RS Aggarwal Class 7 Mathematics Solutions Chapter 14 Properties of Parallel Lines

Question 1. Given: l || m, t is a transversal. ∠5 = 70°. Find all the angles.
Answer: Since l || m and t is a transversal with ∠5 = 70°, we use the properties of parallel lines cut by a transversal.

∠5 = ∠3 = 70° (alternate interior angles)
∠5 + ∠8 = 180° (linear pair)
or 70° + ∠8 = 180°
∠8 = 110°

∠1 = ∠3 = 70° (vertically opposite angles)
∠3 + ∠4 = 180° (linear pair)
or ∠4 = 180° - ∠3 = 180° - 70° = 110°

Therefore: ∠1 = 70°, ∠2 = 110°, ∠3 = 70°, ∠4 = 110°, ∠5 = 70°, ∠6 = 110°, ∠7 = 70°, ∠8 = 110°
In simple words: When two parallel lines are cut by a line called a transversal, angles in matching positions are equal. Also, angles on a straight line add up to 180°.

Exam Tip: Always mark the given angle first, then use alternate interior angles and linear pair properties systematically to find all eight angles.

Question 2. Given: l || m, t is a transversal. ∠1 : ∠2 = 5 : 7. Find all the angles.
Answer: Let the angles measure 5x and 7x respectively. Since ∠1 and ∠2 form a linear pair:
∠1 + ∠2 = 180°
5x + 7x = 180°
12x = 180°
x = 15

Therefore: ∠1 = 5(15) = 75° and ∠2 = 7(15) = 105°

Using parallel line properties:
∠2 + ∠3 = 180° (linear pair)
∠3 = 180° - 105° = 75°

∠3 + ∠6 = 180° (interior angles on the same side of transversal are supplementary)
∠6 = 180° - ∠3 = 105°

and ∠6 = ∠8 = 105° (vertically opposite angles)

Therefore: ∠1 = 75°, ∠2 = 105°, ∠3 = 75°, ∠8 = 105°
In simple words: Set up the angle ratio as 5x and 7x. Since they sit on a straight line, they must sum to 180°. Solve for x, then use parallel line angle relationships to find all remaining angles.

Exam Tip: When angles are given as a ratio, always assume they form a linear pair (unless stated otherwise) and set their sum equal to 180°.

Question 3. Given: l || m, t is a transversal. ∠1 = (2x - 8)°, ∠2 = (3x - 7)°. Find x and the angles.
Answer: We know that consecutive interior angles are supplementary, so:
∠1 + ∠2 = 180°
(2x - 8) + (3x - 7) = 180°
5x - 15 = 180°
5x = 195
x = 39

Therefore: ∠1 = (2 × 39 - 8) = (78 - 8) = 70° and ∠2 = (3 × 39 - 7) = (117 - 7) = 110°
In simple words: When two parallel lines are cut by a transversal, the angles on the same side (inside) add to 180°. Combine the two expressions, solve for x, then substitute back to find each angle.

Exam Tip: Always verify your value of x by substituting back into both angle expressions and checking that they sum to 180°.

Question 4. Given: Two parallel lines cut by two different transversals. ∠1 = 50° (corresponding angles), ∠2 = 65° (corresponding angles). Find x and y.
Answer: From the given figure:

∠1 = ∠3 = 50° (corresponding angles)
and ∠1 + x° = 180° (linear pair)
or x° = 180° - 50° = 130°
or x = 130

∠2 = ∠4 = 65° (corresponding angles)
and ∠2 + y° = 180° (linear pair)
or y° = 180° - 65° = 115°
or y = 115
In simple words: Corresponding angles are angles in the same position at each intersection. They are equal when lines are parallel. Angles on a straight line sum to 180°, so subtract the given angle from 180° to find the supplementary angle.

Exam Tip: Identify which angles are corresponding (same position at different intersections) and which are supplementary (on the same straight line).

Question 5. Given: ∠B = 65°, ∠C = 45°, DAE || BC. Find x and y.
Answer: The given lines are parallel. When AB is taken as the transversal:
x° = ∠B = 65° (alternate angles)

When AC is taken as the transversal:
y° = ∠C = 45° (alternate angles)

Therefore: x = 65 and y = 45
In simple words: When a line crosses two parallel lines, it acts as a transversal. Alternate angles (on opposite sides of the transversal and inside the parallel lines) are always equal, so x matches angle B and y matches angle C.

Exam Tip: In problems involving a triangle and a line parallel to one side, always use alternate angle properties with each side of the triangle acting as a transversal.

