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Question

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

Area of $$\triangle{ABC}=$$area of $$\triangle{OBC}+$$area of $$\triangle{OAC}+$$area of $$\triangle{OAB}$$
We have $$BD=6$$cm, $$BE=6$$cm(equal tangents)
$$DC=9$$cm, $$CF=9$$cm(equal tangents)
Area of $$\triangle{OBC}=\dfrac{1}{2}\times b\times h=\dfrac{1}{2}\times 15\times 3=\dfrac{45}{2}{cm}^{2}$$
Area of $$\triangle{OAC}=\dfrac{1}{2}\times \left(x+9\right)\times 3=\dfrac{3\left(x+9\right)}{2}{cm}^{2}$$
Area of $$\triangle{OAB}=\dfrac{1}{2}\times \left(x+6\right)\times 3=\dfrac{3\left(x+6\right)}{2}{cm}^{2}$$
By Heron's formula,area of $$\triangle{ABC}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}{cm}^{2}$$
$$s=\dfrac{x+9+x+6+15}{2}=\dfrac{2x+30}{2}=x+15$$
$$\therefore \Delta=\sqrt{\left(x+15\right)\left(x+15-15\right)\left(x-15-x-4\right)\left(x+15-x-6\right)}=\sqrt{x\left(x+15\right)\times 6\times 9}$$
$$\Rightarrow \sqrt{54x\left(x+15\right)}=\dfrac{3\left(x+9\right)}{2}+\dfrac{3\left(x+16\right)}{2}+\dfrac{45}{2}$$
$$\Rightarrow \sqrt{54x\left(x+15\right)}=\dfrac{3}{2}\left[x+9+x+6+15\right]$$
$$\Rightarrow \sqrt{54x\left(x+15\right)}=\dfrac{3}{2}\left(2x+30\right)$$
$$\Rightarrow \sqrt{54x\left(x+15\right)}=3x+15$$
Squaring on both sides we get
$$\Rightarrow 54x\left(x+15\right)=9{\left(x+15\right)}^{2}$$
$$\Rightarrow 6x=x+15$$
$$\Rightarrow 6x-x=15$$
$$\Rightarrow 5x=15$$
$$\therefore x=\dfrac{15}{5}=3$$cm
Hence the sides are $$15$$cm, $$12$$cm and $$9$$cm

Mathematics

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