Question

# In a circle of radius 5 cm, AB and AC are two chords. such that $$AB=AC=6\:cm$$. Find the length of chord BC.

Solution

## We know that, if AB and AC are two equal chords of a circle, then the centre of the circle lies on the bisector of $$\angle BAC$$.Here, $$AB=AC=6\:cm$$. So, the bisecor of $$\angle BAC$$ passes through the centre O i.e. OA is the bisector of $$\angle BAC$$.Since, the internal bisector of an angle divides the opposite sides in the ratio of the sides containing the angle. Therefore, M divides BC in the ratio $$1:1$$, i.e. M is the mid-point of BC.$$\Rightarrow OM\perp BC$$. In the right triangle ABM, we have $$\Rightarrow AB^{2}=AM^{2}+BM^{2}$$$$\Rightarrow 36=AM^{2}+BM^{2}$$$$\Rightarrow BM^{2}=36-AM^{2}$$ ...(i)In the right triangle OBM, we have$$\Rightarrow OB^{2}=OM^{2}+BM^{2}$$$$\Rightarrow 25=(OA-AM)^{2}+BM^{2}$$$$\Rightarrow BM^{2}=25-(OA_AM)^{2}$$$$\Rightarrow 25-(5-AM)^{2}$$  .....(ii)From equations (i) and (ii), we get$$36-AM^{2}=25-(5-AM)^{2}$$$$\Rightarrow 11-AM^{2}+(5-AM)^{2}=0$$$$\Rightarrow 11-AM^{2}+25-10AM+AM^{2}=0$$$$\Rightarrow 10AM=36$$$$\Rightarrow AM=3.6$$Putting $$AM=3.6$$ in equation (i), we get$$BM^{2}=36-(3.6)^{2}=36-12.96$$$$\Rightarrow BM=\sqrt{36-12.96}=\sqrt{23.04}=4.8\:cm$$Hence, $$BC=2\:BM=2\times 4.8\:cm=9.6\:cm$$.Mathematics

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