# Curved Surface Area of a Cone

If a perpendicular cut is made from a point on the circumference of the base to the vertex and the cone is opened up, a sector of a circle with radius $l$ is produced. Since the circumference of the base of the cone is $2\pi r$, therefore the arc length of the sector of the circle is $2\pi r$. $\begin{gathered} {\text{Curved surface area}} = {\text{Area of sector }}OAA’ \\ {\text{Curved surface area}} = \left( {\frac{{{\text{Arc length of sector}}}}{{{\text{Circumference of circle}}}}} \right) \times {\text{Area of circle}} \\ {\text{Curved surface area}} = \frac{{2\pi r}}{{2\pi l}} \times \pi {l^2} = \pi rl \\ \end{gathered}$

Where
$l =$ slant height of the cone
$r =$ radius of the base of the cone

Total Surface Area

The total surface area $=$area of curved surface $+$area of base
$\therefore$     $S = \pi rl + \pi {r^2} = \pi r\left( {l + r} \right)$

Rule:

1. The curved surface area of a right circular cone equals the perimeter of the base times one-half slant height.
2. The total surface area equals the curved surface area of the base.

Example:

The slant height of a conical tomb is $10.5$ m. If its diameter is $16.8$ m, find the cost of cleaning it at \$2 per cubic meter and also the cost of whitewashing the curved surface at $50$ cents per square meter.

Solution:
Now, slant height,       $l = 10.5$ m
Perpendicular height $h = \sqrt {{{\left( {10.5} \right)}^2} – {{\left( {8.4} \right)}^2}} = 6.3$ m   (as $h = \sqrt {{l^2} – {r^2}}$)
Volume of the conical tomb $= \frac{1}{3}\pi {r^2}h$
$= \frac{1}{3} \times \frac{{22}}{7} \times 8.4 \times 8.4 \times 6.3$
$= 465.7$ cubic meters
Cost of construction             $= 465.7 \times 2 = 931.39$
Curved surface                     $= \pi rl = \frac{{22}}{7} \times 8.4 \times 10.6 = 277.2$ square meters
Cost of white washing                      $= 277.2 \times \frac{{50}}{{100}} = 138.60$ dollars