Question

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

Answer 1

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, $AB=AC=6cm$. 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

$\Rightarrow A{B}^2=A{M}^2+B{M}^2$

$\Rightarrow 36=A{M}^2+B{M}^2$

$\Rightarrow B{M}^2=36-A{M}^2$ ...(i)

In the right triangle OBM, we have

$\Rightarrow O{B}^2=O{M}^2+B{M}^2$

$\Rightarrow 25=(OA-AM{)}^2+B{M}^2$

$\Rightarrow B{M}^2=25-(O{A}_{A}M{)}^2$

$\Rightarrow 25-(5-AM{)}^2$ .....(ii)

From equations (i) and (ii), we get

$36-A{M}^2=25-(5-AM{)}^2$

$\Rightarrow 11-A{M}^2+(5-AM{)}^2=0$

$\Rightarrow 11-A{M}^2+25-10AM+A{M}^2=0$

$\Rightarrow 10AM=36$

$\Rightarrow AM=3.6$

Putting $AM=3.6$ in equation (i), we get

$B{M}^2=36-(3.6{)}^2=36-12.96$

$\Rightarrow BM=\sqrt{36-12.96}=\sqrt{23.04}=4.8cm$

Hence, $BC=2BM=2\times 4.8cm=9.6cm$.