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CHAPTER 5 — PARALLEL & INTERSECTING LINES
Class VII Mathematics — NCERT Ganita Prakash 2024
Preeti Kushwah Classes — Unit Test | Set 4 — HOTS Challenge
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. All proofs must state Given, To Prove, and each step with the property name.
7. Think carefully before writing — these questions require multi-step reasoning.
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. All proofs must state Given, To Prove, and each step with the property name.
7. Think carefully before writing — these questions require multi-step reasoning.
Section A — (1 Mark Each) [6 × 1 = 6]
Q1.1
Lines l and m are both perpendicular to line n. Without measuring, state whether l ∥ m. Name the geometric property that justifies your answer.
Q2.1
A transversal makes 90° with one of two parallel lines. What angle does it make with the second parallel line? Justify in one line.
Q3.1
Can vertically opposite angles be supplementary? If yes, find the measure of each angle. If no, explain why not.
Q4.1
A transversal cuts two lines. The co-interior angles formed are equal. Find the value of each co-interior angle. Are the lines parallel? Give a one-line justification.
Q5.1
A transversal cuts two lines making alternate interior angles of 75° and 105°. Are the two lines parallel? Give a one-line reason.
Q6.1
True or False (justify in one line): “Alternate interior angles can be supplementary when two lines are parallel.”
Section B — (2 Marks Each) [5 × 2 = 10]
Q7.2
Two lines intersect at O. ∠1 = (x + 30)° and ∠3 = (3x − 10)° are vertically opposite angles.
(a) Find x and ∠1.
(b) Find all four angles at O. Name the property used for each new angle found.
(a) Find x and ∠1.
(b) Find all four angles at O. Name the property used for each new angle found.
Q8.2
Three parallel lines p ∥ q ∥ r are all cut by the same transversal. The transversal makes an angle of 55° with line p.
(a) Find the corresponding angle at q. State the property.
(b) Find the alternate interior angle between p and q. Explain whether it equals the angle at p.
(c) Are all angles at p, q, and r in the same corresponding position equal? Why?
(a) Find the corresponding angle at q. State the property.
(b) Find the alternate interior angle between p and q. Explain whether it equals the angle at p.
(c) Are all angles at p, q, and r in the same corresponding position equal? Why?
Q9.2
Lines l ∥ m. A transversal cuts them. At the upper intersection, ∠1 = 115° (upper-left, exterior).
(a) The angles ∠1 (upper-left at upper intersection) and ∠7 (lower-right at lower intersection) are called alternate exterior angles. Find ∠7. State the property.
(b) Find both co-interior angles on the right side of the transversal. Show your working.
(a) The angles ∠1 (upper-left at upper intersection) and ∠7 (lower-right at lower intersection) are called alternate exterior angles. Find ∠7. State the property.
(b) Find both co-interior angles on the right side of the transversal. Show your working.
Q10.2
A transversal cuts two lines forming corresponding angles of (3x + 15)° and (5x − 25)°.
(a) Assuming the lines are parallel, find x and both angles.
(b) If instead x = 18, are the lines parallel? Justify your answer with a calculation.
(a) Assuming the lines are parallel, find x and both angles.
(b) If instead x = 18, are the lines parallel? Justify your answer with a calculation.
Q11.2
Lines l ∥ m. Transversal t meets l at P and m at Q. The interior angle at P (on the left of t) = 70°. The bisector of this 70° angle is drawn at P. The bisector of the co-interior angle at Q is also drawn. The two bisectors meet at point R.
Prove that ∠PRQ = 90°.
Hint: Let ∠RPQ = α. Find ∠RQP using the co-interior angle. Apply angle sum of triangle.
Prove that ∠PRQ = 90°.
Hint: Let ∠RPQ = α. Find ∠RQP using the co-interior angle. Apply angle sum of triangle.
