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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=6cmBE=BD=8cmAE=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(sa)(sb)(sc)
=(14+x)(14+x14)(14+xx6)(14+xx8)=(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+x48x=16(14+x)48x=224+16x32x=224x=7cm
Hence, AB =x+8 =7 +8=15 cm
CA =6 +x =6 +7 =13 cm

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