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

This is Practice Paper 14 for CBSE Class 7 Maths Ganita Prakash Chapter 5, Parallel and Intersecting Lines: a 40-mark paper of 20 questions in Sections A to E, with a worked answer for each one. Many questions here turn angle facts into small equations, using ratios of complementary and supplementary angles, linear pairs written as expressions in x, and equal corresponding or alternate angles. It also covers finding all eight angles with two parallel lines, angles inside a triangle drawn between parallel lines, and two case studies on a traffic junction and railway tracks.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. Two supplementary angles are such that one angle is 60° more than half the other. Find the measure of the two angles.
Answer:
Let one angle be y°. The other angle is (\( \frac{y}{2} \) + 60)°.
Supplementary angles add up to 180°: y + \( \frac{y}{2} \) + 60 = 180.
\( \frac{3y}{2} \) = 120, so 3y = 240 and y = 80.
Other angle = 40 + 60 = 100.
The two angles are 80° and 100°.
Check: 80° + 100° = 180°, and half of 80° plus 60° is 100°.
In simple words: The angles are 80° and 100°.

Teacher's Note:
1. Supplementary means the sum is 180°.
2. Read carefully: '60° more than half the other'.

Question 12. Two intersecting lines form vertically opposite angles (3x + 10)° and (4x − 30)°. Find the measure of the adjacent angle (the one forming a linear pair).
Answer:
Vertically opposite angles are equal: 3x + 10 = 4x − 30.
10 + 30 = 4x − 3x, so x = 40.
Each of these angles = 3 × 40 + 10 = 130°.
The adjacent angle makes a linear pair with it: 180° − 130° = 50°.
In simple words: Opposite angles are 130° each, so the angle next to them is 50°.

Teacher's Note:
1. First find x, then the angle, then its supplement.
2. The answer is not x; x = 40 is only a number in the expression.

Question 13. Find the measure of an angle that is equal to half its own complement.
Answer:
Let the angle be x°. Its complement is (90 − x)°.
x = \( \frac{1}{2} \) × (90 − x), so 2x = 90 − x.
3x = 90, so x = 30.
The angle is 30°. Check: its complement is 60° and half of 60° is 30°.
In simple words: The angle is 30° because its complement 60° is twice as big.

Teacher's Note:
1. Complement of x is 90 − x.
2. Clear the fraction by multiplying both sides by 2.

Question 14. In the figure below, lines l and m intersect. If ∠a and ∠b are in the ratio 1 : 2, find the measures of ∠b and ∠c.

Answer:
∠a and ∠b form a linear pair, so ∠a + ∠b = 180°.
Let ∠a = x and ∠b = 2x: x + 2x = 180, so 3x = 180 and x = 60.
∠b = 2 × 60° = 120°.
∠c is vertically opposite to ∠a, so ∠c = 60°.
In simple words: ∠a = 60°, so ∠b = 120° and ∠c = ∠a = 60°.

Teacher's Note:
1. Ratio 1 : 2 means 3 equal parts of 180°.
2. The figure is drawn to these measures, so ∠a looks acute.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Find the unknown angles in the given figure if p ∥ q.

Answer:
e and 120° form a linear pair on line p: e = 180° − 120° = 60°.
f is vertically opposite to e: f = 60°.
a corresponds to e (both are above the transversal and to the left of their parallel line): a = 60°.
c is vertically opposite to a: c = 60°.
d corresponds to the 120° angle: d = 120°; b is vertically opposite to d: b = 120°.
So a = c = e = f = 60° and b = d = 120°.
In simple words: All the acute angles are 60° and all the obtuse angles are 120°.

Teacher's Note:
1. Start with the linear pair next to the given angle.
2. Then copy angles across with vertically opposite and corresponding angles.

Question 16. A transversal cuts two parallel lines l and m. A pair of alternate interior angles are given by (x + 20)° and (3x − 40)°. Find the value of x and the measure of the interior angle on the same side of the transversal as (x + 20)°.
Answer:
Alternate interior angles made with parallel lines are equal: x + 20 = 3x − 40.
20 + 40 = 3x − x, so 2x = 60 and x = 30.
The angle is x + 20 = 50°.
Interior angles on the same side add up to 180°, so the required angle = 180° − 50° = 130°.
In simple words: x = 30, the alternate angles are 50°, and the co-interior angle is 130°.

Teacher's Note:
1. 'Consecutive interior angle' in the original means the interior angle on the same side.
2. Check: 3 × 30 − 40 = 50, the same as x + 20.

Question 17. If AB ∥ CD, ∠APQ = 70° and ∠PRD = 130°, find ∠QPR, ∠BPR and ∠PQR.

