CBSE Class 7 Mathematics Parallel and Intersecting Lines Practice Paper Worksheet Set 02 with Answers

CBSE Class 7 Maths Ganita Prakash Chapter 5 Parallel and Intersecting Lines Practice Paper 13 is a 40-mark test with 20 questions in Sections A to E, and every question below has a step-by-step answer. It practises vertically opposite angles and linear pairs, the angles made when a transversal cuts two lines (corresponding, alternate and interior angles), the test for parallel lines, angles in a parallelogram, and multi-step angle chasing with extra parallel lines in a railway-track case study.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. In the adjoining figure, if ∠a = 115°, find ∠b, ∠c and ∠d.

Answer:
∠a and ∠b lie on the straight line l and together make a linear pair, so ∠a + ∠b = 180°.
∠b = 180° − 115° = 65°.
∠c is vertically opposite to ∠a, so ∠c = 115°.
∠d is vertically opposite to ∠b, so ∠d = 65°.
Check: 115° + 65° + 115° + 65° = 360°, the full turn round the point.
In simple words: The angle next to 115° is 65°, and opposite angles copy each other: ∠b = 65°, ∠c = 115°, ∠d = 65°.

Teacher's Note:
1. Find one neighbour by the linear pair rule, then use vertically opposite angles for the rest.
2. The four angles at a point always add up to 360°.

Question 12. A transversal intersects two lines such that the corresponding angles are 78° each. Are the lines parallel? Why?
Answer:
The corresponding angles made by the transversal are equal (78° = 78°).
If a transversal makes equal corresponding angles with two lines, then the two lines are parallel.
So yes, the lines are parallel, because their corresponding angles are equal.
In simple words: Equal corresponding angles prove that the two lines are parallel.

Teacher's Note:
1. This is the converse of the corresponding angles property.
2. Always give the reason, not just 'yes'.

Question 13. Two complementary angles are such that one angle is 30° less than twice the other. Find the measure of the larger angle.
Answer:
Let the smaller angle be x°. Then the other angle is (2x − 30)°.
Complementary angles add up to 90°: x + (2x − 30) = 90.
3x − 30 = 90, so 3x = 120 and x = 40.
Other angle = 2 × 40 − 30 = 50.
The two angles are 40° and 50°, so the larger angle is 50°.
Check: 40° + 50° = 90° and 2 × 40° − 30° = 50°.
In simple words: The angles are 40° and 50°; the larger one is 50°.

Teacher's Note:
1. Complementary means the sum is 90°.
2. Always check both conditions with your answer.

Question 14. Draw two lines intersecting at a point such that one angle is 35°. Find the remaining three angles.
Answer:
Draw a line AB. At a point O on it, use a protractor to draw ∠AOC = 35° and extend CO to D, so that the lines AB and CD cross at O.

∠COB and ∠AOC form a linear pair, so ∠COB = 180° − 35° = 145°.
∠BOD is vertically opposite to ∠AOC, so ∠BOD = 35°.
∠AOD is vertically opposite to ∠COB, so ∠AOD = 145°.
The remaining three angles are 145°, 35° and 145°.
In simple words: Opposite angles are equal and neighbouring angles add to 180°, so the angles are 35°, 145°, 35°, 145°.

Teacher's Note:
1. Two crossing lines give two pairs of equal angles.
2. Neighbouring angles always add to 180°.

Section C: Short Answer Questions (3 Marks Each)

Question 15. In the figure below, l ∥ m and t is a transversal. If ∠6 = 135°, find the measures of ∠1, ∠2, ∠3, ∠4, ∠5, ∠7 and ∠8.

Answer:
∠8 is vertically opposite to ∠6, so ∠8 = 135°.
∠5 and ∠6 form a linear pair, so ∠5 = 180° − 135° = 45°; ∠7 is vertically opposite to ∠5, so ∠7 = 45°.
Since l ∥ m, corresponding angles are equal: ∠2 = ∠6 = 135°, ∠4 = ∠8 = 135°, ∠1 = ∠5 = 45° and ∠3 = ∠7 = 45°.
∠1 = 45°, ∠2 = 135°, ∠3 = 45°, ∠4 = 135°, ∠5 = 45°, ∠7 = 45°, ∠8 = 135°.
In simple words: With parallel lines, the angles are either 135° (the wide ones) or 45° (the narrow ones).

Teacher's Note:
1. The small arrows on l and m mark them as parallel.
2. In such figures, all the obtuse angles are equal and all the acute angles are equal.
3. Any obtuse angle + any acute angle = 180°.

Question 16. In the figure below, ∠a = 120° and ∠f = 70°. Are l and m parallel? Give reason.

Answer:
∠a and ∠b form a linear pair on line l, so ∠b = 180° − 120° = 60°.
∠b (at l) and ∠f (at m) are corresponding angles: both are above their line and to the right of t.
If l ∥ m, these corresponding angles would be equal. But ∠b = 60° and ∠f = 70°, which are not equal.
So l and m are not parallel, because the corresponding angles ∠b and ∠f are unequal.
In simple words: ∠b = 60° but ∠f = 70°; corresponding angles are not equal, so the lines are not parallel.

Teacher's Note:
1. First find the angle at l that matches ∠f.
2. Unequal corresponding angles always mean the lines are not parallel.
3. In the figure, m tilts and will meet l far to the left.

Question 17. In the figure below, AB ∥ CD ∥ EF and EA ⟂ AB. If ∠BEF = 55°, find x and y.

