Exercise 10.1 — Surface Areas and Volumes
Problems based on surface areas and volumes.
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Exercise 10.1 — Mensuration | Class 10 Maths
Exercise 10.1 from Chapter 10 — Mensuration — covers the surface areas and volumes of cones and cylinders, which are among the most frequently tested topics in Class 10 board examinations for CBSE, Telangana, and Andhra Pradesh boards. All 8 questions in this exercise apply the formulas for lateral surface area (CSA), total surface area (TSA), and volume of right circular cones and cylinders to real-world contexts like joker's caps, shuttlecock cylinders, iron rods, and rice heaps.
Cone Surface Area
Cylinder Surface Area
Cone Volume
Cylinder Volume
Slant Height
Key Formulas — Cone and Cylinder at a Glance
Every question in Exercise 10.1 uses one or more of these formulas. Memorise them before attempting the problems.
Right Circular Cone
r = base radius, h = height, l = slant height
Right Circular Cylinder
r = base radius, h = height
| Shape | Measurement | Formula | Unit |
|---|---|---|---|
| Cone | Slant height | l = √(r² + h²) | cm / m |
| Lateral (Curved) Surface Area | CSA = πrl | cm² | |
| Total Surface Area | TSA = πrl + πr² = πr(l + r) | cm² | |
| Volume | V = (1/3)πr²h | cm³ | |
| Cylinder | Lateral (Curved) Surface Area | CSA = 2πrh | cm² |
| Total Surface Area | TSA = 2πr(h + r) | cm² | |
| Volume | V = πr²h | cm³ |
💡 π value tip: Use π = 22/7 when the radius is a multiple of 7. Use π = 3.14 when the problem explicitly says so (like Q8). Using the wrong value of π is a common way to lose marks.
Question 1 — Joker's Cap: Sheet Required for 10 Cones
Question 1
A joker's cap is in the form of a right circular cone whose base radius is 7 cm and height is 24 cm. Find the area of the sheet required to make 10 such caps.
Radius (r)7 cm
Height (h)24 cm
FindSheet for 10 caps
FormulaCSA = πrl
Joker's Cap (Cone)
r=7, h=24, l=25 cm
Step 1: Find slant height l
l = √(r² + h²) = √(7² + 24²)
= √(49 + 576) = √625
∴ l = 25 cm
Step 2: Lateral surface area of 1 cap (= sheet for 1 cap)
CSA = πrl = (22/7) × 7 × 25
= 22 × 25
= 550 cm²
Step 3: Sheet for 10 caps
Area = 10 × 550
= 5500 cm²
∴ Area of sheet required for 10 caps = 5500 cm²
📌 Why CSA, not TSA? A joker's cap is open at the bottom (no base). So only the lateral (curved) surface area = πrl is needed, not the full TSA. Always check whether the solid is open or closed at the base.
Question 2 — Shuttlecock Cylinders: Sheet for 100 Cylinders
Question 2
A sports company was ordered to prepare 100 paper cylinders for packing shuttlecocks. The required dimensions are height 35 cm and radius 7 cm. Find the area of thick paper sheet needed to make 100 cylinders.
Radius (r)7 cm
Height (h)35 cm
FindSheet for 100 cylinders
FormulaTSA = 2πr(h + r)
Step 1: TSA of 1 cylinder (both circular ends + curved surface)
TSA = 2πr(h + r)
= 2 × (22/7) × 7 × (35 + 7)
= 2 × 22 × 42
= 1848 cm²
Step 2: Sheet for 100 cylinders
Area = 100 × 1848
= 1,84,800 cm²
∴ Area of paper sheet for 100 cylinders = 1,84,800 cm²
💡 TSA vs CSA for cylinders: A shuttlecock cylinder is closed at both ends (it needs a lid and a base to hold the shuttlecocks). So we use TSA = 2πr(h + r) which includes the two circular bases. If the cylinder were open at both ends, only CSA = 2πrh would apply.
Question 3 — Volume of a Right Circular Cone
Question 3
Find the volume of a right circular cone with radius 6 cm and height 7 cm.
Radius (r)6 cm
Height (h)7 cm
FormulaV = (1/3)πr²h
V = (1/3) × πr²h
= (1/3) × (22/7) × 6 × 6 × 7
= (1/3) × 22 × 6 × 6 (7 cancels with 7)
= 22 × 2 × 6 (1/3 × 6 = 2)
= 22 × 12
= 264 cm³
∴ Volume of the cone = 264 cm³
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Question 4 — Ratio of Height of Cylinder to Slant Height of Cone
Question 4
The lateral surface area of a cylinder is equal to the curved surface area of a cone. If their bases are the same, find the ratio of the height of the cylinder to the slant height of the cone.
