# Circular Cone

• Nov 20th 2009, 07:52 AM
Ilsa
Circular Cone
Consider a right circular cone with the vertex V and O the center of the base. the radius of the base is 1 and the height VO is square root of 35. consider also the diameter AB of the base and C the midpoint of VA. Comput the length of the shortest path, on the surface of the cone from B to C.
• Nov 20th 2009, 09:33 AM
Soroban
Hello, Ilsa!

Quote:

Consider a right circular cone with the vertex $V$ and the center of the base $O$.
The radius of the base is 1 and the height $VO$ is $\sqrt{35}$
Consider also the diameter $AB$ of the base and $C$ the midpoint of $VA$.

Compute the length of the shortest path, on the surface of the cone from $B$ to $C.$

Code:

            V             *           /|\           / | \         /  |  \       C * __|  \ 6       / √35|    \       /    |    \     /      |      \   A * - - - * - - - * B             O  1
We have a right triangle with legs 1 and $\sqrt{35}.$
The hypotenuse (slant height of the cone) is 6.

If we "unroll" half the cone and press it flat,
. . we have the diagram below.

Code:

                    V                     *                   / \                   /30°\ 3                 /    \  C               6 /      *               /    *  \               /  *      \ 3             / *          \           B * - - - - - - - *                   * * *                     π

The circumference of the base of the cone is: . $2\pi r \:=\:2\pi(1) \:=\:2\pi$
Half the circumference is: $\pi$

We have: . $s \:=\:r\theta \quad\Rightarrow\quad \theta \:=\:\frac{s}{r}$
. . Hence: . $\theta \:=\:\frac{\pi}{6}$
The vertex angle is 30°.

The shortest distance between $B$ and $C$ is a straight line.

In $\Delta BVC$, Law of Cosines:
. . $BC^2 \:=\:6^2 + 3^2 - 2(6)(3)\cos30^o \;=\;45 - 36\left(\frac{\sqrt{3}}{2}\right) \;=\;9(5 - 2\sqrt{3})$

Therefore: . $BC \;=\;3\sqrt{5-2\sqrt{3}} \;\approx\;3.718$