9.2 Properties of Chords

Step 1

Let chord be $AB = 10$ cm. Centre $O$ . Distance is perpendicular $OM = 12$ cm.

Step 2

By Theorem 9.3, $OM$ bisects $AB$ . So $AM = 10/2 = 5$ cm.

Step 3

In right $\Delta OMA$ , apply Pythagoras: $OA^2 = OM^2 + AM^2$ .

Step 4

$r^2 = 12^2 + 5^2 = 144 + 25 = 169$ .

Step 5

$r = \sqrt{169} = 13$ cm.

Final Answer

13 cm

Operation

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Radius (r)

Chord Length (c)

Dist from Center (d)

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