Ex-4.2 Triangles, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

Exercise 4.2

Q.1:  In a Δ ABC, D and E are points on the sides AB and AC respectively such that DE ∥ BC.

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

2.) If ADDB = 34 and AC = 15 cm, Find AE.

3.) If ADDB = 23 and AC = 18 cm, Find AE.

4.) If AD = 4 cm, AE = 8 cm, DB = x – 4 cm and EC = 3x – 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 ADBD = 45 and EC = 2.5 cm, Find AE.

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

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

11.) If AD = 4x – 3, AE = 8x – 7, BD = 3x – 1, and CE = 5x – 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 Δ ABC AND DE ∥ BC

We have to find AC,

Since, AD = 6 cm,

DB = 9 cm and AE = 15 cm.

AB = 15 cm.

So, AD/BD = AE/CE (using Thales Theorem)

Then, 6/9 = 8/x

6x = 72 cm

x = 72/6 cm

x = 12 cm

Hence, AC = 12 + 8 = 20.

2) It is given that AD/BD = 3/4 and AC = 15 cm

We have to find out AE,

Let, AE = x

So,   AD/BD = AE/CE (using Thales Theorem)

Then, 

45 – 3x = 4x

-3x – 4x = – 45

7x = 45

x = 45/7

x = 6.43 cm

3) It is given that AD/BD = 2/3 and AC = 18 cm

We have to find out AE,

Let, AE = x and CE = 18 – x

So,   AD/BD = AE/CE (using Thales Theorem)

Then, 

3x = 36 – 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 – 4 and EC = 3x – 19

We have to find x,

So, AD/BD = AE/CE (using Thales Theorem)

Then, 

4(3x – 19) = 8(x – 4)

12x – 76 = 8(x – 4)

12x – 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, AD/BD = AE/CE (using Thales Theorem)

Then, 8/4 = 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, AD/BD = AE/CE (using Thales Theorem)

Then, 4/4.5 = 8/AC

AC =

AC = 9 cm

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

We have to find out AE

DB = 6 – 2 = 4 cm

So, AD/BD = AE/CE (using Thales Theorem)

Then, 

4x = 18 – 2x

6x = 18

 X = 3 cm

8) It is given that AD/BD = 4/5 and EC = 2.5 cm

We have to find out AE

So, AD/BD = AE/CE (using Thales Theorem)

Then, 4/5 = AE/2.5

AE =  

9) 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, AD/BD = AE/CE (using Thales Theorem)

Then, 

X(x – 1) = (x – 2)(x + 2)

x² – x – x² + 4 = 0

x = 4

10)  It is given that AD = 8x – 7, DB = 5x – 3, AER = 4x – 3 and EC = 3x -1

We have to find the value of x

So, AD/BD = AE/CE (using Thales Theorem)

Then, 

(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

Since the side of triangle can never be negative

Therefore, x = 1.

11)  It is given that AD = 4x – 3, BD = 3x – 1, AE = 8x – 7 and EC = 5x – 3

For finding the value of x

So, AD/BD = AE/CE (using Thales Theorem)

Then, 

(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

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, AD/BD = AE/CE (using Thales Theorem)

Then, 2.5/3 = 3.75/CE

2.5CE = 3.75 x 3

CE = 4.5

Now, AC = 3.75 + 4.5

AC = 8.25 cm.

Q.2)  In a Δ ABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DE ∥ 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)  It is given that D and R are the points on sides AB and AC.

We have to find that DE ∥ BC.

Acc. To Thales Theorem,

AD/DB = AE/CE

8/4 = 12/6

2 = 2  (LHS = RHS)

Hence, DE ∥ BC.

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

We need to prove that DE ∥ BC

Acc. To Thales Theorem,

AD/DB = AE/CE

1.4/4.2 = 1.8/5.4

1/3 = 1/3  (RHS)

Hence, DE ∥ BC.

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

We need to prove DE ∥ BC.

