Right circular cone
A cone has a circular base and a pointed top. Find the slant height l, √(h² + r²), first. Then CSA is πrl and volume is (1/3)πr²h.
Slant height comes first
A right circular cone is formed when a right triangle turns around one of its perpendicular sides. The base radius r, the height h and the slant height l form a right triangle. By Pythagoras theorem, l = √(h² + r²). Most cone sums need l, so find it first and write it on your figure.
Cone formulas
| Measure | Formula | Remember |
|---|---|---|
| Slant height | l = √(h² + r²) | From Pythagoras theorem |
| Curved surface area | πrl | Equal to the area of the sector that folds into the cone |
| Total surface area | πr(l + r) | CSA + base circle πr² |
| Volume | (1/3)πr²h | One third of a cylinder with the same base and height |
Example 1: all four measures
A cone has radius 7 cm and height 24 cm. Find its slant height, curved surface area, total surface area and volume. (Take π = 22/7.)
- Given: r = 7 cm, h = 24 cm.
- Formula: l = √(h² + r²) = √(24² + 7²)
- = √(576 + 49) = √625 = 25 cm
- CSA = πrl = 22/7 × 7 × 25 = 550 cm²
- TSA = πr(l + r) = 22/7 × 7 × (25 + 7) = 22 × 32 = 704 cm²
- Volume = (1/3)πr²h = 1/3 × 22/7 × 7 × 7 × 24
- = 22 × 7 × 8 = 1232 cm³
l = 25 cm, CSA = 550 cm², TSA = 704 cm², volume = 1232 cm³
Example 2: folding a sector into a cone
A sector of angle 252° is cut from a circle of radius 10 cm. Its two straight edges are joined to make a cone, like a paper cap. Find the radius and the curved surface area of the cone. (Take π = 22/7.)
- When the sector is folded, its radius becomes the slant height: l = 10 cm.
- Its arc becomes the circle at the base, so arc length = 2πr.
- Arc length of the sector = (θ/360°) × 2π × (radius of the sector) = (252/360) × 2π × 10.
- So 2πr = (252/360) × 2π × 10, which gives r = (252/360) × 10 = 7 cm.
- CSA = πrl = 22/7 × 7 × 10 = 220 cm²
- Check: area of the sector = (252/360) × 22/7 × 10² = 220 cm². The two agree.
Radius of the cone = 7 cm, CSA = 220 cm²
Example 3: canvas for a tent
A conical tent for a village festival has base radius 3.5 m and height 12 m. How much canvas is needed? Find its cost at an example rate of ₹ 100 per m². (Take π = 22/7.)
- Given: r = 3.5 m, h = 12 m.
- l = √(h² + r²) = √(144 + 12.25) = √156.25 = 12.5 m
- A tent has no canvas floor, so canvas = CSA = πrl.
- CSA = 22/7 × 3.5 × 12.5 = 11 × 12.5 = 137.5 m²
- Cost = 137.5 × 100 = ₹ 13750
Canvas = 137.5 m²; cost = ₹ 13,750 at the example rate
A birthday cap, an ice cream cone, a funnel and a conical tent at a village festival are all cones. The canvas for a tent is its CSA, πrl, because the floor is not covered.
1 marks: “Find the slant height of a cone with radius 5 cm and height 12 cm.” l = √(12² + 5²) = √169. Answer: 13 cm.
2 marks: “Find the curved surface area of a cone of radius 14 cm and slant height 20 cm. (π = 22/7)” CSA = πrl = 22/7 × 14 × 20. Answer: 880 cm².
5 marks: “A sector of angle 216° is cut from a circle of radius 15 cm and bent into a cone. Find the radius, height and curved surface area of the cone. (π = 3.14)” Slant height l = 15 cm. r = (216/360) × 15 = 9 cm. h = √(15² − 9²) = √144 = 12 cm. CSA = πrl = 3.14 × 9 × 15 = 423.9 cm².
Cone rule: find l first. The side is πrl; the volume is one third of the cylinder, (1/3)πr²h.