2. Parallel Lines

Select the correct alternative and fill in the blanks in the following statements:

Click on the question to view the answer

The correct option is: (C) 180°

Solution:

The correct option is: (C) 8

Solution:

The correct option is: (A) 40°

Solution:

In \(\triangle\)ABC,
\(\angle\)A + \(\angle\)B + \(\angle\)C = 180° ... (Theorem)
∴ 76° + 48° + \(\angle\)C = 180°
∴ 124° + \(\angle\)C = 180°
∴ \(\angle\)C = 180° − 124°
∴ \(\angle\)C = 56°

∴ The correct option is: (B) 56°

Solution:

The correct option is: (C) 75°

2*. Ray PQ and ray PR are perpendicular to each other. Points B and A are in the interior and exterior of \(\angle\)QPR respectively. Ray PB and ray PA are perpendicular to each other.
Draw a figure showing all these rays and write:

A pair of complementary angles

A pair of supplementary angles

A pair of congruent angles

Procedure:

From the given information, let’s draw the figure as per the following instructions:

Draw ray PQ and ray PR such that they are perpendicular to each other.
Now, select a point B in the interior of \(\angle\)QPR and a point A in the exterior of \(\angle\)QPR such that the angle between them is 90°.
Now, draw ray PB and ray PA.

Solution:

(i) A pair of complementary angles is \(\angle\)APR and \(\angle\)RPB.

(ii) A pair of supplementary angles is \(\angle\)APB and \(\angle\)RPQ.

(iii) A pair of congruent angles is \(\angle\)QPB and \(\angle\)RPA.

3. Prove that, if a line is perpendicular to one of the two parallel lines, then it is perpendicular to the other line also.

Given:

Line l | | line m and line n is their transversal.
Also, line l \(\perp\) line n.

To Prove:

Line n \(\perp\) line m.

Proof:

Let’s name the angles a and b as shown in the figure.

Now, line l | | line m ... (Given)
∴ \(\angle b\) \(\cong\) \(\angle a\) ... (Corresponding Angles)
But, \(\angle a\) = 90° ... (Given)
∴ \(\angle b\) = 90°
i.e. line n \(\perp\) line m.

4. In figure 2.24, measures of some angles are shown. Using the measures, find \(\angle x\) and \(\angle y\) and hence show that line l | | line m.

Proof:

Problem Set 2 : Problem 4 : Textbook page 23 : Figure 2.24

\(\angle x\) = 130°
... (Vertically Opposite Angles)

and \(\angle y\) = 50°
... (Vertically Opposite Angles)

Now, \(\angle x\) + 50°
= 130° + 50°
= 180°

∴ line l | | line m
... (Interior Angles Test)

5. Line AB | | line CD | | line EF and line QP is their transversal. If y ∶ z = 3 ∶ 7, then find the measure of \(\angle x\). (See figure 2.25)

Solution:

Problem Set 2 : Problem 5 : Textbook page 23 : Figure 2.25

line AB | | line EF ... (Given)
Consider transversal PQ.
∴ \(\angle x\) \(\cong\) \(\angle z\) ... (Alternate angles) ... (i)
But, \(\angle y\) ∶ \(\angle z\) = 3 ∶ 7 ... (Given)
∴ \(\angle y\) ∶ \(\angle x\) = 3 ∶ 7

Now, let the common multiple of these ratios be m.
∴ \(\angle y\) = 3m and \(\angle x\) = 7m ... (ii)

Also, line AB | | line CD ... (Given)
Consider transversal PQ.
∴ \(\angle x\) + \(\angle y\) = 180° ... (Interior angles theorem)
∴ 7m + 3m = 180° ... [From (ii)]
∴ 10m = 180°

∴ m = \(\displaystyle \frac{180°}{10}\)

∴ m = 18° ... (iii)

∴ \(\angle x\) = 7m = 7 × 18°
∴ \(\angle x\) = 126°

6. In figure 2.26, if line q | | line r, line p is their transversal and if a = 80°, find the values of f and g.

Proof:

Problem Set 2 : Problem 6 : Textbook page 23 : Figure 2.26

\(\angle g\) \(\cong\) \(\angle a\) ... (Exterior Alternate Angles)
But, \(\angle a\) = 80° ... (Given)
∴ \(\angle g\) = 80° ... (i)

Also, \(\angle f\) + \(\angle g\) = 180° ... (Angles in Linear Pair)
∴ \(\angle f\) + 80° = 180° ... [From (i)]
∴ \(\angle f\) = 180° − 80°
∴ \(\angle f\) = 100° ... (ii)

7. In figure 2.27, if line AB | | line CF and line BC | | line ED, then prove that: \(\angle\)ABC = \(\angle\)FDE.

Proof:

Problem Set 2 : Problem 7 : Textbook page 23 : Figure 2.27

line AB | | line CF ... (Given)
∴ \(\angle\)ABC \(\cong\) \(\angle\)BCD ... (Alternate Angles) ... (i)

Also,
line BC | | line ED ... (Given)
∴ \(\angle\)BCD \(\cong\) \(\angle\)FDE ... (Corresponding Angles) ... (ii)

From (i) and (ii):
\(\angle\)ABC \(\cong\) \(\angle\)FDE ... (Transitivity)
i.e. \(\angle\)ABC = \(\angle\)FDE

8. In figure 2.28, line PS is a transversal of parallel line AB and line CD. If Ray QX, ray QY, ray RX, ray RY are angle bisectors, then prove that \(\Box\)QXRY is a rectangle.

