A triangle ABC is drawn to 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 (see Fig.) . Find the sides AB and AC.

Let us consider the triangle ABC,

We know that the length of any two tangents which are drawn from the same point to the circle is equal.

So,

(i) CF = CD = 6 cm

(ii) BE = BD = 8 cm

(iii) AE = AF = x

We can see that

(i) AB = EB+AE = 8+x

(ii) CA = CF+FA = 6+x

(iii) BC = DC+BD = 6+8 = 14

The semi perimeter “s” can be calculated as

2s = AB+CA+BC

Substituting the values, we get

2s = 28+2x

s = 14+x

On solving the above equation we get

= √(14+x)48x ……… (i)

The area of △ABC = 2 × area of (△AOF + △COD + △DOB)

= 2×[(½×OF×AF)+(½×CD×OD)+(½×DB×OD)]

= 2×½(4x+24+32) = 56+4…………..(ii)

From equations (i) and (ii) we get,

√(14+x)48= 56+4x

On squaring both the sides, we get

48x(14+x) = (56+4x)2

48x = [4(14+x)]2/(14+x)

48x = 16(14+x)

48x = 224+16x

32x = 224

x = 7 cm

So, AB = 8+x

i.e. AB = 15 cm

And, CA = x+6 =13 cm.

Hence, AB = 15 cm

AC = 13 cm