Chapter 9 of 13

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Andhra Pradesh (BIEAP)Class 9 Mathematics★ AP Board (BSEAP) Class 9 Mathematics · ~10 marks · Board: AP Board

Surface Areas and Volumes

Complete MENSURATION for AP Class 9 — Surface Areas and Volumes of CUBE, CUBOID, CYLINDER, CONE, SPHERE, and HEMISPHERE. All formulas with worked numerical examples. Common exam mistakes. AP Board Class 9 Mathematics.

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

1State and apply CSA, TSA, and Volume formulas for cube and cuboid
2State and apply CSA, TSA, and Volume formulas for right circular cylinder
3State and apply CSA, TSA, and Volume formulas for right circular cone (using slant height l = √(r²+h²))
4State and apply Surface Area and Volume formulas for sphere and hemisphere
5Solve problems involving combination of solids
6Convert between volume units: m³, cm³, litres

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Why this chapter matters

Surface Areas and Volumes (Mensuration) is the most formula-intensive chapter in AP Class 9 Mathematics and consistently the highest-scoring. A cylinder or cone calculation problem appears in almost every exam for 4-5 marks. Sphere and hemisphere questions appear for 3-4 marks. Students must know the difference between Curved Surface Area (CSA), Total Surface Area (TSA), and Volume for each solid. The slant height formula l = √(r² + h²) is critical for cone problems. Unit conversion (1 m³ = 1000 L) is needed for capacity/volume problems. Combination of solids (cone on cylinder, hemisphere on cone) appears in 5-mark questions.

Before you start — revise these

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

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Surface Areas and Volumes — Class 9 Mathematics

"Mensuration is the mathematics of SPACE — how much PAINT to cover a surface, how much WATER to fill a tank."

1. Complete Formula Table

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

2. Key Relationships

Why ⅓ for Cone and Pyramid?

'The volume of a cone is EXACTLY ONE-THIRD the volume of a CYLINDER with the SAME base radius and height. This is because the cone TAPERS to a point — it has less volume near the top.' Similarly, a pyramid = ⅓ × base area × height.

Why 4πr² for Sphere?

'The surface area of a sphere is EXACTLY FOUR TIMES the area of a GREAT CIRCLE (a circle through the centre, area = πr²). So TSA of sphere = 4πr².'

3. Worked Examples

Example 1 — Cuboid

'A room is 8 m long, 6 m wide, and 3 m high. Find the cost of whitewashing the four walls at ₹50 per m².' Area of 4 walls = CSA of cuboid = 2(l+b)h = 2(8+6)×3 = 2×14×3 = 84 m². Cost = 84 × 50 = ₹4,200.

Example 2 — Cylinder

'A cylindrical water tank has radius 1.4 m and height 2 m. Find: (i) CSA, (ii) TSA, (iii) Volume in litres.' (i) CSA = 2πrh = 2×3.14×1.4×2 = 17.584 ≈ 17.6 m². (ii) TSA = 2πr(r+h) = 2×3.14×1.4×3.4 = 29.89 ≈ 29.9 m². (iii) Volume = πr²h = 3.14×1.4²×2 = 12.3088 m³ ≈ 12,309 L. (1 m³ = 1,000 L)

Example 3 — Cone

'A cone has radius 7 cm and slant height 10 cm. Find: (i) CSA, (ii) TSA. (π=22/7)' CSA = πrl = (22/7)×7×10 = 220 cm². Height h = √(l²−r²) = √(100−49) = √51 ≈ 7.14 cm. TSA = πr(r+l) = (22/7)×7×(7+10) = 22×17 = 374 cm². Volume = ⅓πr²h = (1/3)×(22/7)×49×7.14 ≈ 366.67 cm³.

Example 4 — Sphere

'Find the surface area and volume of a sphere of radius 21 cm. (π=22/7)' TSA = 4πr² = 4×(22/7)×21×21 = 4×22×3×21 = 5,544 cm². Volume = (4/3)πr³ = (4/3)×(22/7)×21×21×21 = 4×22×21×21 = 38,808 cm³.