Section C — (3 Marks Each) [4 × 3 = 12]
Q12.3
A transversal cuts two lines. At the upper intersection P, ∠1 = 65° (upper-left). At the lower intersection Q, ∠5 = 70° (also upper-left, the corresponding position).
(a) If the lines were parallel, what would ∠5 need to equal? Using this, determine whether the lines are parallel. [1 mark]
(b) Find ∠4 at P (lower-left interior). [½ mark]
(c) Calculate ∠4 + ∠5. Does this sum confirm your answer in (a)? State the property that this should satisfy for parallel lines. [1½ marks]
(a) If the lines were parallel, what would ∠5 need to equal? Using this, determine whether the lines are parallel. [1 mark]
(b) Find ∠4 at P (lower-left interior). [½ mark]
(c) Calculate ∠4 + ∠5. Does this sum confirm your answer in (a)? State the property that this should satisfy for parallel lines. [1½ marks]
Q13.3
Three parallel lines l ∥ m ∥ n are cut by two transversals t₁ and t₂. Both transversals pass through the same point G on line l. Transversal t₁ makes 60° with l. Transversal t₂ makes 75° with l (both angles measured on the same side, going downward).
(a) Find the angles t₁ and t₂ make with line m at their respective intersections. [1 mark]
(b) Find the angles t₁ and t₂ make with line n. [½ mark]
(c) Find ∠t₁Gt₂ — the angle between the two transversals at G on line l. [1½ marks]
(a) Find the angles t₁ and t₂ make with line m at their respective intersections. [1 mark]
(b) Find the angles t₁ and t₂ make with line n. [½ mark]
(c) Find ∠t₁Gt₂ — the angle between the two transversals at G on line l. [1½ marks]
Q14.3
Lines l ∥ m. Transversal t meets l at P and m at Q. Let the interior angle at P on the left of the transversal = 2α.
(a) Write the co-interior angle at Q in terms of α. [½ mark]
(b) The bisector of the angle 2α at P and the bisector of the co-interior angle at Q meet at a point R. Find ∠RPQ and ∠RQP in terms of α. [1 mark]
(c) Using the angle sum of a triangle, find ∠PRQ. What is special about this result? [1½ marks]
(a) Write the co-interior angle at Q in terms of α. [½ mark]
(b) The bisector of the angle 2α at P and the bisector of the co-interior angle at Q meet at a point R. Find ∠RPQ and ∠RQP in terms of α. [1 mark]
(c) Using the angle sum of a triangle, find ∠PRQ. What is special about this result? [1½ marks]
Q15.3
In the figure, AB ∥ CD. Transversal EF meets AB at P and CD at Q. ∠EPB = 130° (E is above P, B is to the right of P on line AB).
(a) Find ∠EPA. [½ mark]
(b) Find all four angles at P. State the property for each. [1 mark]
(c) Find ∠PQC and ∠PQD (C is to the right of Q, D is to the left of Q on line CD; P is above Q along the transversal). State which angle property gives each answer. [1½ marks]
(a) Find ∠EPA. [½ mark]
(b) Find all four angles at P. State the property for each. [1 mark]
(c) Find ∠PQC and ∠PQD (C is to the right of Q, D is to the left of Q on line CD; P is above Q along the transversal). State which angle property gives each answer. [1½ marks]
Section D — (5 Marks Each) [2 × 5 = 10]
Q16.5
Lines l ∥ m. Transversal t cuts l at P and m at Q. Alternate interior angles: ∠3 (lower-right interior at P) = (4x + 7)° and ∠5 (upper-left interior at Q) = (6x − 13)°.