Condition 1Same base → radius of cylinder = radius of cone = r
Condition 2LSA of cylinder = CSA of cone
LetHeight of cylinder = h; Slant height of cone = l
Findh : l
LSA of cylinder = CSA of cone
2πrh = πrl
(πr cancels from both sides)
2h = l
h/l = 1/2
∴ h : l = 1 : 2
∴ Height of cylinder : Slant height of cone = 1 : 2
📌 Concept: This is a comparison/ratio question — no numbers are given. The key is to set the two area formulas equal, then simplify algebraically. The πr term cancels, leaving a clean relationship between h and l.
Question 5 — How Many Joker's Caps from 1000 cm² Sheet?
Question 5
A self-help group wants to manufacture joker's caps of 3 cm radius and 4 cm height. If the available paper sheet is 1000 cm², how many caps can be manufactured?
Radius (r)3 cm
Height (h)4 cm
Sheet available1000 cm²
FindNumber of caps
3-4-5 right triangle
Pythagorean triple!
Step 1: Slant height (3–4–5 Pythagorean triple!)
l = √(r² + h²) = √(3² + 4²) = √(9 + 16) = √25
∴ l = 5 cm
Step 2: CSA (sheet for 1 cap)
CSA = πrl = (22/7) × 3 × 5 = 330/7 cm²
Step 3: Number of caps
Number = Total sheet ÷ Sheet per cap
= 1000 ÷ (330/7)
= 1000 × (7/330)
= 7000/330
= 21.21...
∴ Number of caps = 21 (only whole caps count)
∴ 21 caps can be manufactured from 1000 cm² of paper sheet.
⚠️ Important: The calculation gives 21.21, but you can only make whole caps — you cannot make 0.21 of a cap. So the answer is always rounded down (floor value), not rounded to the nearest integer. 21.21 → 21 caps, never 22.
Question 6 — Show Volume of Cylinder : Cone = 3 : 1
Question 6
A cylinder and cone have bases of equal radii and are of equal heights. Show that their volumes are in the ratio 3 : 1.
GivenSame radius r and same height h for both
V(cylinder)πr²h
V(cone)(1/3)πr²h
ShowV(cylinder) : V(cone) = 3 : 1
:
= 3 : 1
V(cylinder) : V(cone)
= πr²h : (1/3)πr²h
= 1 : 1/3 (πr²h cancels)
= 1 × 3 : (1/3) × 3 (multiply both by 3)
= 3 : 1
✅ Hence Shown: For the same base and height, a cylinder holds exactly 3 times the volume of a cone. This is why a cone-shaped funnel fills a cylinder 3 times before the cylinder overflows — a beautiful geometric fact!
Question 7 — Total Volume of 50 Cylindrical Iron Rods
Question 7
A solid iron rod is cylindrical in shape. Its height is 11 cm and base diameter is 7 cm. Find the total volume of 50 such rods.
Height (h)11 cm
Diameter (d)7 cm → radius r = 7/2 cm
FormulaV = πr²h
FindVolume of 50 rods
Step 1: Radius r = d/2 = 7/2 cm
Step 2: Volume of 1 rod
V = πr²h = (22/7) × (7/2) × (7/2) × 11
= (22/7) × (49/4) × 11
= (22 × 49 × 11) / (7 × 4)
= (22 × 7 × 11) / 4 (49/7 = 7)
= 1694 / 4
= 423.5 cm³
Step 3: Total volume of 50 rods
Total V = 50 × 423.5
= 21,175 cm³
∴ Total volume of 50 cylindrical rods = 21,175 cm³
💡 Diameter vs Radius: The problem gives diameter = 7 cm, so radius = 7/2 cm. This is a very common trap — always halve the diameter before substituting into the formula. Using 7 instead of 3.5 will quadruple your answer incorrectly.
