#### Exercise 4.2

**Q.1:Â In a \( \Delta \) ABC, D and E are points on the sides AB and AC respectively such that DE \( \parallel \) BC.**

**1.) If AD = 6 cm, DB = 9 cm and AE = 8 cm, Find AC.**

**2.) If \( \frac{AD}{DB} = \frac{3}{4} \) and AC = 15 cm, Find AE.**

**3.) If \( \frac{AD}{DB} = \frac{2}{3} \) and AC = 18 cm, Find AE.**

**4.) If AD = 4 cm, AE = 8 cm, DB = x â€“ 4 cm and EC = 3x â€“ 19, find x.**

**5.) If AD = 8 cm, AB = 12 cm and AE = 12 cm, find CE.**

**6.) If AD = 4 cm, DB = 4.5 cm and AE = 8 cm, find AC.**

**7.) If AD = 2 cm, AB = 6 cm and AC = 9 cm, find AE.**

**8.) If \( \frac{AD}{BD} = \frac{4}{5} \) and EC = 2.5 cm, Find AE.**

**9.) If AD = x cm, DB = x â€“ 2 cm, AE = x + 2 cm, and EC = x â€“ 1 cm, find the value of x.**

**10.) If AD = 8x â€“ 7 cm, DB = 5x â€“ 3 cm, AE = 4x â€“ 3 cm, and EC = (3x – 1) cm, Find the value of x.**

**11.) If AD = 4x â€“ 3, AE = 8x â€“ 7, BD = 3x â€“ 1, and CE = 5x â€“ 3, find the value of x.**

**12.) If AD = 2.5 cm, BD = 3.0 cm, and AE = 3.75 cm, find the length of AC.**

**Sol:**

**1)** It is given that \( \Delta \)

We have to find AC,

Since, AD = 6 cm,

DB = 9 cm and AE = 15 cm.

AB = 15 cm.

So, \( \frac{AD}{BD} = \frac{AE}{CE} \)

Then, \( \frac{6}{9} = \frac{8}{x} \)

6x = 72 cm

x = 72/6 cm

x = 12 cm

Hence, **AC = 12 + 8 = 20.**

**2)**Â It is given that \( \frac{AD}{BD} = \frac{3}{4} \)

We have to find out AE,

Let, AE = x

So,Â Â \( \frac{AD}{BD} = \frac{AE}{CE} \)

Then, \( \frac{3}{4} = \frac{x}{15 – x} \)

45 â€“ 3x = 4x

-3x â€“ 4x = – 45

7x = 45

x = 45/7

**x = 6.43 cm**

**3)** It is given that \( \frac{AD}{BD} = \frac{2}{3} \)

We have to find out AE,

Let, AE = x and CE = 18 – x

So,Â Â \( \frac{AD}{BD} = \frac{AE}{CE} \)

Then, \( \frac{2}{3} = \frac{x}{18 – x} \)

3x = 36 â€“ 2x

5x = 36 cm

X = 36/5 cm

X = 7.2 cm

Hence**, AE = 7.2 cm**

**4)** It is given that AD = 4 cm, AE = 8 cm, DB = x â€“ 4 and EC = 3x â€“ 19

We have to find x,

So, \( \frac{AD}{BD} = \frac{AE}{CE} \)

Then, \( \frac{4}{x – 4} = \frac{8}{3x – 19} \)

4(3x – 19) = 8(x – 4)

12x â€“ 76 = 8(x – 4)

12x â€“ 8x = – 32 + 76

4x = 44 cm

**X = 11 cm**

**5)** It is given that AD = 8 cm, AB = 12 cm, and AE = 12 cm.

We have to find CE,

So, \( \frac{AD}{BD} = \frac{AE}{CE} \)

Then, \( \frac{8}{4} = \frac{12}{CE} \)

8CE = 4 X 12 cm

CE = (4 X 12)/8 cm

CE = 48/8 cm

**CE = 6 cm**

**6)** It is given that AD = 4 cm, DB = 4.5 cm, AE = 8 cm

We have to find out AC

So, \( \frac{AD}{BD} = \frac{AE}{CE} \)

