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# 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 figure). Find the sides AB and AC. Let us join AO, OC, and OB. It is given that BD = 8cm, CD = 6 cm.

$$\therefore$$ Lengths of two tangents drawn from an external point of circle are equal.

$$\therefore$$ $$BF \ = \ BD \ = \ 8 cm \ ,$$
$$\ CE \ = \ CD \ = \ 6 cm \ ,$$
$$\ and \ let \ AF \ = \ AE \ = \ x$$ cm.

Then, the sides of the triangle are 14 cm, (x+6) cm and (x+8) cm.

$$\therefore 2s \ = \ 14 \ + \ (x + 6) \ + \ (x+8) \ = \ 28 \ + \ 2x$$

$$\Rightarrow \ s \ = \ 14 \ + \ x$$

$$s \ - \ a \ = \ 14 \ + \ x \ - \ 14 \ = \ x$$,

$$s \ - \ b \ = \ 14 \ + \ x \ - \ x \ - \ 6 \ = \ 8$$

and, $$s\ - \ c \ = \ 14 \ + \ x \ - \ x \ - \ 8 \ = \ 6$$

$$\therefore$$ Area of $$∆ \ ABC \$$
$$= \ \sqrt{s(s \ - \ a)(s \ - \ b)(s \ - \ c)}$$
$$= \ \sqrt{(14 \ + \ x)(x)(8)(6)}$$

$$= \ \sqrt{48(x^2 \ + \ 14x)}$$

Also, Area of $$∆ \ ABC \$$
$$= \ Area (∆ OBC) \ + \ Area (∆ OCA) \ + \ Area (∆ OAB)$$

$$= \ \frac{1}{2} \ × \ BC \ × \ OD \ + \ \frac{1}{2} \ × \ CA \ × \ OE \ + \ \frac{1}{2} \ × \ AB \ × \ OF$$

$$= \ \frac{1}{2} \ × \ 14 \ × \ 4 \ + \ \frac{1}{2} \ × \ (x+6) \ × \ 4 \ + \ \frac{1}{2} \ × \ (x+8) \ × \ 4$$

$$= \ 2(14 \ + \ x \ + \ 6 \ + \ x \ + \ 8) \ = \ 2(28 \ + \ 2x)$$

$$\therefore \sqrt{(48(x^2 \ + \ 14x)} \ = \ 4(14 \ + \ x)$$

Squaring, we get

$$48(x^2 \ + \ 14x) \ = \ 16(14 \ + \ x)^2$$

$$\Rightarrow \ 3(x^2 \ + \ 14x) \ = \ 196 \ + \ 28x \ + \ x^2$$

$$\Rightarrow 2x^2 \ + \ 14x \ - \ 196 \ = \ 0$$

$$\Rightarrow \ x^2 \ + \ 7x \ - \ 98 \ = \ 0$$

$$\Rightarrow \ (x \ - \ 7)(x \ + \ 14) \ = \ 0$$

$$\Rightarrow \ x \ = \ 7 \ or \ x \ = \ -14$$

But x cannot be negative,

$$\therefore x \ = \ 7$$

Thus, AB = x + 8 = 7 + 8 = 15 cm and AC = x + 6 = 6 + 7 = 13 cm.

Hence, the sides AB and AC are 15 cm and 13 cm respectively.