUTION Volume of cylinder = πr2h = π(60)2(180) = 648000π cm3. Volume of solid = volume of cone + volume of hemisphere = ⅓π(60)2(120) + ⅔π(60)3. = ⅓π(3600)(120) + ⅔π(216000) = 144000π + 144000π = 288000π cm3. Water left = 648000π − 288000π = 360000π cm3. Using π = 22/7: = 360000 × (22/7) = 1131428.57 cm3 ≈ 1.131 m3. ∴ Volume of water left ≈ 1131428.57 cm3 ≈ 1.131 m3.

8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

SOLUTION Neck (cylinder): radius 2/2 = 1 cm, length 8 cm ⇒ Vneck = πr2h = 3.14 × (1)2 × 8 = 25.12 cm3. Spherical part: radius 8.5/2 = 4.25 cm ⇒ Vsphere = ⅔πr3 = (4/3) × 3.14 × (4.25)3 = (4/3) × 3.14 × 76.7656 = 321.39 cm3. Total volume = 25.12 + 321.39 = 346.51 cm3. The child measured 345 cm3, but the correct volume is ≈ 346.51 cm3. ∴ She is not correct; the actual capacity is approximately 346.51 cm3.

Common Mistakes to Avoid

Watch out for these

  • For surface area, adding the full surface areas of the parts — you must drop the faces that get joined or covered and add only the exposed surfaces.
  • Confusing diameter with radius — always halve the diameter first (e.g. d = 14 cm ⇒ r = 7 cm).
  • Forgetting to subtract the part’s length when finding a cylinder’s height in a capsule/gulab jamun (subtract 2r for two hemispherical ends, or 2×cone height for two cones).
  • Using the height of the cone as its slant height — compute l = √(r2 + h2) separately.
  • Mixing up the π value asked for — some questions specify π = 3.14, others 22/7.
  • For volume problems, wrongly removing joined surfaces — volumes of parts are simply added (nothing is lost on joining).

Practice MCQs & Assertion–Reason

1. Two cubes of volume 64 cm3 each are joined end to end. The surface area of the resulting cuboid is:

(a) 128 cm2    (b) 160 cm2    (c) 192 cm2    (d) 96 cm2

2. The total surface area of a hemisphere of radius r is:

(a) 2πr2    (b) 3πr2    (c) 4πr2    (d) πr2

3. When two basic solids are joined to form one solid, the volume of the combination is:

(a) the difference of the volumes    (b) the sum of the volumes    (c) the larger volume only    (d) less than either volume

4. A cone of radius 3.5 cm is mounted on a hemisphere of the same radius; total height of the toy is 15.5 cm. The slant height of the cone is:

(a) 12 cm    (b) 12.5 cm    (c) 13 cm    (d) 11.5 cm

5. The curved surface area of a cylinder of radius r and height h is:

(a) πr2h    (b) 2πr2    (c) 2πrh    (d) 2πr(r + h)

6. A capsule is a cylinder with two hemispherical ends of radius 2.5 mm; if total length is 14 mm, the cylinder length is:

(a) 14 mm    (b) 11.5 mm    (c) 9 mm    (d) 7 mm

7. The number of spheres of radius 0.5 cm whose total volume equals (50/3)π cm3 is:

(a) 50    (b) 75    (c) 100    (d) 150

8. The volume of a cone equals which fraction of the volume of a cylinder of the same base and height?

(a) ½    (b) ⅓    (c) ⅔    (d) ¼

9. While computing the surface area of a combination of solids, the surfaces that are joined together are:

(a) added twice    (b) added once    (c) not included    (d) doubled

10. A hemispherical depression of radius l/2 is cut from one face of a cube of edge l. The surface area of the remaining solid is:

(a) 6l2 − πl2/4    (b) 6l2 + πl2/4    (c) 6l2    (d) 6l2 + πl2/2

Answer key: 1-(b), 2-(b), 3-(b), 4-(b), 5-(c), 6-(c), 7-(c), 8-(b), 9-(c), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: When two solids are joined, the total surface area of the new solid is less than the sum of the surface areas of the two solids.

Reason: The surfaces in contact are hidden and are not counted in the surface area of the combination.

A-R 2. Assertion: The volume of a solid formed by joining two basic solids is the sum of their volumes.

Reason: No part of the volume is lost when two solids are joined together.

A-R 3. Assertion: The slant height of a cone of radius 3.5 cm and height 12 cm is 12.5 cm.

Reason: The slant height of a cone is l = r2 + h2.

A-R 4. Assertion: The total surface area of a hemisphere is twice its curved surface area.

Reason: TSA of a hemisphere = 3πr2 while its CSA = 2πr2.

A-R 5. Assertion: A hemispherical depression cut from a cube increases the surface area of the solid by 2πr2 − πr2 = πr2.

Reason: Cutting the depression removes the flat circular area πr2 and adds the curved area 2πr2.

Answer key: 1-(A), 2-(A), 3-(C), 4-(A), 5-(A).

Quick Revision Summary

  • Real objects are often combinations of a cuboid, cube, cone, cylinder, sphere or hemisphere.
  • Surface area of a combination = sum of the exposed surfaces; surfaces in contact are not counted.
  • Volume of a combination = sum of the volumes of the parts (nothing is lost on joining).
  • Cone: V = ⅓πr2h, CSA = πrl, with l = √(r2 + h2).
  • Cylinder: V = πr2h, CSA = 2πrh. Sphere: V = ⅔πr3, SA = 4πr2.
  • Hemisphere: V = ⅔πr3, CSA = 2πr2, TSA = 3πr2.
  • Capacity of a container is reduced by any raised/hollowed portion inside it.

How to score full marks in this chapter

Start every problem by naming the basic solids and writing their radii and heights (convert diameter to radius first). Decide whether the question asks for surface area (add only exposed parts) or volume (add all parts). Compute the slant height separately for cones, keep the value of π the question specifies, and write your final answer with the correct unit (cm2/cm3/m3). Show each step clearly so every line earns its mark.

Frequently Asked Questions

What is Class 10 Maths Chapter 12 about?

Chapter 12, Surface Areas and Volumes, teaches how to find the surface area, volume and capacity of solids made by combining basic shapes — cuboid, cube, cone, cylinder, sphere and hemisphere — such as capsules, tents, toys and bird-baths.

How many exercises are there in Class 10 Maths Chapter 12?

There are two exercises: Exercise 12.1 (9 questions on surface area of combinations) and Exercise 12.2 (8 questions on volume and capacity). All 17 questions are solved step by step on this page.

Why is the surface area of a combined solid less than the sum of the parts?

When two solids are joined, the faces that touch are hidden inside and are no longer on the outside. So those contact surfaces are not counted, making the combined surface area less than the simple sum of the two separate surface areas.

Are these Class 10 Maths Chapter 12 solutions free?

Yes. All solutions are free and follow the official NCERT Mathematics textbook for the 2026–27 session, with every answer verified.

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