A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of the cylinder is 24 m. The height of the cylindrical portion is 11 m while the vertex of the cone is 16 m above the ground. The area of the canvas required for the tent is
Aptitude
Volume and Surface Area
Difficulty: Hard
Choose an option
-
A1300 m$^2$
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B1310 m$^2$
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C1320 m$^2$
-
D1330 m$^2$
Answer
Correct Answer: 1320 m$^2$
Explanation
### Concept & Surface Area of Composite Figures
The canvas required for a tent equals the sum of the curved surface area of the cylindrical base and the curved surface area of the conical top. The floor is not covered with canvas.
$$ \text{Area} = 2\pi r h_{\text{cyl}} + \pi r l_{\text{cone}} $$
### Step-by-Step Solution
- Diameter of the cylinder = 24 m, so the radius $r = \frac{24}{2} = 12$ m.
- Height of the cylinder $h_1 = 11$ m.
- Total height of the tent is 16 m, so the height of the cone $h_2 = 16 - 11 = 5$ m.
- Find the slant height $l$ of the cone: $l = \sqrt{r^2 + h_2^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = 13$ m.
- Calculate the total canvas area: $A = 2 \pi r h_1 + \pi r l$.
- Factor out $\pi r$: $A = \pi r (2h_1 + l)$.
- Substitute the values: $A = \frac{22}{7} \times 12 \times (2(11) + 13)$.
- Simplify inside the bracket: $22 + 13 = 35$.
- Calculate final area: $A = \frac{22}{7} \times 12 \times 35$.
- Cancel 35 with 7 to get 5: $A = 22 \times 12 \times 5 = 22 \times 60 = 1320$ m$^2$.
### Exam Strategy & Shortcut
Recognize the 5-12-13 Pythagorean triple immediately for the cone's dimensions to save time calculating the slant height. From there, apply the factored formula $\pi r (2h + l)$ for rapid calculation.
### Common Pitfall
Using the total height (16 m) as the height of the cone, rather than subtracting the cylinder's height first.
### Final Answer
Therefore, the correct answer is **1320 m$^2$**.