RD Sharma Solutions Class 10 Triangles Exercise 4.2

RD Sharma Class 10 Solutions Chapter 4 Ex 4.2 PDF Free Download

Exercise 4.2

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

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

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

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

iv) If AD = 4 cm, AE = 8 cm, DB = x – 4 cm and EC = 3x – 19, find x.

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

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

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

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

ix) If AD = x cm, DB = x – 2 cm, AE = x + 2 cm, and EC = x – 1 cm, find the value of x.

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

xi) If AD = 4x – 3, AE = 8x – 7, BD = 3x – 1, and CE = 5x – 3, find the value of x.

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

Solution:

i) Given: \( \Delta \) ABC AND DE \( \parallel \) BC

To find AC,

Given that,

AD = 6 cm, DB = 9 cm and AE = 8 cm.

Hence, Using Thales Theorem

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

Let CE=x. then,

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

6x = 72 cm

x = 72/6 cm

x = 12 cm

∴, AC = 12 + 8 = 20.

 

ii) Given: \( \frac{AD}{BD} = \frac{3}{4} \) and AC = 15 cm

To find: AE,

Hence, Using Thales Theorem,

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

Let, AE = x, then CE= 15-x.

⇒\( \frac{3}{4} = \frac{x}{15 – x} \)

45 – 3x = 4x

-3x – 4x = – 45

7x = 45

x = 45/7

x = 6.43 cm

∴, AE= 6.43cm

 

iii) Given: \( \frac{AD}{BD} = \frac{2}{3} \) and AC = 18 cm

To find: AE,

Hence, Using Thales Theorem,

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

Let, AE = x and CE = 18 – x

⇒\( \frac{2}{3} = \frac{x}{18 – x} \)

3x = 36 – 2x

5x = 36 cm

x = 36/5 cm

x = 7.2 cm

∴, AE = 7.2 cm

 

iv) Given: AD = 4 cm, AE = 8 cm, DB = x – 4 and EC = 3x – 19

To find x,

Hence, Using Thales Theorem,

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

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

4(3x – 19) = 8(x – 4)

12x – 76 = 8(x – 4)

12x – 8x = – 32 + 76

4x = 44 cm

x = 11 cm

 

v) Given: AD = 8 cm, AB = 12 cm, and AE = 12 cm.

To find CE,

Hence, Using Thales Theorem,

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

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

8CE = 4 x 12 cm

CE = (4 x 12)/8 cm

CE = 48/8 cm

CE = 6 cm

 

vi) Given: AD = 4 cm, DB = 4.5 cm, AE = 8 cm

To find = AC

Hence, Using Thales Theorem,

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

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

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

AC = 9 cm

 

vii)  Given: AD = 2 cm, AB = 6 cm, and AC = 9 cm

To find= AE

DB = 6 – 2 = 4 cm

Hence, Using Thales Theorem,

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

\( \frac{2}{4} = \frac{x}{9 – x} \)

4x = 18 – 2x

6x = 18

x = 3 cm

∴, AE= 3cm

 

viii) Given: \( \frac{AD}{BD} = \frac{4}{5} \) and EC = 2.5 cm

To find= AE

Hence, Using Thales Theorem,

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

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

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

 

ix) It is given that AD = x, DB = x – 2, AE = x + 2 and EC = x – 1

We have to find the value of x

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

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

X(x – 1) = (x – 2)(x + 2)

x2 – x – x2 + 4 = 0

x =4

 

x)  Given: AD = 8x – 7, DB = 5x – 3, AER = 4x – 3 and EC = 3x -1

To find= x

Hence, Using Thales Theorem,

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

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

(8x – 7)(3x – 1) = (5x – 3)(4x – 3)

24x2 – 29x + 7 = 20x2 – 27x + 9

4x2 – 2x – 2 = 0

2(2x2 – x – 1) = 0

2x2 – x – 1 = 0

2x2 – 2x + x – 1 = 0

2x(x – 1) + 1(x – 1) = 0

(x – 1)(2x + 1) = 0

x = 1 or x = -1/2

We know that the side of triangle can never be negative. Hence we take the positive value.

∴, x = 1.

 

xi)  Given: AD = 4x – 3, BD = 3x – 1, AE = 8x – 7 and EC = 5x – 3

To find= x

Hence, Using Thales Theorem,

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

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)

20x2 – 12x – 15x + 9 = 24x2 – 29x + 7

20x2 -27x + 9 = 242 -29x + 7

Then,

-4x2 + 2x + 2 = 0

4x2 – 2x – 2 = 0

4x2 – 4x + 2x – 2 = 0

4x(x – 1) + 2(x – 1) = 0

(4x + 2)(x – 1) = 0

X = 1 or x = -2/4

We know that the side of triangle can never be negative. Hence we take the positive value.

∴, x = 1

 

xii) Given: AD = 2.5 cm, AE = 3.75 cm and BD = 3 cm

Hence, Using Thales Theorem,

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

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

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.

 

2. 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.

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

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

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

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

Solution:

i) Given: D and R are the points on sides AB and AC.

To find= DE \( \parallel \) BC.

Hence, Using Thales Theorem,

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

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

2 = 2  (LHS = RHS)

∴, DE \( \parallel \) BC.

