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Andhra Pradesh (BIEAP)Class 10 Mathematics★ AP Board (BSEAP) SSC Mathematics · ~8 marks · Board: AP Board

Mensuration — Surface Areas and Volumes

SURFACE AREAS and VOLUMES of cubes, cuboids, cylinders, cones, spheres, hemispheres, and frustums. Complete formula guide with AP SSC focus. AP Board Class 10 Mathematics.

⏱ 10 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Calculate curved surface area (CSA), total surface area (TSA), and volume of cylinders, cones, spheres, and hemispheres
2Calculate the volume and CSA of a frustum of a cone
3Solve combination problems (e.g., cone + cylinder, hemisphere + cone)
4Apply the conservation of volume principle when one solid is melted and recast into another
5Calculate the volume of water displaced when a solid is immersed

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Why this chapter matters

Mensuration is one of the MOST RELIABLE scoring chapters in AP SSC — the questions are entirely formula-based, with no proof or deduction required. The CSA, TSA, and Volume formulas for cylinder, cone, sphere, hemisphere, and frustum must be memorised. Combination problems (e.g., a cone on top of a cylinder, or melting one solid to form another) are standard 4-mark questions. Students who memorise the formula table score full marks in this chapter with practice.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Mensuration — Surface Areas and Volumes

Complete Formula Table

Solid	CSA	TSA	Volume
Cube (side s)	4s²	6s²	s³
Cuboid (l,b,h)	2(l+b)h	2(lb+bh+hl)	lbh
Cylinder	2πrh	2πr(r+h)	πr²h
Cone (l=√(r²+h²))	πrl	πr(r+l)	⅓πr²h
Sphere	4πr²	4πr²	(4/3)πr³
Hemisphere	2πr²	3πr²	(2/3)πr³

Frustum of a Cone: Volume = ⅓πh(r₁² + r₂² + r₁r₂). CSA = π(r₁+r₂)l.

Key Memory Tricks

Cone volume = ⅓ of cylinder (same base, same height). 'The ⅓ factor comes from tapering.'
Sphere surface area = 4πr² (exactly 4 times the area of a great circle).

Common Mistakes

Forgetting UNITS — area in cm², volume in cm³.
Mixing l (slant height) and h (vertical height) in cone formulas.

Worked Examples

Cube

Example 1: Find TSA, CSA, and volume of a cube with side 10 cm. TSA = 6s² = 6×100 = 600 cm². CSA = 4s² = 400 cm². Volume = s³ = 1000 cm³.

Example 2: TSA of a cube is 384 cm². Find its volume. 6s² = 384 → s² = 64 → s = 8 cm. Volume = 8³ = 512 cm³.

Cuboid

Example 3: A cuboid measures 12 cm × 8 cm × 5 cm. Find its TSA and volume. TSA = 2(lb+bh+hl) = 2(96+40+60) = 2×196 = 392 cm². Volume = 12×8×5 = 480 cm³.

Cylinder

Example 4: A cylinder has radius 7 cm and height 15 cm. Find CSA, TSA, and volume. (π = 22/7) CSA = 2πrh = 2×(22/7)×7×15 = 660 cm². TSA = 2πr(r+h) = 44×(7+15) = 44×22 = 968 cm². Volume = πr²h = (22/7)×49×15 = 2310 cm³.

Example 5: Volume of a cylinder is 1540 cm³ and its radius is 7 cm. Find its height. πr²h = 1540 → (22/7)×49×h = 1540 → 154h = 1540 → h = 10 cm.

Cone

Example 6: Find slant height, CSA, TSA, and volume of a cone with r = 6 cm and h = 8 cm. l = √(36+64) = √100 = 10 cm. CSA = πrl = 3.14×6×10 = 188.4 cm². TSA = πr(r+l) = 3.14×6×16 = 301.44 cm². Volume = ⅓πr²h = ⅓×3.14×36×8 = 301.44 cm³.

Example 7: CSA of a cone is 308 cm² and its slant height is 14 cm. Find its radius. (π = 22/7) πrl = 308 → (22/7)×r×14 = 308 → 44r = 308 → r = 7 cm.

