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Question

In the given figure, a triangel ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point o f contant D, are of lengths 6 cm and 9 cm respectively. If the area of Δ = 54 cm2 then find the lengths of sides AB and AC.

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Solution

Consider the above figure. We assume that the circle touches AB in F, BC in D and AC in E. Also given is BD = 9cm and DC = 3cm.Let AF = x.

For ΔABC, AF = AE = x (∵ tangents drawn from an external point to a circle are congruent i.e. AE and AF are tangent drawn from external point A.)

Similarly we have, BE = BD = 3cm (∵ congruent tangents from point B)

And CF = CD = 9cm (∵ congruent tangents from point C)

Now, AB = AE + EB = x + 3

BC = BD + DC = 12

AC = AF + FC = x + 9

Then,

2s = AB + BC +CA = x + 3 + 12 + 1 + x + 9 = 2x + 24

s = x + 12

Using Heron’s formula,

Area of ΔABC=s(sa)(sb)(sc)

=(12+x)(12+x12)(12+xx9)(12+xx3)

=(12+x)(x)(3)(9)

=27(12x+x2)

=33(12x+x2) -----(1)

Area of ΔOBC=12×12×3=18

Area of ΔOAB=12×(3+x)×3=9+3x2

Area of ΔOAC=12×(9+x)×3=27+3x2

Total Area = Area of ΔOBC + Area of ΔOAB + Area of ΔOAC

=8+9+3x2+27+3x2

33(12x+x2)=8+9+3x2+27+3x2

33(12x+x2)=36+9+3x+27+3x2

33(12x+x2)=72+6x2=12+x

Squaring both sides,

[3(12x+x2)]2=(12+x)2

3(12x+x2)=(12+x)2

36x+3x2=144+24x+x2

2x2+12x144=0

x2+6x72=0

(x+12)(x6)=0

(x=6

Hence AB = x + 3 = 9 + 3 = 12 cm

BC = 12 cm

AC = x + 9 = 15 cm


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