Question:
The radius of a circle is 8 cm and the length of one of its chords is 12 cm. Find the distance of the chord from the centre.
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
