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CHAPTER 5 — PARALLEL & INTERSECTING LINES
Class VII Mathematics — NCERT Ganita Prakash 2024
Preeti Kushwah Classes — Unit Test | Set 3 — Advanced (Reasoning)
General Instructions:
1. All questions are compulsory. One optional Bonus question (2 marks) at the end.
2. Section A: 6 questions × 1 mark = 6 marks.
3. Section B: 5 questions × 2 marks = 10 marks.
4. Section C: 4 questions × 3 marks = 12 marks.
5. Section D: 2 questions × 5 marks = 10 marks.
6. Show all steps and state the angle property used for every calculation.
7. Use correct notation: ∠ for angle, ∥ for parallel, ⊥ for perpendicular.
1. All questions are compulsory. One optional Bonus question (2 marks) at the end.
2. Section A: 6 questions × 1 mark = 6 marks.
3. Section B: 5 questions × 2 marks = 10 marks.
4. Section C: 4 questions × 3 marks = 12 marks.
5. Section D: 2 questions × 5 marks = 10 marks.
6. Show all steps and state the angle property used for every calculation.
7. Use correct notation: ∠ for angle, ∥ for parallel, ⊥ for perpendicular.
Section A — (1 Mark Each) [6 × 1 = 6]
Q1.1
One angle of a linear pair is twice the other. Find both angles.
Q2.1
A transversal cuts two non-parallel lines at two points. At most, how many distinct angle values can appear among the 8 angles formed? (Hint: think about what happens at each intersection due to VOA.)
Q3.1
Write the meaning of: line p ⊥ line q. State the measure of the angle they form.
Q4.1
If alternate interior angles formed by a transversal on two lines are not equal, what can you conclude about the two lines?
Q5.1
Angles that are at the same relative position at each of two intersections made by a transversal are called _______ angles.
Q6.1
A transversal cuts two lines forming co-interior angles of 82° and 98°. Are the two lines parallel? Give your reason in one sentence.
Section B — (2 Marks Each) [5 × 2 = 10]
Q7.2
Two lines intersect at O. Among the four angles formed, ∠1 = (5x − 10)° and ∠3 = (3x + 20)° are vertically opposite angles.
(a) Find the value of x.
(b) Find the measure of ∠1. Name the property used.
(a) Find the value of x.
(b) Find the measure of ∠1. Name the property used.
Q8.2
Lines l ∥ m. A transversal t cuts them. At the upper intersection P, ∠1 = 110° (upper-left, exterior angle).
Using the linear pair property and then the alternate interior angle property, find the alternate interior angle at the lower intersection Q. Show each step clearly.
Using the linear pair property and then the alternate interior angle property, find the alternate interior angle at the lower intersection Q. Show each step clearly.
Q9.2
Lines l ∥ m. A transversal forms corresponding angles of (4y − 10)° and (2y + 30)° at the two intersections.
(a) Find y.
(b) Find both corresponding angles and verify they are equal.
(a) Find y.
(b) Find both corresponding angles and verify they are equal.
Q10.2
Lines l ∥ m. A transversal cuts them. The interior angle at the upper intersection P = (2x + 35)°. Its co-interior angle at the lower intersection Q = (4x − 5)°.
(a) Find x.
(b) Find both co-interior angles and verify they sum to 180°.
(a) Find x.
(b) Find both co-interior angles and verify they sum to 180°.
Q11.2
Prove: If a transversal cuts two lines such that the corresponding angles are equal, then the alternate interior angles are also equal.
Hint: Use the vertically opposite angle property and the given corresponding angle condition.
Hint: Use the vertically opposite angle property and the given corresponding angle condition.
Section C — (3 Marks Each) [4 × 3 = 12]
Q12.3
Lines l ∥ m. A transversal cuts l at P and m at Q. At P, ∠1 = (3x + 5)°. The corresponding angle ∠5 at Q = (5x − 25)°.
