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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!
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
In circle O with radius OA = OB = 10, chord AB is drawn; OC = 6.Code:* * *
* *
* *
* *
* O *
* * *
* / | \ *
/ |6 \ 10
* / | \ *
A* - - - + - - - *B
* C 8 *
* * *
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:In right triangle ORQ, we have: .d² + 4² .= .10² . → . d² .= .84Code:* * *
* *
* *
* *
* O *
* * *
* / | \ *
/ |d \ 10
* / | \ *
P* - - - + - - - *Q
* 4 R 4 *
* * *
Therefore: .d .= .√84