Answer:
AB ∥ CD and PR is a transversal, so ∠APR = ∠PRD = 130° (alternate interior angles).
∠APR = ∠APQ + ∠QPR, so ∠QPR = 130° − 70° = 60°.
A, P, B lie on a straight line: ∠APQ + ∠QPR + ∠BPR = 180°, so ∠BPR = 180° − 70° − 60° = 50°.
AB ∥ CD and PQ is a transversal, so ∠PQR = ∠APQ = 70° (alternate interior angles).
Check: in triangle PQR, 60° + 70° + 50° = 180°.
In simple words: ∠QPR = 60°, ∠BPR = 50° and ∠PQR = 70°.

Teacher's Note:
1. Look for Z-shapes between the parallel lines to spot alternate angles.
2. Angles on the straight line AB add up to 180°.

Section D: Long Answer Question (5 Marks)

Question 18. In the figures below, the lines marked with arrows are parallel. What angles do x and y stand for?

Answer:
Figure (i): The vertical line is perpendicular to the top line (right-angle mark). The two horizontal lines are parallel, so it is perpendicular to the bottom line too: the angle between the vertical line and the bottom line is 90°.
At the lower crossing, the angles on the upper side of the bottom line are 65° and the angle between the slant line and the right part of the bottom line. These two make 90°, so that angle = 90° − 65° = 25°.
x° is vertically opposite to this 25° angle, so x = 25.
The slant line is a transversal of the two parallel lines. At the top line, the angle between the slant line (going down-left) and the left part of the top line is 25°, because it is an alternate angle to the 25° angle at the bottom line.
y° is the angle below the top line between the slant line and the right part of the top line, so it makes a linear pair with that 25° angle: y° = 180° − 25° = 155°.
So y = 155 (equivalently y° = 90° + 65°, alternate angles).
Figure (ii): The two slant lines meet at a point on the top line. Line t makes 53° and line s makes 78° with the bottom line.
By corresponding angles, at the top line t makes 53° and s makes 78° with the right part of the top line.
The angle between the two lines there = 78° − 53° = 25°, and x° is vertically opposite to it, so x = 25.
Answer: in (i) x = 25°, y = 155°; in (ii) x = 25°.
In simple words: In (i) x = 90 − 65 = 25 and y = 180 − 25 = 155; in (ii) x = 78 − 53 = 25.

Teacher's Note:
1. A line perpendicular to one of two parallel lines is perpendicular to the other too.
2. Move angles from one parallel line to the other with corresponding angles, then subtract.
3. The words saying which lines are parallel were added; the original only showed arrow marks.

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

Question 19. Case Study 1: Traffic Junction and Signals At a busy traffic junction, two parallel roads l and m are crossed by a slant road t (a transversal). The traffic police place digital signals at the junction angles. One of the angles formed is 65°. The design ensures equal alternate interior angles for safe road crossing.
(i) Identify the type of the pair ∠3 and ∠7 in the figure.
(ii) If ∠3 = 65°, find ∠7.
(iii) What is the vertically opposite angle to ∠3?
(iv) An angle measures 35°. What is the measure of the supplement of its complement?
Answer:
(i) ∠3 and ∠7 are both below their road and to the right of t, so they are corresponding angles.
(ii) The roads are parallel, so corresponding angles are equal: ∠7 = ∠3 = 65°.
(iii) ∠1 is opposite to ∠3 at the same crossing, so the vertically opposite angle is ∠1 (also 65°).
(iv) Complement of 35° = 90° − 35° = 55°; supplement of 55° = 180° − 55° = 125°.
In simple words: ∠3 and ∠7 are corresponding, so ∠7 = 65°; ∠1 is opposite ∠3; the answer to (iv) is 125°.

Teacher's Note:
1. Corresponding angles sit in the same corner at each crossing.
2. The figure is drawn with ∠3 = 65°.
3. The traffic photo is decorative and not needed.

Question 20. Case Study 2: Railway Tracks and Ladder Two parallel railway tracks l and m are shown with a ladder t crossing them (a transversal). Angles ∠f and ∠d are formed at the crossing points. An engineer checks if ∠f = ∠d to confirm that the tracks are parallel. He measures ∠f = 120°.
(i) Name the type of angles ∠f and ∠d.
(ii) Find ∠d if ∠f = 120°.
(iii) Which property ensures the tracks are parallel?
(iv) A certain angle is 14° more than its own complement. Find the measure of the angle.
Answer:
(i) ∠d (below l, left of t) and ∠f (above m, right of t) are between the tracks on opposite sides of the ladder, so they are alternate interior angles.
(ii) For parallel lines, alternate interior angles are equal: ∠d = ∠f = 120°.
(iii) If a transversal makes equal alternate interior angles with two lines, the lines are parallel. Since ∠f = ∠d, the tracks are parallel.
(iv) Let the angle be x°. Then x = (90 − x) + 14, so 2x = 104 and x = 52° (its complement is 38°, and 52 − 38 = 14).
In simple words: ∠d and ∠f are alternate interior angles, so ∠d = 120°; the angle in (iv) is 52°.

Teacher's Note:
1. Alternate interior angles form a Z-shape between the lines.
2. Equal alternate angles are a test for parallel lines.
3. The railway photo is decorative.