Chapter 4: Triangles Exercise – 4.2
Question: 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 AD/DB = 3/4 and AC = 15 cm, Find AE.
3.) If AD/DB = 2/3 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 AD/BD = 4/5 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.
Solution:
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.
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
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
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,
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,
8CE = 4 × 12 cm
CE = (4 × 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
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
4x = 18 – 2x
6x = 18
x = 3 cm
8. It is given that
We have to find out 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
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
(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
(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
2.5CE = 3.75 × 3
CE = 4.5
Now, AC = 3.75 + 4.5
AC = 8.25 cm.
Question: 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.
Solution:
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,
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,
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 = AB – DB = 10.8 – 4.5 = 6.3
And,
EC = AC – AE = 4.8 – 2.8 = 2
Now,
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,
Hence, DE ∥ BC.
Question: 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.
Solution:
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,
2PB = 2.4 × 3 cm
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 similarity)
Since, the corresponding sides of similar triangles are proportional,
Therefore, PQ = 2.4 cm.
Question: 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.
Solution:
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,
2.4 + DB = 6
DB = 6 – 2.4
DB = 3.6 cm
3.2 + EC = 8
EC = 8 – 3.2
EC = 4.8 cm
Therefore, BD = 3.6 cm and CE = 4.8 cm.
Question: 5
In figure given below, state PQ ∥ EF.
Solution:
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,
As we can see it is not prortional.
So, PQ is not parallel to EF.
Question: 6
M and N are the points on the sides PQ and PR respectively, of a ΔPQR. 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.
Solution:
(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,
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,
Hence, MN ∥ QR.
Question: 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.
Solution:
In ΔOAB, Since, LM ∥ AB,
In Δ OBC, Since, MN ∥ BC,
From the above equations,
In a Δ OCA,
LN ∥ AC (by converse BPT)
Question: 8
If D and E are the points on sides AB and AC respectively of a ΔABC such that DE ∥ BC and BD = CE. Prove that Δ ABC is isosceles.
Solution:
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 = AE
Now, BD = CE and AD = AE.
So, AD + BD = AE + CE.
Therefore, AB = AC.
Therefore, Δ ABC is isosceles.