(a) Find the value of x. [1 mark]
(b) Find all 8 angles at P and Q. Label clearly and state all properties used. [2 marks]
(c) A ray from P bisects ∠3. Find the angle this bisector makes with line l. [1 mark]
(d) This bisector meets line m at a new point R. Using the alternate interior angle property (since l ∥ m), find the angle the bisector makes with m at R. Is this the same as the angle at l? Why? [1 mark]
(a) Find the value of x. [1 mark]
(b) Find all 8 angles at P and Q. Label clearly and state all properties used. [2 marks]
(c) A ray from P bisects ∠3. Find the angle this bisector makes with line l. [1 mark]
(d) This bisector meets line m at a new point R. Using the alternate interior angle property (since l ∥ m), find the angle the bisector makes with m at R. Is this the same as the angle at l? Why? [1 mark]
Q17.5
(a) In trapezoid ABCD, AB ∥ CD. Diagonal AC is drawn (acting as a transversal). ∠BAC = 35° (the angle at A between side AB and diagonal AC). Find ∠ACD (the angle at C between diagonal CA and side CD). Name the angle property that relates ∠BAC and ∠ACD. [2 marks]
(b) In parallelogram PQRS, PQ ∥ SR. A transversal cuts PQ at X and SR at Y. ∠PXY = 65° (angle at X between XP going left and XY going down toward Y). Find ∠XYR (angle at Y between YX going up and YR going right). State the property used. [2 marks]
(c) At point P on line l, transversal t makes a 55° angle with l. A line n passing through P is perpendicular to t (i.e., n ⊥ t). Line m is parallel to l. Find the angle that n makes with line m. [1 mark]
(b) In parallelogram PQRS, PQ ∥ SR. A transversal cuts PQ at X and SR at Y. ∠PXY = 65° (angle at X between XP going left and XY going down toward Y). Find ∠XYR (angle at Y between YX going up and YR going right). State the property used. [2 marks]
(c) At point P on line l, transversal t makes a 55° angle with l. A line n passing through P is perpendicular to t (i.e., n ⊥ t). Line m is parallel to l. Find the angle that n makes with line m. [1 mark]
Bonus Question (Optional) [2 Marks]
Q18.2
★ Three parallel lines l ∥ m ∥ n are cut by a transversal. The transversal makes an angle of (2x + 5)° with line l and (x + 25)° with line m. Find x. Then find the angle the transversal makes with line n. Explain, using angle properties, why the angle with n is the same as with m.
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Answer Key & Detailed Solutions — Set 4
Q1. [1 Mark]
Yes, l ∥ m.
Property: Two lines perpendicular to the same line are parallel to each other.
Since l ⊥ n and m ⊥ n, both make 90° with n. The corresponding angles (both 90°) are equal → l ∥ m.
Property: Two lines perpendicular to the same line are parallel to each other.
Since l ⊥ n and m ⊥ n, both make 90° with n. The corresponding angles (both 90°) are equal → l ∥ m.
Q2. [1 Mark]
90°.
Corresponding angles are equal when lines are parallel. Since the transversal makes 90° with the first line, it makes 90° with the second line too.
Corresponding angles are equal when lines are parallel. Since the transversal makes 90° with the first line, it makes 90° with the second line too.
Q3. [1 Mark]
Yes.
VOA are equal, so let each = β. Supplementary means sum = 180°: β + β = 180° → β = 90°.
VOA can be supplementary when both equal 90° (the two lines are perpendicular).
VOA are equal, so let each = β. Supplementary means sum = 180°: β + β = 180° → β = 90°.
VOA can be supplementary when both equal 90° (the two lines are perpendicular).
Q4. [1 Mark]
Let each co-interior angle = θ. Equal and supplementary: θ + θ = 180° → θ = 90°.
Yes, lines are parallel because co-interior angles summing to 180° is the condition for parallel lines (90° + 90° = 180° ✓).
Yes, lines are parallel because co-interior angles summing to 180° is the condition for parallel lines (90° + 90° = 180° ✓).
Q5. [1 Mark]
75° ≠ 105°, so alternate interior angles are not equal.
No, the lines are not parallel — alternate interior angles must be equal for lines to be parallel.
No, the lines are not parallel — alternate interior angles must be equal for lines to be parallel.
Q6. [1 Mark]
True — but only in a special case.
