RS Aggarwal Class 7 Solutions Chapter 14 Properties of Parallel Lines

In this chapter, we provide RS Aggarwal Solutions for Class 7 Chapter 14 Properties of Parallel Lines Ex 14 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 7 Chapter 14 Properties of Parallel Lines Ex 14 Maths pdf, free RS Aggarwal Solutions Class 7 Chapter 14 Properties of Parallel Lines Ex 14 Maths book pdf download. Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 7
SubjectMaths
ChapterChapter 12
Chapter NameProperties of Parallel Lines
Exercise14

RS Aggarwal Solutions for Class 7 Chapter 14 Properties of Parallel Lines Download PDF

Question 1.
Solution:
A transversal t intersects two parallel lines l and m.
∠ 1 = ∠ 5 (corresponding angles)
But ∠ 5 = 70° (given)
∠ 1 = 70°
RS Aggarwal Class 7 Solutions Chapter 14 Properties of Parallel Lines Ex 14 1
But ∠ 3 = ∠ 5 (Alternate angles)
∠ 3 = 70°
∠4 + ∠5 = 180° (Sum of co-interior angles)
⇒ ∠4 + 70° = 180°
⇒ ∠4 = 180° – 70°
⇒ ∠4 = 110°
But ∠ 4 = ∠ 8 (corresponding angles)
∠ 8 = 110°
Hence ∠ 1 = 70°, ∠3 = 70°, ∠4 = 110° and ∠ 8 = 110°

Question 2.
Solution:
A transversal t intersects two parallel lines l and m
∠1 : ∠2 = 5 : 7
But ∠ 1 + ∠ 2 = 180° (Linear pair)
RS Aggarwal Class 7 Solutions Chapter 14 Properties of Parallel Lines Ex 14 2
RS Aggarwal Class 7 Solutions Chapter 14 Properties of Parallel Lines Ex 14 3
But ∠ 3 = ∠ 1 (vertically opposite angles)
∠ 3 = 75°
∠ 8 = ∠ 4 (corresponding angles)
and ∠ 4 = ∠ 2 (vertically opposite angles)
∠8 = ∠2 = 105°
Hence ∠ 1 = 75°, ∠2 = 105°, ∠3 = 75° and ∠ 8 = 105°

Question 3.
Solution:
A transversal t intersects two parallel lines l and m interior angles of the same side of t are (2x – 8)° and (3x – 7)°
(2x – 8)° + (3x – 7)° = 180° (sum of co-interior angles)
⇒ 2x – 8 + 3x – 7 = 180°
⇒ 5x – 15° = 180°
⇒ 5x = 180° + 15°
⇒ 5x = 195°
RS Aggarwal Class 7 Solutions Chapter 14 Properties of Parallel Lines Ex 14 4
⇒ x = 1955 = 39°
First angle = 2x – 8° = 2 x 39° – 8° = 78° – 8° = 70°
Second angle = 3x – 7 = 3 x 39° – 7° = 117° – 7° = 110°

Question 4.
Solution:
l || m and two transversals intersect these lines but s is not parallel to t.
RS Aggarwal Class 7 Solutions Chapter 14 Properties of Parallel Lines Ex 14 5
∠ 5 = ∠ 1 (vertically opposite angles)
∠ 5 = 50°
But l || m and s the transversal
∠ 5 + ∠ 2 = 180° (sum of co-interior angles)
⇒ 50° + x = 180°
⇒ x = 180° – 50° – 130°
x = 130°
∠ 4 = ∠ 6 (vertically opposite angles)
∠ 6 = y
But l || m and t is the transversal
∠ 6 + ∠ 3 = 180° (sum of co-interior angles)
⇒ y + 65° = 180°
⇒ y = 180° – 65° = 115°
y = 115°
Hence x = 130° and y = 115°

Question 5.
Solution:
In the figure, ABC is a triangle, DAE || BC
∠B = 65°, ∠C = 45°
∠ DAB = x° and ∠ EAC = y°
DAE || BC and AB is transversal
∠ DAB = ∠ B (Alternate angles)
RS Aggarwal Class 7 Solutions Chapter 14 Properties of Parallel Lines Ex 14 6
⇒ x° = 65°
Similarly ∠ EAC = ∠ C (Alternate angles)
y° = 45°
Hence x = 65° and y = 45°

