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Question

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

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Solution

## CE=CD=6cm[∵ length of tangent drawn from the external point are equal]BF=BD=8cmAE=AF=xCB=14cm=cAC=AE+EC=(x+6)cm=aAB=AF+FC=(x+8)cm=bArea of ΔABC by Heron's formula√s(s−a)(s−b)(s−c)s=a+b+c2=x+6+x+8+142=2(x+14)2=x+14Area =√(x+14)[x+14−(x+6)][x+14−(x+8)][x+14−14]=√(x+14)(8)(6)(x)=√48x2+672xArea of ΔABC = area of ΔAOC+ area of ΔAOB+ area of ΔBOC=12×OE×AC+12×OF×AB+12×OD×BC=12×4(x+6)+12×4(x+8)+12×4×14=2x×12+2x+16+28=4x+56√48x2+672x=4x+5648x2+672x=(4x+56)248x2+672x=16x2+448x+313632x2+224x−3136=032(x2+7x+98)=0x2+7x−98=0x2+14x−7x−98=0x(x+14)−7(x+14)=0(x−7)(x+14)=0x=7 and x=−14x can't be negative∴AC=x+6=7+6=13cmAB=x+8=7+8=15cm

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