Sphere

Example 8: Find SA and volume of a sphere of radius 10.5 cm. SA = 4πr² = 4×(22/7)×110.25 = 4×346.5 = 1386 cm². Volume = (4/3)πr³ = (4/3)×(22/7)×1157.625 = (4/3)×3638.25 = 4851 cm³.

Hemisphere

Example 9: Find CSA, TSA, and volume of a hemisphere of radius 7 cm. CSA = 2πr² = 2×(22/7)×49 = 308 cm². TSA = 3πr² = 3×(22/7)×49 = 462 cm². Volume = (2/3)πr³ = (2/3)×(22/7)×343 = (2/3)×1078 = 718.67 cm³.

Frustum of a Cone

When a cone is cut by a plane parallel to its base, the lower part is a FRUSTUM.

Formulas:

Volume = ⅓πh(r₁² + r₂² + r₁r₂)
CSA = π(r₁+r₂)l where l = √[(r₁−r₂)² + h²]
TSA = π(r₁+r₂)l + πr₁² + πr₂²

Example 10 — Bucket: A bucket has diameters 20 cm (top) and 12 cm (bottom), height 10 cm. Find its volume. r₁ = 10 cm, r₂ = 6 cm, h = 10 cm. Volume = ⅓π×10×(100+36+60) = ⅓π×10×196 = (1960π)/3 = (1960×22)/(3×7) ≈ 2053.33 cm³.

Combination of Solids

AP exams frequently combine two or more solids. Strategy: Identify each shape. Compute visible SA (ignore shared surfaces). Add volumes.

Example 11 — Cylinder with Hemispherical Ends: A solid has a cylinder with hemispheres on both ends. r = 3.5 cm, total length = 19 cm. Find TSA. Cylinder height = 19 − 2×3.5 = 12 cm. TSA = CSA of cylinder + 2×CSA of hemisphere = 2πrh + 2×2πr² = 2πr(h+2r) = 2×(22/7)×3.5×(12+7) = 22×19 = 418 cm².

Example 12 — Cone on Cylinder: A solid has a cone on a cylinder. r = 5 cm, cylinder height = 8 cm, total height = 20 cm. Find volume. Cone height = 20−8 = 12 cm. Volume = πr²h_cyl + ⅓πr²h_cone = π×25×8 + ⅓π×25×12 = 200π + 100π = 300π = 942.48 cm³.

Unit Conversions

Given	Convert To	Operation
m → cm	cm	× 100
cm → mm	mm	× 10
m² → cm²	cm²	× 10000
m³ → cm³	cm³	× 1000000
litre → cm³	cm³	× 1000
cm³ → litres	litres	÷ 1000

'Volume in cm³ divided by 1000 gives litres. Very common in water tank problems.'

AP Exam Focus

Topic	Marks	Frequency
Surface area of cube/cuboid/cylinder	4	Very Common
Volume of cone/sphere/hemisphere	4	Very Common
Frustum of cone	5	Common
Combination of solids	5	Very Common
Unit conversion (volume to capacity)	2-3	Moderate

Self-Test Questions

Find the volume of a sphere whose surface area is 616 cm². (Answer: 4πr²=616 → r=7 → V=(4/3)π×343 ≈ 1437.33 cm³)
TSA of a cuboid is 332 cm², length 10 cm, height 6 cm. Find breadth. (Answer: 2(10b+6b+60)=332 → 16b=106 → b=6.625 cm)
Find CSA of a cylinder with height 21 cm and base radius 5 cm. (Answer: 2×3.14×5×21 = 659.4 cm²)
A cone (r=8, h=15) is melted and recast into a sphere. Find sphere radius. (Answer: ⅓π×64×15 = (4/3)πr³ → r³ = 240 → r ≈ 6.21 cm)
Hemisphere diameter = 14 cm. Find CSA and TSA. (Answer: CSA = 308 cm², TSA = 462 cm²)
A metallic sphere of radius 4.2 cm is melted and recast into a cylinder of radius 6 cm. Find cylinder height. (Answer: (4/3)π(4.2)³ = π×36×h → h = 2.744 cm)
How many litres can a cylindrical tank of radius 3.5 m and height 10 m hold? (Answer: V = 385 m³ = 385,000 litres)

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Mensuration Formulas

CYLINDER (r=radius, h=height): CSA = 2πrh. TSA = 2πr(r+h). Volume = πr²h. CONE (r=radius, h=height, l=slant height=√(r²+h²)): CSA = πrl. TSA = πr(r+l). Volume = (1/3)πr²h. SPHERE (r=radius): SA = 4πr². Volume = (4/3)πr³. HEMISPHERE (r=radius): CSA = 2πr². TSA = 3πr². Volume = (2/3)πr³. FRUSTUM (r₁=top radius, r₂=bottom radius, h=height, l=slant height=√[h²+(r₂−r₁)²]): Volume = (1/3)πh(r₁²+r₂²+r₁r₂). CSA = π(r₁+r₂)l.

