**DAV Class 8 Maths Solution Chapter 10 Parallel Lines**

DAV School Books Class 8 Maths Solution Chapter 10 parallel lines all Question Answer.

DAV Class 8 10th chapter parallel lines full Chapter explanation is provided by expert teacher. DAV class 8 maths chapter wise solution can be found here.

DAV School Books Class 8 Maths Solution Chapter 10 parallel lines

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

**Ans:**

∠ 1 + ∠ x= 180° (Supplementary angles as they form a linear pair)

∴ ∠ x = 180° – 110° = 70° (∵∠ 1= 110°)

∠ x = ∠ y (they form alternate exterior alternate angles)

∴ y = 70°

∠y = ∠z (they form alternate interior angles)

∴ z = 70°

∠w + ∠z = 180° (Supplementary angles as they form a linear pair)

∴ ∠ w = 180° – 70° = 110° (∵∠ z = 70)

Thus ∠w , ∠x, ∠y and ∠z are 110°, 70°,70° and 70° respectively.

**2) In the quadrilateral ABCD shown in the fig**

**∠1= ∠2 = 90°**

**Is AD** **∥ ****BC? Justify your answer.**

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**Ans:**

AD and BC are two lines that are cut by a transversal line AB.

∠1 and ∠2 form the consecutive interior angles.

According to the problem ∠1= ∠2 = 90°

We know that if the consecutive interior angles are equal to each other, then the two lines will be parallel to each other.

In this case ∠1 and ∠2 are equal to each other. Hence, AD and BC are parallel to each other.

**3) Draw a line segment AB = 6 cm. Mark two points P and Q in 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.**

**Ans:**

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PR and QS are both perpendicular to the line segment AB.

∴ The angle PR and QS form with AB is 90°.

∴ ∠a= ∠b = 90°

Both ∠a and ∠b are consecutive interior angles.

We know that if the consecutive interior angles are equal to each other, then the two lines will be parallel to each other.

In this case ∠a and ∠b are equal to each other. Hence, PR and QS are parallel to each other.

** 4) In this figure l** **∥ m** **∥ n. ****Find ****∠x ****and ****∠y.**

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**Ans:**

From the figure given in the problem we can see that ∠A = 50°.

∠A+ ∠x = 180° (As they form consecutive internal angles)

∴ ∠x = 180° – 50° = 130°

On the other hand ∠C and ∠y form a pair of alternative interior angles.

∴ ∠C = ∠y = 30 ° (∵∠C = 30°)

∴ ∠x = 130° and ∠C= 30°

**5) ABCD is a quadrilateral in which all the four angles ****∠A = ****∠B = ****∠C = ****∠D= 90****°. Show that AB ****∥ ****CD and AD** **∥ ****BC.**

**Ans:**

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AB and CD are two lines that are cut by a transversal line BC.

∠B and ∠C form the consecutive interior angles.

According to the problem ∠B= ∠C = 90°

We know that if the consecutive interior angles are equal to each other, then the two lines will be parallel to each other.

In this case ∠B and ∠C are equal to each other. Hence, AB and CD are parallel to each other.

∴ AB ∥ CD

AD and BC are two lines that are cut by a transversal line AB.

∠A and ∠B form the consecutive interior angles.

According to the problem ∠A= ∠B = 90°

In this case ∠A and ∠B are equal to each other. Hence, AD and BC are parallel to each other.

∴ AD ∥ BC

**6) In this figure show that AB** **∥ ****CD if ****∠****BOA = ****∠****BAO and ****∠****COD = ****∠****CDO**

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**Ans:**

AB and CD are two lines that are cut by two transversal lines AD and BC.

∠BOA and ∠COD form vertical angles

∴ ∠BOA = ∠COD

According to the problem ∠BOA = ∠BAO and ∠COD = ∠CDO

∴ ∠BAO = ∠CDO

∠BAO and ∠CDO form the alternate interior angles.

We know that if the alternate interior angles are equal to each other, then the two lines will be parallel to each other.

In this case ∠BAO and ∠CDO are equal to each other. Hence, AB and CD are parallel to each other.

∴ AB ∥ CD

**7) In this figure ****∠****A = 75° and CE ****∥ ****AB. If ****∠****ECD = 40°, Find the other two angles of the triangle.**

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**Ans:**

In the △ABC

∠BCA + ∠ACD = 180°

∴ ∠BCA = 180° – ∠ACD

Now, ∠ACD = ∠ACE + ∠ECD = ∠ACE + 40°

According to the problem CE ∥ AB and AC is the transversal line passing through them.

∴ ∠CAB and ∠ACE form alternate interior angles.

∴ ∠CAB = ∠ACE = 75°

∴ ∠ACD = 75° + 40° = 115°

Now, ∠BCA = 180° – ∠ACD

Or, ∠BCA = 180° – 115° = 65°

Now for △ABC,

∠ABC+∠BCA+∠CAB = 180°

Or, ∠ABC = 180 – ∠BCA – ∠CAB

Or, ∠ABC = 180° – 65° – 75° = 40°

∴ ∠ ABC = 40° and ∠BCA = 65°