4. Hollow Solids and Combinations

Hollow Cylinder: Volume of material = π(R²−r²)h (R=outer radius, r=inner radius). Hemispherical Bowl (Hollow) : Inner CSA = 2πr². Outer CSA = 2πR². Total surface area to paint = inner CSA + outer CSA + area of the RIM = 2πr² + 2πR² + π(R²−r²).

5. Common Mistakes to Avoid

Confusing slant height (l) with vertical height (h) in cone — l = √(r²+h²). In CSA = πrl and TSA = πr(r+l), use SLANT height. In Volume = ⅓πr²h, use VERTICAL height.
Forgetting the ⅓ factor — Cone and pyramid volumes = ⅓ of corresponding cylinder/prism.
Unit errors — If dimensions are in cm, area is in cm², volume is in cm³. 1 m³ = 1,000 L (not 100 L!).
Using diameter instead of radius — Formulas use RADIUS (r). If diameter is given, halve it first.

6. AP Exam Focus

Topic	Marks
Cylinder and cone problems	4-5
Sphere and hemisphere	3-4
Combined solids	4-5

More Worked Examples

Example 5 — Hemisphere: Find the TSA and volume of a solid hemisphere of radius 7 cm. TSA = 3πr² = 3×(22/7)×49 = 3×22×7 = 462 cm². Volume = (2/3)πr³ = (2/3)×(22/7)×343 = (2×22×343)/(3×7) = 15092/21 ≈ 718.67 cm³.

Example 6 — Combined Solid: A toy is a cone mounted on a hemisphere. Radius = 3.5 cm. Total height = 15.5 cm. Find TSA. Hemisphere radius = 3.5 cm. Cone radius = 3.5 cm. Cone height = 15.5 − 3.5 = 12 cm. Slant height l = √(12² + 3.5²) = √(144 + 12.25) = √156.25 = 12.5 cm. TSA = CSA of cone + CSA of hemisphere = πrl + 2πr² = (22/7)(3.5)(12.5) + 2(22/7)(3.5)² = 137.5 + 77 = 214.5 cm².

Quick Formula Memory

'TSA of hemisphere = 3πr² (NOT 2πr² — that's only the CURVED surface. The flat circular face adds another πr²).' 'For frustum of a cone: Volume = ⅓πh(r₁² + r₂² + r₁r₂). CSA = π(r₁+r₂)l (l = √(h²+(r₁−r₂)²)).'

AP Exam Strategy for Mensuration: 1. Draw a CLEAR diagram. 2. Label all given dimensions. 3. Identify the CORRECT formula. 4. Substitute CAREFULLY. 5. Check UNITS (cm² for area, cm³ for volume). 6. Round to the specified decimal places. 'Mensuration is the HIGHEST SCORING section in AP Math — formulas are given (mostly), you just need to apply them correctly.'

More Worked Examples — Exam Pattern

Example 7 — Cost of Plastering: A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the CSA at ₹12.50 per m². Diameter = 50 cm → r = 25 cm = 0.25 m. h = 3.5 m. CSA = 2πrh = 2 × (22/7) × 0.25 × 3.5 = 2 × 22 × 0.25 × 0.5 = 5.5 m². Cost = 5.5 × 12.50 = ₹68.75. 'Always convert ALL measurements to the SAME unit before calculating. Mixing cm and m is a common error.'

Example 8 — Sphere Packing: How many spheres of radius 1 cm can be made from a solid sphere of radius 8 cm? Volume of big sphere = (4/3)πR³. Volume of small sphere = (4/3)πr³. Number = R³/r³ = 8³/1³ = 512. 'When melting and recasting, VOLUME is CONSERVED. Number of small solids = Volume(big)/Volume(small).'

Example 9 — Water Flow Problem: Water flows through a cylindrical pipe of internal diameter 7 cm at 2 m/s. How much water is discharged per minute? r = 3.5 cm. Area of cross-section = πr² = (22/7)×3.5² = 38.5 cm². Water flowing per second = area × speed = 38.5 × 200 cm³ (2 m/s = 200 cm/s) = 7,700 cm³/s. Per minute = 7,700 × 60 = 462,000 cm³ = 462 L.

