Subject: General Aptitude · Chapter: Volume and Surface Area · Exam: 2006 · Marks: · Difficulty:
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
(a)$1300 \mathrm{m}^{2}$
(b)$1310 \mathrm{m}^{2}$
(c)$1320 \mathrm{m}^{2}$
(d)$1330 \mathrm{m}^{2}$
Answer
Answer (as printed): C
Explanation
Radius, $r=12 \mathrm{m}$. Height of conical part, $h=(16-11) \mathrm{m}=5 \mathrm{m}$. Slant height of conical part, $$l=\sqrt{r^{2}+h^{2}}=\sqrt{(12)^{2}+5^{2}}=\sqrt{169}=13 \mathrm{m} .$$ Height of cylindrical part, $H=11 \mathrm{m}$. Area of canvas required = Curved surface area of cylinder + Curved surface area of cone $$\begin{aligned} & =2 \pi r H+\pi r l \\ & =\left[\frac{22}{7}(2 \times 12 \times 11+12 \times 13)\right] \mathrm{m}^{2} \\ & =\left[\frac{22}{7}(264+156)\right] \mathrm{m}^{2}=\left(\frac{22}{7} \times 420\right) \mathrm{m}^{2}=1320 \mathrm{m}^{2} . \end{aligned}$$