Question 6. Given: CE || BA, ∠BAC = 80°, ∠ECD = 35°. Find ∠ABC and ∠ACB.
Answer: Given: CE || BA, ∠BAC = 80°, ∠ECD = 35°

(i) ∠BAC = ∠ACE = 80° (alternate angles with AC as transversal)

(ii) ∠ACB + ∠ACD = 180° (linear pair)
or ∠ACB + ∠ACE + ∠ECD = 180°
or ∠ACB + 80° + 35° = 180°
or ∠ACB = 65°

(iii) In triangle ABC:
∠BAC + ∠ACB + ∠ABC = 180° (angle sum property)
80° + 65° + ∠ABC = 180°
∠ABC = 35°
In simple words: Since CE is parallel to BA, angle ACE equals angle BAC by the alternate angle rule. Then use the straight line property to find angle ACB. Finally, apply the triangle angle sum rule to get angle ABC.

Exam Tip: Always apply alternate angle properties first when two lines are parallel, then use linear pair and triangle angle sum properties in sequence.

Question 7. Given: From the figure, find ∠ABC.
Answer: From the figure, we observe that line BD extended is parallel to line AC. Using the properties of parallel lines, alternate interior angles are equal. The angle that line AC makes with line AB equals the angle that the extended line BD makes with AB. Through calculation of the angles shown and applying the angle sum property of triangles, we determine ∠ABC = 35°.
In simple words: When a line is parallel to one side of a triangle, the angle it forms with another side equals an alternate angle. Use this to find the unknown angle in the triangle.

Exam Tip: Look for parallel lines in geometry problems and use alternate interior angles to connect unknown angles to known ones.

Question 8. Given: AO || CD, OB || CE, ∠AOB = 50°, ∠CDO = 40°. Find ∠BOD.
Answer: Given: AO || CD, OB || CE, ∠AOB = 50°, ∠CDO = 40°

Since AO || CD and OB is the transversal:
∠AOB = ∠CDB = 50° (when AO || CD and OB is transversal)

Since CD || OB and CD is the transversal:
∠FOD = ∠ODC = 40° (when CD || OB and OD is transversal)

Therefore: ∠BOD = ∠BOF + ∠FOD = 50° + 40° = 90°
In simple words: Draw a line through O parallel to both CD and AB to help break down the angle. Apply alternate angle properties twice - once for each pair of parallel lines - then add the two resulting angles together.

Exam Tip: When multiple pairs of parallel lines are given, construct an auxiliary line parallel to all of them to simplify the angle relationships.

Question 9. Given: AB || CD. GL and HM are angle bisectors of ∠AGH and ∠GHD, respectively. Show that GL || HM.
Answer: Given: AB || CD

GL and HM are angle bisectors of ∠AGH and ∠GHD respectively.

Since AB || CD, we know that ∠AGH = ∠GHD (alternate angles)

or \( \frac{1}{2} \angle AGH = \frac{1}{2} \angle GHD \)

or ∠LGH = ∠GHM (given)

Therefore, GL || HM as we know that if the angles of any pair of alternate interior angles are equal, then the lines are parallel.
In simple words: Two parallel lines make equal alternate angles with a transversal. When you bisect (cut in half) those equal angles, the bisectors themselves become alternate angles that are also equal. Equal alternate angles mean the bisectors must be parallel to each other.

Exam Tip: To prove two lines are parallel, always show that a pair of alternate interior angles formed by them are equal.

Question 10. Given: AB || CD. ∠ABE = 120°, ∠ECD = 100°, ∠BEC = x°. Find x.
Answer: Given: AB || CD, ∠ABE = 120°, ∠ECD = 100°, ∠BEC = x°

Construction: FEG || AB

Now, since AB || FEG and AB || CD, FEG || AB || CD

∠ABE = ∠BEG = 120° (alternate angles, when AB || FEG and BE is transversal)

∠FOD = ∠ODC = 40° (alternate angles, when CD || FEG and EG is transversal)

∠BOD = ∠BOF + ∠FOD = 50° + 40° = 90°
In simple words: Draw a helper line through E parallel to both AB and CD. This splits the angle at E into two parts. Use alternate angle properties to find each part, then add them to get the total angle BEC.

Exam Tip: When finding an angle between two rays from a point, and parallel lines are involved, construct an auxiliary line through that point parallel to the given parallel lines.

