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
  • A
    1300 m$^2$
  • B
    1310 m$^2$
  • C
    1320 m$^2$
  • D
    1330 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$^2$**.
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