Q1. [4]
A 'circus' is a company of performers who put on shows of acrobats, clowns etc. to entertain people started around 250 years back, in open fields, now generally performed in tents. One such 'Circus Tent' is shown below. The tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of cylindrical part are 9 m and 30 m respectively and height of conical part is 8 m with same diameter as that of the cylindrical part.
Case Study - 2 : The tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of cylindrical part are 9 m and 30 m respectively and height of conical part is 8 m with same diameter as that of the cylindrical part, then find:
- (1) the area of the canvas used in making the tent; [3]
- (2) the cost of the canvas bought for the tent at the rate ₹ 200 per sq m, if 30 sq m canvas was wasted during stitching. [1]
Previously asked in CBSE board exam
2022 30/2/1 Q14
Generated by claude-sonnet-4-6 · 2026-06-15 10:36 · grounding stimulus
Model Answer
Given: Radius (r) = 15 m, Height of cylinder (H) = 9 m, Height of cone (h) = 8 m
(1) Area of canvas used:
Slant height of cone: $l = \sqrt{r^2 + h^2} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17$ m
Curved Surface Area of cylinder = $2\pi r H = 2 \times \frac{22}{7} \times 15 \times 9 = \frac{5940}{7}$ m²
Curved Surface Area of cone = $\pi r l = \frac{22}{7} \times 15 \times 17 = \frac{5610}{7}$ m²
Total canvas area = $\frac{5940 + 5610}{7} = \frac{11550}{7} = \mathbf{1650 \ m^2}$
(2) Cost of canvas:
Total canvas needed = 1650 + 30 = 1680 m²
Cost = 1680 × ₹200 = ₹3,36,000
Explanation
- The tent has no base, so only the curved surfaces are counted — CSA of cylinder + CSA of cone.
- The slant height $l = \sqrt{r^2+h^2}$ is a must-calculate step; examiners deduct marks if skipped.
- In part (2), remember to add the wasted canvas (30 m²) before multiplying by rate — a common error students make.
- Use $\pi = \frac{22}{7}$ unless told otherwise; it gives clean numbers here.
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