Question

In a circle with center O, a perpendicular is dropped from the center to a chord PQ. Find the length of PO, if length of PQ = 24 cm and KO = 5 cm.

Answer 1

The perpendicular from the centre of a circle to a chord bisects the chord.
$\therefore PK=KQ=12cm$
Given, KO = 5 cm.
Using Pythagoras theorem in $\Delta PKO$,
we get, $(PO{)}^2=(KO{)}^2+(PK{)}^2$
$\Rightarrow (PO{)}^2=5^2+{12}^2$
$\Rightarrow (PO{)}^2=169$
$\Rightarrow PO=13cm$.