(a) Find the value of x. [1 mark]
(b) Find the measure of ∠1. [½ mark]
(c) Find all four angles at P (∠1, ∠2, ∠3, ∠4). State the property used for each. [1½ marks]
(a) Find the value of x. [1 mark]
(b) Find the measure of ∠1. [½ mark]
(c) Find all four angles at P (∠1, ∠2, ∠3, ∠4). State the property used for each. [1½ marks]
Q13.3
Three parallel lines a ∥ b ∥ c are all cut by the same transversal. The transversal makes an angle of 70° with line a at the upper-left position.
(a) Find the angle at the corresponding position with line b. State the property used. [1 mark]
(b) Find the angle at the corresponding position with line c. State the property used. [1 mark]
(c) Find the co-interior angle between lines a and b on the same side as the 70° angle. [1 mark]
(a) Find the angle at the corresponding position with line b. State the property used. [1 mark]
(b) Find the angle at the corresponding position with line c. State the property used. [1 mark]
(c) Find the co-interior angle between lines a and b on the same side as the 70° angle. [1 mark]
Q14.3
Lines PQ and RS intersect at O. ∠POR = (2x + 15)° and ∠POS = (3x − 5)°.
(a) ∠POR and ∠POS form a linear pair. Write the equation and find x. [1½ marks]
(b) Find all four angles at O. Name the property used for each. [1½ marks]
(a) ∠POR and ∠POS form a linear pair. Write the equation and find x. [1½ marks]
(b) Find all four angles at O. Name the property used for each. [1½ marks]
Q15.3
Prove: When a transversal cuts two parallel lines, the co-interior angles (same-side interior angles) are supplementary (sum = 180°).
Hint: Let ∠1 be an exterior angle at P and ∠4, ∠5 be the co-interior pair. Use: (i) corresponding angles property, (ii) linear pair property.
Hint: Let ∠1 be an exterior angle at P and ∠4, ∠5 be the co-interior pair. Use: (i) corresponding angles property, (ii) linear pair property.
Section D — (5 Marks Each) [2 × 5 = 10]
Q16.5
Lines l ∥ m. A transversal t cuts l at A and m at B. At A, ∠1 = (3x + 22)° and ∠2 = (5x − 10)° form a linear pair (∠1 is upper-left, ∠2 is upper-right).
(a) Use the linear pair property to find x. [1 mark]
(b) Find all four angles at A (∠1, ∠2, ∠3, ∠4). [1 mark]
(c) Find all four angles at B (∠5, ∠6, ∠7, ∠8). Use all three properties: corresponding angles, alternate interior angles, and co-interior angles — state which property gives which angle. [2 marks]
(d) Verify: the co-interior angles on the left side of the transversal sum to 180°. [1 mark]
(a) Use the linear pair property to find x. [1 mark]
(b) Find all four angles at A (∠1, ∠2, ∠3, ∠4). [1 mark]
(c) Find all four angles at B (∠5, ∠6, ∠7, ∠8). Use all three properties: corresponding angles, alternate interior angles, and co-interior angles — state which property gives which angle. [2 marks]
(d) Verify: the co-interior angles on the left side of the transversal sum to 180°. [1 mark]
Q17.5
(a) A surveyor is mapping two parallel boundary lines crossed by a path. The path makes angles of (2y + 12)° and (4y − 8)° at corresponding positions with the two boundary lines. Find y and the actual angle the path makes. [2 marks]
(b) A diagonal road crosses two parallel streets at points P (upper street) and Q (lower street). The interior angle of the road with the upper street at P = 62°.
(i) Find the other interior angle at P. [½ mark]
(ii) Find both interior angles at Q using alternate interior angles and co-interior angles. State the property used for each. [1½ marks]
(c) Draw a rough figure showing a transversal t crossing two parallel lines l and m. Label all 8 angles (∠1 to ∠8). Circle one pair of alternate interior angles and put a box around one pair of co-interior angles. [1 mark]
(b) A diagonal road crosses two parallel streets at points P (upper street) and Q (lower street). The interior angle of the road with the upper street at P = 62°.
(i) Find the other interior angle at P. [½ mark]
(ii) Find both interior angles at Q using alternate interior angles and co-interior angles. State the property used for each. [1½ marks]
(c) Draw a rough figure showing a transversal t crossing two parallel lines l and m. Label all 8 angles (∠1 to ∠8). Circle one pair of alternate interior angles and put a box around one pair of co-interior angles. [1 mark]
Bonus Question (Optional) [2 Marks]
Q18.2
★ A transversal cuts two lines forming co-interior angles of (2x + 10)° and (3x − 20)°. For what value of x will the two lines be parallel? Find both angles and verify your answer.