Then, \( \frac{4}{4.5} = \frac{8}{AC} \)

\(AC = \frac{4.5 \times 8 }{4}\)

**AC = 9 cm**

**Â **

**7)**Â It is given that AD = 2 cm, AB = 6 cm, and AC = 9 cm

We have to find out AE

DB = 6 â€“ 2 = 4 cm

So, \( \frac{AD}{BD} = \frac{AE}{CE} \)

Then, \( \frac{2}{4} = \frac{x}{9 – x} \)

4x = 18 â€“ 2x

6x = 18

**Â X = 3 cm**

**8)** It is given that \( \frac{AD}{BD} = \frac{4}{5} \)

We have to find out AE

So, \( \frac{AD}{BD} = \frac{AE}{CE} \)

Then, \( \frac{4}{5} = \frac{AE}{2.5} \)

**AE = \(\frac{4 \times 2.5}{5}\) = 2 cm**

**9)** It is given that AD = x, DB = x â€“ 2, AE = x + 2 and EC = x â€“ 1

We have to find the value of x

So, \( \frac{AD}{BD} = \frac{AE}{CE} \)

Then, \( \frac{x}{x – 2} = \frac{x + 2}{x – 1} \)

X(x – 1) = (x – 2)(x + 2)

x^{2} â€“ x â€“ x^{2} + 4 = 0

**x =4**

**10)**Â It is given that AD = 8x â€“ 7, DB = 5x â€“ 3, AER = 4x â€“ 3 and EC = 3x -1

We have to find the value of x

So, \( \frac{AD}{BD} = \frac{AE}{CE} \)

Then, \( \frac{8x – 7}{5x – 3} = \frac{4x – 3}{3x – 1} \)

(8x – 7)(3x – 1) = (5x â€“ 3)(4x – 3)

24x^{2} â€“ 29x + 7 = 20x^{2 }â€“ 27x + 9

4x^{2} â€“ 2x â€“ 2 = 0

2(2x^{2} â€“ x – 1) = 0

2x^{2} â€“ x – 1 = 0

2x^{2} â€“ 2x + x â€“ 1 = 0

2x(x – 1) + 1(x – 1) = 0

(x – 1)(2x + 1) = 0

X = 1 or x = -1/2

Since the side of triangle can never be negative

**Therefore, x = 1**.

**11)**Â It is given that AD = 4x â€“ 3, BD = 3x â€“ 1, AE = 8x â€“ 7 and EC = 5x â€“ 3

For finding the value of x

So, \( \frac{AD}{BD} = \frac{AE}{CE} \)

Then, \( \frac{4x – 3}{3x – 1} = \frac{8x – 7}{5x – 3} \)

(4x – 3)(5x – 3) = (3x – 1)(8x – 7)

4x(5x – 3) -3(5x – 3) = 3x(8x – 7) -1(8x – 7)

20x^{2} â€“ 12x â€“ 15x + 9 = 24x^{2} â€“ 29x + 7

20x^{2} -27x + 9 = 24^{2} -29x + 7

Then,

-4x^{2} + 2x + 2 = 0

4x^{2} – 2x â€“ 2 = 0

4x^{2} – 4x + 2x â€“ 2 = 0

4x(x â€“ 1) + 2(x – 1) = 0

(4x + 2)(x â€“ 1) = 0

X = 1 or x = -2/4

Since, side of triangle can never be negative

**Therefore x = 1**

**12)** It is given that, AD = 2.5 cm, AE = 3.75 cm and BD = 3 cm

So, \( \frac{AD}{BD} = \frac{AE}{CE} \)

Then, \( \frac{2.5}{3} = \frac{3.75}{CE} \)

2.5CE = 3.75 x 3

\(CE = \frac{3.75\times 3}{2.5}\)

\(CE = \frac{11.25}{2.5}\)