Example 10 — Cone from Sector: A sector of a circle of radius 15 cm and angle 216° is folded to form a cone. Find the base radius of the cone. Arc length of sector = (θ/360)×2πR = (216/360)×2π×15 = 0.6×30π = 18π cm. This arc becomes the circumference of the base: 2πr = 18π → r = 9 cm. Slant height of cone = radius of sector = 15 cm. Height h = √(l²−r²) = √(225−81) = √144 = 12 cm.

Unit Conversion Quick Reference

From	To	Multiply by
cm³	mL	1 (IDENTICAL)
m³	L	1000
L	cm³	1000
cm	m	0.01
m²	cm²	10000
hectare	m²	10000

' 1 cm³ = 1 mL and 1 m³ = 1000 L — these two conversions appear in almost every AP exam.'

Word Problem Patterns in AP Exam

Cost problems: Area calculated → cost = area × rate per unit area.
Melting/Recasting: Volume conserved → number = V₁/V₂.
Embankment/Road roller: CSA of cylinder × number of revolutions.
Tent/Cap problems: Usually need CSA of cone (the base is on the ground).
Combined solids: Break into individual parts → add CSAs (NOT TSA — hidden surfaces that are joined don't count).

Self-Test — Quick Numericals

CSA of cube of side 5 cm? (Answer: 4s² = 100 cm².)
Volume of cylinder with r=7cm, h=10cm? (Answer: πr²h = 1540 cm³ with π=22/7.)
TSA of hemisphere of radius 7 cm? (Answer: 3πr² = 462 cm².)
Slant height of cone with r=3 cm, h=4 cm? (Answer: l = 5 cm.)
Volume of sphere of radius 6 cm, in terms of π? (Answer: (4/3)π×216 = 288π cm³.)
How many litres in 5 m³? (Answer: 5000 L.)

Key formulas & results

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

Surface Areas and Volumes — All Solids

CUBE (side = s): CSA (4 faces) = 4s². TSA (6 faces) = 6s². Volume = s³. Diagonal = s√3. CUBOID (length l, breadth b, height h): CSA (4 side faces) = 2h(l+b). TSA = 2(lb + bh + hl). Volume = lbh. Diagonal = √(l²+b²+h²). RIGHT CIRCULAR CYLINDER (radius r, height h): CSA (curved surface only) = 2πrh. TSA = 2πrh + 2πr² = 2πr(r+h). Volume = πr²h. Note: 2πr = circumference of base, so CSA = circumference × height. HOLLOW CYLINDER (outer radius R, inner radius r, height h): CSA = 2πh(R+r). TSA = 2πh(R+r) + 2π(R²−r²). Volume = πh(R²−r²). RIGHT CIRCULAR CONE (radius r, height h, slant height l): SLANT HEIGHT: l = √(r²+h²). (From Pythagoras: slant is hypotenuse of right triangle with legs r and h). CSA (curved surface only) = πrl. TSA = πrl + πr² = πr(l+r). Volume = (1/3)πr²h. SPHERE (radius r): Surface Area = 4πr². Volume = (4/3)πr³. Note: Surface area = 4 times the area of a great circle. HEMISPHERE (radius r): CSA (curved part only) = 2πr². TSA = 2πr² + πr² = 3πr² (flat circular base included). Volume = (2/3)πr³. UNIT CONVERSIONS: 1 m = 100 cm. 1 m² = 10,000 cm². 1 m³ = 1,000,000 cm³ = 1,000 litres. 1 litre = 1000 cm³ = 1 dm³. 1 cm³ = 1 mL. COMBINATION PROBLEMS: Find total surface area by adding exposed surfaces only (subtract hidden surfaces at junctions). Volume = sum of individual volumes. EXAMPLE — Hemisphere on cylinder: TSA = 2πr² (hemisphere CSA) + 2πrh (cylinder CSA) + πr² (cylinder base) = πr(3r+2h). Note: the circular junction is hidden on both pieces — neither is part of the external surface.

