NCERT Class 10 - Surface Areas and Volumes (Chapter 12)

🎯 Key Concepts from Chapter 12

Master these formulas before attempting the exercises. π = 22/7 unless stated otherwise.

📦 Cuboid

Surface Area = 2(lb + bh + hl)
Volume = l × b × h
Diagonal = √(l² + b² + h²)

🧊 Cube

Surface Area = 6a²
Volume = a³
Side from volume: a = ∛V

⭕ Cylinder

CSA = 2πrh
TSA = 2πr(r + h)
Volume = πr²h

🔺 Cone

CSA = πrl (l = slant height)
l = √(r² + h²)
Volume = (1/3)πr²h

🌍 Sphere & Hemisphere

Sphere: Surface = 4πr², Volume = (4/3)πr³
Hemisphere: CSA = 2πr², Volume = (2/3)πr³
Hemisphere TSA = 3πr² (including base)

⚡ Combination Solids Strategy

Surface Area: Add visible curved surfaces only. Don't double-count joined surfaces.
Volume: Simply add volumes of individual solids.
For depressions/hollows: Subtract volume, but ADD curved surface area.

📚 Essential Formulas

Surface Area of Combined Solids:
TSA = Sum of all visible curved surface areas (exclude joined/base surfaces)
Example: Cone on hemisphere = πrl + 2πr² (not 3πr² + πr²)

Volume of Combined Solids:
Volume = Sum of volumes of individual components
Example: Cylinder + 2 hemispheres = πr²h + (4/3)πr³

Key Identities:
π = 22/7 ≈ 3.1429 | When diameter given: r = d/2
Slant height: l = √(r² + h²) | For cone on hemisphere: total height = h_cone + r

💡 Problem-Solving Strategy

1 Visualize the solid — Draw or identify the shape. Is it a combination?

2 Find dimensions — Extract r, h, l from given data. Watch for diameter vs radius!

3 Identify formula — Surface area or volume? Combined or single solid?

4 Calculate step-by-step — Use π = 22/7. Keep fractions when possible.

5 Check units — cm² for area, cm³ for volume. Convert if needed.

📘 Exercise 12.1 — Surface Areas

9 Questions | Combined Solids Surface Area

Surface Area

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

Fig. 12.1 — Two 4 cm cubes joined form an 8×4×4 cuboid

✅ Step-by-Step Solution

1Volume of one cube = 64 cm³

2Side of cube = ∛64 = 4 cm

3Two cubes joined end to end form a cuboid: length = 4+4 = 8 cm, breadth = 4 cm, height = 4 cm

4Surface area = 2(lb + bh + hl) = 2(8×4 + 4×4 + 4×8)

5= 2(32 + 16 + 32) = 2×80 = 160 cm²

🎯 Final Answer: 160 cm²

Combined Solid

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.

Fig. 12.2 — Hollow hemisphere (r=7) + hollow cylinder (h=6)

✅ Step-by-Step Solution

1Radius of hemisphere = 14/2 = 7 cm

2Height of cylindrical part = Total height - Hemisphere radius = 13 - 7 = 6 cm

3Inner surface area = CSA of hemisphere + CSA of cylinder

4CSA of hemisphere = 2πr² = 2×(22/7)×7² = 2×22×7 = 308 cm²

5CSA of cylinder = 2πrh = 2×(22/7)×7×6 = 264 cm²

6Total inner surface area = 308 + 264 = 572 cm²

🎯 Final Answer: 572 cm²

Combined Solid

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.

Fig. 12.3 — Cone (r=3.5, h=12) on hemisphere (r=3.5), l=12.5

✅ Step-by-Step Solution

1Radius r = 3.5 cm (same for cone and hemisphere)

2Height of cone = Total height - Radius of hemisphere = 15.5 - 3.5 = 12 cm

3Slant height of cone: l = √(r² + h²) = √(3.5² + 12²) = √(12.25 + 144) = √156.25 = 12.5 cm

4Total surface area = CSA of cone + CSA of hemisphere

5= πrl + 2πr² = πr(l + 2r)

6= (22/7)×3.5×(12.5 + 7) = 11×19.5 = 214.5 cm²

🎯 Final Answer: 214.5 cm²

Combined Solid

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.