For parallel lines, alternate interior angles are equal. If they are also supplementary (sum = 180°), then each equals 90°. This happens when the transversal is perpendicular to both parallel lines. So alternate interior angles can be supplementary when lines are parallel, provided the transversal is perpendicular.
For parallel lines, alternate interior angles are equal. If they are also supplementary (sum = 180°), then each equals 90°. This happens when the transversal is perpendicular to both parallel lines. So alternate interior angles can be supplementary when lines are parallel, provided the transversal is perpendicular.
Q7. [2 Marks]
Property: Vertically Opposite Angles are equal.
∠1 = ∠3 → x + 30 = 3x − 10
40 = 2x → x = 20 [½ mark]
∠1 = 20 + 30 = 50° [½ mark]
All four angles: [1 mark]
∠1 = ∠3 = 50° (Vertically Opposite Angles)
∠2 = 180° − 50° = 130° (Linear Pair with ∠1)
∠4 = 130° (Vertically Opposite to ∠2)
∠1 = ∠3 → x + 30 = 3x − 10
40 = 2x → x = 20 [½ mark]
∠1 = 20 + 30 = 50° [½ mark]
All four angles: [1 mark]
∠1 = ∠3 = 50° (Vertically Opposite Angles)
∠2 = 180° − 50° = 130° (Linear Pair with ∠1)
∠4 = 130° (Vertically Opposite to ∠2)
Q8. [2 Marks]
Transversal makes 55° with p.
(a) Corresponding angle at q = 55° [½ mark]
Property: Corresponding angles are equal when p ∥ q.
(b) Alternate interior angle between p and q: [½ mark]
The interior angle at p (on the side facing q) = 180° − 55° = 125° (if 55° is exterior-side).
Alternatively, the 55° angle is an interior angle on one side → its alternate interior = 55° (equal when parallel). The alternate interior angle equals 55° — same as the angle at p on that side.
(c) Yes, all corresponding angles at p, q, and r are equal (= 55°). [1 mark]
Since p ∥ q ∥ r, corresponding angles with any transversal are equal across all three parallel lines.
(a) Corresponding angle at q = 55° [½ mark]
Property: Corresponding angles are equal when p ∥ q.
(b) Alternate interior angle between p and q: [½ mark]
The interior angle at p (on the side facing q) = 180° − 55° = 125° (if 55° is exterior-side).
Alternatively, the 55° angle is an interior angle on one side → its alternate interior = 55° (equal when parallel). The alternate interior angle equals 55° — same as the angle at p on that side.
(c) Yes, all corresponding angles at p, q, and r are equal (= 55°). [1 mark]
Since p ∥ q ∥ r, corresponding angles with any transversal are equal across all three parallel lines.
Q9. [2 Marks]
Given: l ∥ m, ∠1 = 115° (upper-left exterior at upper intersection P).
(a) Alternate exterior angle ∠7: [1 mark]
∠1 (upper-left at P) and ∠7 (lower-right at Q) are on opposite sides of the transversal → alternate exterior angles.
∠7 = 115° (Alternate exterior angles are equal when lines are parallel.)
(b) Co-interior angles on the right side: [1 mark]
At P: ∠3 (lower-right interior) = ∠1 = 115° (VOA at P)
At Q: ∠6 (upper-right interior) = 180° − ∠7 = 180° − 115° = 65° (Linear Pair at Q)
Co-interior pair: ∠3 = 115° and ∠6 = 65°
Verification: 115° + 65° = 180° ✓
(a) Alternate exterior angle ∠7: [1 mark]
∠1 (upper-left at P) and ∠7 (lower-right at Q) are on opposite sides of the transversal → alternate exterior angles.
∠7 = 115° (Alternate exterior angles are equal when lines are parallel.)
