DAV Class 8 Maths Chapter 10 Worksheet 1 Solutions

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:

1. Interior angles: ∠2 and ∠3, ∠6 and ∠7
2. Pair of alternate angles: ∠2 and ∠7, ∠3 and ∠6
3. Pair of vertically opposite angles: ∠1 and ∠6, ∠2 and ∠5, ∠3 and ∠8, ∠4 and ∠7.
4. 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, A₁, A₂, A₃, A₄ and A₅ on AP at equal distances and B₁, B₂, B₃, B₄ and B₅ on BQ.
(v) Join A to B₅ and B to A₅.
(vi) Join A₁B₄, A₂B₃, A₃B₂ and A₄B₁.
(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.