CE = 4.5

Now, AC = 3.75 + 4.5

**AC = 8.25 cm.**

**Q.2)Â In a \( \Delta \) ABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DE \( \parallel \) BC.**

**1.) AB = 12 cm, AD = 8 cm, AE = 12 cm, and AC = 18 cm.**

**2.) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm, and AE = 1.8 cm.**

**3.) AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm, and AE = 2.8 cm.**

**4.) AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm, and EC = 5.5 cm.**

**Sol: **

**1)**Â It is given that D and R are the points on sides AB and AC.

We have to find that DE \( \parallel \)

Acc. To Thales Theorem,

\( \frac{AD}{DB} = \frac{AE}{CE} \)

\( \frac{8}{4} = \frac{12}{6} \)

2 = 2Â (LHS = RHS)

**Hence, DE \( \parallel \) BC.**

**2)**Â It is given that D and E are the points on sides AB and AC

We need to prove that DE \( \parallel \)

Acc. To Thales Theorem,

\( \frac{AD}{DB} = \frac{AE}{CE} \)

\( \frac{1.4}{4.2} = \frac{1.8}{5.4} \)

\( \frac{1}{3} = \frac{1}{3} \)

**Hence, DE \( \parallel \) BC.**

**3)**Â It is given that D and E are the points on sides AB and AC.

We need to prove DE \( \parallel \)

Acc. To Thales Theorem,

\( \frac{AD}{DB} = \frac{AE}{CE} \)

AD = AB â€“ DB = 10.8 â€“ 4.5 = 6.3

And,

EC = AC â€“ AE = 4.8Â – 2.8 = 2

Now,

\( \frac{6.3}{4.5} = \frac{2.8}{2.0} \)

**Hence, DE \( \parallel \) BC.**

**4)** It is given that D and E are the points on sides AB and Ac.

We need to prove that DE \( \parallel \)

Acc. To Thales Theorem,

\( \frac{AD}{DB} = \frac{AE}{CE} \)

\( \frac{5.7}{9.5} = \frac{3.3}{5.5} \)

\( \frac{3}{5} = \frac{3}{5} \)

**Hence, DE \( \parallel \) BC.**

**Q.3)Â In a \( \Delta \) ABC, P and Q are the points on sides AB and AC respectively, such that PQ \( \parallel \) BC. If AP = 2.4 cm, AQ = 2 cm, QC = 3 cm, and BC = 6 cm, Find AB and PQ.**

**Sol:**

It is given that AP = 2.4 cm, AQ = 2 cm, QC = 3 cm, and BC = 6 cm.

We need to find AB and PQ.

Using Thales Theorem,

\( \frac{AP}{PB} = \frac{AQ}{QC} \)

\( \frac{2.4}{PB} = \frac{2}{3} \)

2PB = 2.4 x 3 cm

\(PB = \frac{2.4 \times 3}{2}\)

PB = 3.6 cm

Now, AB = AP + PB

AB = 2.4 + 3.6

AB = 6 cm

Since, PQ \( \parallel \)

\( \Delta \)

Since, PQ \( \parallel \)

\( \Delta \)

In \( \Delta \)

\(\angle APQ = \angle ABC\)

\(\angle AQP = \angle ACB\)

Therefore, \( \Delta \)

Since, the corresponding sides of similar triangles are proportional,

Therefore,Â Â \(\frac{AP}{AB} = \frac{PQ}{BC} = \frac{AQ}{AC}\)

\(\frac{AP}{AB} = \frac{PQ}{BC}\)

\(\frac{2.4}{6} = \frac{PQ}{6}\)

**Therefore, PQ = 2.4 cm.**

**Q.4)Â In a \( \Delta \) ABC, D and E are points on AB and AC respectively, such that DE \( \parallel \) BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm, and BC = 5 cm. Find BD and CE.**

**Sol:**Â It is given that AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BE = 5 cm.

We need to find BD and CE.

