GSEB Solutions for Class 7 Mathematics – Parallel Lines

GSEB Solutions for Class 7 Mathematics – Parallel Lines (English Medium)

GSEB SolutionsMathsScience
Exercise 

Solution 1:

1. Line t1 and t2 are parallel lines. We write in symbol as t1 || t2
2. In all eight angles are formed by a transversal of two lines.
3. Two pairs of alternate angles are formed by a transversal of two lines.
4. Four pairs of corresponding angles are formed by a transversal of two lines.
5. The sum of the measures of both the angles in every pair of interior angles on the same side of the transversal is 180°.

Solution 2:

1. Line l is the transversal of line n and p.
2. Line l is the transversal of line m and n.
3. Line l is the transversal of line m and p.
4. Line m is the transversal of line n and l.
5. Line m is the transversal of line l and p.
6. Line m is the transversal of line n and p.
7. Line n is the transversal of line m and l.
8. Line p is the transversal of line m and l.

Solution 3:

Pairs of alternate angles:
∠JOP and ∠OPN
∠KOP and ∠OPM
Pairs of corresponding angles:
∠AOK and ∠OPN
∠AOJ and ∠OPM
Pairs of interior angles on the same side of the transversal:
∠KOP and ∠OPN
∠JOP and ∠OPM

Solution 4:

1. ∠PAY and ∠ABN are corresponding angles.
∴ m∠PAY = m∠ABN
But m∠ABN = m∠PBN = 55° (given)
∴ m∠PAY = 55°
2. ∠XAB and ∠ABN are alternate angles.
∴ m∠XAB = m∠ABN
∴ m∠ABN = m∠XAB = 55°
3. ∠YAB and ∠ABN are interior angles on the same side of the transversal.
(Interior angles on the same side of the transversal are supplementary)
∴ m∠YAB + m∠ABN = 180°˚
m∠ABN = 55°
∴ 55° + m∠YAB = 180°
∴ m∠YAB = 180° – 55° = 125°
4. ∠PAX and ∠ABM are corresponding angles.
∴ m∠PAX = m∠ABM
∴ m∠PAX = 125°
5. ∠XAB and ∠MBQ are corresponding angles.
∴ m∠XAB = m∠MBQ
m∠XAB = 55°
∴ m∠MBQ = 55°
6. ∠ABM and ∠YAB are alternate angles.
∴ m∠ABM = m∠YAB
But m∠YAB = 125°
∴ m∠ABM = 125°
7. ∠NBQ and ∠YAB are corresponding angles.
∴ m∠NBQ = m∠YAB
m∠YAB = 125°
∴ m∠NBQ = 125°

Solution 5:

∠XPA and ∠BPQ are vertically opposite angles.
∴ m∠XPA = m∠BPQ
∴ m∠BPQ = 120°
∠PQD and ∠BPQ are interior angles on the same side of the transversal XY.
(Interior angles on the same side of the transversal are supplementary)
∴ m∠PQD + m∠BPQ = 180°
But m∠BPQ = 120°
∴ m∠PQD + 120° = 180°
∴ m∠PQD = 180° – 120° = 60°
∴ m∠PQD = 60°
Hence, m∠BPQ = 120° and m∠PQD = 60°

Solution 6:

∠XPA and ∠PQC are corresponding angles.
∴ m∠XPA = m∠PQC
But m∠XPA = 110° (given)
∴ m∠PQC = 110° …..(1)
Since, ∠BPQ and ∠PQC are alternate angles.
∴ m∠BPQ = m∠PQC
But m∠PQC = 110°°
∴ m∠BPQ = 110˚
∠APQ and ∠PQC are interior angles on the same side of the transversal.
∴ m∠APQ + m∠PQC = 180°
(Interior angles on the same side of the transversal are supplementary)
But m∠PQC = 110°
∴ m∠APQ + 110° = 180°
∴ m∠APQ = (180 – 110)°
∴ m∠APQ = 70°
∠APQ and ∠PQD are alternate angles.
∴ m∠APQ = m∠PQD
∴ m∠APQ = m∠PQD = 70°

Activity – 1

Solution 1:

1. No, AB and CD do not interest each other.
2. Yes, a straight line forms the sides of a scale.
3. Yes, the distance between both the lines is the same.
4. The distance between two straight lines is 1 cm.
5. No, the lines do not intersect each other at any one place on extending.

Solution 2:

1. On extending lines l and m in figures (1), (2) and (3), they will intersect each other as shown.
The point of intersection of figure (1) is point O.
The point of intersection of figure (2) is point A.
The point of intersection of figure (3) is point B.
2. By extending lines l and m in figure (4), they will never intersect each other because lines l and m are parallel to each other.

Activity – 2

Solution 1:

1. Books
2. Scale or ruler
3. Door
4. Currency notes
5. Eraser
6. Screen of computer

Activity – 3

Solution 1:

The distance between the opposite sides of the page is the same every time.
Hence, the opposite sides of a page are parallel.

