  Question

In Figure $$5$$, a $$\triangle ABC$$ is drawn to circumscribe a circle of radius $$3\ cm$$, such that the segments $$BD$$ and $$DC$$ are respectively of lengths $$6\ cm$$ and $$9\ cm$$. If the area of $$\triangle ABC$$ is $$54\ cm^2$$, then find the lengths of sides $$AB$$ and $$AC$$. Solution

Given $$OD =3\ cm$$$$BD$$$$= BE = 6\ cm$$ (equal tangent)$$DC = CF=9\ cm$$ (equal tangent)Area of triangle $$ABC = 54\ {cm}^2$$ (Given)Area of triangle $$OBC =$$ $$\cfrac{1}{2}\times 15\times 3=\dfrac{45}{2} cm^{2}$$Area of triangle $$OAC=$$$$\cfrac{1}{2}(x+9)\times 3=\dfrac{3(x+9)}{2} cm^{2}$$Area of triangle $$OAB=$$$$\cfrac{1}{2}(x+6)\times 3=\dfrac{3(x+6)}{2} cm^{2}$$$$\displaystyle 54 = \cfrac{3}{2}\left [ (x+9)+(x+6)+15 \right ]$$$$\displaystyle \Rightarrow 54=\cfrac{3}{2}(2x+30)$$$$\displaystyle \Rightarrow 54=3(x+15)$$$$\displaystyle \Rightarrow 17=x+15$$$$\displaystyle \Rightarrow x = 2$$Then side are $$x+9=2+9=11,x+6=2+6=8$$So sides of triangle $$ABC$$ are $$11,8$$ and $$15$$. Mathematics

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