AP EXAM KEY TRAPS: (1) ALWAYS compute slant height l for cones: l = √(r²+h²). A common exam setup gives r and h, and expects CSA = πrl — students who skip slant height get wrong answers. (2) CSA vs TSA: CSA for cylinder = 2πrh (no top or bottom circles). TSA = CSA + 2πr² (adds two circles). Many questions ask for just CSA (e.g., painting the curved part of a pillar) or just TSA (wrapping a tin). Read carefully. (3) HEMISPHERE TSA = 3πr² (curved 2πr² + flat base πr²). If it's a BOWL (open hemisphere), TSA = 2πr² only (no base). (4) Volume ALWAYS uses (1/3) for CONE: (1/3)πr²h. Cylinder = πr²h (no 1/3). (5) 1 m³ = 1000 litres (NOT 100). This comes up in water tank problems. (6) π ≈ 22/7 when r is a multiple of 7, and π ≈ 3.14 otherwise — follow the problem's instruction.

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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

✗ Forgetting to calculate slant height for cone problems, or using height instead of slant height in CSA = πrl

✓ For a CONE, there are THREE measurements: radius r (base), height h (perpendicular height from apex to center of base), and slant height l (distance along the slanted surface from apex to base rim). CSA = πrl uses the SLANT HEIGHT l, NOT the perpendicular height h. FORMULA: l = √(r² + h²) — this comes from Pythagoras' theorem applied to the right triangle formed by r (horizontal), h (vertical), and l (hypotenuse). PROCEDURE: Always compute l first. Given r = 5 cm, h = 12 cm: l = √(5² + 12²) = √(25 + 144) = √169 = 13 cm. Then CSA = π × 5 × 13 = 65π cm². If you used h=12 instead: CSA = 60π — WRONG. Write l = √(r²+h²) as the first step in every cone problem.

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 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.

•CUBE (side s): CSA = 4s² (4 lateral faces). TSA = 6s² (all 6 faces). Volume = s³. Diagonal = s√3.
•CUBOID (length l, breadth b, height h): CSA = 2h(l+b) (4 side faces). TSA = 2(lb+bh+hl). Volume = lbh. Diagonal = √(l²+b²+h²).
•RIGHT CIRCULAR CYLINDER (radius r, height h): CSA = 2πrh (curved surface only — the 'tin can wrap'). TSA = 2πr(r+h) = CSA + 2 circles. Volume = πr²h. Note: 2πr = circumference of base.
•HOLLOW CYLINDER (outer R, inner r, height h): CSA = 2πh(R+r) (outer + inner curved). TSA = 2πh(R+r) + 2π(R²−r²) (adds two annular rings). Volume = πh(R²−r²).
•RIGHT CIRCULAR CONE (radius r, height h, slant height l): SLANT HEIGHT l = √(r²+h²) — ALWAYS COMPUTE FIRST. CSA = πrl. TSA = πr(l+r) = πrl + πr² (curved + base circle). Volume = (1/3)πr²h. The (1/3) factor comes from the cone being 1/3 the volume of a cylinder with same base and height.
•SPHERE (radius r): SURFACE AREA = 4πr² (no CSA/TSA distinction — only one surface). Volume = (4/3)πr³. Note: surface area = 4 × area of a great circle.
•HEMISPHERE (radius r): CSA = 2πr² (curved part only, the 'dome'). TSA = 3πr² (includes flat circular base). Volume = (2/3)πr³ (half a sphere's volume).
•UNIT CONVERSIONS: 1 m = 100 cm. 1 m² = 10,000 cm². 1 m³ = 1,000,000 cm³ = 1,000 litres = 1 kilolitre. 1 litre = 1000 cm³ = 1 dm³. 1 cm³ = 1 mL.
•RELATIONSHIPS BETWEEN SOLIDS: Cone volume = (1/3) × Cylinder volume (same r, h). Hemisphere volume = (1/2) × Sphere volume. Surface area of sphere = 4 × area of great circle = 4πr².
•COMBINATION OF SOLIDS — TSA: count only the EXPOSED surfaces. When two solids are joined, the JUNCTION FACES are hidden and NOT counted. Example: hemisphere on cylinder: TSA = 2πr² (hemisphere CSA) + 2πrh (cylinder CSA) + πr² (cylinder bottom). The flat face where hemisphere meets cylinder is hidden.
•COMBINATION OF SOLIDS — VOLUME: simply ADD the individual volumes. No subtraction needed for volume (the junction has no volume itself).

