A circle has a radius of 5 cm. The chord EF is 7 cm.
How far is the midpoint of the chord from the centre of the circle?
Add the radii, OE and OF, to make two right-angled triangles.
FM is half of the length of chord EF.
FM = 3.5 cm
Use Pythagoras' theorem to calculate the length OM.
OF2 = FM2 + OM2
OM2 = 12.75
OM = 3.6 cm (to 1 decimal place)
In the diagram below, AB is the chord of a circle with centre O.
OM is the perpendicular from the centre to the chord.
Angles OMA and OMB are both right-angles.
OA is the hypotenuse of triangle OAM.
OB is the hypotenuse of triangle OBM.
OA = OB as both are radii of the circle.
OM is common to both triangles.
Therefore, triangles OAM and OMB are congruent (RHS – right-angle, hypotenuse, side).
Therefore, the remaining sides of the triangles are equal, AM = MB.
So, M must be the mid-point of AB, and the chord has been bisected.