Since, DE \( \parallel \)

\(\angle APQ = \angle ABC\)

Since, DE \( \parallel \)

\(\angle AED = \angle ACB\)

In \( \Delta \)

\( \angle ADE = \angle ABC \)

\( \angle AED = \angle ACB \)

So, \( \Delta \)

Since, the corresponding sides of similar triangles are proportional, then,

Therefore,Â Â \( \frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC} \)

\( \frac{AD}{AB} = \frac{DE}{BC} \)

\( \frac{2.4}{2.4 + DB} = \frac{2}{5} \)

2.4 + DB = 6

DB = 6 â€“ 2.4

DB = 3.6 cm

Similarly, \( \frac{AE}{AC} = \frac{DE}{BC} \)

\( \frac{3.2}{3.2 + EC} = \frac{2}{5} \)

3.2 + EC = 8

EC = 8 â€“ 3.2

EC = 4.8 cm

**Therefore, BD = 3.6 cm and CE = 4.8 cm.**

**Q.5)**Â In figure given below, state PQ \( \parallel \)

**Sol:Â **

It is given that EP = 3 cm, PG = 3.9 cm, FQ = 3.6 cm and QG = 2.4 cm

We have to check that PQ \( \parallel \)

Acc. to Thales Theorem,

\( \frac{PG}{GE} = \frac{GQ}{FQ} \)

Now,

\(\frac{3.9}{3} \neq \frac{3.6}{2.4}\)

As we can see it is not prortional.

**So, PQ is not parallel to EF.**

**Q.6)Â M and N are the points on the sides PQ and PR respectively, of a \( \Delta \) PQR. For each of the following cases, state whether MN \( \parallel \) QR.**

**(i)Â PM = 4 cm, QM = 4.5 cm, PN = 4 cm, NR = 4.5 cm.**

**(ii) PQ = 1.28 cm, PR = 2.56 cm, PM = 0.16 cm, PN = 0.32 cm.**

**Sol:Â **

**(i)** It is given that PM = 4 cm, QM = 4.5 cm, PN = 4 cm, and NR = 4.5 cm.

We have to check that MN \( \parallel \)

Acc. to Thales Theorem,

\( \frac{PM}{QM} = \frac{PN}{NR} \)

\( \frac{4}{4.5} = \frac{4}{4.5} \)

**Hence, MN \( \parallel \) QR.**

**Â **

**(ii)** It is given that PQ = 1.28 cm, PR = 2.56 cm, PM = 0.16 cm, and PN = 0.32 cm.

We have to check that MN \( \parallel \)

Acc. to Thales Theorem,

\( \frac{PM}{QM} = \frac{PN}{NR} \)

Now,

\( \frac{PM}{MQ} = \frac{0.16}{1.12} \)

\( \frac{PN}{NR} = \frac{0.32}{2.24} \)

Since,

\( \frac{0.16}{1.12} = \frac{0.32}{2.24} \)

**Hence, MN \( \parallel \) QR.**

**Q.7)Â In three line segments OA, OB, and OC, points L, M, N respectively are so chosen that LM \( \parallel \) AB and MN \( \parallel \) BC but neither of L, M, and N nor A, B, C are collinear. Show that LN \( \parallel \) AC.Â **

**Sol:**

In \( \Delta \)

Then,Â \( \frac{OL}{LA} = \frac{OM}{MB} \)

In \( \Delta \)

Then,Â \( \frac{OM}{MB} = \frac{ON}{NC} \)

Therefore,Â \( \frac{ON}{NC} = \frac{OM}{MB} \)

From the above equations,

We get,Â \( \frac{OL}{LA} = \frac{ON}{NC} \)

In a \( \Delta \)

\( \frac{OL}{LA} = \frac{ON}{NC} \)

LN \( \parallel \)

**Q.8)Â If D and E are the points on sides AB and AC respectively of a \( \Delta \) ABC such that DE \( \parallel \) BC and BD = CE. Prove that \( \Delta \) ABC is isosceles.**

**Sol:**

It is given that in \( \Delta \)

We need to prove that \( \Delta \)

Acc. to Thales Theorem,

\( \frac{AD}{BD} = \frac{AE}{EC} \)

AD = AE

Now, BD = CE and AD = AE.

So, AD + BD = AE + CE.

Therefore, AB = AC.

**Therefore, \( \Delta \) ABC is isosceles.**