(b) Co-interior angles on the right side: [1 mark]
At P: ∠3 (lower-right interior) = ∠1 = 115° (VOA at P)
At Q: ∠6 (upper-right interior) = 180° − ∠7 = 180° − 115° = 65° (Linear Pair at Q)
Co-interior pair: ∠3 = 115° and ∠6 = 65°
Verification: 115° + 65° = 180° ✓
Q10. [2 Marks]
(a) Lines parallel → corresponding angles equal: [1 mark]
3x + 15 = 5x − 25
40 = 2x
x = 20
Both angles = 3(20) + 15 = 75°
(b) If x = 18: [1 mark]
First angle = 3(18) + 15 = 54 + 15 = 69°
Second angle = 5(18) − 25 = 90 − 25 = 65°
69° ≠ 65° → Corresponding angles are not equal.
Lines are NOT parallel (corresponding angles must be equal for parallel lines).
3x + 15 = 5x − 25
40 = 2x
x = 20
Both angles = 3(20) + 15 = 75°
(b) If x = 18: [1 mark]
First angle = 3(18) + 15 = 54 + 15 = 69°
Second angle = 5(18) − 25 = 90 − 25 = 65°
69° ≠ 65° → Corresponding angles are not equal.
Lines are NOT parallel (corresponding angles must be equal for parallel lines).
Q11. [2 Marks]
Given: l ∥ m. Interior angle at P (left of transversal) = 70°.
Co-interior angle at Q = 180° − 70° = 110° (co-interior angles on parallel lines sum to 180°). [½ mark]
Let ∠RPQ = α (bisector of 70°): α = 70° ÷ 2 = 35°
∠RQP = bisector of 110°: ∠RQP = 110° ÷ 2 = 55° [½ mark]
In △PRQ:
∠PRQ + ∠RPQ + ∠RQP = 180°
∠PRQ + 35° + 55° = 180°
∠PRQ = 180° − 90° = 90° [1 mark]
∴ The bisectors of co-interior angles on parallel lines meet at right angles. ■
Co-interior angle at Q = 180° − 70° = 110° (co-interior angles on parallel lines sum to 180°). [½ mark]
Let ∠RPQ = α (bisector of 70°): α = 70° ÷ 2 = 35°
∠RQP = bisector of 110°: ∠RQP = 110° ÷ 2 = 55° [½ mark]
In △PRQ:
∠PRQ + ∠RPQ + ∠RQP = 180°
∠PRQ + 35° + 55° = 180°
∠PRQ = 180° − 90° = 90° [1 mark]
∴ The bisectors of co-interior angles on parallel lines meet at right angles. ■
Q12. [3 Marks]
(a) Testing for parallel lines: [1 mark]
For parallel lines, corresponding angles must be equal → ∠5 should equal ∠1 = 65°.
But ∠5 = 70° ≠ 65°.
The lines are NOT parallel.
Property used: Corresponding Angles Criterion (equal iff lines parallel).
(b) ∠4 at P (lower-left interior): [½ mark]
∠4 + ∠1 = 180° (Linear Pair on line at P)
∠4 = 180° − 65° = 115°
(c) Check ∠4 + ∠5: [1½ marks]
∠4 + ∠5 = 115° + 70° = 185°
This does NOT equal 180°. For parallel lines, co-interior angles must sum to 180° (Co-interior Angle Property).
Since 185° ≠ 180°, this confirms the lines are not parallel.
For parallel lines, corresponding angles must be equal → ∠5 should equal ∠1 = 65°.
But ∠5 = 70° ≠ 65°.
The lines are NOT parallel.
Property used: Corresponding Angles Criterion (equal iff lines parallel).
(b) ∠4 at P (lower-left interior): [½ mark]
∠4 + ∠1 = 180° (Linear Pair on line at P)
∠4 = 180° − 65° = 115°
(c) Check ∠4 + ∠5: [1½ marks]
∠4 + ∠5 = 115° + 70° = 185°
This does NOT equal 180°. For parallel lines, co-interior angles must sum to 180° (Co-interior Angle Property).
Since 185° ≠ 180°, this confirms the lines are not parallel.