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Answer Key & Detailed Solutions — Set 3
Q1. [1 Mark]
Let the smaller angle = y°. The other angle = 2y°.
Linear pair: y + 2y = 180° → 3y = 180° → y = 60°
Angles: 60° and 120°.
Linear pair: y + 2y = 180° → 3y = 180° → y = 60°
Angles: 60° and 120°.
Q2. [1 Mark]
At each intersection, VOA gives 2 pairs → only 2 distinct values per intersection.
With two non-parallel lines, the two intersections can have different angle measures.
Maximum distinct values = 4 (2 distinct values at each of 2 intersections, all four different).
With two non-parallel lines, the two intersections can have different angle measures.
Maximum distinct values = 4 (2 distinct values at each of 2 intersections, all four different).
Q3. [1 Mark]
p ⊥ q means line p is perpendicular to line q.
They meet at a right angle: each of the four angles = 90°.
They meet at a right angle: each of the four angles = 90°.
Q4. [1 Mark]
If alternate interior angles are not equal, the two lines are not parallel.
(The converse of the alternate interior angle property: equal alt. interior ⇔ parallel lines.)
(The converse of the alternate interior angle property: equal alt. interior ⇔ parallel lines.)
Q5. [1 Mark]
Angles at the same relative position at each intersection are called corresponding angles.
Q6. [1 Mark]
82° + 98° = 180°.
Yes, the lines are parallel because co-interior angles sum to 180° iff the lines are parallel.
Yes, the lines are parallel because co-interior angles sum to 180° iff the lines are parallel.
Q7. [2 Marks]
Property used: Vertically Opposite Angles are equal.
∠1 = ∠3 → 5x − 10 = 3x + 20
5x − 3x = 20 + 10
2x = 30
x = 15 [1 mark]
∠1 = 5(15) − 10 = 75 − 10 = 65° [1 mark]
Verification: ∠3 = 3(15) + 20 = 65° ✓
∠1 = ∠3 → 5x − 10 = 3x + 20
5x − 3x = 20 + 10
2x = 30
x = 15 [1 mark]
∠1 = 5(15) − 10 = 75 − 10 = 65° [1 mark]
Verification: ∠3 = 3(15) + 20 = 65° ✓
Q8. [2 Marks]
Given: l ∥ m, ∠1 = 110° (upper-left exterior at P).
Step 1 — Linear Pair at P:
∠4 (lower-left interior at P) + ∠1 = 180° (angles on a straight line)
∠4 = 180° − 110° = 70° [1 mark]
Step 2 — Alternate Interior Angles (l ∥ m):
∠4 (lower-left interior at P) and ∠6 (upper-right interior at Q) are alternate interior angles.
∠6 = ∠4 = 70° [1 mark]
The alternate interior angle at Q = 70°.
Step 1 — Linear Pair at P:
∠4 (lower-left interior at P) + ∠1 = 180° (angles on a straight line)
∠4 = 180° − 110° = 70° [1 mark]
Step 2 — Alternate Interior Angles (l ∥ m):
∠4 (lower-left interior at P) and ∠6 (upper-right interior at Q) are alternate interior angles.
∠6 = ∠4 = 70° [1 mark]
The alternate interior angle at Q = 70°.
Q9. [2 Marks]
Property: Corresponding angles are equal when lines are parallel.
4y − 10 = 2y + 30
4y − 2y = 30 + 10
2y = 40
y = 20 [1 mark]
First angle = 4(20) − 10 = 70°
Second angle = 2(20) + 30 = 70°
Verification: 70° = 70° ✓ [1 mark]
4y − 10 = 2y + 30
4y − 2y = 30 + 10
2y = 40
y = 20 [1 mark]
First angle = 4(20) − 10 = 70°
Second angle = 2(20) + 30 = 70°
Verification: 70° = 70° ✓ [1 mark]
Q10. [2 Marks]
Property: Co-interior angles on parallel lines are supplementary (sum = 180°).
