Exercise 10.3 — Cone

Surface area and volume of cone.

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Chapter 10 · Surface Areas & Volumes

Exercise 10.3 — Right Circular Cone

Full step-by-step solutions for all 12 questions covering volume, curved surface area, total surface area, slant height and real-life applications of cones.

Class 9 Maths CBSE Telangana Board Andhra Pradesh Board 12 Questions Solved

What is a Right Circular Cone?

A right circular cone is a three-dimensional solid formed when a right-angled triangle is rotated about one of its legs. The fixed leg becomes the height (h) of the cone, and the rotating leg traces out the circular base of radius r. The slanting side sweeps out the lateral (curved) surface and its length is called the slant height (l).

You encounter cones every day — ice-cream cones, traffic cones, party hats, funnels, carrot shapes and pine cones all approximate this solid. Understanding how to measure their surface and volume is a key skill tested in Class 9 board exams for CBSE, Telangana and Andhra Pradesh.

🍦 Ice-cream cone 🚧 Traffic cone 🎉 Party hat 🌲 Christmas tree 🥕 Carrot 🔺 Funnel
h r l V
Right Circular Cone
h = height  |  r = radius  |  l = slant height
Right Oblique
Right vs Oblique Cone
In a right cone, the apex is directly above the centre of the base.

Key Terms and Formulas — Right Circular Cone

Before solving problems, make sure you know the three measurements of a cone and how they relate:

SymbolNameMeaning
rBase radiusRadius of the circular base
hHeightPerpendicular distance from apex to base centre
lSlant heightDistance from apex to any point on the base circle (along the surface)

The three measurements are connected by the Pythagorean relationship:

l² = r² + h²
Base Circumference
2πr
Base Area
πr²
Curved (Lateral) Surface Area
πrl
Total Surface Area
πr(l + r)
Volume
(1/3)πr²h
Slant Height
l = √(r² + h²)
💡 Memory tip: A cone's volume is exactly one-third of the volume of a cylinder with the same radius and height. This is why the formula has a ⅓ factor. Fill a cone three times to fill the matching cylinder!

Questions 1 & 2 — Finding the Height from Given Volume

These questions teach you to rearrange the volume formula to find the height when the base area or radius is already given. The key move is to substitute what you know directly into the volume formula.

Question 1
The base area of a cone is 38.5 cm². Its volume is 77 cm³. Find its height.

Given: Base area = πr² = 38.5 cm², Volume = 77 cm³

Strategy: The volume formula is (1/3) × πr² × h. We already know πr² = 38.5, so we can substitute directly without needing r separately.

Volume = (1/3) × πr² × h 77 = (1/3) × 38.5 × h h = (77 × 3) ÷ 38.5 h = 231 ÷ 38.5 Height of the cone = 6 cm
Answer: Height = 6 cm
Question 2
The volume of a cone is 462 m³. Its base radius is 7 m. Find its height.

Given: r = 7 m, Volume = 462 m³, π = 22/7

Volume = (1/3) × πr²h (1/3) × (22/7) × 7 × 7 × h = 462 (1/3) × 22 × 7 × h = 462 (7 in numerator and denominator cancel) (22 × 7 / 3) × h = 462 h = (462 × 3) ÷ (22 × 7) h = 1386 ÷ 154 ∴ Height of the cone = 9 m
Answer: Height = 9 m

Question 3 — Finding Radius and Total Surface Area from Curved Surface Area

This question is a two-part problem. First, use the curved surface area formula (πrl) to find the radius, then apply the total surface area formula πr(l + r).

Question 3
Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find: (i) radius of the base (ii) total surface area.