Fig. 12.4 — Hemisphere on cube, greatest diameter = side of cube = 7 cm

✅ Step-by-Step Solution

1Greatest diameter = side of cube = 7 cm (so radius = 3.5 cm)

2Surface area = TSA of cube - base area of hemisphere + CSA of hemisphere

3= 6a² - πr² + 2πr² = 6a² + πr²

4= 6×7² + (22/7)×3.5²

5= 6×49 + (22/7)×12.25 = 294 + 38.5 = 332.5 cm²

🎯 Final Answer: d = 7 cm, SA = 332.5 cm²

Depression

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.

Fig. 12.5 — Hemispherical depression in cube face

✅ Step-by-Step Solution

1Let edge of cube = l, so radius of hemisphere = l/2

2Surface area of remaining solid = TSA of cube - area of circular base of hemisphere + CSA of hemisphere

3= 6l² - π(l/2)² + 2π(l/2)²

4= 6l² - πl²/4 + πl²/2 = 6l² + πl²/4

5Using π = 22/7: = 6l² + (22/7)×l²/4 = 6l² + 11l²/14 = (84l² + 11l²)/14 = 95l²/14

🎯 Final Answer: 6l² + πl²/4 (or 95l²/14 using π=22/7)

Capsule

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

Fig. 12.10 — Capsule = Cylinder + 2 hemispheres

✅ Step-by-Step Solution

1Radius r = 5/2 = 2.5 mm

2Length of cylindrical part = Total length - 2×radius = 14 - 5 = 9 mm

3Surface area = CSA of cylinder + 2 × CSA of hemisphere

4= 2πrh + 2×2πr² = 2πr(h + 2r)

5= 2×(22/7)×2.5×(9 + 5) = (110/7)×14 = 110×2 = 220 mm²

🎯 Final Answer: 220 mm²

Tent

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.

Fig. 12.6 — Tent = Cylinder (h=2.1m, d=4m) + Cone (l=2.8m)

✅ Step-by-Step Solution

1Radius r = 4/2 = 2 m

2Canvas area = CSA of cylinder + CSA of cone (base not covered)

3= 2πrh + πrl

4= 2×(22/7)×2×2.1 + (22/7)×2×2.8

5= (88/7)×2.1 + (44/7)×2.8 = 26.4 + 17.6 = 44 m²

6Cost = 44 × 500 = ₹22,000

🎯 Final Answer: Area = 44 m², Cost = ₹22,000

Cavity

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.

Fig. 12.7 — Conical cavity hollowed from cylinder

✅ Step-by-Step Solution

1Radius r = 1.4/2 = 0.7 cm, Height h = 2.4 cm

2Slant height of cone: l = √(r² + h²) = √(0.49 + 5.76) = √6.25 = 2.5 cm

3TSA of remaining solid = CSA of cylinder + Area of base + CSA of cone

4= 2πrh + πr² + πrl

5= 2×(22/7)×0.7×2.4 + (22/7)×0.49 + (22/7)×0.7×2.5

6= 10.56 + 1.54 + 5.5 = 17.6 cm²

7Rounded to nearest cm² = 18 cm²

🎯 Final Answer: 18 cm²

Scooped

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.

Fig. 12.11 — Cylinder with hemispheres scooped from each end

✅ Step-by-Step Solution

1Radius r = 3.5 cm, Height of cylinder h = 10 cm

2Total surface area = CSA of cylinder + 2 × CSA of hemisphere

3= 2πrh + 2×2πr² = 2πr(h + 2r)

4= 2×(22/7)×3.5×(10 + 7)

5= 22×17 = 374 cm²

🎯 Final Answer: 374 cm²

📗 Exercise 12.2 — Volumes

8 Questions | Combined Solids Volume

Volume

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.