Question 8 — Rice Heap (Cone): Volume and Canvas Required
Question 8
A heap of rice is in the form of a cone with diameter 12 m and height 8 m. Find its volume. How much canvas cloth is required to cover the heap? (Use π = 3.14)
Diameter (d)12 m → radius r = 6 m
Height (h)8 m
π3.14 (use 22/7 gives ≈ same)
FindVolume + Canvas area (CSA)
Conical Rice Heap
6-8-10 triple (3-4-5 × 2)
Step 1: Radius r = 12/2 = 6 m
Step 2: Volume of conical heap
V = (1/3)πr²h
= (1/3) × (22/7) × 6 × 6 × 8
= (1/3) × (22/7) × 288
= (22 × 288) / 21
= 6336 / 21
≈ 301.7 m³
Step 3: Slant height (6-8-10 triple)
l = √(r² + h²) = √(6² + 8²) = √(36 + 64) = √100
∴ l = 10 m
Step 4: Canvas cloth = Lateral surface area of cone
CSA = πrl = (22/7) × 6 × 10
= 1320/7
≈ 188.57 m² ≈ 188.5 m²
∴ Volume = 301.7 m³ | Canvas required = 188.5 m²
💡 Pythagorean triples save time: In Q1, r=7, h=24 → l=25 (7–24–25 triple). In Q5, r=3, h=4 → l=5 (3–4–5 triple). In Q8, r=6, h=8 → l=10 (6–8–10 = 2× the 3–4–5 triple). Recognising these triples means you don't need a calculator for the square root.
Common Mistakes to Avoid in Exercise 10.1
- Using TSA when CSA is needed (and vice versa): For open-base objects like joker's caps or rice heaps → use CSA = πrl. For closed cylinders like shuttlecock containers → use TSA = 2πr(h + r). Reading the question carefully is essential.
- Forgetting to halve the diameter: Q7 and Q8 give the diameter, not the radius. Always divide diameter by 2 before substituting into any formula.
- Not computing slant height first: Many students substitute h into πrl directly. l ≠ h. Slant height must be found using l = √(r² + h²) first.
- Rounding up when counting items: In Q5, the calculation gives 21.21 caps. Since you cannot make a fraction of a cap, you must round down to 21, not up to 22.
- Wrong π value: Q8 says "use π = 3.14". In all other questions, π = 22/7 is appropriate (radius is a multiple or factor of 7). Using the wrong π value leads to a wrong answer even if all other steps are correct.
⛔ Board Exam Alert (Telangana & AP SSC): Q1, Q5, and Q8 are the most commonly picked from Exercise 10.1 for 4–5 mark questions. Q6 (prove 3:1 ratio) and Q4 (ratio h:l = 1:2) are popular 2-mark questions. For any multi-part question like Q8, ensure you answer both parts (volume AND canvas area) — missing the second part costs 2 marks.
Quick Reference — All Answers at a Glance
| Q# | Problem | Key Formula | Answer |
|---|---|---|---|
| Q1 | Sheet for 10 joker's caps (r=7, h=24) | 10 × πrl; l=25 | 5500 cm² |
| Q2 | Sheet for 100 shuttlecock cylinders (r=7, h=35) | 100 × 2πr(h+r) | 1,84,800 cm² |
| Q3 | Volume of cone (r=6, h=7) | (1/3)πr²h | 264 cm³ |
| Q4 | h(cylinder) : l(cone) if LSA = CSA, same base | 2πrh = πrl → h/l = 1/2 | 1 : 2 |
| Q5 | Caps from 1000 cm² (r=3, h=4) | 1000 ÷ (πrl); l=5 | 21 caps |
| Q6 | Show V(cylinder) : V(cone) = 3:1 | πr²h : (1/3)πr²h | 3 : 1 ✓ |
| Q7 | Volume of 50 cylindrical rods (d=7, h=11) | 50 × πr²h; r=3.5 | 21,175 cm³ |
| Q8 | Rice heap cone (d=12, h=8); volume + canvas | (1/3)πr²h; πrl; l=10 | 301.7 m³; 188.5 m² |
What Exercise 10.1 Prepares You For
The surface area and volume skills from Exercise 10.1 are the foundation for the entire Mensuration chapter. The next exercises extend these ideas to spheres and hemispheres, and then to combinations of solids — such as a cone placed on top of a cylinder, or a hemisphere scooped out of a cylinder. The slant height calculation using l = √(r² + h²) and the understanding of when to use CSA vs TSA will be needed in all subsequent exercises.
For earlier foundation work on 2D areas that leads into this chapter, you can review Surface Areas and Volumes from Class 9. The ratio and proportion thinking used in Q4 and Q6 connects directly to Similar Triangles Chapter 8.
📐 Exam Tips — Telangana & AP SSC Board:
- Always write Given, Formula, Substitution, Calculation, Answer as separate lines — this earns step marks even if your final answer has an arithmetic error.
- Memorise the three Pythagorean triples that appear in this exercise: 7–24–25, 3–4–5, and 6–8–10. They save time and avoid square-root errors.
- For Q6-type "show that" questions, write the full ratio setup and simplification — don't just state the answer.
- Units are compulsory: area answers in cm² or m², volume in cm³ or m³. Missing units costs half a mark per answer in most boards.