Discussion
The radius of a circle is 6 cm. The perpendicular distance from the centre of the circle to the chord which is 8 cm in length is?
Answer: B
The distance is \(2\sqrt{5}\) cm.
The perpendicular from the centre bisects the chord, so it meets a radius and half the chord (4 cm) in a right triangle:
\[ d = \sqrt{6^{2} - 4^{2}} = \sqrt{20} = 2\sqrt{5} \]
\(\sqrt{5}\) would come from halving the result. \(2\sqrt{7}\) is \(\sqrt{36 - 8}\), using the whole chord without squaring, and \(\sqrt{7}\) is half of that.
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