Acc. To Thales Theorem,

AD/DB = AE/CE

AD = AB – DB = 10.8 – 4.5 = 6.3

And,

EC = AC – AE = 4.8  – 2.8 = 2

Now,

6.34.5 = 2.82.0

Hence, DE ∥ BC.

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

We need to prove that DE ∥ BC.

Acc. To Thales Theorem,

AD/DB = AE/CE

5.7/9.5 = 3.3/5.5

3/5 = 3/5   (LHS = RHS)

Hence, DE ∥ BC.

Q.3)  In a Δ ABC, P and Q are the points on sides AB and AC respectively, such that PQ ∥ 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,

AP/PB = AQ/QC

2.4/PB = 2/3

2PB = 2.4 x 3 cm

PB =

PB = 3.6 cm

Now, AB = AP + PB

AB = 2.4 + 3.6

AB = 6 cm

Since, PQ ∥ BC, AB is transversal, then,

Δ APQ  =  Δ ABC    (by corresponding angles)

Since, PQ ∥ BC, AC is transversal, then,

Δ APQ  =  Δ ABC    (by corresponding angles)

In Δ ABQ and Δ ABC,

∠APQ = ∠ABC

∠AQP = ∠ACB

Therefore, Δ APQ =  Δ ABC (angle angle similarity)

Since, the corresponding sides of similar triangles are proportional,

Therefore,   AP/AB = PQ/BC = AQ/AC

AP/AB = PQ/BC

2.4/6 = PQ/6

Therefore, PQ = 2.4 cm.

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

Sol:  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 ∥ BC, AB is transversal, then,

∠APQ = ∠ABC

Since, DE ∥ BC, AC is transversal, then,

∠AED = ∠ACB

In Δ ADE and Δ ABC,

∠ADE = ∠ABC

∠AED = ∠ACB

So, Δ ADE =  Δ ABC (angle angle similarity)

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

Therefore,   AD/AB = AE/AC = DE/BC

AD/AB = DE/BC

2.4 + DB = 6

DB = 6 – 2.4

DB = 3.6 cm

Similarly, AE/AC = DE/BC

3.2 + EC = 8

EC = 8 – 3.2

EC = 4.8 cm

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

Q.5)  In figure given below, state PQ ∥ EF.

Sol: 

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 ∥ EF or not.

Acc. to Thales Theorem,

PG/GE = GQ/FQ

Now,

3.9/3 ≠ 3.6/2.4

As we can see it is not prortional.

So, PQ is not parallel to EF.

Q.6)  M and N are the points on the sides PQ and PR respectively, of a Δ PQR. For each of the following cases, state whether MN ∥ 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.

Sol: (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 ∥ QR or not.

Acc. to Thales Theorem,

PM/QM = PN/NR

44.5 = 44.5

Hence, MN ∥ QR.

(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 ∥ QR or not.

Acc. to Thales Theorem,

PM/QM = PN/NR

Now,

PM/MQ = 0.161.12  = 1/7

PN/NR = 0.322.24  = 1/7

Since,

0.161.12 = 0.322.24

Hence, MN ∥ QR.

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

Sol: In Δ OAB, Since, LM ∥ AB,

Then,  OL/LA = OM/MB     (using BPT)

In Δ OBC, Since, MN ∥ BC,

Then,  OM/MB = ON/NC     (using BPT)

Therefore,  ON/NC = OM/MB

From the above equations,

We get,  OL/LA = ON/NC

In a Δ OCA,

OL/LA = ON/NC

LN ∥ AC (by converse BPT)

Q.8)  If D and E are the points on sides AB and AC respectively of a Δ ABC such that DE ∥ BC and BD = CE. Prove that Δ ABC is isosceles.

Sol: It is given that in Δ ABC, DE ∥ BC and BD = CE.

We need to prove that Δ ABC is isosceles.

Acc. to Thales Theorem,

AD/BD = AE/EC

AD = AE

Now, BD = CE and AD = AE.

So, AD + BD = AE + CE.

Therefore, AB = AC.

Therefore, Δ ABC is isosceles.