Q13. [3 Marks]
(a) Angles with line m: [1 mark]
Since l ∥ m, corresponding angles are equal.
t₁ makes 60° with m (corresponding angle = 60°, same as at l).
t₂ makes 75° with m.
(b) Angles with line n: [½ mark]
Since l ∥ n (and m ∥ n), corresponding angles remain the same.
t₁ makes 60° with n. t₂ makes 75° with n.
(c) Angle between t₁ and t₂ at G on line l: [1½ marks]
At G on line l, t₁ makes 60° and t₂ makes 75° on the same side of l, both going downward.
The angle ∠t₁Gt₂ = 75° − 60° = 15°.
(Both transversals are on the same side of G; the angle between them is the difference of their angles with l.)
Since l ∥ m, corresponding angles are equal.
t₁ makes 60° with m (corresponding angle = 60°, same as at l).
t₂ makes 75° with m.
(b) Angles with line n: [½ mark]
Since l ∥ n (and m ∥ n), corresponding angles remain the same.
t₁ makes 60° with n. t₂ makes 75° with n.
(c) Angle between t₁ and t₂ at G on line l: [1½ marks]
At G on line l, t₁ makes 60° and t₂ makes 75° on the same side of l, both going downward.
The angle ∠t₁Gt₂ = 75° − 60° = 15°.
(Both transversals are on the same side of G; the angle between them is the difference of their angles with l.)
Q14. [3 Marks]
Given: Interior angle at P = 2α. l ∥ m.
(a) Co-interior angle at Q: [½ mark]
Co-interior angles sum to 180° (parallel lines).
Co-interior angle at Q = 180° − 2α
(b) Angles in triangle PRQ: [1 mark]
Bisector of 2α at P: ∠RPQ = 2α ÷ 2 = α
Bisector of (180° − 2α) at Q: ∠RQP = (180° − 2α) ÷ 2 = 90° − α
(c) Finding ∠PRQ: [1½ marks]
In △PRQ: ∠PRQ + ∠RPQ + ∠RQP = 180°
∠PRQ + α + (90° − α) = 180°
∠PRQ + 90° = 180°
∠PRQ = 90°
Special result: The bisectors of co-interior angles formed by a transversal on two parallel lines are always perpendicular to each other — regardless of the angle of the transversal. ■
(a) Co-interior angle at Q: [½ mark]
Co-interior angles sum to 180° (parallel lines).
Co-interior angle at Q = 180° − 2α
(b) Angles in triangle PRQ: [1 mark]
Bisector of 2α at P: ∠RPQ = 2α ÷ 2 = α
Bisector of (180° − 2α) at Q: ∠RQP = (180° − 2α) ÷ 2 = 90° − α
(c) Finding ∠PRQ: [1½ marks]
In △PRQ: ∠PRQ + ∠RPQ + ∠RQP = 180°
∠PRQ + α + (90° − α) = 180°
∠PRQ + 90° = 180°
∠PRQ = 90°
Special result: The bisectors of co-interior angles formed by a transversal on two parallel lines are always perpendicular to each other — regardless of the angle of the transversal. ■
Q15. [3 Marks]
Given: AB ∥ CD, ∠EPB = 130° (upper-right at P).
(a) ∠EPA: [½ mark]
∠EPA + ∠EPB = 180° (Linear Pair on AB)
∠EPA = 180° − 130° = 50°
(b) All four angles at P: [1 mark]
∠EPB = 130° (given) — upper-right
∠EPA = 50° (Linear Pair) — upper-left
∠FPB = 50° (Vertically Opposite to ∠EPA) — lower-right
∠FPA = 130° (Vertically Opposite to ∠EPB) — lower-left
(c) Angles at Q: [1½ marks]
Since AB ∥ CD and PQ is the transversal, P is above Q, C is to the right of Q.
Ray QP = ray QE (both point upward from Q along the transversal).
∠PQC = ∠EQC corresponds to ∠EPB (corresponding positions, both upper-right).