Fig. 12.8 — Cone (r=1, h=1) on hemisphere (r=1)

✅ Step-by-Step Solution

1Radius r = 1 cm (same for cone and hemisphere)

2Height of cone h = 1 cm (equal to radius)

3Volume = Volume of cone + Volume of hemisphere

4= (1/3)πr²h + (2/3)πr³

5= (1/3)π×1²×1 + (2/3)π×1³

6= π/3 + 2π/3 = 3π/3 = π cm³

🎯 Final Answer: π cm³

Volume

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.

Fig. 12.9 — Cylinder with two cones at ends, like an aeroplane model

✅ Step-by-Step Solution

1Radius r = 3/2 = 1.5 cm

2Height of each cone = 2 cm

3Height of cylinder = Total length - 2×cone height = 12 - 4 = 8 cm

4Volume of air = Volume of cylinder + 2×Volume of cone

5= πr²h + 2×(1/3)πr²h_cone

6= π×(1.5)²×8 + 2×(1/3)×π×(1.5)²×2

7= π×2.25×8 + (4/3)×π×2.25

8= 18π + 3π = 21π cm³ = 66 cm³ (using π=22/7)

🎯 Final Answer: 66 cm³ (or 21π cm³)

Volume

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.

Fig. 12.15 — Gulab jamun = Cylinder + 2 hemispheres

✅ Step-by-Step Solution

1Radius r = 2.8/2 = 1.4 cm

2Height of cylindrical part = 5 - 2×1.4 = 2.2 cm

3Volume of one gulab jamun = πr²h + (4/3)πr³

4= (22/7)×1.4²×2.2 + (4/3)×(22/7)×1.4³

5= (22/7)×1.96×2.2 + (88/21)×2.744

6= 6.16×2.2 + 11.498 ≈ 13.552 + 11.498 ≈ 25.05 cm³

7Syrup in one = 30% of 25.05 ≈ 7.515 cm³

8Syrup in 45 = 45 × 7.515 ≈ 338.2 ≈ 338 cm³

🎯 Final Answer: 338 cm³

Volume

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.

Fig. 12.16 — Cuboid with 4 conical depressions for pens

✅ Step-by-Step Solution

1Volume of cuboid = 15×10×3.5 = 525 cm³

2Radius of each cone r = 0.5 cm, height (depth) = 1.4 cm

3Volume of one conical depression = (1/3)πr²h

4= (1/3)×(22/7)×0.5²×1.4

5= (1/3)×(22/7)×0.25×1.4 = (22×0.35)/21 = 7.7/21 ≈ 0.3667 cm³

6Volume of 4 depressions = 4×0.3667 ≈ 1.467 cm³

7Volume of wood = 525 - 1.467 ≈ 523.53 cm³ ≈ 523.5 cm³

🎯 Final Answer: 523.5 cm³

Volume

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.

Fig. 12.12 — Lead shots displace water in inverted cone

✅ Step-by-Step Solution

1Volume of cone = (1/3)πr²h = (1/3)×(22/7)×5²×8 = (22×200)/21 = 4400/21 cm³

2Water that flows out = 1/4 × 4400/21 = 1100/21 cm³

3Volume of one lead shot (sphere) = (4/3)πr³

4= (4/3)×(22/7)×(0.5)³ = (4/3)×(22/7)×0.125 = 11/21 cm³

5Number of lead shots = (1100/21) ÷ (11/21) = 1100/11 = 100

6Therefore, 100 lead shots were dropped.

