Question

In a $\Delta ABC$, AD is the bisector of $\angle A$.

(i) If AB = 6.4 cm, AC = 8 cm and BD = 5.6 cm, find DC.

(ii) If AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC.

(iii) If AB = 5.6 cm, BD = 3.2 cm and BC = 6 cm, find AC.

(iv) If AB = 5.6 cm, AC = 4 cm and DC = 3 cm, find BC.

Answer 1

(i) AB = 6.4 cm, AC = 8 cm, BD = 5.6 cm

Let BC = x

Now,
DC = (BC-BD) = (x-5.6) cm

In ΔABC, AD is the base of ∠A

So, by the angle bisector theorem, we have

BD/DC=AB/AC

⇒5.6/x−5.6=6.4/8

⇒⇒ 6.4x - 35.84 = 44.8

⇒⇒ 6.4x = 80.64

⇒⇒ x = 12.6

Hence, BC = 12.6 cm and DC = (12.6 - 5.6) cm = 7 cm

(ii) AB = 10 cm, AC = 14 cm, BC = 6 cm

Let BD = x,

DC = (BC - BD) = (6-x) cm

In ΔABC, AD is the bisector of ∠A

So, By angle bisector theorem,

We have, BD/DC=AB/AC

⇒ x/6−x=10/14

⇒ 14x = 60 - 10x

⇒ 24x = 60

x = 2.5

Hence, BD = 2.5 cm and DC = (6 - 2.5) cm = 3.5 cm

(iii) AB = 5.6 cm, BD = 3.2 cm and BC = 6 cm

DC = BC - BD = (6-3.2) cm = 2.8 cm

Let, AC = x,

In ΔABC, AD is the bisector of ∠A

So, by the angle bisector theorem we have

Therefore, BD/DC=AB/AC

⇒ 3.2/2.8=5.6/x

⇒ x= 5.6×2.8/3.2=4.95cm

Hence, AC = 4.9 cm

(iv) AB = 5.6 cm, AC = 4 cm, DC = 3 cm

Let BD = x,

In ΔABC, AD is the bisector of ∠A

So, by the angle bisector theorem we have

Therefore, BD/DC=AB/AC

⇒ x/3=5.6/4

⇒ x= 5.6×3/4=4.2 cm

Hence, BD = 4.2 cm

So BC = BD + AC = (4.2 + 3) cm

BC = 7.2 cm