(2x + 35) + (4x − 5) = 180°
6x + 30 = 180°
6x = 150°
x = 25 [1 mark]
First angle = 2(25) + 35 = 85°
Second angle = 4(25) − 5 = 95°
Verification: 85° + 95° = 180° ✓ [1 mark]
(2x + 35) + (4x − 5) = 180°
6x + 30 = 180°
6x = 150°
x = 25 [1 mark]
First angle = 2(25) + 35 = 85°
Second angle = 4(25) − 5 = 95°
Verification: 85° + 95° = 180° ✓ [1 mark]
Q11. [2 Marks]
Given: Transversal cuts two lines. Corresponding angles are equal: ∠1 = ∠5 (where ∠1 is upper-left at P and ∠5 is upper-left at Q). [Setup: ½ mark]
To prove: Alternate interior angles are equal (∠3 = ∠5, where ∠3 is lower-right interior at P and ∠5 is upper-left interior at Q).
Proof:
Step 1: ∠3 = ∠1 [Vertically Opposite Angles at P] …(i) [½ mark]
Step 2: ∠1 = ∠5 [Corresponding Angles, given] …(ii)
From (i) and (ii): ∠3 = ∠5 [1 mark]
∴ Alternate interior angles are equal. ■
To prove: Alternate interior angles are equal (∠3 = ∠5, where ∠3 is lower-right interior at P and ∠5 is upper-left interior at Q).
Proof:
Step 1: ∠3 = ∠1 [Vertically Opposite Angles at P] …(i) [½ mark]
Step 2: ∠1 = ∠5 [Corresponding Angles, given] …(ii)
From (i) and (ii): ∠3 = ∠5 [1 mark]
∴ Alternate interior angles are equal. ■
Q12. [3 Marks]
(a) Finding x: [1 mark]
Since l ∥ m, corresponding angles ∠1 = ∠5:
3x + 5 = 5x − 25
30 = 2x
x = 15
(b) ∠1 = 3(15) + 5 = 50° [½ mark]
(c) All four angles at P: [1½ marks]
∠1 = 50° (given)
∠3 = 50° — Vertically Opposite to ∠1 [½ mark]
∠2 = 130° — Linear Pair with ∠1: 180° − 50° = 130° [½ mark]
∠4 = 130° — Vertically Opposite to ∠2 [½ mark]
Since l ∥ m, corresponding angles ∠1 = ∠5:
3x + 5 = 5x − 25
30 = 2x
x = 15
(b) ∠1 = 3(15) + 5 = 50° [½ mark]
(c) All four angles at P: [1½ marks]
∠1 = 50° (given)
∠3 = 50° — Vertically Opposite to ∠1 [½ mark]
∠2 = 130° — Linear Pair with ∠1: 180° − 50° = 130° [½ mark]
∠4 = 130° — Vertically Opposite to ∠2 [½ mark]
Q13. [3 Marks]
Transversal makes 70° (upper-left) with line a.
(a) Angle at corresponding position with b: [1 mark]
Since a ∥ b, corresponding angles are equal.
Angle = 70°
(b) Angle at corresponding position with c: [1 mark]
Since b ∥ c (and therefore a ∥ c), corresponding angles are equal.
Angle = 70°
(c) Co-interior angle between a and b: [1 mark]
Co-interior angles on the same side of the transversal are supplementary (since a ∥ b).
Co-interior angle = 180° − 70° = 110°
(a) Angle at corresponding position with b: [1 mark]
Since a ∥ b, corresponding angles are equal.
Angle = 70°
(b) Angle at corresponding position with c: [1 mark]
Since b ∥ c (and therefore a ∥ c), corresponding angles are equal.
Angle = 70°
(c) Co-interior angle between a and b: [1 mark]
Co-interior angles on the same side of the transversal are supplementary (since a ∥ b).