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 tank capacity calculations

AP's village water supply schemes use cylindrical and spherical tanks. A typical overhead tank in AP villages is cylindrical, 3 m radius × 5 m height → πr²h × 1000 L/m³ ≈ 141,000 litres — enough for several hundred people daily. Engineers at AP Water Resources Department use these formulas every day to design rural and urban water infrastructure. Understanding cylinder volume and unit conversion is a literal job skill.

Manufacturing — cans, cones, and packaging

Every cylindrical food can, ice cream cone, ball bearing, and sealed bottle is designed using the formulas from this chapter. Manufacturers compute material requirements (TSA × cost per cm²) to minimise production cost. AP's manufacturing sector (food processing, beverages, chemicals) employs thousands of engineers who use surface area and volume calculations daily.

Construction — concrete and material estimation

Civil engineers building columns, beams, foundations, and tanks in AP's infrastructure projects (Amaravati Capital, Polavaram dam, new highways) must calculate concrete volumes (cylinders, cuboids) and steel surface areas (for paint/protective coating). The volume formula × density gives weight; the surface area × paint coverage gives paint required. Class 9 mensuration is the foundation of construction estimation.

Exam strategy

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

ALWAYS list given values first: 'Given: r = ___, h = ___' before any calculation. This visual organisation prevents errors.
Slant height FIRST for cones: write l = √(r²+h²) as the first step. Compute and label clearly: 'Slant height = ___ cm.' Then use l in subsequent CSA/TSA formulas.
Formula table (before exam): write all CSA/TSA/Volume formulas for cube, cuboid, cylinder, cone, sphere, hemisphere on a single sheet. Memorise. In exam, write them at top of working space.
Read CSA vs TSA carefully: 'curved surface area' (CSA) vs 'total surface area' (TSA) vs 'area to be painted' (depends on context — usually CSA if it's a side, or sum of relevant faces). Highlight the asked quantity.
Unit conversions: m³ → L (×1000). cm³ → L (÷1000). Always show the conversion step explicitly. State final units in your answer.

Going beyond the textbook

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

Research Archimedes' calculation of the sphere volume (~250 BCE) — using his method of exhaustion (a precursor to integral calculus), Archimedes proved that the volume of a sphere is exactly 2/3 the volume of the smallest cylinder containing it. This was so important to him that he requested a sphere-in-cylinder diagram on his tombstone. Research how Archimedes' methods compare to modern calculus.
Investigate Cavalieri's Principle (1635) — if two solids have the same height and the same cross-sectional area at every height, they have the same volume. This generalises to arbitrary solid shapes. Cavalieri's Principle is used to derive cone and sphere volume formulas elegantly.
Explore packing efficiency — when you pack spheres into a box (like oranges in a crate), what fraction of the box is filled? For random packing: about 64%. For dense regular packing (face-centred cubic): π/(3√2) ≈ 74.05%. Kepler conjectured (1611) that this is the maximum; Thomas Hales proved it (1998). Research the Kepler conjecture.
Research surface area to volume ratio in biology — SA/V = 3/r for a sphere. As animals get LARGER, surface area grows as r² but volume grows as r³, so SA/V DECREASES. Small animals lose heat faster (large SA/V → more heat loss per gram). Cells must stay small to exchange nutrients efficiently. The Class 9 mensuration formulas explain biological size limits.