∠PQC = ∠EPB = 130° [Property: Corresponding Angles, AB ∥ CD] [½ mark]
∠PQD: D is to the left of Q. ∠PQD = ∠EQD corresponds to ∠EPA = 50°.
∠PQD = 50° [Property: Corresponding Angles; or Alternate Interior Angles with ∠EPA] [1 mark]
Check: ∠PQC + ∠PQD = 130° + 50° = 180° (Linear Pair on CD) ✓
(a) ∠EPA: [½ mark]
∠EPA + ∠EPB = 180° (Linear Pair on AB)
∠EPA = 180° − 130° = 50°
(b) All four angles at P: [1 mark]
∠EPB = 130° (given) — upper-right
∠EPA = 50° (Linear Pair) — upper-left
∠FPB = 50° (Vertically Opposite to ∠EPA) — lower-right
∠FPA = 130° (Vertically Opposite to ∠EPB) — lower-left
(c) Angles at Q: [1½ marks]
Since AB ∥ CD and PQ is the transversal, P is above Q, C is to the right of Q.
Ray QP = ray QE (both point upward from Q along the transversal).
∠PQC = ∠EQC corresponds to ∠EPB (corresponding positions, both upper-right).
∠PQC = ∠EPB = 130° [Property: Corresponding Angles, AB ∥ CD] [½ mark]
∠PQD: D is to the left of Q. ∠PQD = ∠EQD corresponds to ∠EPA = 50°.
∠PQD = 50° [Property: Corresponding Angles; or Alternate Interior Angles with ∠EPA] [1 mark]
Check: ∠PQC + ∠PQD = 130° + 50° = 180° (Linear Pair on CD) ✓
Q16. [5 Marks]
(a) Finding x — Alternate interior angles are equal: [1 mark]
4x + 7 = 6x − 13
20 = 2x
x = 10
∠3 = 4(10) + 7 = 47°
(b) All 8 angles: [2 marks]
At P:
∠3 = 47° (lower-right interior, given)
∠1 = 47° (Vertically Opposite to ∠3)
∠2 = 133° (Linear Pair with ∠1: 180° − 47°)
∠4 = 133° (Vertically Opposite to ∠2)
At Q (using Corresponding Angles, l ∥ m):
∠5 = 47° (corresponds to ∠1; also = ∠3 by alternate interior)
∠6 = 133° (corresponds to ∠2)
∠7 = 47° (corresponds to ∠3)
∠8 = 133° (corresponds to ∠4)
(c) Bisector of ∠3: [1 mark]
∠3 = 47°. The bisector divides it into 47° ÷ 2 = 23.5°.
∠3 is the angle between line l and transversal t at P (lower-right interior).
The bisector makes 23.5° with line l (half of ∠3).
(d) Bisector meets m at R: [1 mark]
The bisector line is a new transversal cutting l (at P) and m (at R).
Since l ∥ m, the bisector makes equal corresponding angles: 23.5° with m at R.
Yes, same angle (23.5°) — because corresponding angles are equal when lines are parallel.
4x + 7 = 6x − 13
20 = 2x
x = 10
∠3 = 4(10) + 7 = 47°
(b) All 8 angles: [2 marks]
At P:
∠3 = 47° (lower-right interior, given)
∠1 = 47° (Vertically Opposite to ∠3)
∠2 = 133° (Linear Pair with ∠1: 180° − 47°)
∠4 = 133° (Vertically Opposite to ∠2)
At Q (using Corresponding Angles, l ∥ m):
∠5 = 47° (corresponds to ∠1; also = ∠3 by alternate interior)
∠6 = 133° (corresponds to ∠2)
∠7 = 47° (corresponds to ∠3)
∠8 = 133° (corresponds to ∠4)
(c) Bisector of ∠3: [1 mark]
∠3 = 47°. The bisector divides it into 47° ÷ 2 = 23.5°.
