Question

Find the surface area of a frustum of cone, whose larger and smaller radius is $8$ m and $4$ m. The height of the cone is $3$ cm. (Use $\pi =3$)

• A. $420m^2$
• B. $421m^2$
• C. $430m^2$
• D. $440m^2$

Answer 1

The correct option is A $420m^2$

TSA of frustum = $\pi (R+r)s+\pi (R^2+r^2)$

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

= $3\times (8+4)\sqrt{4^2+3^2}+3(8^2+4^2)$

= $3\times 12\times 5+3\times 80$

= $420$