Question 6.
Solution:
In ∆ABC, AB || CE
∠BAC = 80°, ∠ECD = 35°
AB || CE and BCD is the transversal
∠ABC = ∠ECD (corresponding angles)
RS Aggarwal Class 7 Solutions Chapter 14 Properties of Parallel Lines Ex 14 7
⇒ ∠ABC = 35° (∠ECD = 35°)
Again AB || CE and AC is the transversal
∠ BAC = ∠ ACE (alternate angles)
∠ACE = 80° (∠BAC = 80°)
In ∆ABC
∠A + ∠B + ∠ACB = 180° (Sum of angles of a triangle)
∠ 80° + ∠ 35° + ∠ACB = 180°
⇒ ∠ACB + ∠ 115° = 180°
⇒ ∠ACB = 180° – 115° = 65°
Hence ∠ ACE = 80°, ∠ ACB = 65° and ∠ ABC = 35°

Question 7.
Solution:
In the figure,
AO || CD, DB || CE and ∠AOB = 50°
AO || CD and CD is the transversal
∠ AOB = ∠ CDB (corresponding angles)
RS Aggarwal Class 7 Solutions Chapter 14 Properties of Parallel Lines Ex 14 8
∠ CDB = 50° (∠ AOB = 50°)
Similarly CE || OB and CD in transversal
∠ECD + ∠CEB = 180° (sum of co-interior angles)
⇒ ∠ECD + 50° = 180°
⇒ ∠ECD = 180° – 50° = 130°
∠ECD = 130°

Question 8.
Solution:
In the fig, AB || CD
∠ABO = 50° and ∠CDO = 40°
From O, draw EOF || AB or CD
AB || EF and BO is the transversal
RS Aggarwal Class 7 Solutions Chapter 14 Properties of Parallel Lines Ex 14 9
∠ABO = ∠ 1 (Alternate angles) …(i)
∠ CDO = ∠ 2 (Alternate angles) …(ii)
Similarly, EF || CD and OD is the transversal
Adding (i) and (ii),
∠ 1 + ∠ 2 = ∠ABO + ∠CDO
⇒ ∠BOD = 50° + 40° = 90°
Hence ∠ BOD = 90°

Question 9.
Solution:
Given : In the figure, AB || CD and EF is a transversal which intersects them at G and H respectively
GL and HM are the angle bisectors or ∠ AGH and ∠ GHD respectively.
To prove : GL || HM.
Proof : AB || CD and EF is a transversal
∠ AGH = ∠ CHD (Alternate angles)
GL is the bisector of ∠ AGH
∠ 1 = ∠2 = 12 ∠ AGH
RS Aggarwal Class 7 Solutions Chapter 14 Properties of Parallel Lines Ex 14 10
Similarly, HM is the bisectors of ∠ GHD
∠3 = ∠4 = 12 ∠ GHD
∠ AGH = ∠ GHD (proved)
∠ 1 = ∠3
But, these are alternate angles
BL || HM
Hence proved.

Question 10.
Solution:
In the given figure,
AB || CD
∠ ABE = 120° and ∠ECD = 100° ∠ BEC = x°
From E, draw FG || AB or CD.
AB || EF
RS Aggarwal Class 7 Solutions Chapter 14 Properties of Parallel Lines Ex 14 11
∠ABE + ∠1 = 180° (sum of co-interior angles)
⇒ 120° + ∠1 = 180°
⇒ ∠1 = 180°- 120° = 60°
Similarly CD || EG
∠ECD + ∠2 = 180°
⇒ 100° + ∠2 = 180°
⇒ ∠2 = 180° – 100°
∠ 2 = 80°
But ∠1 + ∠x + ∠2 = 180° (Angles on one side of a straight line)
⇒ 60° + x + 80° = 180°
⇒ x + 140° = 180°
⇒ x = 180° – 140° = 40°
x = 40°

Question 11.
Solution:
Given : In the figure, ABCD is a quadrilateral in which AB || DC and AD || BC
To prove : ∠ADC = ∠ABC
Proof : AB || DC and DA is the transversal
∠ADC + ∠ DAB = 180° (co-interior angles)
RS Aggarwal Class 7 Solutions Chapter 14 Properties of Parallel Lines Ex 14 12
Similarly, AD || BC and AB is the transversal
∠DAB + ∠ABC = 180° …(ii)
from (i) and (ii),
∠ ADC + ∠ DAB = ∠DAB + ∠ABC
⇒ ∠ADC = ∠ABC
Hence ∠ ADC = ∠ ABC
Hence proved.