Question 11. Given: AB || CD, AD || BC. Prove: ∠1 = ∠3.
Answer: Given: AB || CD, AD || BC

∠1 + ∠2 = 180° (AB || CD and AD is the transversal) ... (i)
∠2 + ∠3 = 180° (AD || BC and AB is the transversal) ... (ii)

From (i) and (ii):
∠1 + ∠2 = 180° = ∠2 + ∠3
∠1 = ∠3

Therefore, ∠ADC = ∠ABC
In simple words: When you have two pairs of parallel lines, angles formed at one corner relate to angles at another corner through linear pairs. Since both pairs equal 180°, the angles on opposite corners must be equal.

Exam Tip: In a parallelogram (where opposite sides are parallel), always use linear pair and transversal properties to connect angles at different corners.

Question 12. Given: The figure shows a quadrilateral ABCD with AB || CD and AD || BC. Using the properties of parallel lines, establish angle relationships.
Answer: When AB || CD with transversal AD, we have ∠DAB + ∠ADC = 180° (co-interior angles).

Similarly, when AD || BC with transversal AB, we have ∠DAB + ∠ABC = 180° (co-interior angles).

This shows that ∠ADC = ∠ABC. Likewise, applying the same logic to the other pair of sides proves that ∠DAB = ∠BCD.

In quadrilateral ABCD where both pairs of opposite sides are parallel, opposite angles are equal and consecutive angles are supplementary.
In simple words: In a quadrilateral where opposite sides are parallel (a parallelogram), opposite angles must be equal because consecutive angles always sum to 180°.

Exam Tip: Remember that in any parallelogram, opposite angles are equal and adjacent angles are supplementary - this is a fundamental property that speeds up many geometry proofs.

Question 13. Given: l || m, p || q. ∠1 = 65°. Find all the angles.
Answer: Given: l || m, p || q, ∠1 = 65°

∠1 = ∠a = 65° (vertically opposite angles)
∠a + ∠d = 180° (consecutive interior angles on the same side of a transversal are supplementary)
or ∠d = 180° - 65° = 115°

∠c + ∠d = 180° (consecutive interior angles on the same side of a transversal are supplementary)
or ∠c = 180° - 115° = 65°

∠c + ∠b = 180° (consecutive interior angles on the same side of a transversal are supplementary)
or ∠b = 180° - 65° = 115°

Therefore: ∠a = 65°, ∠b = 115°, ∠c = 65°, ∠d = 115°
In simple words: When two pairs of parallel lines intersect, they create a pattern. The vertically opposite angles are equal, and angles on a straight line sum to 180°. Use these two rules to find all angles.

Exam Tip: In a grid formed by two pairs of parallel lines, the angles repeat in a regular pattern - typically alternating between two values that are supplementary.

Question 14. Given: AB || DC, AD || BC. ∠BAC = 35°, ∠CAD = 40°. Find x and y.
Answer: Given: AB || DC, AD || BC, ∠BAC = 35°, ∠CAD = 40°

Since AB || DC with transversal AC:
∠BAC = y = 35° (alternate angles when AB || DC)
∠CAD = x = 40° (alternate angles when AD || BC)

Therefore: x = 40, y = 35
In simple words: In a parallelogram (opposite sides parallel), a diagonal creates alternate angles with each pair of parallel sides. The angle the diagonal makes with one side equals the angle it makes with the opposite side.

Exam Tip: When a diagonal is drawn in a parallelogram, use alternate angle properties with each pair of parallel sides to find the angles the diagonal makes.

Question 15. Given: AB || CD. ∠BAE = 125°. Find ∠AED and ∠CAB + ∠BAE.
Answer: Given: AB || CD, ∠BAE = 125°

(i) ∠1 + ∠2 = 180° (linear pair)
or 130° + ∠2 = 180°
or ∠2 = 50° \( \therefore \) 40° = ∠3

∠2 = 50° \( \therefore \) ∠2 ≠ ∠3

(ii) ∠2 + ∠3 = 180° (linear pair)
35° + ∠3 = 180°
∠3 = 145° = 145° = ∠1

(iii) ∠2 + ∠3 = 180° (linear pair)
∠3 = 180° - 125° = 55°
∠3 = 55° \( \therefore \) 60° = ∠1

∠2 = 180° - ∠1 = 180° - 55° = 125°
In simple words: Use the properties that angles on a straight line sum to 180° and apply parallel line rules with alternate angles to find each unknown angle systematically.

Exam Tip: Always work through such multi-part problems step by step, identifying linear pairs and alternate angles at each stage to avoid calculation errors.