Co-interior angle = 180° − 70° = 110°
Q14. [3 Marks]
(a) Equation and value of x: [1½ marks]
∠POR and ∠POS are a linear pair (they lie on straight line RS at O):
∠POR + ∠POS = 180°
(2x + 15) + (3x − 5) = 180°
5x + 10 = 180°
5x = 170°
x = 34
(b) All four angles at O: [1½ marks]
∠POR = 2(34) + 15 = 68 + 15 = 83°
∠POS = 3(34) − 5 = 102 − 5 = 97°
∠QOS = 83° — Vertically Opposite Angles (∠QOS = ∠POR)
∠QOR = 97° — Vertically Opposite Angles (∠QOR = ∠POS)
Verification: 83° + 97° + 83° + 97° = 360° ✓
∠POR and ∠POS are a linear pair (they lie on straight line RS at O):
∠POR + ∠POS = 180°
(2x + 15) + (3x − 5) = 180°
5x + 10 = 180°
5x = 170°
x = 34
(b) All four angles at O: [1½ marks]
∠POR = 2(34) + 15 = 68 + 15 = 83°
∠POS = 3(34) − 5 = 102 − 5 = 97°
∠QOS = 83° — Vertically Opposite Angles (∠QOS = ∠POR)
∠QOR = 97° — Vertically Opposite Angles (∠QOR = ∠POS)
Verification: 83° + 97° + 83° + 97° = 360° ✓
Q15. [3 Marks]
Given: Lines l ∥ m. Transversal t meets l at P and m at Q.
Label: ∠1 = upper-left at P (exterior); ∠4 = lower-left at P (interior); ∠5 = upper-left at Q (interior).
∠4 and ∠5 are co-interior angles (both on the left of the transversal).
To prove: ∠4 + ∠5 = 180°
Proof:
Step 1: ∠1 = ∠5 [Corresponding Angles, l ∥ m] …(i) [1 mark]
Step 2: ∠1 + ∠4 = 180° [Linear Pair on line l at P] …(ii) [1 mark]
Step 3: Substitute (i) into (ii):
∠5 + ∠4 = 180°
∴ Co-interior angles ∠4 and ∠5 are supplementary. ■ [1 mark]
Label: ∠1 = upper-left at P (exterior); ∠4 = lower-left at P (interior); ∠5 = upper-left at Q (interior).
∠4 and ∠5 are co-interior angles (both on the left of the transversal).
To prove: ∠4 + ∠5 = 180°
Proof:
Step 1: ∠1 = ∠5 [Corresponding Angles, l ∥ m] …(i) [1 mark]
Step 2: ∠1 + ∠4 = 180° [Linear Pair on line l at P] …(ii) [1 mark]
Step 3: Substitute (i) into (ii):
∠5 + ∠4 = 180°
∴ Co-interior angles ∠4 and ∠5 are supplementary. ■ [1 mark]
Q16. [5 Marks]
(a) Finding x: [1 mark]
∠1 + ∠2 = 180° (Linear Pair on line l)
(3x + 22) + (5x − 10) = 180°
8x + 12 = 180°
8x = 168°
x = 21
(b) Angles at A: [1 mark]
∠1 = 3(21) + 22 = 63 + 22 = 85°
∠2 = 5(21) − 10 = 105 − 10 = 95° (Check: 85 + 95 = 180° ✓)
∠3 = 85° (VOA of ∠1) ∠4 = 95° (VOA of ∠2)
(c) Angles at B: [2 marks]
Using Corresponding Angles (l ∥ m):
∠5 = 85° (corresponds to ∠1) [½ mark]
∠6 = 95° (corresponds to ∠2) [½ mark]
Using Alternate Interior Angles: ∠7 = ∠3 = 85° [½ mark]
Using Co-interior Angles: ∠8 = 180° − ∠7 = 95° (or VOA of ∠6) [½ mark]
(d) Verification — co-interior on left: [1 mark]
Left co-interior pair: ∠4 (lower-left interior at A) and ∠5 (upper-left interior at B)
∠4 + ∠5 = 95° + 85° = 180° ✓
∠1 + ∠2 = 180° (Linear Pair on line l)
(3x + 22) + (5x − 10) = 180°
8x + 12 = 180°
8x = 168°
x = 21
(b) Angles at A: [1 mark]
∠1 = 3(21) + 22 = 63 + 22 = 85°
∠2 = 5(21) − 10 = 105 − 10 = 95° (Check: 85 + 95 = 180° ✓)
∠3 = 85° (VOA of ∠1) ∠4 = 95° (VOA of ∠2)
(c) Angles at B: [2 marks]
Using Corresponding Angles (l ∥ m):
∠5 = 85° (corresponds to ∠1) [½ mark]
