Class 9 Mathematics
Chapter 11: Surface Areas and Volumes
Exercise 11.1 – Stepwise Solutions
Important Formula Used
Curved Surface Area of Cone: πrl
Total Surface Area of Cone: πr(l + r)
Slant Height of Cone: l = √(r² + h²)
Note: Assume π = 22/7 unless stated otherwise.
Q1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Given:
Diameter = 10.5 cm
So, radius r = 10.5 / 2 = 5.25 cm
Slant height l = 10 cm
Formula: Curved surface area = πrl
CSA = (22/7) × 5.25 × 10
= (22/7) × (21/4) × 10
= 165 cm²
Answer: 165 cm²
Q2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Given:
Slant height l = 21 m
Diameter = 24 m
Radius r = 12 m
Formula: Total surface area = πr(l + r)
TSA = (22/7) × 12 × (21 + 12)
= (22/7) × 12 × 33
= 1244.57 m²
Answer: 1244.57 m²
Q3. 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 of the cone
Given:
CSA = 308 cm²
l = 14 cm
(i) Find radius
Using formula:
πrl = 308
(22/7) × r × 14 = 308
44r = 308
r = 7 cm
Radius = 7 cm
(ii) Total surface area
TSA = πr(l + r)
= (22/7) × 7 × (14 + 7)
= 22 × 21
= 462 cm²
Answer: Radius = 7 cm, TSA = 462 cm²
Q4. A conical tent is 10 m high and the radius of its base is 24 m. Find:
(i) slant height of the tent
(ii) cost of the canvas required to make the tent, if the cost of 1 m² canvas is ₹ 70
Given:
Height h = 10 m
Radius r = 24 m
(i) Slant height
l = √(r² + h²)
= √(24² + 10²)
= √(576 + 100)
= √676
= 26 m
(ii) Cost of canvas
Canvas needed = curved surface area of cone
CSA = πrl
= (22/7) × 24 × 26
= 1961.14 m²
Cost = 1961.14 × 70
= ₹ 1,37,280
Answer: (i) 26 m (ii) ₹ 1,37,280
Q5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material required for stitching margins and wastage in cutting is approximately 20 cm. (Use π = 3.14)
Given:
Height h = 8 m
Radius r = 6 m
Width of tarpaulin = 3 m
Step 1: Slant height
l = √(r² + h²)
= √(6² + 8²)
= √(36 + 64)
= √100 = 10 m
Step 2: Curved surface area
CSA = πrl
= 3.14 × 6 × 10
= 188.4 m²
Step 3: Length of tarpaulin
Length = Area / Width
= 188.4 / 3
= 62.8 m
Add extra 20 cm = 0.2 m
Required length = 62.8 + 0.2 = 63.0 m
Answer: 63 m
Q6. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of ₹ 210 per 100 m².
Given:
Slant height l = 25 m
Diameter = 14 m
Radius r = 7 m
Step 1: Curved surface area
CSA = πrl
= (22/7) × 7 × 25
= 550 m²
Step 2: Cost
Cost for 100 m² = ₹ 210
Cost for 550 m² = (210/100) × 550
= ₹ 1155
Answer: ₹ 1155
Q7. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Given:
Radius r = 7 cm
Height h = 24 cm
Step 1: Slant height
l = √(r² + h²)
= √(7² + 24²)
= √(49 + 576)
= √625 = 25 cm
Step 2: Area of one cap
Since cap is open at base, only curved surface area is needed.
CSA = πrl
= (22/7) × 7 × 25
= 550 cm²
Step 3: Area for 10 caps
= 10 × 550 = 5500 cm²
Answer: 5500 cm²
Q8. A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each cone is to be painted and the cost of painting is ₹ 12 per m², what will be the cost of painting all these cones? (Use π = 3.14 and take √1.04 = 1.02)
Given:
Number of cones = 50
Diameter = 40 cm = 0.4 m
Radius r = 0.2 m
Height h = 1 m
Step 1: Slant height
l = √(r² + h²)
= √(0.2² + 1²)
= √(0.04 + 1)
= √1.04
= 1.02 m
Step 2: Curved surface area of one cone
CSA = πrl
= 3.14 × 0.2 × 1.02
= 0.64056 m²
Step 3: Area of 50 cones
= 50 × 0.64056
= 32.028 m²
Step 4: Cost of painting
Cost = 32.028 × 12
= ₹ 384.336
Answer: ₹ 384.34 (approx.)