The figure below shows a circle centered at O and of radius 5 cm. AB and AC are two chords such that AB = AC = 6 cm. AP is perpendicular to BC. Find the length of the chord BC. [4 Marks]
In △APC,
⇒AC2=PC2+AP2
⇒PC2=AC2−AP2
⇒PC2=62−AP2
⇒PC2=36−AP2 ---(i) [1 Mark]
In △POC,
PO=AO−AP
PO=5−AP
⇒PC2=OC2−OP2
⇒PC2=25−(5−AP)2
⇒PC2=25−(25+AP2−10AP)
⇒PC2=10AP−AP2 ------- (ii) [1 Mark]
Substitute (i) in (ii)
36−AP2=10AP−AP2
⇒AP=3.6 cm
PC2=36−AP2
PC2=36−12.96
PC2=23.04
PC=4.8 cm [1 Mark]
Since a perpendicular from the centre of a circle to a chord bisects the chord,
BC=2×PC
=2×4.8
=9.6 cm [1 Mark]
∴ The length of chord BC is 9.6 cm.