Question

Find the surface area of a frustum of cone, whose larger and smaller radius is $12$ cm and $8$ cm. The height of the cone is $3$ cm. (Use $\pi =3$)

• A. $914c{m}^2$
• B. $924c{m}^2$
• C. $934c{m}^2$
• D. $944c{m}^2$

Answer 1

Answer: (B)

$924c{m}^2$

Surface area of frustum cone

$=\pi (r+R)\sqrt{(R-r{)}^2+h^2}+\pi r^2+\pi R^2$

$=\pi (8+12)\sqrt{(12-8{)}^2+3^2}+\pi \ast 8^2+\pi {12}^2$

$=3\times 20\sqrt{16+9}+3\times 64+3\times 144$

$=60\times 5+192+432$

$=924c{m}^2$