Activity – 4

Solution 1:

Activity – 5

Solution 1:

In all, eight angles are formed.
There are two points of intersection. i.e. A and B.
The names of the angles formed are as follows:
∠RBY, ∠SBY, ∠ABS, ∠ABR, ∠PAB, ∠QAB, ∠XAQ, ∠XAP

Activity – 6

Solution 1:

Conclusion:

1. The measures of the angles of the pair of alternate angles are equal.
2. The measures of the angles of the pair of corresponding angles are equal.
3. The sum of the measures of both angles in every pair of interior angles on the same side of the transversal is 180°.

Practice – 1

Solution 1:

Figure (1):
Lines AB and CD are parallel lines.

Figure (2):
Lines m and n intersect each other and hence, they are not parallel lines.
Figure (3):
On extending lines XYand line CD, they will intersect each other.
Hence, lines XY and CD are not parallel lines.
Figure (4):
Lines EP and FQ are parallel lines.

Solution 2:

Transversal is a line that intersects two other lines in the same plane at two distinct points.

Figure (1):
Line l₃is the transversal of lines l₁ and l₂.
Figure (2):
Line t is the transversal of lines m and n.
Line n is the transversal of lines m and t.
Line m is the transversal of lines t and n.

Solution 3:

1. Line PQ is a transversal of the other two lines, AB and CD.
2.  In all, eight angles are formed.
   They are ∠CYQ, ∠DYQ, ∠AXY, ∠BXY, ∠XYC, ∠XYD, ∠PXA, ∠PXB.
3.  Two pairs of alternate angles are formed.
   ∠AXY and ∠XYD,
   ∠BXY and ∠XYC
4.  Four pairs of corresponding angles are formed.
   ∠AXY and ∠CYQ,
   ∠BXY and ∠DYQ,
   ∠PXA and ∠XYC,
   ∠PXB and ∠XYD
5. Two pairs of interior angles on the same side of the transversal are formed.
   ∠BXY and ∠XYD,
   ∠AXY and ∠XYC.

Practice – 2

Solution 1:

2. ∠CPX and ∠EQP are a pair of corresponding angles.
3. ∠DPQ and ∠PQE are a pair of alternate angles.
4. ∠DPQ and ∠PQF are a pair of interior angles on the same side of the transversal.

Solution 2:

1.  ∠CGP and ∠GKX are corresponding angles.
   ∴ m∠CGP = m∠GKX
   But m∠CGP = m∠PGC = 65° (given)
   ∴ m∠GKX = 65°
2. ∠CGR and ∠GKY are corresponding angles.
   ∴ m∠GKY = m∠CGR
   m∠CGP + m∠CGR = 180° (Supplementary angles)
   ∴ 65° + m∠CGR = 180°
   ∴ m∠CGR = 180° – 65° = 115°
   ∴ m∠GKY = m∠CGR = 115°
3. ∠PGK and ∠XKD are corresponding angles.
   ∴ m∠PGK = m∠XKD
   m∠PGK + m∠PGC = 180° (Supplementary angles)
   ∴ m∠PGK = 180° – 65° = 115°
   ∴ m∠XKD = 115°
4.  ∠RGK and ∠YKD are corresponding angles.
   ∴ m∠RGK = m∠YKD
   m∠RGK + m∠CGR = 180°
   ∴ m∠RGK + 115° = 180°
   ∴ m∠RGK = 180° – 115° = 65°
   ∴ m∠YKD = m∠RGK = 65°
5.  ∠GKX and ∠RGK are alternate angles.
   ∴ m∠GKX = m∠RGK
   But m∠RGK = 65°
   ∴ m∠GKX = 65°
6.  ∠PGK and ∠GKY are alternate angles.
   ∴ m∠PGK = m∠GKY
   But m∠PGK = 115°
   ∴ m∠GKY = 115°
7. ∠RGK and ∠GKY are interior angles on the same side of the transversal.
   ∴ m∠RGK + m∠GKY = 180°
   m∠RGK = 65°
   ∴ 65° + m∠GKY = 180°
   ∴ m∠GKY = 180° – 65° = 115°

Solution 3:

1.  ∠PXA and ∠XYC are corresponding angles.
   ∴ m∠PXA = m∠XYC
   But m∠XYC = 85° (given)
   ∴ m∠PXA = 85°
2. ∠BXY and ∠XYC are alternate angles.
   ∴ m∠BXY = m∠XYC
   But m∠XYC = 85°
   ∴ m∠BXY = 85°
3. ∠BXY and ∠XYD are interior angles on the same side of the transversal.
   ∴ m∠BXY + m∠XYD = 180°
   m∠BXY = 85°
   ∴ 85° + m∠XYD = 180°
   ∴ m∠XYD = 180° – 85° = 95°
   ∴ m∠XYD = 95°
4.  ∠PXB and ∠XYD are corresponding angles.
   ∴ m∠PXB = m∠XYD
   But m∠XYD = 95°
   ∴ m∠PXB = 95°
5.  ∠BXY and ∠DYQ are corresponding angles.
   ∴ m∠BXY = m∠DYQ
   But m∠BXY = 85°
   ∴ m∠DYQ = 85°
6.  ∠AXY and ∠XYD are alternate angles.
   ∴ m∠AXY = m∠XYD
   But m∠XYD = 95°
   ∴ m∠AXY = 95°
7.  ∠CYQ and ∠AXY are corresponding angles.
   ∴ m∠CYQ = m∠AXY
   But m∠AXY = 95°
   ∴ m∠CYQ = 95°