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Question

# Question 12 A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the tangents BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.

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Solution

## Length of two tangents drawn from the same point to the circle are equal. ∴ CF = CD = 6 cm ∴ BE = BD = 8 cm ∴ AE = AF = x We observed that, AB = AE + EB = x + 8 BC = BD + DC = 8 + 6 = 14 CA = CF + FA = 6 + x Let perimeter of triangle ABC be s. ⇒ 2s = AB + BC + CA = x + 8 + 14 + 6 + x = 28 + 2x ⇒ s = 14 + x Area of Δ ABC = √s(s−a)(s−b)(s−c) =√(14+x)(14+x−14)(14+x−x−6)(14+x−x−8) =√(14+x)(x)(8)(6) =√(14+x)48x...(i) Also, area of Δ ABC = 2 × (area of ΔAOF+ΔCOD+ΔDOB) =2×[(12×OF×AF)+(12×CD×OD)+(12×DB×OD)] =2×12(4x+24+32)=56+4x...(ii) Equating equations (i) and (ii) we get, √(14+x)48x=56+4x Squaring both sides, 48x(14+x)=(56+4x)2 ⇒48x=[4(14+x)]2(14+x) ⇒48x=16(14+x) ⇒48x=224+16x ⇒32x=224 ⇒ x = 7 cm Hence, AB = x + 8 = 7 + 8 = 15 cm CA = 6 + x = 6 + 7 = 13 cm.

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