 

ii)  Given: D and E are the points on sides AB and AC

To prove= DE \( \parallel \) BC

Hence, Using Thales Theorem,

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

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

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

∴, DE \( \parallel \) BC.

 

iii)  Given: D and E are the points on sides AB and AC.

To prove= DE \( \parallel \) BC.

Hence, Using Thales Theorem,

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

AD = AB – DB = 10.8 – 4.5 = 6.3

And,

EC = AC – AE = 4.8  – 2.8 = 2

Now,

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

∴, DE \( \parallel \) BC.

 

iv) Given: D and E are the points on sides AB and Ac.

To prove: DE \( \parallel \) BC.

Hence, Using Thales Theorem,

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

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

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

Hence, DE \( \parallel \) BC.

 

3. 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.

Solution:

Given: AP = 2.4 cm, AQ = 2 cm, QC = 3 cm, and BC = 6 cm.

To find= AB and PQ.

Class 10 maths chapter 4 triangles exercise 4.2-1

Hence, Using Thales Theorem,

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

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

2PB = 2.4 x 3 cm

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

PB = 3.6 cm

Now, AB = AP + PB

AB = 2.4 + 3.6

AB = 6 cm

As PQ \( \parallel \) BC, AB is transversal,

\( \Delta \) APQ  = \( \Delta \) ABC    (by corresponding angles)

As PQ \( \parallel \) BC, AC is transversal,

\( \Delta \) APQ  = \( \Delta \) ABC    (by corresponding angles)

In \( \Delta \) ABQ and \( \Delta \) ABC,

\(\angle APQ = \angle ABC\)

\(\angle AQP = \angle ACB\)

Hence, \( \Delta \) APQ = \( \Delta \) ABC (angle angle similarity)

We know that the corresponding sides of similar triangles are proportional,

⇒\(\frac{AP}{AB} = \frac{PQ}{BC} = \frac{AQ}{AC}\)

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

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

∴, PQ = 2.4 cm.

 

4.  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.

Solution:

Given: AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BE = 5 cm.

To find= BD and CE.

As DE \( \parallel \) BC, AB is transversal,

\(\angle APQ = \angle ABC\)

As DE \( \parallel \) BC, AC is transversal,

\(\angle AED = \angle ACB\)

In \( \Delta \) ADE and \( \Delta \) ABC,

\( \angle ADE = \angle ABC \)

\( \angle AED = \angle ACB \)

∴, \( \Delta \) ADE = \( \Delta \) ABC (angle angle similarity)

We know that the corresponding sides of similar triangles are proportional,

⇒  \( \frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC} \)

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

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

2.4 + DB = 6

DB = 6 – 2.4

DB = 3.6 cm

Same way,

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

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

3.2 + EC = 8

EC = 8 – 3.2

EC = 4.8 cm

∴, BD = 3.6 cm and CE = 4.8 cm.

 

5.  In figure 4.35 given below, state PQ \( \parallel \) EF.

Class 10 maths chapter 4 triangles exercise 4.2-2

Fig 4.35

Solution:

Class 10 maths chapter 4 triangles exercise 4.2-3

Given: EP = 3 cm, PG = 3.9 cm, FQ = 3.6 cm and QG = 2.4 cm

To check: PQ \( \parallel \) EF or not.

Hence, Using Thales Theorem,

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

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

Hence, it is not proportional.

∴, PQ is not parallel to EF.

 

6. 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)  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.

Solution:

(i) Given: PM = 4 cm, QM = 4.5 cm, PN = 4 cm, and NR = 4.5 cm.

To check= MN \( \parallel \) QR or not.

Hence, Using Thales Theorem,

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

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

∴, MN \( \parallel \) QR.

 

(ii) Given: PQ = 1.28 cm, PR = 2.56 cm, PM = 0.16 cm, and PN = 0.32 cm.

To check= MN \( \parallel \) QR or not.

Hence, Using Thales Theorem,

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

\( \frac{PM}{MQ} = \frac{0.16}{1.12} \) = 1/7

\( \frac{PN}{NR} = \frac{0.32}{2.24} \) = 1/7

But, since,

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

∴, MN \( \parallel \) QR.

 

7. 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. 

Solution:

In \( \Delta \) OAB, Given that, LM \( \parallel \) AB,

Then, using BPT,

\( \frac{OL}{LA} = \frac{OM}{MB} \)

Class 10 maths chapter 4 triangles exercise 4.2-5

Now,

In \( \Delta \) OBC, Given that, MN \( \parallel \) BC,

Then, using BPT,

\( \frac{OM}{MB} = \frac{ON}{NC} \)

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

We can infer from the above equations that,

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

Now, In \( \Delta \) OCA,

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

By converse BPT

LN \( \parallel \) AC

 

8. 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.

Solution:

Given: \( \Delta \) ABC, DE \( \parallel \) BC and BD = CE.

To prove= \( \Delta \) ABC is isosceles.

Hence, Using Thales Theorem,

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

AD = AE

Now, BD = CE and AD = AE.

So, AD + BD = AE + CE.

i.e., AB = AC.

∴, \( \Delta \) ABC is isosceles.

Leave a Comment

Your email address will not be published. Required fields are marked *