Question

The radius of a circle is 6 cm and the length of one of its chords is 6 cm, the distance of the chords from the center is:

Answer 1

$O$ is centre.

Since we know that perpendicular from centre on a chord bisects it.

So, $AP=BP$

and in $△OPA$,

$r^2$$=(OA{)}^2=(OP{)}^2+\left(AP{)}^2\right(6{)}^2=(OP{)}^2+{\left(\frac{AB}{2}\right)}^{2}36=(OP{)}^2+3^2\left(OP\right)=36-9OP=\sqrt{27}=3\sqrt{3}$