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=8cm
AE=AF=x
CB=14cm=c
AC=AE+EC=(x+6)cm=a
AB=AF+FC=(x+8)cm=b
Area 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+14
Area =√(x+14)[x+14−(x+6)][x+14−(x+8)][x+14−14]
=√(x+14)(8)(6)(x)
=√48x2+672x
Area of ΔABC = area of ΔAOC+ area of ΔAOB+ area of ΔBOC