📐 NCERT Class 10 Mathematics

Chapter 12: Surface Areas and Volumes

Complete theory, formulas for surface areas and volumes of cubes, cuboids, spheres, hemispheres, cones, cylinders, frustums, and combinations of solids. Prepared strictly according to CBSE board exam pattern 2025-26.

📖 Introduction 🧊 Cube & Cuboid 🥫 Cylinder 🍦 Cone ⚽ Sphere & Hemisphere 📐 Frustum of Cone 🔷 Combinations of Solids 📝 Solved Examples ⭐ Key Points

📖 Introduction to Surface Areas and Volumes

In this chapter, we will learn how to calculate the surface areas (total surface area and curved/lateral surface area) and volumes of various 3D shapes — cubes, cuboids, cylinders, cones, spheres, hemispheres, and frustums. We will also learn how to find surface areas and volumes of combinations of solids (like a cone mounted on a cylinder, a hemisphere on a cube, etc.).

💡 Important Distinction:

Total Surface Area (TSA): Sum of areas of all surfaces (including top and bottom).
Curved/Lateral Surface Area (CSA/LSA): Area of only the curved surface (excluding top and bottom).
Volume: The amount of space occupied by the solid.

🧊 Cube and Cuboid

📌 Cuboid

Length = l, Breadth = b, Height = h
Lateral Surface Area = 2h(l + b)
Total Surface Area = 2(lb + bh + hl)
Volume = l × b × h
Diagonal = √(l² + b² + h²)

📌 Cube

Side = a
Lateral Surface Area = 4a²
Total Surface Area = 6a²
Volume = a³
Diagonal = a√3

📝 Example

Find the total surface area and volume of a cube of side 7 cm.
Solution: TSA = 6 × 7² = 6 × 49 = 294 cm². Volume = 7³ = 343 cm³.

🥫 Cylinder

📌 Right Circular Cylinder

Radius = r, Height = h
Curved Surface Area = 2πrh
Total Surface Area = 2πr(r + h)
Volume = πr²h

📌 Hollow Cylinder

External radius = R, Internal radius = r, Height = h
CSA = 2πh(R + r)
TSA = 2π(R + r)(R - r + h)
Volume = πh(R² - r²)

📝 Example

Find the curved surface area and volume of a cylinder of radius 7 cm and height 10 cm.
Solution: CSA = 2 × 22/7 × 7 × 10 = 2 × 22 × 10 = 440 cm².
Volume = 22/7 × 7² × 10 = 22 × 7 × 10 = 1540 cm³.

🍦 Cone

📌 Right Circular Cone

Radius = r, Height = h, Slant Height = l
l = √(r² + h²)
Curved Surface Area = πrl
Total Surface Area = πr(r + l)
Volume = (1/3)πr²h

📝 Example

Find the curved surface area of a cone with radius 7 cm and height 24 cm.
Solution: l = √(7² + 24²) = √(49 + 576) = √625 = 25 cm.
CSA = π × 7 × 25 = 22/7 × 7 × 25 = 22 × 25 = 550 cm².

⚽ Sphere and Hemisphere

📌 Sphere

Radius = r
Surface Area = 4πr²
Volume = (4/3)πr³

📌 Hemisphere (Half Sphere)

Radius = r
Curved Surface Area = 2πr²
Total Surface Area = 3πr²
Volume = (2/3)πr³

📝 Example

Find the surface area and volume of a sphere of radius 21 cm.
Solution: SA = 4 × 22/7 × 21² = 4 × 22/7 × 441 = 4 × 22 × 63 = 5544 cm².
Volume = (4/3) × 22/7 × 21³ = (4/3) × 22/7 × 9261 = (4/3) × 22 × 1323 = 4 × 22 × 441 = 38808 cm³.

📐 Frustum of a Cone

A frustum is the portion of a cone that remains after cutting off the top portion by a plane parallel to the base.

