The radius of a circle is 8 cm and the length of one of its chords is 12 cm.

Solution:

Given that,

Radius of circle (OA) = 8 cm

Chord (AB) = 12 cm

Draw OC⊥AB

We know that

The perpendicular from centre to chord bisects the chord

$\mathrm{AC}^{2}+\mathrm{OC}^{2}=\mathrm{OA}^{2}$

$\Rightarrow 6^{2}+O C^{2}=8^{2}$

$\Rightarrow 36+O C^{2}=64$

$\Rightarrow O C^{2}=64-36$

 

$\Rightarrow O C^{2}=28$

$\Rightarrow O C=\sqrt{28}$

$\Rightarrow O C=5.291 \mathrm{~cm}$

∴ AC = BC = 12/2 = 6 cm

Now in ΔOCA, by Pythagoras theorem