COMBINATION PROBLEMS: (1) Identify each solid in the combination. (2) For TSA: only count EXPOSED surfaces (the flat circular face where two solids join is hidden — not counted). (3) For Volume: simply ADD the volumes. RECASTING: Melting and recasting → Volume₁ = Volume₂ (volume is conserved). WATER DISPLACED: Volume of water displaced = Volume of solid immersed (Archimedes principle).

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Using diameter instead of radius in mensuration formulas

✓ ALL mensuration formulas use RADIUS. If the problem gives diameter, HALVE it before substituting. Example: 'A cylinder with diameter 14 cm' → r = 7 cm. Using diameter 14 directly gives 4× the correct volume and 2× the correct area. This is the most common mensuration error.

WATCH OUT

✗ Counting the internal joining face in combination TSA

✓ When two solids are joined (e.g., a cone placed on top of a cylinder), the circular face at the junction is NOT exposed — it is shared/hidden. Do NOT add the area of this face to the TSA. For a cone on a cylinder: TSA = CSA of cylinder + base circle of cylinder + CSA of cone (but NOT the circular base of the cone, which is attached to the cylinder top).

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Mensuration — Surface Areas and Volumes?

1 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

1 questions~2 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•CYLINDER: CSA = 2πrh (just the curved surface, no circles). TSA = 2πr(r+h) = 2πrh + 2πr² (add both circular ends). Volume = πr²h. MNEMONIC: 'Cylinder is a can — 2 circles + rectangle wrapped around.'
•CONE: l = slant height = √(r²+h²) (ALWAYS compute l first if not given). CSA = πrl (just the curved surface, no base circle). TSA = πr(r+l) = πrl + πr² (add base circle). Volume = (1/3)πr²h. MNEMONIC: 'Cone is 1/3 of cylinder with same r and h.'
•SPHERE: SA = 4πr² (no TSA vs CSA distinction — there's only one surface). Volume = (4/3)πr³. HEMISPHERE: CSA = 2πr² (curved surface only). TSA = 3πr² (add flat circular base). Volume = (2/3)πr³ = half sphere. MNEMONIC: 'Hemisphere TSA = half sphere SA + base circle = 2πr² + πr² = 3πr².'
•FRUSTUM (truncated cone): l = slant height = √[h²+(r₂−r₁)²]. Volume = (πh/3)(r₁²+r₂²+r₁r₂). CSA = π(r₁+r₂)l. TSA = π(r₁+r₂)l + πr₁² + πr₂². r₁ = top radius, r₂ = bottom radius. Always identify which radius is larger (the base).
•COMBINATION TSA RULE: Count ONLY exposed surfaces. When two solids are joined, the FLAT FACE at the junction is hidden — DON'T ADD IT. Example: cone on cylinder. TSA = CSA(cylinder) + base circle(cylinder bottom) + CSA(cone). The top circle of cylinder and base circle of cone are hidden — not counted.
•COMBINATION VOLUME RULE: Simply ADD the volumes of individual solids. No adjustment needed for the junction — volumes add directly. Volume(compound solid) = V₁ + V₂ + ...
•RECASTING/MELTING PRINCIPLE: Volume is CONSERVED. Volume of original solid = Total volume of new solids. Example: large sphere melted into small spheres → (4/3)πR³ = n × (4/3)πr³ → n = R³/r³. This ratio only involves the cube of the radius ratio.
•WATER DISPLACED: Volume of object submerged = Volume of water displaced (Archimedes). If a solid is dropped into a full cylinder, water spilled = Volume of solid. If cylinder is NOT full, rise in water level × base area of cylinder = Volume of solid.
•π VALUE: Use π = 22/7 when the radius is a multiple of 7 or when the problem says 'use π = 22/7'. Use π = 3.14 when the radius is a decimal. Use symbolic π when exact answer is required. AP SSC typically uses π = 22/7.
•UNITS AND CONVERSIONS: Always check units are consistent. If radius is in cm and height in m — convert first. 1 litre = 1000 cm³. Volume in cm³ ÷ 1000 = volume in litres. This conversion appears in 'tank capacity in litres' questions.

