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Question

# In the given figure, a triangel ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point o f contant D, are of lengths 6 cm and 9 cm respectively. If the area of Δ = 54 cm2 then find the lengths of sides AB and AC.

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Solution

## Consider the above figure. We assume that the circle touches AB in F, BC in D and AC in E. Also given is BD = 9cm and DC = 3cm.Let AF = x. For ΔABC, AF = AE = x (∵ tangents drawn from an external point to a circle are congruent i.e. AE and AF are tangent drawn from external point A.) Similarly we have, BE = BD = 3cm (∵ congruent tangents from point B) And CF = CD = 9cm (∵ congruent tangents from point C) Now, AB = AE + EB = x + 3 BC = BD + DC = 12 AC = AF + FC = x + 9 Then, 2s = AB + BC +CA = x + 3 + 12 + 1 + x + 9 = 2x + 24 ∴ s = x + 12 Using Heron’s formula, Area of ΔABC=√s(s−a)(s−b)(s−c) =√(12+x)(12+x−12)(12+x−x−9)(12+x−x−3) =√(12+x)(x)(3)(9) =√27(12x+x2) =3√3(12x+x2) -----(1) Area of ΔOBC=12×12×3=18 Area of ΔOAB=12×(3+x)×3=9+3x2 Area of ΔOAC=12×(9+x)×3=27+3x2 Total Area = Area of ΔOBC + Area of ΔOAB + Area of ΔOAC =8+9+3x2+27+3x2 ⇒3√3(12x+x2)=8+9+3x2+27+3x2 ⇒3√3(12x+x2)=36+9+3x+27+3x2 ⇒3√3(12x+x2)=72+6x2=12+x Squaring both sides, ⇒[√3(12x+x2)]2=(12+x)2 ⇒3(12x+x2)=(12+x)2 ⇒36x+3x2=144+24x+x2 ⇒2x2+12x−144=0 ⇒x2+6x−72=0 ⇒(x+12)(x−6)=0 ⇒(x=6 Hence AB = x + 3 = 9 + 3 = 12 cm BC = 12 cm AC = x + 9 = 15 cm

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