∠6 = 95° (corresponds to ∠2) [½ mark]
Using Alternate Interior Angles: ∠7 = ∠3 = 85° [½ mark]
Using Co-interior Angles: ∠8 = 180° − ∠7 = 95° (or VOA of ∠6) [½ mark]
(d) Verification — co-interior on left: [1 mark]
Left co-interior pair: ∠4 (lower-left interior at A) and ∠5 (upper-left interior at B)
∠4 + ∠5 = 95° + 85° = 180° ✓
Q17. [5 Marks]
(a) Surveyor problem: [2 marks]
Corresponding angles are equal when lines are parallel:
2y + 12 = 4y − 8
20 = 2y
y = 10 [1 mark]
Actual angle = 2(10) + 12 = 32° [1 mark]
(b) Diagonal road:
(i) The other interior angle at P = 180° − 62° = 118° [½ mark]
(Linear Pair on the upper street at P)
(ii) Interior angles at Q: [1½ marks]
Using Alternate Interior Angles: The interior angle on the same side of the road as 62° (at P) → its alternate = 62° at Q.
Using Co-interior Angles: Co-interior pair with 62° at P → 180° − 62° = 118° at Q.
Both interior angles at Q: 62° and 118°.
(c) Figure description: [1 mark]
Draw two horizontal parallel lines l (top) and m (bottom). Draw a diagonal line t cutting both. At the upper intersection (P) label ∠1 (upper-left), ∠2 (upper-right), ∠3 (lower-right interior), ∠4 (lower-left interior). At the lower intersection (Q) label ∠5 (upper-left interior), ∠6 (upper-right interior), ∠7 (lower-right), ∠8 (lower-left). Circle ∠3 & ∠5 (alt. interior), box ∠4 & ∠5 (co-interior).
Corresponding angles are equal when lines are parallel:
2y + 12 = 4y − 8
20 = 2y
y = 10 [1 mark]
Actual angle = 2(10) + 12 = 32° [1 mark]
(b) Diagonal road:
(i) The other interior angle at P = 180° − 62° = 118° [½ mark]
(Linear Pair on the upper street at P)
(ii) Interior angles at Q: [1½ marks]
Using Alternate Interior Angles: The interior angle on the same side of the road as 62° (at P) → its alternate = 62° at Q.
Using Co-interior Angles: Co-interior pair with 62° at P → 180° − 62° = 118° at Q.
Both interior angles at Q: 62° and 118°.
(c) Figure description: [1 mark]
Draw two horizontal parallel lines l (top) and m (bottom). Draw a diagonal line t cutting both. At the upper intersection (P) label ∠1 (upper-left), ∠2 (upper-right), ∠3 (lower-right interior), ∠4 (lower-left interior). At the lower intersection (Q) label ∠5 (upper-left interior), ∠6 (upper-right interior), ∠7 (lower-right), ∠8 (lower-left). Circle ∠3 & ∠5 (alt. interior), box ∠4 & ∠5 (co-interior).
Q18. Bonus [2 Marks]
Condition for parallel lines: Co-interior angles must sum to 180°. [½ mark]
(2x + 10) + (3x − 20) = 180°
5x − 10 = 180°
5x = 190°
x = 38 [½ mark]
First angle = 2(38) + 10 = 76 + 10 = 86° [½ mark]
Second angle = 3(38) − 20 = 114 − 20 = 94° [½ mark]
Verification: 86° + 94° = 180° ✓
The lines will be parallel when x = 38.
(2x + 10) + (3x − 20) = 180°
5x − 10 = 180°
5x = 190°
x = 38 [½ mark]
First angle = 2(38) + 10 = 76 + 10 = 86° [½ mark]
Second angle = 3(38) − 20 = 114 − 20 = 94° [½ mark]
Verification: 86° + 94° = 180° ✓
The lines will be parallel when x = 38.