The DAV Maths Class 8 Solutions and **DAV Class 8 Maths Chapter 10 Worksheet 1** Solutions of Parallel Lines offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 10 WS 1 Solutions

Question 1.

In the given figure, l ∥ m ∥ n and p is a transversal. If ∠1 = 110°, find angles x, y, z and w.

Answer:

Given that l ∥ m ∥ n and p is the transversal.

∴ ∠1 + ∠x = 180° [Supplementary angles]

⇒ 110° + ∠x = 180°

⇒ ∠x = ∠y [Corresponding angles]

70° = ∠y

∠y = 70°

∠y = ∠z [Alternate angles]

70° = ∠z

Hence, ∠z = 70°

Question 2.

In the quadrilateral ABCD shown in the figure, ∠1 = ∠2 = 90°. Is AD ∥ BC? Justify your answer.

Answer:

∠1 = 90°

∠2 = 90°

AB is a transversal.

∴∠1 + ∠2 = 180°.

∴ ∠1 and ∠2 are supplementary angles.

Hence, AB ∥ CD.

Question 3.

Draw a line segment \(\overline{\mathrm{AB}}\) = 6 cm. Mark two points P and Q on it. Draw lines perpendicular to AB through P and Q (PR and QS). What can you say about PR and QS? Are these parallel? Justify your answer.

Answer:

In the given figure,

PR ⊥ AB and QS ⊥ AB

∴ ∠1 = ∠2 = 90°

But ∠1 and ∠2 are correspondings angles

∴ \(\overleftrightarrow{\mathrm{PR}} \| \overrightarrow{\mathrm{QS}}\)

Question 4.

In the given figure, l∥m∥n. Find ∠x and ∠y.

Answer:

l ∥ m ∥ n (given)

∠x + 50° = 180° [Sum of interior angles]

⇒ ∠x = 180° – 50°

∠x = 130°

∠y = 30° [Pair of alternate angles]

Question 5.

ABCD is a quadrilateral in which all the four angles are equal. Show that AB∥CD and AD ∥BC.

Answer:

Let the measure of each angle of the quadrilateral ABCD be x.

Then ∠A = ∠B = ∠C = ∠D = x.

We know that,

the sum of all the angles of a quadrilateral = 360°

⇒ ∠A + ∠B + ∠C + ∠D = 360°

⇒ x + x + x + x = 360°

⇒ 4x = 360°

⇒ x = 90°

So, ∠A = ∠B = ∠C = ∠D = 90°

Now, ∠A + ∠B = 90° + 90° = 180°.

They form a pair of interior angles on the same side of a transversal.

Hence, \(\overline{\mathrm{AD}} \| \overline{\mathrm{BC}}\).

Similarly, ∠B + ∠C = 90° + 90° = 180°.

∴ They form a pair of interior angles on the same side of a transversal.

Hence, \(\overline{\mathrm{AB}} \| \overline{\mathrm{DC}}\).

Question 6.

In the given figure, if ∠BAO = ∠DCO and OC = OD, show that \(\overline{\mathrm{AB}} \| \overline{\mathrm{CD}}\).

Answer:

Given that ∠BAO = ∠DCO .. .(i)

and OC = OD …..(ii)

Consider ΔOCD.

Since OC = OD

⇒ ∠DCO = ∠CDO (Angles opposite to equal sides of a triangle are equal)

⇒ ∠BAO = ∠CDO [From (i]

But these are alternate angles.

∴ AB ∥ CD.

Question 7.

In the given figure, ∠A =75° and \(\overline{\mathrm{CE}} \| \overline{\mathrm{AB}}\). If ∠ECD = 40°, find the other two angles of the triangle ABC.

Answer:

\(\overline{\mathrm{AB}} \| \overline{\mathrm{CE}}\) [Given]

∠BAC = ∠ACE

75° = ∠ACE

∠ACD = ∠ACE + ∠ECD

= 75° + 40° = 115°

∠ACB + ∠ACD = 180° [Corresponding angles]

∠ACB + 115° = 180°

∠ACB = 180° – 115° = 65°

∠ABC = ∠ECD [Corresponding angles]

∴ ∠ABC = 40°

Hence, ∠ACB = 65° and ∠ABC = 40°.

Question 8.

In the given figure, l∥ m, find ∠x, ∠y and ∠z.