Question 12.
Solution:
In the figure,
l || m and p || q.
∠1 = 65°
∠ 2 = ∠ 1 (vertically opposite angles)
RS Aggarwal Class 7 Solutions Chapter 14 Properties of Parallel Lines Ex 14 13
∠ 2 = 65°
⇒ a = 65°
p || q and l is the transversal
∠ 2 + ∠ 3 = 180° (co-interior angles)
⇒ a + b= 180°
⇒ 65° + b = 180°
⇒ b = 180° – 65° = 115°
Again l || m and p is the transversal
∠ 3 + ∠4 = 180°
⇒ b + c = 180°
⇒ 115° + c = 180°
⇒ c = 180° – 115° = 65°
l || m and q is the transversal
∠ 2 + ∠ 5 = 180°
⇒ a + d = 180°
⇒ 65° + d = 180°
⇒ d = 180° – 65° = 115°
Hence a = 65°, b = 115°, c = 65° and d = 115°

Question 13.
Solution:
In the given figure, AB || DC and AD || BC and AC is the diagonal of parallelogram ABCD.
RS Aggarwal Class 7 Solutions Chapter 14 Properties of Parallel Lines Ex 14 14
∠BAC = 35°, ∠CAD = 40°, ∠ACB = x° and ∠ ACD = y°. .
AB || DC and CA is the transversal
∠ DCA = ∠ CAB (Alternate angles)
⇒ y = 35°
and similarly AD || BC and AC is the transversal
∠ CAD = ∠ ACB (Alternate angles)
⇒ 40° = x°
x = 40° and y = 35°

Question 14.
Solution:
In the figure, AB || CD and CD has been produced to E so that
∠ BAE = 125° ∠ BAC = x°, ∠ ABD = x°, ∠ BDC = y° and ∠ ACD = z°
DAE is a straight line and AB stands on it.
RS Aggarwal Class 7 Solutions Chapter 14 Properties of Parallel Lines Ex 14 15
∠ BAD + ∠ BAE = 180° (Linear pair)
⇒ x + 125° = 180°
⇒ x = 180° – 125° = 55°
But ∠ABC = x = 55°
DC || AB and CB is the transversal
∠ABC + ∠ BCD = 180° (co-interior angles)
⇒ x + y = 180°
⇒ 55° + y = 180°
⇒ y = 180° – 55° = 125°
Again DC || AB and DAE is its transversal
∠ CDA = ∠ BAE (corresponding angles).
z = 125°
Hence x = 55°, y = 125° and z = 125°

Question 15.
Solution:
Given : In each figure,
l and m are two lines and t is the transversal
To prove : l || m or not
Proof:
(i) fig. (i)
A transversal t intersects two lines l and m
RS Aggarwal Class 7 Solutions Chapter 14 Properties of Parallel Lines Ex 14 16
and ∠ 1 = 40°, ∠2 = 130°
But ∠ 1 + ∠3 = 180° (Linear pair)
⇒ 40° + ∠ 3 = 180°
⇒ ∠3 = 180° – 40° = 140°
l || m,
If ∠ 3 = ∠ 2
⇒ 140° = 130°
Which is not possible.
l is not parallel to m.
(ii) fig. (ii)
Transversal t, intersects l and m and ∠ 1 = 35°, ∠2 = 145°
But ∠ 1 = ∠ 3 (vertically opposite angles).
∠3 = 35°
l || m,
if ∠3 + ∠2 = 180°
if 35° + 145° = 180°
RS Aggarwal Class 7 Solutions Chapter 14 Properties of Parallel Lines Ex 14 17
if 180°= 180°
which is true
l || m
(iii) Transversal t, intersects l and m.
∠ 1 = 125°, ∠ 2 = 60°
RS Aggarwal Class 7 Solutions Chapter 14 Properties of Parallel Lines Ex 14 18
But ∠ 1 = ∠ 3 (vertically opposite angles)
∠ 3 = 125°
l || m
If ∠3 + ∠2 = 180° (co-interior angles)
If 125° + 60° = 180°
If 185° =180°
which is not possible.
Hence l is not parallel to m.

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