Lesson 3, Topic 2
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# Surface Area of a Cone

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Consider the diagram 1 and 2 above, to show that a sector of a circle can be bent to form the curved surface of an open cone. In diagram 1, the sector OAXB is of radius l and arc AXB subtends angle at Î¸. This sector is bent to form a cone of base radius r and slant height â€“ see Diagram 2.

Hint:

(a) The area of the sector is equal to the area of the curved surface of the cone.

(b) The length of arc AXB in Diagram 1 is the same as the circumference of the circular base of the cone in Diagram 2.

Curve Surface Area of Cone = $$\frac {\theta}{360} \scriptsize \: \times \: \pi l^2$$ â€¦………(1)

Also, Circumference of circular base of the cone = Length of Arc AXB

i.e.Â  2Ï€r = $$\frac {\theta}{360} \scriptsize \: \times \: 2 \pi l$$ â€¦â€¦â€¦â€¦â€¦(2)

From equation (2) $$\frac {r}{l} = \frac {\theta}{360}$$

Substitue for $$\frac {\theta}{360}$$ in equation 1

i.e. Curved surface area of cone =$$\frac {r}{l}\: \times \: \pi l^2$$

Therefore, Curved surface area of cone = Ï€rl

Therefore, Total Surface area of cone = $$\scriptsize Ï€rl+Ï€r^2$$

= $$\scriptsize Ï€r(l\:+\:r)$$

error: