Introduction to Mensuration

Formulae for surface areas and volumes of 3D objects.

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Mensuration — Introduction to Surface Areas and Volumes

Mensuration is the branch of Class 10 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) that deals with measuring the surface area and volume of solid shapes. While earlier classes focused on flat, two-dimensional figures like triangles and circles, this chapter takes the next step into three dimensions — boxes, cylinders, cones, spheres, and hemispheres.

This introductory lesson lays the groundwork for the entire chapter by presenting the standard formulas for six fundamental solids. Every exercise that follows — including combinations of two or more solids joined together — builds directly on these six formula sets, so getting comfortable with them now will make the rest of the chapter far easier.

Cuboid Cube Cylinder Cone Sphere Hemisphere
💡 Foundation fact: For every solid in this chapter, you will calculate up to three quantities — lateral (or curved) surface area (the area of the side faces only), total surface area (every face, including top and bottom), and volume (the space enclosed inside). Always check which one a question is asking for before substituting numbers.
📌 Units to remember: Surface area is always measured in square units (cm², m²) since it measures a flat region, while volume is always measured in cubic units (cm³, m³) since it measures enclosed space. Writing the wrong unit is one of the most common mark-losing mistakes in mensuration answers.

1. Cuboid

A cuboid is a solid with six rectangular faces — think of a matchbox, a brick, or a book. It has three different measurements: length (l), breadth (b) and height (h).

Solid Shape
l h b
Lateral surface area = 2h(l + b)
Total surface area = 2(lh + bh + lb)
Volume = l × b × h
Quick example: l = 8 cm, b = 5 cm, h = 4 cm LSA = 2 × 4 × (8+5) = 104 cm² TSA = 2 × (32+20+40) = 184 cm² Volume = 8 × 5 × 4 = 160 cm³

2. Cube

A cube is simply a special cuboid where all three dimensions are equal — length, breadth and height are all the same value, l. A dice or a Rubik's cube is a perfect everyday example.

Solid Shape
l l l
Lateral surface area = 4l²
Total surface area = 6l²
Volume = l³
Quick example: l = 6 cm LSA = 4 × 36 = 144 cm² TSA = 6 × 36 = 216 cm² Volume = 6³ = 216 cm³
📌 Cuboid vs Cube: Every formula for the cube can be obtained from the cuboid's formula by simply replacing both b and h with l. This is a useful memory shortcut — you don't need to memorise two completely separate sets of formulas.

3. Cylinder

A cylinder has two flat circular ends (radius r) joined by a curved surface, with height h between them — like a tin can, a pipe, or a water tank.

Solid Shape
r h
Lateral surface area = 2πrh
Total surface area = 2πr(h + r)
Volume = πr²h
Quick example: r = 7 cm, h = 10 cm (π = 22/7) LSA = 2 × 22/7 × 7 × 10 = 440 cm² TSA = 2 × 22/7 × 7 × 17 = 748 cm² Volume = 22/7 × 49 × 10 = 1540 cm³
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4. Cone

A cone has a single circular base (radius r) that narrows smoothly up to a single point called the apex, at height h above the base — like an ice-cream cone or a birthday party hat. The distance from the apex to any point on the base's edge is called the slant height, l.

Solid Shape
h l r
Lateral surface area = πrl
Total surface area = πr(l + r)
Volume = ⅓ πr²h
Slant height: l² = r² + h²
Quick example: r = 7 cm, h = 24 cm (π = 22/7) l = √(7² + 24²) = √625 = 25 cm LSA = 22/7 × 7 × 25 = 550 cm² TSA = 22/7 × 7 × 32 = 704 cm² Volume = ⅓ × 22/7 × 49 × 24 = 1232 cm³
💡 Don't confuse h and l: The height h is the straight vertical distance from the apex to the centre of the base, while the slant height l runs along the outer surface from the apex to the edge of the base. They are only equal to each other when r = 0, so always check the question is giving you the right one — or use l² = r² + h² to convert between them.

5. Sphere

A sphere is a perfectly round solid, like a ball — every single point on its surface is the same distance, r, from its centre. Unlike the other solids, a sphere has no flat faces or edges at all, so it has only one surface area formula (there is no separate "lateral" and "total" version).