Answer:
Line BE is a transversal of the parallel lines CD and EF.
∠y (at D) and ∠BEF = 55° (at E) lie between CD and EF on the same side of the transversal, so they are interior angles on the same side: y + 55 = 180.
y = 125, so y° = 125°.
∠x (at B) and ∠y (at D) are corresponding angles for the parallel lines AB and CD, so x = y = 125.
Answer: x = 125°, y = 125°.
In simple words: y = 180° − 55° = 125°, and x equals y because they are corresponding angles.

Teacher's Note:
1. The source answer key gave 145°; it wrongly subtracted 35° instead of 55°.
2. EA ⟂ AB is extra information and is not needed here.
3. Same-side interior angles add to 180°; corresponding angles are equal.

Section D: Long Answer Question (5 Marks)

Question 18. In the figure below, ∠ABC = 45° and ∠IKJ = 78°. Find ∠GEH, ∠HEF and ∠FED with reasoning.

Answer:
The arrow marks show that the lines IA and GD are parallel. Lines CH and JF are transversals that meet at E on GD.
Lines IA and CH cross at B, so ∠KBE = ∠ABC = 45° (vertically opposite angles).
Lines IA and JF cross at K, so ∠BKE = ∠IKJ = 78° (vertically opposite angles).
∠GEH and ∠KBE are corresponding angles (IA ∥ GD, transversal CH), so ∠GEH = 45°.
∠FED and ∠BKE are corresponding angles (IA ∥ GD, transversal JF), so ∠FED = 78°.
G, E, D lie on a straight line, so ∠GEH + ∠HEF + ∠FED = 180°.
∠HEF = 180° − 45° − 78° = 57°.
In simple words: ∠GEH = 45°, ∠FED = 78° (corresponding angles), and the middle angle ∠HEF = 180° − 45° − 78° = 57°.

Teacher's Note:
1. First move the given angles across the crossing with vertically opposite angles.
2. Then move them down to E with corresponding angles.
3. Angles on a straight line add up to 180°.

Section E: Case Study Based Questions (4 Marks Each)

Question 19. Case Study 1: To ease rush-hour congestion, city planners are designing a new flyover over two major, parallel avenues, Elm Street and Oak Avenue. A connector road is being added to link them, acting as a transversal line. For smooth traffic flow and driver safety, the engineers must measure the corresponding angles where the connector road meets the main avenues and make sure they are equal, so that the transition between the parallel roads is as smooth as possible. In the figure, l and m are the two avenues and t is the connector road.
(i) Which angles are corresponding angles in the above figure?
(ii) If ∠a = 65°, what is ∠b?
(iii) Why must ∠a = ∠b for parallelism?
(iv) Write one real-life importance of parallel roads in city planning.
Answer:
(i) ∠a and ∠b are corresponding angles: both lie above their avenue and to the right of the connector road t.
(ii) l ∥ m, so corresponding angles are equal: ∠b = ∠a = 65°.
(iii) Two lines cut by a transversal are parallel exactly when their corresponding angles are equal. If ∠a ≠ ∠b, the two avenues would not be parallel and would meet somewhere.
(iv) Parallel roads keep the same distance apart all the way, so lanes stay of equal width, roads never cross by accident and traffic flows smoothly (any one valid reason).
In simple words: Corresponding angles a and b are equal (65°) because the avenues are parallel.

Teacher's Note:
1. Corresponding angles sit in matching corners at the two crossings.
2. Equal corresponding angles are the test for parallel lines.
3. The aerial photo of the flyover is not needed to answer.

Question 20. Case Study 2: At the busy New Delhi Railway Junction, two main tracks, LM and PQ, run perfectly parallel. A new high-speed connecting track MNOP cuts across them, creating the turning points N and O where engineers must manage traffic flow. In the first figure ∠LMN = 40°, ∠MNO = 96°, ∠OPQ = 52° and ∠NOP = a°. To test safety and signalling, the head engineer sketched two virtual test lines: line RS through point N, parallel to LM, and line TU through point O, parallel to PQ (second figure).
(i) If ∠NOP = a°, what type of angle relation must be checked to confirm the parallelism of the tracks?
(ii) Find the angle x°.
(iii) Find the angle z°.
(iv) Find the angle a°.
Answer:
(i) Alternate angles: each crossing track acts as a transversal, and the tracks (and test lines) are parallel only if the alternate angles are equal, that is ∠LMN = ∠MNS, ∠SNO = ∠NOT and ∠TOP = ∠OPQ (equivalently, interior angles on the same side add up to 180°).
(ii) RS ∥ LM and MN is a transversal, so w° = ∠MNS = ∠LMN = 40° (alternate angles). Since w° + x° = ∠MNO = 96°, x° = 96° − 40° = 56°.
(iii) TU ∥ PQ and OP is a transversal, so z° = ∠TOP = ∠OPQ = 52° (alternate angles).
(iv) RS ∥ TU (both are parallel to the parallel tracks) and NO is a transversal, so y° = ∠NOT = ∠SNO = x° = 56° (alternate angles).
a° = y° + z° = 56° + 52° = 108°.
In simple words: x = 96 − 40 = 56, z = 52, y = x = 56, so a = 56 + 52 = 108.

Teacher's Note:
1. Drawing a helper line parallel to the tracks through each bend splits a bend angle into two alternate angles.
2. The source key wrote w° = 56° by mistake; w° is 40°.
3. The photo of the junction is only for context.