Question

What is the surface area of a frustum of cone, whose larger and smaller radius is $6$ in. and $2$ in. The height of the cone is $3$ in. (Use $\pi =3.14$)

• A. $231.2i{n}^2$
• B. $241.2i{n}^2$
• C. $221.2i{n}^2$
• D. $251.2i{n}^2$

Answer 1

The correct option is D $251.2i{n}^2$

Surface area of frustum cone = $\pi (r+R)\sqrt{(R-r{)}^2+h^2}+\pi r^2+\pi R^2$

$=\pi (6+2)\sqrt{(6-2{)}^2+3^2}+\pi \ast 6^2+\pi \ast 2^2$

$=3.14\times 8\sqrt{16+9}+3.14\times 36+3.14\times 4$

$=25.12\times 5+113.04+12.56$

$=251.2i{n}^2$