Answer:

l ∥ m (given)

and q is transversal.

∴ ∠x = 110° [Corresponding angles]

∠x + ∠z = 180° (Supplementary angles)

110° + ∠z = 180°

∴ ∠z = 180° – 110° = 70°

40° + ∠1 = 180°

∠1 = 180° – 40° = 140°

But ∠1 = ∠y (Alternate angles)

∴ ∠y = 140°

Hence, ∠x = 110°, ∠y = 140° and ∠z = 70°.

Question 9.

In the given figure, show that:

(i) \(\overline{\mathrm{AB}} \| \overline{\mathrm{CD}}\)

Answer:

∠ABC = 40° (Given)

∠BCD = 20° + 20° = 40° (Given)

∴ ∠ABC = ∠BCD

But they are alternate angles.

∴ \(\overline{\mathrm{AB}} \| \overline{\mathrm{CD}}\)

(ii) \(\overline{\mathrm{CD}} \| \overline{\mathrm{EF}}\)

Answer:

∠ECD = 20° (Given)

∠CEF = 160° (Given)

∠ECD + ∠CEF = 20° + 160°

= 180°

∴ They are interior angles on the same side of transversal CE.

∴ \(\overline{\mathrm{CD}} \| \overline{\mathrm{EF}}\)

(iii) \(\overline{\mathrm{AB}} \| \overline{\mathrm{EF}}\)

Justify your answer.

Answer:

\(\overline{\mathrm{AB}} \| \overline{\mathrm{CD}}\) [from (i)]

\(\overline{\mathrm{AB}} \| \overline{\mathrm{CD}}\) [from (ii)]

∴ \(\overline{\mathrm{AB}} \| \overline{\mathrm{CD}}\)

Hence proved.

### DAV Class 8 Maths Chapter 10 Worksheet 1 Notes

If two lines in a two dimensional plane do not intersect each other, they are called parallel lines.

**Transversal line:**

Any line which intersects the parallel lines is called transversal line.

Here in the alongside figure, l || m and t is the transversal intersecting l and m at A and B respectively.

**Some pair of angles:**

- Interior angles: ∠2 and ∠3, ∠6 and ∠7
- Pair of alternate angles: ∠2 and ∠7, ∠3 and ∠6
- Pair of vertically opposite angles: ∠1 and ∠6, ∠2 and ∠5, ∠3 and ∠8, ∠4 and ∠7.
- Pair of corresponding angles: ∠1 and ∠3, ∠5 and ∠7, ∠2 and ∠4, ∠6 and ∠8.

**Supplementary angles:**

Any two angles whose sum is 180° are called supplementary angles.

Example 1.

In the given figure, ∠2 = 65°, find ∠1, ∠4 and ∠3.

Solution:

(i) ∠1 + ∠2 = 180° [Supplementary angles)

∠ + 65° = 180°

∴ ∠1 = 180° – 65° = 115

(ii) ∠1 = ∠4 [Corresponding angles]

115° = ∠4

∴ ∠4 = 115°

(iii) ∠3 + ∠4 = 180° [Supplementary angles]

∠3 + 115° = 180°

∠3 = 180° – 115°

= 65°

Example 2.

Divide a given line AB = 4 cm into 5 equal parts.

Solution:

Steps of Construction:

(i) Draw AB = 4 cm.

(ii) At A, draw a ray AP making an acute angle with \(\overline{\mathrm{AB}}\).

(iii) At B, draw AP || BQ on opposite side.

(iv) With the help of compass, mark 5 points, A_{1}, A_{2}, A_{3}, A_{4} and A_{5} on AP at equal distances and B_{1}, B_{2}, B_{3}, B_{4} and B_{5} on BQ.

(v) Join A to B_{5} and B to A_{5}.

(vi) Join A_{1}B_{4}, A_{2}B_{3}, A_{3}B_{2} and A_{4}B_{1}.

(vii) Measure \(\overline{\mathrm{AC}_1}\), \(\overline{\mathrm{C}_1 \mathrm{C}_2}\), \(\overline{\mathrm{C}_2 \mathrm{C}_3}\), \(\overline{\mathrm{C}_3 \mathrm{C}_4}\) and \(\overline{\mathrm{C}_4 \mathrm{~B}}\) which are equal.