Where else this chapter is tested

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

AP Board SSC (Class 10) — MensurationVery High — Class 9 surface areas and volumes directly extend to Class 10 combination of solids and frustum

JEE Main and AdvancedMedium — 3D mensuration appears in JEE often combined with coordinate geometry

NTSE (Mathematics)High — mensuration problems are standard NTSE topics

AP EAPCETHigh — 3D solids and mensuration are part of EAPCET mathematics syllabus

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Questions students ask

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CURVED SURFACE AREA (CSA) / LATERAL SURFACE AREA: The area of the SIDE surfaces only — excluding the flat circular ends (for cylinders, cones) or top/bottom (for cuboids). Example: the wrapping label on a soup can (excluding the lid) = CSA of cylinder. TOTAL SURFACE AREA (TSA): The area of ALL surfaces — including the flat top, bottom, and sides. Example: the area of metal sheet needed to make a closed cylindrical box including the top and bottom = TSA. WHICH TO USE: depends on what's being asked. 'Paint the walls of a circular pillar' → CSA (just the curved part). 'Cost of metal to make a closed water tank' → TSA (all 6 surfaces of cuboid OR cylinder including top and bottom). 'Wallpaper for the inside of a room' → CSA of cuboid + ceiling (no floor) = 2h(l+b) + lb. Read the question carefully — students often compute TSA when CSA is needed.

Because the CSA formula for a cone uses SLANT HEIGHT (l), NOT the perpendicular height (h). CSA = πrl, NOT πrh. The slant height is the distance along the SLANTING SURFACE from the APEX to the EDGE of the base circle. The perpendicular height is the straight-up distance from the apex to the CENTRE of the base. These are different! In the right triangle formed by r (base radius), h (perpendicular height), and l (slant height), the slant is the hypotenuse: l² = r² + h² → l = √(r² + h²). EXAMPLE: r = 6 cm, h = 8 cm. l = √(36+64) = √100 = 10 cm. CSA = π × 6 × 10 = 60π cm² (NOT π × 6 × 8 = 48π). For the VOLUME of a cone, we use the perpendicular height h (not l): Volume = (1/3)πr²h. Different parts of the cone formula use different heights — be careful which one!

Imagine a cylinder of radius r and height h — its volume is πr²h. Now imagine a cone with the SAME radius r at the base and the SAME height h, with its apex (point) at the top of the cylinder. The cone clearly takes up LESS volume than the cylinder. Specifically, the cone occupies exactly 1/3 of the cylinder's volume — the rest is empty space. This 1/3 ratio is a deep geometric fact, proven rigorously using calculus (integration) — Archimedes proved it 2300 years ago without calculus by an ingenious method. SAME RELATIONSHIP holds for pyramids and prisms: volume of pyramid = (1/3) × volume of prism with same base and height. This 1/3 factor is essential — students who forget it get answers three times too large for cone volumes.

USE UNIT CONVERSION: 1 m³ = 1,000 LITRES (1 kilolitre). PROCEDURE: (1) Compute volume in m³ using V = πr²h with r and h in metres. (2) Multiply by 1000 to get litres. EXAMPLE: cylinder with r = 2 m, h = 5 m. Volume = π × 4 × 5 = 20π m³ ≈ 62.83 m³. In litres: 62.83 × 1000 = 62,830 litres ≈ 63 kL. ALTERNATIVE METHOD: convert dimensions to dm first (1 m = 10 dm). r = 20 dm, h = 50 dm. Volume = π × 400 × 50 = 20,000π dm³ = 62,830 dm³ = 62,830 litres (since 1 dm³ = 1 L). COMMON ERROR: thinking 1 m³ = 100 L (this is WRONG — 1 m³ = 1000 L). Always double-check the conversion factor before submitting an answer.

DECISION RULE: count only the SURFACES THAT ARE EXPOSED in the final compound figure. Don't include the surfaces that are HIDDEN at the junction. EXAMPLE: cone (radius r, slant l) placed on top of cylinder (same r, height h). EXPOSED surfaces: (1) Bottom of cylinder (πr² — the cylinder sits on the table). (2) Curved surface of cylinder (2πrh). (3) Curved surface of cone (πrl). HIDDEN: top of cylinder and bottom of cone (both = πr², but they coincide). TSA = πr² + 2πrh + πrl = πr(r + 2h + l). Note: the cone's base and the cylinder's top are FUSED — neither is exposed. Be very careful not to add them. SIMILARLY for hemisphere-on-cylinder: hemisphere CSA + cylinder CSA + cylinder bottom (the flat base of hemisphere coincides with cylinder top — hidden).

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