Given: Curved surface area (CSA) = 308 cm², Slant height l = 14 cm

Part (i) — Find radius r:
CSA = πrl = 308 (22/7) × r × 14 = 308 22 × r × 2 = 308 44r = 308 r = 7 cm
Part (ii) — Total surface area:
TSA = πr(l + r) = (22/7) × 7 × (14 + 7) = 22 × 21 = 462 cm²
Answers: (i) Radius = 7 cm  |  (ii) Total Surface Area = 462 cm²
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Questions 4 & 5 — Cost of Painting and Sector-to-Cone Conversion

These are the most challenging questions in the exercise. Question 4 uses a cost problem to find the total surface area, then works backwards to find radius (requiring a quadratic equation). Question 5 involves an interesting real-world construction: bending a circular sector into a cone.

Question 4
The cost of painting the total surface area of a cone at 25 paise per cm² is ₹176. Find the volume, if slant height is 25 cm.

Step 1 — Find total surface area from cost:

Area painted at ₹0.25 per cm² costs ₹176 Total surface area = 176 ÷ 0.25 = 176 × 100 ÷ 25 Total surface area = 704 cm²

Step 2 — Find radius using Total Surface Area = 704 cm²:

πr(l + r) = 704, where l = 25 cm (22/7) × r(25 + r) = 704 r(25 + r) = 704 × 7/22 = 224 r² + 25r - 224 = 0 r² + 32r - 7r - 224 = 0 (factorising: 32 × (-7) = -224, 32 + (-7) = 25) r(r + 32) - 7(r + 32) = 0 (r + 32)(r - 7) = 0 r = -32 (rejected, radius can't be negative) or r = 7 cm ✓

Step 3 — Find height using Pythagoras:

l² = r² + h² 25² = 7² + h² 625 = 49 + h² h² = 576 → h = 24 cm

Step 4 — Find volume:

V = (1/3)πr²h = (1/3) × (22/7) × 7 × 7 × 24 = (1/3) × 22 × 7 × 24 ∴ Volume = 1232 cm³
Answer: Volume = 1232 cm³  (r = 7 cm, h = 24 cm)
📐 Why does a quadratic appear in Q4? The total surface area formula πr(l + r) expands to πrl + πr², which has both an r and an r² term. When you set this equal to a number, you get a quadratic equation in r. This is exactly why Chapter 4 (Quadratic Equations) in Class 10 matters — the skill connects across chapters!
Question 5
From a circle of radius 15 cm, a sector with angle 216° is cut out and its bounding radii are bent to form a cone. Find its volume.

Key insight: When you roll a sector into a cone, the radius of the circle becomes the slant height of the cone, and the arc length of the sector becomes the circumference of the cone's base.

Slant height of cone l = radius of circle = 15 cm Sector arc = circumference of cone base Lateral surface area of cone = Area of sector πrl = (x/360) × πR² r × 15 = (216/360) × 15 × 15 r × 15 = (3/5) × 225 = 135 r = 9 cm Find height: l² = r² + h² 15² = 9² + h² → 225 = 81 + h² h² = 144 → h = 12 cm Volume = (1/3) × π × 9² × 12 = (1/3) × 3.14 × 81 × 12 ∴ Volume = 1017.36 cm³
Answer: Volume = 1017.36 cm³  (r = 9 cm, h = 12 cm, l = 15 cm)

Questions 6 & 7 — Conical Tent Problems

These questions apply cone geometry to practical real-world scenarios — finding the cost of canvas for a tent, and finding the volume when both CSA and base area are given.

Question 6
Height of a tent = 9 m, base diameter = 24 m. Find slant height and cost of canvas cloth at ₹14 per sq.m.
Find slant height:
r = 24/2 = 12 m, h = 9 m l² = r² + h² = 12² + 9² l² = 144 + 81 = 225 l = 15 m
Find cost of canvas:
CSA = πrl = (22/7) × 12 × 15 = 3960/7 m² Cost = (3960/7) × 14 = ₹7920
Answers: Slant height = 15 m  |  Cost of canvas = ₹7,920
Question 7
Curved surface area = 1159 5/7 cm², base area = 254 4/7 cm². Find the volume.
Base area: πr² = 254 4/7 = 1782/7 (22/7) × r² = 1782/7 → r² = 81 → r = 9 cm CSA: πrl = 1159 5/7 = 8118/7 (22/7) × 9 × l = 8118/7 → l = 41 cm Height: l² = r² + h² 41² = 9² + h² → 1681 = 81 + h² h² = 1600 → h = 40 cm Volume = (1/3) × π × 9² × 40 = (1/3) × 3.14 × 81 × 40 ∴ Volume = 3391.2 cm³
Answer: Volume = 3391.2 cm³

