Arcs and Chords | MathHelp.com - By MathHelp.com
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| 00:0-1 | Find the length of the radius of a circle . | |
| 00:03 | If a chord of the circle has a length of | |
| 00:05 | 12 cm and is four cm from the center of | |
| 00:10 | the circle . Let's start this problem by drawing a | |
| 00:14 | picture . We have a circle , wow ! Yeah | |
| 00:21 | . Which will call circle . Oh and we have | |
| 00:25 | a cord . Yeah Which will call a B . | |
| 00:35 | We know that the cord has a length of 12 | |
| 00:38 | cm . Mhm And is four cm from the center | |
| 00:47 | of the circle . This means that we can draw | |
| 00:50 | a perpendicular segment from the center of the circle to | |
| 00:54 | the cord Which has a length of four which will | |
| 00:58 | call O . C . Remember that the distance from | |
| 01:03 | a point to a segment is represented by the perpendicular | |
| 01:07 | to the segment . Now notice that segment oc is | |
| 01:12 | part of a diameter of the circle . And remember | |
| 01:18 | that if a diameter of a circle is perpendicular to | |
| 01:22 | accord , then it bisects the cord . So segment | |
| 01:26 | A . B . Which has a length of 12 | |
| 01:29 | , is split into two congruent segments , each with | |
| 01:33 | a length of six . Now to find the radius | |
| 01:39 | of the circle , let's draw radius Obe and give | |
| 01:44 | it a length of X . Notice that right triangle | |
| 01:50 | O . C . B is formed , so we | |
| 01:53 | can use the pythagorean theorem to set up the equation | |
| 01:57 | four squared plus six squared equals x squared . Yeah | |
| 02:09 | . which simplifies to 16 plus 36 equals X squared | |
| 02:17 | , or 52 equals x squared and square rooting both | |
| 02:24 | sides . The square root of 52 simplifies to two | |
| 02:31 | . Route 13 . So we have to route 13 | |
| 02:35 | equals x . So the length of the radius of | |
| 02:41 | the circle Is to Route 13 cm . |
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