Proof:

Problem Set 2 : Problem 8 : Figure 2.28 : Textbook Page 23

Line AB | | line CD ... (Given)
Consider transversal PS.
∴ \(\angle\)AQR = \(\angle\)DRQ ... (Alternate Angles)
∴ \(\displaystyle \frac{1}{2}\) \(\angle\)AQR = \(\displaystyle \frac{1}{2}\) \(\angle\)DRQ ... (i)

Now, ray QX is the bisector of \(\angle\)AQR ... (Given)
∴ \(\angle\)XQR = \(\displaystyle \frac{1}{2}\) \(\angle\)AQR ... (ii)

Also, ray RY is the bisector of \(\angle\)DRQ ... (Given)
∴ \(\angle\)YRQ = \(\displaystyle \frac{1}{2}\) \(\angle\)DRQ ... (iii)

From (i), (ii) and (iii):
∴ \(\angle\)XQR = \(\angle\)YRQ
∴ QX | | RY ... (Alternate Angles Test) ... (iv)

Similarly, line AB | | line CD ... (Given)
Consider transversal PS.
∴ \(\angle\)BQR = \(\angle\)CRQ ... (Alternate Angles)
∴ \(\displaystyle \frac{1}{2}\) \(\angle\)BQR = \(\displaystyle \frac{1}{2}\) \(\angle\)CRQ ... (v)

Now, ray QY is the bisector of \(\angle\)BQR ... (Given)
∴ \(\angle\)YQR = \(\displaystyle \frac{1}{2}\) \(\angle\)BQR ... (vi)

Also, ray RX is the bisector of \(\angle\)CRQ ... (Given)
∴ \(\angle\)XRQ = \(\displaystyle \frac{1}{2}\) \(\angle\)CRQ ... (vii)

From (v), (vi) and (vii):
∴ \(\angle\)YQR = \(\angle\)XRQ
∴ QY | | RX ... (Alternate Angles Test) ... (viii)

From (iv) and (viii),
\(\Box\)QXRY is a parallelogram. ... (Definition of a Parallelogram) ... (ix)
∴ \(\angle\)BQR + \(\angle\)DRQ = 180° ... (Interior Angles)
Dividing both sides by 2,
\(\displaystyle \frac{1}{2}\) \(\angle\)BQR + \(\displaystyle \frac{1}{2}\) \(\angle\)DRQ = \(\displaystyle \frac{180°}{2}\)
∴ \(\angle\)YQR + \(\angle\)YRQ = 90° ... [From (vi) and (iii)] ... (x)

Now, in \(\triangle\)YQR,
∴ \(\angle\)YQR + \(\angle\)YRQ + \(\angle\)QYR = 180° ... (Theorem)
∴ 90° + \(\angle\)QYR = 180° ... [From (x)]
∴ \(\angle\)QYR = 180° − 90°
∴ \(\angle\)QYR = 90° ... (xi)

A parallelogram is a rectangle if one of its angles is 90°.
∴ \(\Box\)QXRY is a rectangle. ... [From (ix) and (xi)]

Select the correct alternative and fill in the blanks in the following statements: Click on the question to view the answer

If a transversal intersects two parallel lines, then the sum of the interior angles on the same side of the transversal is ____. 0° 90° 180° 360°

The number of angles formed by a transversal of two lines is _____. 2 4 8 16

Solution:

A transversal intersects two parallel lines. If the measure of one of the angles is 40º, then the measure of its corresponding angle is _____. 40° 140° 50° 180°

Solution:

In \(\triangle\)ABC, \(\angle\)A = 76°, \(\angle\)B = 48º, ∴ \(\angle\)C = _____. 66° 56° 124° 28°

Solution:

Two parallel lines are intersected by a transversal. If the measure of one of the alternate interior angles is 75º, then the measure of the other angle is _____. 105° 15° 75° 45°

Solution:

Procedure:

Solution:

3. Prove that, if a line is perpendicular to one of the two parallel lines, then it is perpendicular to the other line also.

Given:

To Prove:

Proof:

4. In figure 2.24, measures of some angles are shown. Using the measures, find \(\angle x\) and \(\angle y\) and hence show that line l | | line m.

Proof:

5. Line AB | | line CD | | line EF and line QP is their transversal. If y ∶ z = 3 ∶ 7, then find the measure of \(\angle x\). (See figure 2.25)

Solution:

6. In figure 2.26, if line q | | line r, line p is their transversal and if a = 80°, find the values of f and g.

Proof:

7. In figure 2.27, if line AB | | line CF and line BC | | line ED, then prove that: \(\angle\)ABC = \(\angle\)FDE.

Proof:

8. In figure 2.28, line PS is a transversal of parallel line AB and line CD. If Ray QX, ray QY, ray RX, ray RY are angle bisectors, then prove that \(\Box\)QXRY is a rectangle.

Proof:

Select the correct alternative and fill in the blanks in the following statements:

Click on the question to view the answer