Thread: 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.

Hello, DivideBy0!

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 . → . d² .= .84

Therefore: .d .= .√84