Chapter 4: Lines and Angles (Exercise 4.2

Chapter 4: Lines and Angles (Exercise 4.2)

Question 1:
In the given figure, find the values of x and y and then show that AB ∥ CD.

Solution:
* On the straight line (transversal):
50° + x = 180° (Linear pair)
x = 180° - 50° = 130°

* For lines CD and transversal:
y = 130° (Vertically opposite angles)

* Here, we observe that x = y = 130°.
* Since the alternate interior angles are equal, the line AB is parallel to CD (AB ∥ CD).
(Hence proved.)

Question 2:
In the given figure, if AB ∥ CD, CD ∥ EF and y : z = 3 : 7, find the value of x.

Solution:
* Given: AB ∥ CD and CD ∥ EF ⇒ AB ∥ EF.
* Since AB ∥ EF, alternate interior angles are equal:
x = z --- (Equation 1)

* Since AB ∥ CD, interior angles on the same side of the transversal sum to 180°:
x + y = 180°
z + y = 180° (Substituting x = z)

* Given ratio y : z = 3 : 7. Let y = 3k and z = 7k:
3k + 7k = 180°
10k = 180° ⇒ k = 18°

* Calculating z:
z = 7 × 18° = 126°
* Therefore, x = z = 126°.

Question 3:
In the given figure, if AB ∥ CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF, and ∠FGE.

Solution:
* Given: AB ∥ CD, EF ⊥ CD (∠FED = 90°), and ∠GED = 126°.

1. Finding ∠AGE:
Since AB ∥ CD and GE is a transversal, alternate interior angles are equal:
∠AGE = ∠GED = 126°

2. Finding ∠GEF:
∠GED = ∠GEF + ∠FED
126° = ∠GEF + 90°
∠GEF = 126° - 90° = 36°

3. Finding ∠FGE:
∠AGE + ∠FGE = 180° (Linear Pair on line AB)
126° + ∠FGE = 180°
∠FGE = 180° - 126° = 54°

Question 4:
In the given figure, if PQ ∥ ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.

Solution:
* Construction: Draw a line RX through point R parallel to line ST (and thus parallel to PQ).

* Since ST ∥ RX, consecutive interior angles sum to 180°:
∠RST + ∠SRX = 180°
130° + ∠SRX = 180° ⇒ ∠SRX = 50°

* Since PQ ∥ RX, alternate interior angles are equal:
∠PQR = ∠QRX
110° = ∠QRS + ∠SRX
110° = ∠QRS + 50°
∠QRS = 110° - 50° = 60°

Chapter 4: Lines and Angles (Exercise 4.2 - Remaining Questions)

Question 5:
In the given figure, if AB ∥ CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

Solution:
* Since AB ∥ CD and PQ is a transversal:
∠APQ = ∠PQR (Alternate interior angles)
x = 50°

* Since AB ∥ CD and PR is a transversal:
∠APR = ∠PRD (Alternate interior angles)
∠APQ + ∠QPR = 127°
50° + y = 127°
y = 127° - 50° = 77°

Question 6:
In the given figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes mirror PQ at B, reflects along BC to strike RS at C, and reflects back along CD. Prove that AB ∥ CD.

Solution:
* Draw normals BM ⊥ PQ and CN ⊥ RS.
* By law of reflection, angle of incidence = angle of reflection:
∠1 = ∠2 and ∠3 = ∠4.

* Since PQ ∥ RS, the normals BM ∥ CN.
* Thus, ∠2 = ∠3 (Alternate interior angles).
* Multiplying by 2:
2(∠2) = 2(∠3) ⇒ ∠1 + ∠2 = ∠3 + ∠4 ⇒ ∠ABC = ∠BCD.

* Since alternate interior angles are equal, AB ∥ CD.
(Hence proved.)