Question

The radius of a circle with centre O is 25 cm. Find the distance of a chord from the centre if the length of the chord is 48 cm

Answer 1

Given that OP ⊥ chord CD and length of CD = 48 cm
Radius of circle = 25 cm
Thus, OD = 25 cm
Now CD = 48 cm
Length of PD = $\frac{1}{2}$ (CD) (Perpendicular drawn from the centre of a circle to its chord bisects the chord)
PD = 24 cm
In $\Delta$OPD, ∠OPD = 90°
By Pythagoras theorem,
$(OD{)}^2=(OP{)}^2+(PD{)}^2$
$(25{)}^2–(24{)}^2=(OP{)}^2$
$(OP{)}^2=49$
$\therefore$ (OP) = $\sqrt{49}=7$ cm
Thus, the distance of the chord from the centre of the circle is 7 cm.