Andhra Pradesh (BIEAP) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Water tanks and storage capacity

AP's rural water supply relies on overhead cylindrical tanks and ground-level sumps. Engineers use Volume = πr²h to calculate tank capacity in litres. A cylindrical tank of radius 3.5 m and height 10 m has volume = π × 12.25 × 10 ≈ 385 m³ = 385,000 litres — enough for ~1,000 people for a day. The CSA formula (2πrh) determines how much sheet metal is needed to build the cylindrical walls. These calculations are performed daily by engineers at AP Water Resources Board.

Ice cream cones and food packaging

An ice cream cone is exactly the combination solid from this chapter: a hemisphere of ice cream (2/3 × πr³) on top of a cone (1/3 × πr²h). Food manufacturers calculate the volume of ice cream that fits using both formulas. The waffle cone's shape is a frustum (wider at top). Frustum volume determines the size of single-serving cups, funnels, and buckets. Every time a new product is packaged, a mensuration calculation determines material requirements and capacity.

Tarpaulin and tent fabric (CSA applications)

A tent shaped like a cylinder topped by a cone requires fabric only for the curved surfaces — the floor is the ground. Fabric area = CSA(cylinder) + CSA(cone) = 2πrh₁ + πrl₂. Every camping equipment manufacturer, circus tent supplier, and military tent producer uses exactly this calculation to determine fabric cost and cutting requirements. AP's Corrugated Sheets factories and tent manufacturers use CSA formulas for bidding on contracts.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Formula table: write all formulas at the top of the mensuration answer before starting any problem — list cylinder, cone, sphere, hemisphere, frustum with CSA, TSA, Volume. This reference prevents mid-problem formula confusion.
Identify each solid: before computing anything, list all solids in the combination and what formula you'll use for each. 'Solid 1: cylinder (CSA), Solid 2: hemisphere (CSA), Solid 3: cylinder (base circle).' This explicit identification earns structure marks.
Frustum slant height: ALWAYS compute l = √[h²+(r₂−r₁)²] first before applying any frustum formula. Write this step explicitly — examiners look for it.
Recasting problems: write Volume₁ = n × Volume₂. Then show cancellation of π (and any other common factors). Then write the clean equation and solve for n. State the final answer: 'Number of [small solids] = n.'
Units: state units throughout — write '= 22/7 cm² per unit area' or '= 154 cm²' not just '= 154'. And convert volume to litres in capacity problems: 'Volume = 1540 cm³ = 1540/1000 = 1.54 litres.'

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Research the Cavalieri's Principle — if two solids have the same height and equal cross-sectional areas at every height, they have equal volumes. This principle explains WHY the cone volume is 1/3 of the cylinder: at height y from base, a cone of base radius r and height h has cross-section π(r(h−y)/h)² = πr²(h−y)²/h². When integrated (summed) from 0 to h, this gives (1/3)πr²h. Research how Cavalieri's Principle relates to integral calculus and why Archimedes discovered the sphere volume formula (4/3 πr³) 1800 years before calculus was invented.
Investigate Archimedes' method of exhaustion — Archimedes computed the volume and surface area of a sphere by approximating it with inscribed and circumscribed cylinders and cones, then taking the limit as the approximations improved. He inscribed a cone + cylinder (and circumscribed) and showed the sphere volume lies between them, approaching 4/3 πr³. This was essentially integral calculus 1900 years before Newton. Archimedes was so proud of this discovery that he requested a sphere-in-cylinder diagram on his tombstone.
Explore packing efficiency — when you pack spheres (like oranges in a crate) as efficiently as possible, what fraction of the volume is actually occupied? For random packing: ~64%. For the densest known packing (face-centred cubic or hexagonal close packing): π/(3√2) ≈ 74.05%. Kepler conjectured this in 1611; it was finally proved in 1998 by Thomas Hales using computer-assisted proof. Research the Kepler conjecture and what 'proof by computer' means for mathematics.
Research the surface area-to-volume ratio and its biological implications — as a cell grows, its volume increases as r³ but its surface area increases as r². The ratio SA/V = 3/r decreases as the cell gets bigger. Beyond a certain size, a cell cannot exchange nutrients and waste through its surface fast enough for its volume — this is WHY cells divide rather than just growing larger. The same principle explains why small animals lose heat faster (higher SA/V ratio), why ice melts faster when crushed (increased surface area), and why powder burns faster than a solid block. Class 10 mensuration formulas directly explain fundamental biology and chemistry.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

