WAEC Mathematics 2014 Paper

Concept:The perpendicular from the centre of a circle to a chord bisects the chord.Using Pythagoras' theorem in the resulting right triangle gives the required distance.Explanation:Let the chord be $AB = 7\text{ cm}$ and the centre be $O$.Let $OM$ be the perpendicular distance from $O$ to the chord $AB$.Since $OM$ bisects $AB$, we have:$$AM = \frac{7}{2} = 3.5\text{ cm}$$The radius is given as:$$OA = 3.7\text{ cm}$$In right-angled triangle $OAM$, apply Pythagoras' theorem:$$OA^2 = AM^2 + OM^2$$Substitute the known values:$$3.7^2 = 3.5^2 + OM^2$$$$13.69 = 12.25 + OM^2$$$$OM^2 = 13.69 - 12.25 = 1.44$$$$OM = \sqrt{1.44} = 1.2\text{ cm}$$Answer:The distance of the chord from the centre of the circle is $1.2\text{ cm}$.Correct option: B. $1.2\text{ cm}$