Solid Shape
r
Surface area = 4πr²
Volume = (4/3) πr³
Quick example: r = 7 cm (π = 22/7) Surface area = 4 × 22/7 × 49 = 616 cm² Volume = (4/3) × 22/7 × 343 ≈ 1437.3 cm³

6. Hemisphere

A hemisphere is exactly half of a sphere, cut straight through its centre — like a bowl, a dome, or half an orange. It has one flat circular face (the cut) and one curved face, both with radius r.

Solid Shape
r r
Lateral (curved) surface area = 2πr²
Total surface area = 3πr²
Volume = (2/3) πr³
Quick example: r = 7 cm (π = 22/7) LSA = 2 × 22/7 × 49 = 308 cm² TSA = 3 × 22/7 × 49 = 462 cm² Volume = (2/3) × 22/7 × 343 ≈ 718.7 cm³
📌 Why is TSA = 3πr² for a hemisphere? A hemisphere's total surface area is the curved part (2πr²) plus the flat circular base (πr²) — added together, 2πr² + πr² = 3πr². Deriving it this way is safer in an exam than trying to recall "3πr²" from memory alone.

All Six Formulas at a Glance

Keeping all the formulas side by side makes it much easier to spot the patterns — notice how every curved-surface formula contains π, while the cuboid and cube (made only of flat rectangular faces) never do.

SolidLateral / Curved S.A.Total Surface AreaVolume
Cuboid2h(l + b)2(lh + bh + lb)l × b × h
Cube4l²6l²
Cylinder2πrh2πr(h + r)πr²h
Coneπrlπr(l + r)⅓ πr²h
Sphere4πr²(4/3) πr³
Hemisphere2πr²3πr²(2/3) πr³
💡 Spot the family resemblance: A hemisphere's volume is exactly half a sphere's volume of the same radius — (2/3)πr³ is half of (4/3)πr³. This kind of relationship is a great way to double-check your answer in an exam.

Common Mistakes to Avoid

  • Mixing up lateral and total surface area: Lateral (or curved) surface area excludes the flat top/bottom faces — always re-read the question to check whether it asks for "curved surface area," "lateral surface area," or "total surface area."
  • Using height instead of slant height in a cone (or vice versa): πrl (using slant height l) gives the curved surface area of a cone, while the volume formula ⅓πr²h uses the vertical height h — these are not interchangeable.
  • Forgetting the flat face area for a hemisphere: A hemisphere has a circular flat face in addition to its curved surface — when a question asks for total surface area, both 2πr² and πr² must be added.
  • Using the wrong value of π: Use π = 22/7 when the radius is a multiple of 7 (since it cancels neatly), and π = 3.14 otherwise, unless the question specifies which one to use.
  • Mismatched units: If the radius is given in cm but the height is given in m, convert both to the same unit before substituting into any formula — this is an easy mark to lose.
High-value exam tip: Many board exam questions combine two solids — for example, a cone sitting on top of a cylinder (like an ice-cream cone with a scoop), or a cylinder with hemispherical ends. In these combined-solid problems, you simply add or subtract the relevant surface area or volume formulas from this page — there is no separate formula to memorise for combined shapes.

What This Lesson Prepares You For

Every exercise in the Mensuration chapter builds on the six formula sets covered here, especially when two or more solids are joined together to form a single combined shape. The circle-related formulas you have already studied — particularly the area of a circle used inside the cylinder, cone, sphere and hemisphere formulas above — come directly from the sector and segment area work in Exercise 9.3, so revisiting that lesson is useful if the πr² term ever feels unfamiliar.

Going forward, look out for questions that ask you to find the surface area or volume of a single solid directly from these formulas, as well as more advanced questions that combine two solids (such as a hemispherical bowl with a cylindrical rim, or a cone mounted on a cylinder) — these combined problems simply require adding or subtracting the appropriate formulas from the table above.

📐 Board Exam Tip (CBSE, Telangana & AP): Examiners frequently test whether students can correctly identify which formula a real-life object needs — for example, recognising that a test tube is a cylinder with one hemispherical end, or that a funnel is a cone. Practising this identification step, before any calculation, is just as important as memorising the formulas themselves.
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