∠3 is the angle between line l and transversal t at P (lower-right interior).
The bisector makes 23.5° with line l (half of ∠3).
(d) Bisector meets m at R: [1 mark]
The bisector line is a new transversal cutting l (at P) and m (at R).
Since l ∥ m, the bisector makes equal corresponding angles: 23.5° with m at R.
Yes, same angle (23.5°) — because corresponding angles are equal when lines are parallel.
Q17. [5 Marks]
(a) Trapezoid ABCD, AB ∥ CD, diagonal AC: [2 marks]
AC acts as a transversal cutting AB at A and CD at C.
∠BAC is the interior angle at A on the right of transversal AC (angle between AB going right and AC going diagonally to C).
∠ACD is the interior angle at C on the left of transversal CA (angle between CA going back to A and CD going right).
These are on opposite sides of the transversal → Alternate Interior Angles. [1 mark]
∴ ∠ACD = ∠BAC = 35° [1 mark]
(b) Parallelogram PQRS, transversal at X and Y: [2 marks]
∠PXY = 65°: at X, between XP (left along PQ) and XY (downward to Y).
∠XYR: at Y, between YX (upward to X) and YR (right along SR).
∠PXY is interior upper-left at X; ∠XYR is interior lower-right at Y → opposite sides of transversal → Alternate Interior Angles. [1 mark]
Since PQ ∥ SR: ∠XYR = ∠PXY = 65° [1 mark]
(c) Line n ⊥ t at P, l ∥ m: [1 mark]
At P: transversal t makes 55° with line l.
Line n ⊥ t → angle between n and t = 90°.
Angle between n and l = 90° − 55° = 35°.
Since l ∥ m, line n makes equal corresponding angles: 35° with line m.
AC acts as a transversal cutting AB at A and CD at C.
∠BAC is the interior angle at A on the right of transversal AC (angle between AB going right and AC going diagonally to C).
∠ACD is the interior angle at C on the left of transversal CA (angle between CA going back to A and CD going right).
These are on opposite sides of the transversal → Alternate Interior Angles. [1 mark]
∴ ∠ACD = ∠BAC = 35° [1 mark]
(b) Parallelogram PQRS, transversal at X and Y: [2 marks]
∠PXY = 65°: at X, between XP (left along PQ) and XY (downward to Y).
∠XYR: at Y, between YX (upward to X) and YR (right along SR).
∠PXY is interior upper-left at X; ∠XYR is interior lower-right at Y → opposite sides of transversal → Alternate Interior Angles. [1 mark]
Since PQ ∥ SR: ∠XYR = ∠PXY = 65° [1 mark]
(c) Line n ⊥ t at P, l ∥ m: [1 mark]
At P: transversal t makes 55° with line l.
Line n ⊥ t → angle between n and t = 90°.
Angle between n and l = 90° − 55° = 35°.
Since l ∥ m, line n makes equal corresponding angles: 35° with line m.
Q18. Bonus [2 Marks]
Since l ∥ m, corresponding angles are equal: [½ mark]
2x + 5 = x + 25
x = 20 [½ mark]
Angle with l = 2(20) + 5 = 45°
Angle with m = 20 + 25 = 45° ✓ [½ mark]
Angle with n: Since m ∥ n, corresponding angles are equal → angle with n = 45°.
Explanation: Because all three lines are parallel (l ∥ m ∥ n), a single transversal makes equal corresponding angles with all of them. The angle the transversal makes does not change as it crosses parallel lines. [½ mark]
2x + 5 = x + 25
x = 20 [½ mark]
Angle with l = 2(20) + 5 = 45°
Angle with m = 20 + 25 = 45° ✓ [½ mark]
Angle with n: Since m ∥ n, corresponding angles are equal → angle with n = 45°.
Explanation: Because all three lines are parallel (l ∥ m ∥ n), a single transversal makes equal corresponding angles with all of them. The angle the transversal makes does not change as it crosses parallel lines. [½ mark]