AP Board SSC (Class 10)Very High — Mensuration combination and recasting problems are standard 6–10 mark questions in every AP SSC Mathematics paper

JEE Main (Mathematics)Medium — 3D mensuration appears in JEE but is usually combined with coordinate geometry (equations of spheres and cylinders)

NTSE (Mathematics)High — Surface area and volume problems are tested in NTSE Stage I

AP EAPCET and AP ECETHigh — 3D solids and mensuration appear in both engineering entrance exams; Class 10 formulas are directly tested

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CURVED SURFACE AREA (CSA): Only the curved/lateral surface — like peeling the outer skin off a shape. For a cylinder, it is the rectangle you get when you unroll the tube. For a cone, it is the slanted surface (not the base). For a sphere, there is only one surface so SA=4πr² = both CSA and TSA. TOTAL SURFACE AREA (TSA): Everything — curved surface PLUS all flat circular ends. Cylinder TSA = CSA + 2 circles (top+bottom). Cone TSA = CSA + 1 circle (base). Hemisphere TSA = CSA + 1 circle (flat face). RULE OF THUMB: If the solid is sitting on a flat surface or has a flat opening, the flat face is a circle of area πr². Add it to CSA to get TSA.

The faces NOT counted are the JOINED faces — the faces that are in contact with each other. When a cone sits on top of a cylinder: the BASE CIRCLE OF THE CONE (area πr²) and the TOP CIRCLE OF THE CYLINDER (area πr²) are touching each other — both are hidden. So you DO NOT add these faces. TSA of (cone + cylinder) = CSA(cylinder) + BOTTOM CIRCLE of cylinder + CSA(cone). What you count: the outer tube of the cylinder + the bottom (it's sitting on something or is the base) + the slanted surface of the cone. What you don't count: the top of the cylinder (under the cone) + the base of the cone (on the cylinder).

Melting conserves volume: Volume of large sphere = n × Volume of small sphere. (4/3)πR³ = n × (4/3)πr³. The π and 4/3 cancel: R³ = n × r³. Therefore n = R³/r³ = (R/r)³. If large radius = 6 cm and small radius = 2 cm: n = (6/2)³ = 3³ = 27 small spheres. If large radius = 12 cm, small radius = 3 cm: n = (12/3)³ = 4³ = 64. The ratio of volumes is the CUBE of the radius ratio — this is why a sphere 3 times bigger contains 27 times the material.

A FRUSTUM is what you get when you cut a cone parallel to its base — removing the top part. The shape is like a bucket or a drinking glass (wider at top, narrower at bottom). The frustum has TWO circular ends: top radius r₁ (smaller) and bottom radius r₂ (larger). It has a slant height l = √[h²+(r₂−r₁)²]. Volume = (πh/3)(r₁²+r₂²+r₁r₂). The frustum formula appears in AP SSC almost every year. It is tested as either (1) a standalone calculation problem, or (2) as the shape of a bucket/glass/hopper — 'find the volume of water a bucket can hold.' Practical contexts: buckets, lampshades, frustum-shaped traffic cones.

Volume of water overflowing = Volume of sphere (since the cylinder is full, any immersed volume displaces an equal volume of water which overflows). If sphere radius = cylinder radius = r, then Volume of sphere = (4/3)πr³. Water overflow = (4/3)πr³. Note: this requires that the SPHERE FITS inside the cylinder, which means the sphere diameter (2r) ≤ cylinder height. If the sphere is too tall to fit, it sits partially submerged — only the submerged volume displaces water. For AP SSC problems, assume the sphere fully fits unless stated otherwise.

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