# In the given figu

Given: and OD = 6cm

Let OB = x cm

In ΔBOD, By Pythagoras theorem

OB2 = BD2 + OD2

x2 = BD2 + 62

x2 = BD2 + 36

BD2 = x2 – 36

Now consider Δ ABC

Here BC = 2x

By Pythagoras theorem

BC2 = AB2 + AC2

(2x)2 = 4(x2 – 36)+ AC2

4x2 = 4x2 –144 + AC2

AC2 = 144

AC = 12 cm

AC = 12 cm

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