# Circle and Chord

• Mar 22nd 2007, 11:55 PM
DivideBy0
Circle and Chord
In a circle of radius 10cm, a chord is drawn 6cm from the centre. If a chord half the length of the original chord were drawn, its distance, in centimetres, from the centre would be

(A) sqrt(96) (B) sqrt(84) (C) 9 (D) 8 (E) 3pi

I'm not so familiar with the properties of chords in circles, so please don't explain too fast, thanks.
• Mar 23rd 2007, 05:10 AM
Soroban
Hello, DivideBy0!

Quote:

In a circle of radius 10cm, a chord is drawn 6cm from the centre.
If a chord half the length of the original chord were drawn,
its distance, in centimetres, from the centre would be:

. . (A) sqrt(96) . . (B) sqrt(84) . . (C) 9 . . (D) 8 . . (E) 3pi

Code:

```              * * *           *          *         *              *       *                *       *        O        *       *        *        *       *      / | \      *             /  |6  \ 10       *  /    |    \  *       A* - - - + - - - *B           *    C  8  *               * * *```
In circle O with radius OA = OB = 10, chord AB is drawn; OC = 6.

Using Pythagorus on right triangle OCB, we find that: CB = 8
. . Hence: AB = 16.

If we have a chord only 8 units long, it looks like this:
Code:

```              * * *           *          *         *              *       *                *       *        O        *       *        *        *       *      / | \      *             /  |d  \ 10       *  /    |    \  *       P* - - - + - - - *Q           *  4  R  4  *               * * *```
In right triangle ORQ, we have: .d² + 4² .= .10² . . .= .84

Therefore: .d .= .√84