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Question

# A triangle ABC is drawn to circumscribe a circumscribe a circle of radius 4 cm such that the segments 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

## In ΔABC, Length of two tangents drawn from the same point to the circle are equal, ∴CF=CD=6cm∴BE=BD=8cm∴AE=AF=x We observed taht AB +AE +EB =x +8 BC =BD +DC =8+6 =14 CA =CF +FA =6+x Now semiperime of circle s, ⇒2s=AB+BC+CA =x+8+14+6+x =28 +2x ⇒s=14+x Area of ΔABC=√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 ΔABC=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 equation (i) and (ii) we get, √(14+x)48x=56+4x Squaring both sides, 48x(14+x=(56+4x)2 ⇒48x=[(14+x)]214+x⇒48x=16(14+x)⇒48x=224+16x⇒32x=224⇒x=7cm Hence, AB =x+8 =7 +8=15 cm CA =6 +x =6 +7 =13 cm

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