question_answer

In figure 4, 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 \[\Delta \,ABC\] is \[54\text{ }c{{m}^{2}}\], then find the lengths of sides AB and AC.

Answer:

Given, in \[\Delta \text{ }ABC\], circle touch the triangle at point D, F and E respectively and let the lengths of the segment AF be x.

So,       \[BF=BD=6\text{ }cm\] [Tangent from point B]

\[CE=CD=9\text{ }cm\]            [Tangent from point C]

and       \[AE=AF=x\text{ }cm\] [Tangent from point A]

Now,     Area of \[\Delta \,OBC=\frac{1}{2}\times BC\times OD\]

\[=\frac{1}{2}\times (6+9)\times 3\]

\[=\frac{45}{2}c{{m}^{2}}\]

Area of \[\Delta \text{ }OCA=\frac{1}{2}\times AC\times OE\]

\[=\frac{1}{2}\times (9+x)\times 3\]

\[=\frac{3}{2}(9+x)c{{m}^{2}}\]

Area of \[\Delta \,BOA=\frac{1}{2}\times AB\times OF\]

\[=\frac{1}{2}\times (6+x)\times 3\]

\[=\frac{3}{2}\times (6+x)\times c{{m}^{2}}\]

Area of \[\Delta \text{ }ABC=54\text{ }c{{m}^{2}}\]                 [Given]

\[\because \]       Area of \[\Delta \text{ }ABC=\] Area of \[\Delta \,OBC+\] Area of \[\Delta \text{ }OCA+\] Area of \[\Delta \,BOA\]

\[54=\frac{45}{2}+\frac{3}{2}(9+x)+\frac{3}{2}(6+x)\]

\[\Rightarrow 54\times 2=45+27+3x+18+3x\]

\[\Rightarrow 108-45-27-18=6x\]

\[\Rightarrow 6x=18\]

\[\Rightarrow x=3\]

So,       \[AB=AF+FB=x+6=3+6=9\text{ }cm\]

and       \[AC=AE+EC=x+9=3+9=12\text{ }cm\]

Hence, lengths of AB and AC are 9 cm and 12 cm respectively.