Question

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 the figure given). Find the sides AB and AC.

Answer 1

In ΔABC,

Length of two tangents drawn from the same point to the circle are equal,

∴ CF = CD = 6cm

∴ BE = BD = 8cm

∴ AE = AF = x

We observed that

AB + AE + EB = x + 8

BC = BD + DC = 8 + 6 = 14

CA = CF + FA = 6 + x

Now semi perimeter of circle s,

⇒ 2s = AB + BC + CA

= x + 8 + 14 + 6 + x

= 28 + 2x

⇒ s = 14 + x

Area of ΔABC = \(\sqrt{s(s - a) (s - b)(s-c)}\)

\(=\sqrt{(14+x) (14 +x -14)(14 + x - x -6)(14 +x -x-8)}\)

\(= \sqrt{(14 + x)(x) (8)(6)}\)

\(= \sqrt{(14 + x)48x}\) ..... (i)

Also, Area of ΔABC=2× area of (ΔAOF+ΔCOD+ΔDOB)

= 2 × [(1/2 × OF × AF) + (12 × CD × OD) + (12 × DB × OD)] 

= 2 × 12(4x + 24 + 32)

= 56 + 4x ....(ii)

Equating equation (i) and (ii) we get,

\(\sqrt{(14 +x)}48x\) = 56 + 4x

Squaring both sides,

48x(14 + x) = (56 + 4x)²

⇒ 48x = \(\frac{[(14 + x)]^2}{14 + x}\)

⇒ 48x = 16(14 + x)

⇒ 48x = 224 + 16x

⇒ 32x = 224

⇒ x = 7cm

Hence,

AB = x + 8 

= 7 + 8 

= 15 cm

CA = 6 + x 

= 6 + 7 

= 13 cm

Answer 2

To Prove: Side AB = ? and Side AC = ? 

BD = 8 cm. DC = 6 cm. 

BE = 8 cm. CF = 6 cm. 

AE = AF = x cm. 

In ∆ABC, 

a = BC = BD + DC =8 + 6 =14 cm. 

b = CA = (x + 6) cm. 

c = AB = (x + 8) cm.

48x × (x + 14) = 16 x (x + 14)² 

3x = x + 14 

∴ x = 7 cm.

 AB = c = x + 8 = 7 + 8 = 15 cm. 

AC = b = x + 6 = 7 + 6 = 13 cm.