Building blocks

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## Appendix

This appendix reviews various formulae for finding the curved surface area and volume of a cone.

### Surface area of a cone

Here’s a diagram of a cone. We’ve drawn some lengths and an angle on it.

When we cut the (slanted) surface of the cone in a straight line from the vertex to the base and then unwrap it, we end up with a sector of a circle, like this:

The radius of this sector is the slant length of the cone, which is $l$. The outer arc of the sector was the circumference of the base circle of the cone, so its length is $2\pi r$.

Now the circumference of the whole of this circle is $2\pi l$, and therefore the sector is a fraction $\dfrac{2\pi r}{2\pi l}$ of the whole circle. The area of the sector, and hence of the slanted surface of the cone, is therefore $\dfrac{2\pi r}{2\pi l}\times \pi l^2=\pi rl.$

Alternatively, we could have worked out $\phi$ first, using the formula for the arc length of a sector “$l=r\theta$”, giving in this case $2\pi r=l\phi$, so $\phi=\frac{2\pi r}{l}$. Then the area of the sector is given by the formula “$A=\frac12 r^2\theta$”, so calculating $\frac12 l^2 .\frac{2\pi r}{l}=\pi rl$ gives us the area as before.

It’s always good to check that our answers are sensible before going too far in a problem!

How could we do this here?

Sometimes we will know some measurements more readily than others. So we could try rewriting this answer using other variables.

Since $l^2=r^2+h^2$ using Pythagoras, we could also write the surface area as $\pi r\sqrt{r^2+h^2}$ or $\pi l\sqrt{l^2-h^2}$ if we wish.

If we draw a cross-section of the cone, we see that we have a right-angled triangle:

From this diagram, we see that $h=l\cos\theta$ and $r=l\sin\theta$, so we can substitute for $l$ in our earlier answer $\pi rl$ to find that we can also write the surface area as $\pi rl=\frac{\pi rh}{\cos\theta}=\frac{\pi r^2}{\sin\theta}.$

Using the reciprocal trigonometric functions, we could also write these as $A=\pi rh\sec\theta=\pi r^2\cosec\theta$.

### Volume of a cone

The area of the base of the cone is $\pi r^2$. The volume of any pyramidal shape is given by $\tfrac13\times\text{base area}\times\text{perpendicular height}$ so in this case, the volume is $\frac13\pi r^2h$. This result can be proved using calculus.