Questions 8 & 9 — Composite Tent and Tarpaulin Length

Question 8
A tent is cylindrical to height 4.8 m and conical above it. Base radius = 4.5 m, total height = 10.8 m. Find the canvas required.

Key idea: Total canvas = lateral surface of cylinder + lateral surface of cone.

Cylindrical height H = 4.8 m, r = 4.5 m Conical height h = 10.8 - 4.8 = 6 m Slant height of cone: l² = 4.5² + 6² = 20.25 + 36 = 56.25 l = 7.5 m Canvas = 2πrH + πrl = πr(2H + l) = 3.14 × 4.5 × (2 × 4.8 + 7.5) = 14.13 × (9.6 + 7.5) = 14.13 × 17.1 ∴ Canvas required = 241.6 m²
Answer: Canvas required = 241.6 m²
Question 9
A conical tent has height 8 m and base radius 6 m. What length of 3 m wide tarpaulin is needed? (Add 20 cm for wastage, π = 3.14)
r = 6 m, h = 8 m l² = 6² + 8² = 36 + 64 = 100 → l = 10 m Lateral surface area = πrl = 3.14 × 6 × 10 = 188.4 m² Area of tarpaulin = Length × Width = L × 3 L × 3 = 188.4 L = 188.4 ÷ 3 = 62.8 m Add wastage: 62.8 + 0.20 = 63 m
Answer: Length of tarpaulin = 63 m

Questions 10, 11 & 12 — Joker's Cap, Vessel Filling and Similar Cones

Question 10
A joker's cap is a right circular cone of base radius 7 cm and height 27 cm. Find the sheet required for 10 such caps.
r = 7 cm, h = 27 cm l² = 7² + 27² = 49 + 729 = 778 l = √778 ≈ 27.9 cm Lateral surface area of 1 cap = πrl = (22/7) × 7 × 27.9 = 22 × 27.9 = 613.8 cm² For 10 caps: 10 × 613.8 ∴ Sheet required = 6138 cm²
Answer: Sheet required for 10 caps = 6138 cm²
Question 11
Water fills a conical vessel (diameter 5.2 m, slant height 6.8 m) at 1.8 m³/min. How long to fill it?
r = 5.2/2 = 2.6 m, l = 6.8 m l² = r² + h² → (6.8)² = (2.6)² + h² 46.24 = 6.76 + h² → h² = 39.48 h = 6.28 m Volume = (1/3) × π × (2.6)² × 6.28 = (1/3) × 3.14 × 6.76 × 6.28 Volume ≈ 44.43 m³ Time = Volume ÷ Rate = 44.43 ÷ 1.8 ∴ Time ≈ 24.7 minutes
Answer: Time to fill = ≈ 24.7 minutes
Question 12
Two similar cones have volumes 12π and 96π cu. units. If CSA of smaller cone is 15π sq. units, find the CSA of the larger cone.

For similar cones, the ratio of curved surface areas is equal to the volume ratio raised to the power 2/3:

A₁/A₂ = (V₁/V₂)^(2/3)
V₁ = 12π, V₂ = 96π, A₁ = 15π V₁/V₂ = 12π/96π = 1/8 A₁/A₂ = (1/8)^(2/3) = [(1/2)³]^(2/3) = (1/2)² = 1/4 15π / A₂ = 1/4 A₂ = 15π × 4 ∴ CSA of larger cone = 60π sq. units
Answer: Curved surface area of larger cone = 60π sq. units
📐 Formula for similar cones: r₁/r₂ = h₁/h₂ = l₁/l₂ = (A₁/A₂)^(1/2) = (V₁/V₂)^(1/3). This scaling relationship is powerful for comparing similar solids.

Quick Reference — All 12 Answers at a Glance

Q#ProblemKey Answer
Q1Base area 38.5 cm², Volume 77 cm³ → Find heighth = 6 cm
Q2Volume 462 m³, radius 7 m → Find heighth = 9 m
Q3iCSA 308 cm², slant 14 cm → radiusr = 7 cm
Q3iiTotal surface area of Q3 coneTSA = 462 cm²
Q4Painting cost ₹176 → Volume (slant 25 cm)V = 1232 cm³
Q5Sector 216° from circle r=15 cm → Volume of coneV = 1017.36 cm³
Q6Tent h=9 m, d=24 m → Slant height & canvas costl = 15 m; ₹7,920
Q7CSA = 1159 5/7 cm², base area = 254 4/7 cm² → VolumeV = 3391.2 cm³
Q8Composite tent (cylinder + cone) → Canvas area241.6 m²
Q9Tarpaulin 3 m wide for cone tent → Length63 m
Q10Joker's cap r=7, h=27 → Sheet for 10 caps6138 cm²
Q11Conical vessel fills at 1.8 m³/min → Time≈ 24.7 min
Q12Similar cones, CSA ratio from volume ratio60π sq. units

Common Mistakes to Avoid

  • Using diameter instead of radius: When the problem gives diameter, always halve it first before substituting into any formula. This is the most frequent calculation error in Q6 and Q11.
  • Confusing slant height and vertical height: The formulas for CSA (πrl) and Volume (⅓πr²h) use different heights — slant height l for surface area, vertical height h for volume. Mixing them up loses marks.
  • Not rejecting the negative root in Q4: The quadratic gives r = 7 and r = −32. Always state that r = −32 is rejected because a radius cannot be negative, then proceed with r = 7.
  • Forgetting to add waste in Q9: The question specifically mentions 20 cm of extra material. Always read for such conditions before finalising your answer.
  • Wrong formula for similar cones (Q12): Surface areas scale as the square of linear dimensions (ratio²), and volumes scale as the cube (ratio³). The key formula is A₁/A₂ = (V₁/V₂)^(2/3).
  • Incorrect π value: Use π = 22/7 unless the problem says "use π = 3.14." Both appear in this exercise — switching them gives a wrong answer.
Board exam alert (Telangana & AP): In Questions 4 and 5, partial credit is given for each correct step. Write each calculation step clearly — finding TSA, setting up the quadratic, rejecting the negative root, finding h, then finding volume. Missing any step can reduce marks even if the final answer is right.

What This Exercise Prepares You For

Exercise 10.3 is the cone-specific component of Chapter 10. After this, Exercise 10.4 and 10.5 extend the same ideas to spheres and hemispheres, and later to combined solids where a cone is attached to a cylinder or hemisphere. The quadratic equation technique used in Q4 directly revisits skills from Class 9 Polynomials and will reappear in Class 10 Quadratic Equations.

The ratio-of-similar-solids concept in Q12 connects to Similar Triangles in Class 10, where the same surface-to-volume scaling applies. For students revising earlier work, this topic connects to Exponents and Powers when evaluating fractional powers like (1/8)^(2/3).

📐 Surface Areas & Volumes 📊 Similar Solids 📝 Quadratic Equations 🔺 Chapter 10 — Class 9
📌 Exam tip for CBSE, Telangana and AP boards: Exercise 10.3 problems — especially Q4 (quadratic approach) and Q5 (sector-to-cone) — are high-value questions often set for 4–5 marks. Practice these without a calculator to match actual exam conditions. Knowing the 3-4-5 and 5-12-13 Pythagorean triples helps quickly spot slant heights: in Q6, (9, 12, 15) is a multiple of (3, 4, 5); in Q9, (6, 8, 10) is a multiple of (3, 4, 5).
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