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.
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