📌 Frustum Formulas

Radius of top = r₁, Radius of base = r₂, Height = h, Slant height = l
l = √[h² + (r₂ - r₁)²]
Curved Surface Area = πl(r₁ + r₂)
Total Surface Area = π[r₁² + r₂² + l(r₁ + r₂)]
Volume = (1/3)πh(r₁² + r₂² + r₁r₂)

📝 Example

A bucket is in the shape of a frustum with top radius 14 cm, bottom radius 7 cm, and height 24 cm. Find its volume.
Solution: Volume = (1/3) × 22/7 × 24 × (14² + 7² + 14×7) = (1/3) × 22/7 × 24 × (196 + 49 + 98) = (1/3) × 22/7 × 24 × 343 = 22/7 × 8 × 343 = 22 × 8 × 49 = 8624 cm³.

🔷 Combinations of Solids

Many real-life objects are combinations of two or more solids — a cone on a cylinder (tent), a hemisphere on a cube (trophy), a cylinder with hemispherical ends (capsule), etc.

💡 Important Points to Remember:

Total Surface Area: Sum of surface areas of all visible parts. Do not include areas where two solids join (the base of one solid attached to the other is not counted).
Volume: Sum of volumes of individual solids (no subtraction).
When a solid is melted and recast into another shape → Volume remains constant.
When a solid is cut into smaller pieces → Total volume remains the same.

📝 Example

A solid is in the shape of a cone mounted on a cylinder. The height of cylinder is 10 cm, radius of cylinder = 7 cm, and height of cone = 6 cm. Find total volume and TSA.

Solution: Volume = πr²h(cylinder) + (1/3)πr²h(cone) = (22/7)×49×10 + (1/3)×(22/7)×49×6 = 1540 + 308 = 1848 cm³.
TSA = CSA of cylinder + CSA of cone + base area of cylinder = 2πrh + πrl + πr². (l = √(7²+6²)=√85≈9.22cm).

📝 NCERT Solved Examples (Detailed)

🔹 Example 1 (Conversion of Solid)

Question: A solid iron sphere of radius 6 cm is melted and recast into a cylinder of radius 4 cm. Find the height of the cylinder.

Solution: Volume of sphere = (4/3)πr³ = (4/3)π × 216 = 288π cm³.
Volume of cylinder = πr²h = π × 16 × h = 16πh.
Equating: 288π = 16πh → h = 288/16 = 18 cm.

🔹 Example 2 (Combination of Solids)

Question: A wooden article is made by scooping out a hemisphere from one end of a cylinder. The cylinder has height 10 cm and radius 3.5 cm. Find the total surface area.

Solution: TSA = CSA of cylinder + CSA of hemisphere + base area of cylinder = 2πrh + 2πr² + πr² = 2πrh + 3πr².
= 2 × 22/7 × 3.5 × 10 + 3 × 22/7 × 12.25 = 220 + 115.5 = 335.5 cm².

🔹 Example 3 (Frustum - Board Exam PYQ)

Question: The slant height of a frustum of a cone is 10 cm. The radii of its circular ends are 8 cm and 4 cm. Find its curved surface area.

Solution: CSA = πl(r₁ + r₂) = 22/7 × 10 × (8 + 4) = 22/7 × 10 × 12 = 22/7 × 120 = 2640/7 = 377.14 cm².

⭐ Key Points for Board Exam (Quick Revision)

📌 Formula Summary Table

Shape	Curved/Lateral Surface Area	Total Surface Area	Volume
Cuboid	2h(l+b)	2(lb+bh+hl)	l×b×h
Cube	4a²	6a²	a³
Cylinder	2πrh	2πr(r+h)	πr²h
Cone	πrl	πr(r+l)	(1/3)πr²h
Sphere	—	4πr²	(4/3)πr³
Hemisphere	2πr²	3πr²	(2/3)πr³
Frustum	πl(r₁+r₂)	π[r₁²+r₂²+l(r₁+r₂)]	(1/3)πh(r₁²+r₂²+r₁r₂)

⚠️ Important Observations:

When a solid is melted and recast into another solid → Volume remains constant.
When a solid is cut into smaller pieces → Total volume remains the same.
For combination of solids, TSA = sum of visible surface areas (areas of joining surfaces are not included).
For a cylinder with hemispherical ends, total length = height of cylinder + 2 × radius.
l (slant height of cone/frustum) = √(h² + (r₂ - r₁)²).
π = 22/7 or 3.14 as per question requirement.

📌 When solid is melted → Volume remains constant | For combination solids → Add only visible areas

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