🎯 Final Answer: 100

Volume

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)

Fig. 12.13 — Iron pole = Large cylinder + Small cylinder

✅ Step-by-Step Solution

1First cylinder: r₁ = 12 cm, h₁ = 220 cm

2Volume₁ = πr₁²h₁ = 3.14×12²×220 = 3.14×144×220 = 99,532.8 cm³

3Second cylinder: r₂ = 8 cm, h₂ = 60 cm

4Volume₂ = πr₂²h₂ = 3.14×8²×60 = 3.14×64×60 = 12,057.6 cm³

5Total volume = 99,532.8 + 12,057.6 = 111,590.4 cm³

6Mass = 111,590.4 × 8 = 892,723.2 g = 892.72 kg ≈ 892.3 kg

🎯 Final Answer: 892.3 kg

Volume

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.

Fig. 12.14 — Solid displaces water in full cylinder

✅ Step-by-Step Solution

1Volume of cone = (1/3)πr²h = (1/3)π×60²×120 = 144,000π cm³

2Volume of hemisphere = (2/3)πr³ = (2/3)π×60³ = 144,000π cm³

3Volume of solid = 144,000π + 144,000π = 288,000π cm³

4Volume of cylinder = πr²h = π×60²×180 = 648,000π cm³

5Water left = 648,000π - 288,000π = 360,000π cm³

6= 360,000×(22/7) = 7,920,000/7 ≈ 1,131,428.6 cm³ ≈ 1,131,400 cm³

🎯 Final Answer: 1,131,400 cm³ (or 360,000π cm³)

Verification

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.

Fig. 12.17 — Spherical vessel with cylindrical neck

✅ Step-by-Step Solution

1Radius of spherical part = 8.5/2 = 4.25 cm

2Volume of sphere = (4/3)πr³ = (4/3)×3.14×4.25³

3= (4/3)×3.14×76.765625 = 321.39 cm³

4Radius of cylindrical neck = 2/2 = 1 cm, height = 8 cm

5Volume of cylinder = πr²h = 3.14×1²×8 = 25.12 cm³

6Total volume = 321.39 + 25.12 = 346.51 cm³

7Child's measurement = 345 cm³

8Difference = 346.51 - 345 = 1.51 cm³

9The child is approximately correct (within measurement error), but more precisely actual volume ≈ 346.5 cm³

🎯 Final Answer: No, actual volume is about 346.5 cm³ (close but not exactly 345)

🧮 Surface Area & Volume Calculators

Interactive tools for all solid geometry calculations

📦 Cuboid Calculator

Length (cm)

Breadth (cm)

Height (cm)

Surface Area: 160.00 cm²

Volume: 128.00 cm³

⭕ Cylinder Calculator

Radius (cm)

Height (cm)

CSA: 264.00 cm²

TSA: 572.00 cm²

Volume: 924.00 cm³

🔺 Cone Calculator

Radius (cm)

Height (cm)

Slant Height: 12.50 cm

CSA: 137.50 cm²

Volume: 154.00 cm³

🌍 Sphere & Hemisphere

Radius (cm)

Sphere SA: 616.00 cm²

Sphere Vol: 1437.33 cm³

Hemisphere CSA: 308.00 cm²

Hemisphere Vol: 718.67 cm³

💊 Capsule Calculator (Q6 type)

Total Length (mm)

Diameter (mm)

Surface Area: 220.00 mm²

Volume: 179.59 mm³

🏕️ Tent Calculator (Q7 type)

Cylinder Height (m)

Diameter (m)

Cone Slant Height (m)

Canvas Rate (₹/m²)

Canvas Area: 44.00 m²

Cost: ₹22,000

🍬 Gulab Jamun Calculator (Q3 type)

Length (cm)

Diameter (cm)

Syrup %

Number of Jamuns

Total Syrup: 338.05 cm³

⚖️ Mass Calculator (Q6 type)

Solid Type

Dimension 1 (cm)

Dimension 2 (cm)

Density (g/cm³)

Mass: 796.18 kg

📦 Surface Areas and Volumes

📦 Cuboid

🧊 Cube

⭕ Cylinder

🔺 Cone

🌍 Sphere & Hemisphere

⚡ Combination Solids Strategy

📘 Exercise 12.1 — Surface Areas

📗 Exercise 12.2 — Volumes

🧮 Surface Area & Volume Calculators