# [Arcs, sectors and segments] Finding the arc

• Jul 26th 2010, 02:08 AM
Cthul
[Arcs, sectors and segments] Finding the arc
"A circle is modified so that it will measure a total of 30m around the major arc AB and the chord AB"
I don't know where to start.
• Jul 26th 2010, 04:34 AM
Soroban
Hello, Cthul

Quote:

The perimeter of the shape formed by major arc $AB$ and chord $AB$ is 30 m.

Code:

              * * *           *          *         *              *       *                *       *        O        *       *        ♥        *       *    r  *120*  r    *             *      *       *  *          *  *       A ♥ - - - - - - - ♥ B           *          *               * * *

The circumference of the circle is: $2\pi r.$

The major arc has length: $\frac{2}{3}\times 2\pi r \:=\:\frac{4\pi}{3}r$

Using the Law of Cosines on $\Delta AOB\!:$

. . $AB^2 \:=\:r^2 + r^2 - 2(r)(r)\cos120^o \:=\:3r^2$

Hence: . $AB \:=\:r\sqrt{3}$

The perimeter is: . $\frac{4\pi}{3}r + r\sqrt{3} \:=\:30$

. . . . . . . . . . . . $\left(\frac{4\pi}{3} + \sqrt{3}\right)r \:=\:30$

Therefore: . $r \;=\;\dfrac{30}{\frac{4\pi}{3} + \sqrt{3}} \;=\;\dfrac{90}{4\pi + 3\sqrt{3}}\text{ meters}$

• Jul 26th 2010, 04:49 AM
Quote:

Originally Posted by Cthul
"A circle is modified so that it will measure a total of 30m around the major arc AB and the chord AB"
I don't know where to start.

Alternatively, split the triangle OAB into two identical right-angled triangles,
where O is the circle centre.

Then $|\angle OAB|=30^o$

$|AB|=2rcos30^o=r\sqrt{3}$

The length of major arc AB is $\frac{240}{360}{2\pi}r$

$30m=r\sqrt{3}+\frac{4{\pi}r}{3}$
• Jul 27th 2010, 02:34 AM
Cthul
Thanks to both, very helpful steps.