# Finding central angle after slicing cone

• Nov 19th 2011, 08:43 AM
benny92000
Finding central angle after slicing cone
A cone made of cardboard has a vertical height of 8 cm and a radius of 6 cm. If this cone is cut along the slanted height to make a sector, what is the central angle, in degrees, of the sector?

I drew a diagram. I thought the central angle was 90 degrees because of the 6,8,10 triangle that included the slant height. The answer says 216 degrees. (Thinking)
• Nov 19th 2011, 10:15 AM
Soroban
Re: Finding central angle after slicing cone
Hello, benny92000!

The problem is not stated clearly.

Quote:

A cone made of cardboard has a vertical height of 8 cm and a radius of 6 cm.
If this cone is cut along the slanted height to make a sector,
what is the central angle, in degrees, of the sector?

The side view of the cone looks like this:
Code:

              B               *             /:\             / : \           /  :  \           /  :  \ 10         /    :8  \         /    :    \       /      :      \     A *-------*-------* C       : - 6 - : - 6 - :
We see that the slant height is 10 cm.
The circumference of the base is: . $2\pi r = 2\pi(6) = 12\pi\text{ cm}$

Now we cut the cone and lay it flat.
We have a large sector of a circle of radius 10.
Code:

                B               * * *           *          *         *              *       *                *       *        O        *       *        *        *       *        / \        *           10 /  \ 10       *    /    \    *         *  /      \  *         A *          * C               * * *
Length-of-arc Formula: . $s \,=\,r\theta$ . where:. $\begin{Bmatrix}{s &=& \text{length of arc} \\ r &=& \text{radius} \\ \theta &=& \text{central angle}\\[-2mm] && \text{in radians} \end{Bmatrix}$

We have: . $\begin{Bmatrix}s \,=\,\text{arc}(ABC) \,=\, 12\pi \\ r \:=\: 10\end{Bmatrix}$

Therefore: . $12\pi \,=\,10\theta$

. . . . . . . . . . $\theta \:=\:\frac{12\pi}{10}\text{ radians} \:=\:216^o$