Question

If a ∆ ABC, AD is the bisector of ∠A, meeting side BC at D.

(i) If BD = 2.5 cm, AB = 5 cm and AC = 4.2 cm, find DC.
(ii) If BD = 2 cm, AB = 5 cm and DC = 3 cm, find AC.
(iii) If AB = 3.5 cm, AC = 4.2 cm and DC = 2.8 cm, find BD.
(iv) If AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC.
(v) If AC = 4.2 cm, DC = 6 cm and BC = 10 cm. find AB.
(vi) If AB = 5.6 cm, AC = 6 cm and DC = 3 cm, find BC.
(vii) If AD = 5.6 cm, BC = 6 cm and BD = 3.2 cm, find AC.
(viii) If AB = 10 cm, AC = 6 cm and BC = 12 cm, find BD and DC.

Answer 1

(i) It is given that , and .

In , is the bisector of, meeting side at .

We have to find .

Since is bisector

Then

Hence

(ii) It is given that , and.

In , is the bisector of, meeting side at .

We have to find .

Since is bisector

So (is bisector of and side)

Then

Hence

(iii) It is given that , and .

In , is the bisector of, meeting side at .

We have to find BD.

Since is bisector

So (is bisector of and side)

Then

Hence

(iv) It is given that , and .

In , is the bisector of, meeting side at .

We have to find BD and .

Since is bisector

So (is bisector of and side)

Let BD = x cm. Then CD = (6 − x) cm

Then,

$$\begin{gathered} \frac{10}{14}=\frac{x}{6-x} \\ \Rightarrow 14x=60-10x \\ \Rightarrow 24x=60 \\ \Rightarrow x=\frac{60}{24}=2.5 \end{gathered}$$

Hence, BD = 2.5 cm and DC = 6 − 2.5 = 3.5 cm.

(v) It is given that , and .

In , is the bisector of, meeting side at .

We have to find .

Since is bisector

So

Then

Hence

(vi) It is given that , and .

In , is the bisector of, meeting side at .

We have to find .

Since is bisector

So

Then

So

Hence

(vii) If it is given that AB = 5.6 cm, and _.

In, is the bisector of, meeting side at

(viii) It is given that , and .

In , is the bisector of, meeting side at .

We have to find BD and .

Since is bisector

So

Let BD = x cm

Then

⇒

Now

Hence and