Key Formulas & Concepts to Remember

• Slant Height (l): l = √(r² + h²), where r is the base radius and h is the vertical height.
• Curved Surface Area (CSA) of Cone: CSA = πrl
• Total Surface Area (TSA) of Cone: TSA = πr(l + r)
• Standard Value of π: Take π = 22/7 unless specified as 3.14.

Q1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

Solution:

Given: Base Diameter (d) = 10.5 cm
Radius (r) = 10.5 / 2 = 5.25 cm = 21/4 cm
Slant height (l) = 10 cm

Curved Surface Area (CSA) = πrl

CSA = (22/7) × (21/4) × 10 = (22 × 3 × 10) / 4 = 660 / 4 = 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.

Solution:

Given: Slant height (l) = 21 m
Base Diameter (d) = 24 m ⇒ Radius (r) = 24 / 2 = 12 m

Total Surface Area (TSA) = πr(l + r)

TSA = (22/7) × 12 × (21 + 12)
TSA = (22/7) × 12 × 33 = 8712 / 7 = 1244.57 m² (approx.)

Q3. Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find (i) radius of the base and (ii) total surface area of the cone.

Solution:

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

(i) To find radius (r):
CSA = πrl = 308
(22/7) × r × 14 = 308
44 × r = 308 ⇒ r = 308 / 44 = 7 cm

(ii) To find Total Surface Area (TSA):
TSA = πr(l + r)
TSA = (22/7) × 7 × (14 + 7) = 22 × 21 = 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.

Solution:

Given: Height (h) = 10 m, Radius (r) = 24 m

(i) Slant height (l):
l = √(r² + h²) = √(24² + 10²) = √(576 + 100) = √676 = 26 m

(ii) Cost of canvas required:
Area of canvas needed = CSA = πrl = (22/7) × 24 × 26 = 13728 / 7 m²
Cost of 1 m² canvas = ₹ 70
Total Cost = (13728 / 7) × 70 = 13728 × 10 = ₹ 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 that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use π = 3.14).

Solution:

Given: h = 8 m, r = 6 m, Width of tarpaulin = 3 m, Wastage = 20 cm = 0.2 m

Slant height (l) = √(r² + h²) = √(6² + 8²) = √(36 + 64) = √100 = 10 m
CSA of conical tent = πrl = 3.14 × 6 × 10 = 188.4 m²
Length of tarpaulin required = Area / Width = 188.4 / 3 = 62.8 m
Total length required (including extra margin) = 62.8 m + 0.2 m = 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.

Solution:

Given: Slant height (l) = 25 m, Diameter = 14 m ⇒ Radius (r) = 7 m
CSA of tomb = πrl = (22/7) × 7 × 25 = 550 m²
Cost of white-washing 100 m² = ₹ 210
Cost of white-washing 1 m² = ₹ 210 / 100 = ₹ 2.10
Total cost for 550 m² = 550 × 2.10 = ₹ 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.

Solution:

Given: Radius (r) = 7 cm, Height (h) = 24 cm
l = √(r² + h²) = √(7² + 24²) = √(49 + 576) = √625 = 25 cm
CSA of 1 cap = πrl = (22/7) × 7 × 25 = 550 cm²
Area of sheet for 10 caps = 10 × 550 = 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 of the cones 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 √1.04 = 1.02)

Solution:

Given: Diameter = 40 cm ⇒ Radius (r) = 20 cm = 0.2 m
Height (h) = 1 m
Slant height (l) = √(r² + h²) = √(0.2² + 1²) = √(0.04 + 1) = √1.04 = 1.02 m
CSA of 1 cone = πrl = 3.14 × 0.2 × 1.02 = 0.64056 m²
CSA of 50 cones = 50 × 0.64056 = 32.028 m²
Total Cost = 32.028 × ₹ 12 = ₹ 384.34 (approx.)

Aptitude Corner: Quick Concept Check

Question: If both the radius and height of a right circular cone are doubled, how many times will its Curved Surface Area become?

Quick Logic: Since l = √(r² + h²), doubling both r and h doubles l as well (new l' = 2l). Therefore, new CSA = π × (2r) × (2l) = 4 × (πrl). The Curved Surface Area becomes 4 times the original area!