Question

Find the surface area of a frustum of cone, whose larger and smaller radius is 10 and 2 mm. The slant height of the cone is 15 mm.

• A. 274$\pi$ mm${}^2$
• B. 264$\pi$ mm${}^2$
• C. 284$\pi$ mm${}^2$
• D. 254$\pi$ mm${}^2$

Answer 1

Answer: (C)

284$\pi$ mm${}^2$

Surface area of a frustum of cone = $\pi$$(r+R)s+$$\pi$$r$${}^2$$+$$\pi$$R$${}^2$

$R=10mm$, $r=2mm$, $s=15mm$

$SA=(2+10)\ast 15+2+10$

$=\pi \left(12\right)15+4\pi +100\pi$

$=\pi (180+4+100)$

$=284\pi$ mm${}^2$