# Finding the radius of a circle inscribed in a sector.

• Apr 13th 2012, 02:32 PM
Furyan
Finding the radius of a circle inscribed in a sector.
Hello

I have a question where I need to find the area of a circle inscribed, I think that's the right word, in a sector. The circle touches both straight edges of the sector and the arc. I know the angle of the sector, $\dfrac{\pi}{3}$ and the lengths of the two straight edges, 6 cm. I figure I need to find the radius of the circle.

I'm just after a hint. Thank you.
• Apr 13th 2012, 03:39 PM
skeeter
Re: Finding the radius of a circle inscribed in a sector.
think properties of 30-60-90 triangles ...
• Apr 13th 2012, 03:44 PM
emakarov
Re: Finding the radius of a circle inscribed in a sector.
Draw the bisector of the angle and connect the circle center with the points where the circle touches the angle's sides.Then use the fact that the resulting triangles are right-angled.
• Apr 13th 2012, 05:45 PM
Soroban
Re: Finding the radius of a circle inscribed in a sector.
Hello, Furyan!

Quote:

I need to find the area of a circle inscribed in a sector.
The circle touches both straight edges of the sector and the arc.
Angle of the sector is $\tfrac{\pi}{3}$ and the lengths of the two straight edges, 6 cm.
I figure I need to find the radius of the circle. . Right!

Code:

                          B                         * o *                 *  *    |    *  *             *    *      |      *    *         A o    *        |r      *    o C             \            |            /             \  *        |O        *  /               \ *        o        * /               \*        | *  r    */                 \        |  *    /                 *        |    *  *               6  *      |      o                   \*    |    */ Q                     \  * * *  /                     \    |    /                       \  |  /                       \30|30/                         \ | /                         \|/                           o                           P
We have sector $ABCP:\;\angle APC = 60^o,\;AP = BP = CP = 6.$

Inscribed circle: center $O$, radius $OB = OQ = r.$

In right triangle $OQP\!:\;\angle OPQ = 30^o,\;OQ = r.$
. . Hence: . $OP = 2r$

Since $BP = 6r$, we have: . $r + 2r \,=\,6$

• Apr 14th 2012, 09:59 AM
Furyan
Re: Finding the radius of a circle inscribed in a sector.
Quote:

Originally Posted by skeeter
think properties of 30-60-90 triangles ...

Thank you Skeeter. I hadn't spotted it was a 'special' triangle and the 2:1 connection. The graph was very helpful.

Thank you.
• Apr 14th 2012, 10:03 AM
Furyan
Re: Finding the radius of a circle inscribed in a sector.
Quote:

Originally Posted by emakarov
Draw the bisector of the angle and connect the circle center with the points where the circle touches the angle's sides.Then use the fact that the resulting triangles are right-angled.

Thank you emakarov, I had drawn the angle bisector and was wrestling with those right triangles.

Thank you.
• Apr 14th 2012, 10:17 AM
Furyan
Re: Finding the radius of a circle inscribed in a sector.
Quote:

Originally Posted by Soroban
Hello, Furyan!

Code:

                          B                         * o *                 *  *    |    *  *             *    *      |      *    *         A o    *        |r      *    o C             \            |            /             \  *        |O        *  /               \ *        o        * /               \*        | *  r    */                 \        |  *    /                 *        |    *  *               6  *      |      o                   \*    |    */ Q                     \  * * *  /                     \    |    /                       \  |  /                       \30|30/                         \ | /                         \|/                           o                           P
We have sector $ABCP:\;\angle APC = 60^o,\;AP = BP = CP = 6.$

Inscribed circle: center $O$, radius $OB = OQ = r.$

In right triangle $OQP\!:\;\angle OPQ = 30^o,\;OQ = r.$
. . Hence: . $OP = 2r$

Since $BP = 6r$, we have: . $r + 2r \,=\,6$

Thank you Soroban, I have to admit that I hadn't quite got there.

$BP = 3r$, though. Right?

Thank you.