Question

Calculate the total surface area of a frustum of cone whose upper base radius is 2m and lower base radius is 4 m and its height is 12 m. (Use $\pi$ = $\pi$).

• A. 4$\pi$(3$\sqrt{37}$ + 5) m${}^2$
• B. 3$\pi$(3$\sqrt{37}$ + 5) m${}^2$
• C. 2$\pi$(3$\sqrt{37}$ + 5) m${}^2$
• D. $\pi$(3$\sqrt{37}$ + 5) m${}^2$

Answer 1

Answer: (A), (C)

4$\pi$(3$\sqrt{37}$ + 5) m${}^2$
2$\pi$(3$\sqrt{37}$ + 5) m${}^2$
Total surface area, SA = (r + R)$\sqrt{(R-r{)}^2+h^2}+\pi r^2+\pi R^2$

R = 4m, r = 2m, h = 12m

SA = (2 + 4)π$\sqrt{(4-2{)}^2+{12}^2}+\pi 2^2+\pi 4^2$

= 6$\pi \sqrt{(2{)}^2+{12}^2}+4\pi +16\pi$

= 6$\pi \sqrt{4+144}+20\pi$

= 6$\pi \sqrt{148}+20\pi$

= 4$\pi (3\sqrt{37}+5)$ m