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.
Answer: l = √(242 + 72) = 25 cm; one cap = πrl = 227 × 7 × 25 = 550 cm2; 10 caps = 5500 cm2.
Step-by-step solution
Given: r = 7 cm; h = 24 cm; 10 caps
To find: Sheet needed for 10 caps
To find: Sheet needed for 10 caps
Idea: A cap is open at the bottom, so it needs only the curved surface πrl. Find l first.
- l = √(242 + 72) = √(576 + 49) = √625 = 25 cm.1 mark
- Sheet for one cap = πrl = 227 × 7 × 25 = 550 cm2.1 mark
- For 10 caps: 10 × 550 = 5500 cm2.1 mark
Area of sheet = 5500 cm².
Check: 5500 cm2 = 0.55 m2, a sensible amount of card for ten caps ✓.
Answer to write in the exam
l = √(242 + 72) = √625 = 25 cm
One cap = πrl = 227 × 7 × 25 = 550 cm2
∴ 10 caps need 10 × 550 = 5500 cm2
Common mistakes that cost marks
- Using the height 24 cm in πrl.
- Adding the base πr2; a cap has no base.
- Forgetting to multiply by 10.
How this can come in the exam
MCQ (1 mark)
A conical tent has radius 3.5 m and slant height 6 m. The canvas needed for its curved surface is (π = 227)
- 66 m2
- 33 m2
- 132 m2
- 77 m2
Show answer
(A) 66 m2
227 × 3.5 × 6 = 11 × 6 = 66 m2.
Try one yourself
A birthday cap is a cone of radius 5 cm and height 12 cm. Find the paper needed for 20 caps, in terms of π.
Show answer
l = 13 cm; one cap 65π cm2; 20 caps 1300π cm2 (≈ 4082 cm2 with π = 3.14).
More questions like this
- What length of tarpaulin 3 m wide is required to make a conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that is required for stitching margins and wastage in cutting is 20 cm.
- A right triangle with sides 6 cm, 8 cm and 10 cm is rotated through 360° about the side of 8 cm. Find the volume and the curved surface area of the solid so formed.
- Suppose you have a cup in the shape of a right circular cone. Fill it with water to half the depth of the cone. What fraction of the volume of the cup is occupied by the water?
- But you may be able to find your own derivation of the formula by connecting it with the formula for volume of a cone; namely, by writing it as
V = 13 × (4πr2) × r,
i.e., Volume of sphere = 13 × surface area of sphere × radius of sphere.
Try to work out the connecting links on your own. - Take a rubber ball and drive a nail into it. Using the nail for support, wind a string tightly around the ball. When you have reached the ‘fullest’ part of the ball, use pins to keep the string in place, and continue to wind the string around the remaining part of the ball, till you have completely covered it; see the figure. Mark the starting and finishing points on the string, and unwind the string from the surface of the ball. Now, measure the diameter of the ball and get its radius. On a sheet of paper, draw four circles with a radius equal to the radius of the ball. Fill the circles with the string you had wound around the ball (as shown in the figure).