The document RD Sharma Solutions Ex-8.4, (Part -2), Lines And Angles, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.

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**Q 15 : In the below fig, ∠1 = 60 ^{∘} and ∠2 = (2/3)rd of a right angle. Prove that l|| m.**

Ans : Given :

∠1 = 60^{∘} and ∠2 = (2/3)rd of a right angle

To prove : parallel Drawn to m

Proof ∠1 = 60

∠2 = (23)×90 = 60

Since ∠1 =∠1 = 60^{∘}

Therefore, Parallel to m as pair of corresponding angles are equal.

**16. In the below fig, if l||m||n and ∠1 = 60 ^{∘}. Find ∠2.**

**Ans :** Since l parallel to m and p is the transversal

Therefore, Given: l||m||n

∠1 = 60^{∘}

To find ∠2

∠1=∠3 = 60^{∘ } [Corresponding angles]

Now, ∠3and∠4 are linear pair of angles

∠3+∠4 = 180^{∘}

60 + ∠4 = 180

∠4 = 180 — 60

⇒ 120

Also, m||n and P is the transversal

Therefore ∠4 = ∠2 = 120 (Alternative interior angle]

Hence 2 ∠2 = 120

**Q 17 : Prove that the straight lines perpendicular to the same straight line are parallel to one another.**

**Ans :** Let AB and CD be drawn perpendicular to the Line MN

∠ABD = 90^{∘} [ AB is perpendicular to MN ] —–(i)

∠CON = 90^{∘} [CD is perpendicular to MN ] —–(ii)

Now,

∠ABD = ∠CDN = 90^{∘} [From (i) and (ii)]

Therefore, AB||CD, Since corresponding angles are equal.

**Q 18 : The opposite sides of a quadrilateral are parallel. If one angle of the quadrilateral is 601. Find the other angles.**

**Ans :** Given AB || CD

AD|| BC

Since AB || CD and AD is the transversal

Therefore, A + D = 180 (Co-interior angles are supplementary)

60 + D = 180

D = 180 – 60

D = 120

Now. AD || BC and AB is the transversal

A + B = 180 (Co-interior angles are supplementary)

60 +B = 180

B = 180 — 60

= 120

Hence, ∠A = ∠C = 60^{∘} and ∠B =∠D = 120∘

**Q 19 : Two lines AB and CD intersect at O. If ∠AOC+∠COB+∠BOD = 270 ^{∘}, find the measures of ∠AOC,∠COB,∠BOD,∠DOA**

Ans :

Given : ∠AOC+∠COB+∠BOD=270∘

To find : ∠AOC,∠COB,∠BOD,∠DOA

Here, ∠AOC+∠COB+∠BOD = 270^{∘} [ Complete angle]

⇒ 270 + AOD = 360

⇒ AOD = 360 — 270

⇒ AOD = 90

Now, AOD + BOD = 180 [Linear pair]

90 + BOD = 180

⇒ BOD = 180 – 90

⇒ BOD = 90

AOD = BOC = 90 [Vertically opposite angles]

BOD = AOC = 90 [Vertically opposite angles]

**Q 20. In the below figure, p is a transversal to lines m and n, ∠2 = 120 ^{∘} and ∠5 = 60^{∘}. Prove that m|| n.**

Ans :

Given that

∠2 = 120^{∘} and ∠5 = 60^{∘}

To prove,

∠2+∠1 = 180^{∘} [ Linear pair ]

120+∠1 = 180

∠1 = 180−120

∠1 = 60^{∘}

Since ∠1 = ∠5 = 60^{∘}

Therefore, m||n [As pair of corresponding angles are equal]

**Q 21 : In the below fig. transversal t intersects two lines m and n, ∠4 =110 ^{∘} and ∠7 = 65^{∘} Is m||n ?**

Ans : Given :

∠4 = 110^{∘} and ∠7 = 65^{∘}

To find : Is m||n

Here. ∠7 = ∠5 = 65^{∘} [Vertically opposite angle]

Now. ∠4+∠5 = 110+65 = 175^{∘}

Therefore, m is not parallel to n as the pair of co interior angles is not supplementary.

**Q 22 : Which pair of lines in the below fig. is parallel ? give reasons.**

Ans : ∠A+∠B = 115+65 = 180^{∘}

Therefore, AB || BC [ As sum of co interior angles are supplementary]

∠B+∠C = 65+115 = 180^{∘}

Therefore, AB || CD (As sum of interior angles are supplementary]

**Q 23 : If I, m, n are three lines such that I|| m and n perpendicular to l, prove that n perpendicular to m.**

Ans :

Given, l||m, n perpendicular to I

To prove: n perpendicular to m

Since l||m and n intersects

∴ ∠1 = ∠2 [Corresponding angles]

But, U = 90

⇒ ∠2 = 90^{∘}

Hence n is perpendicular to m

**Q 24 : In the below fig, arms BA and BC of ∠ABC are respectively parallel to arms ED and EF of∠DEF. Prove that ∠ABC = ∠DEF.**

**Ans :**

Given

AB || DE and BC || EF

To prove : ∠ABC=∠DEF

Construction: Produce BC to x such that it intersects DE at M.

Proof : Since AB || DE and BX is the transversal

ABC = DMX [Corresponding angle] —–(i)

Also, BX || EF and DE Is the transversal

DMX = DEF [Corresponding angles] —–(ii)

From (i) and (ii)

∠ABC =∠DEF

**Q 25: In the below fig, arms BA and BC of ABC are respectively parallel to arms ED and EF of DEF Prove that ∠ABC+∠DEP = 180 ^{∘}**

**Ans :**

Given:

AB II DE, BC II EF

To prove: ∠ABC+∠DEF = 180^{∘}

Construction: Produce BC to intersect DE at M

Proof :

Since AB || EM and BL is the transversal

∠ABC = ∠EML [Corresponding angle] —–(i)

Also,

EF || ML and EM is the transversal

By the property of co-interior angles are supplementary

∠DEF+∠EML = 180^{∘} (ii)

From (i) and (ii) we have

Therefore ∠DEF+∠ABC = 180^{∘}

**Q 26 : With of the following statements are true (T) and which are false (F)? Give reasons.**

**(1) If two lines are intersected by a transversal, then corresponding angles are equal.**

**(ii) If two parallel lines are intersected by a transversal, then alternate interior angles are equal.**

**(ii) Two lines perpendicular to the same line are perpendicular to each other.**

**(iv) Two lines parallel to the same line are parallel to each other.**

**(v) If two parallel lines are intersected by a transversal, then the interior angles on the same side of the transversal are equal.**

**Ans :**

(i) False

(ii)True

(iii) False

(iv) True

(v) False

**Q 27: Fill in the blanks in each of the following to make the statement true:**

**(i) If two parallel lines are intersected by a transversal, then each pair of corresponding angles are ____________ **

**(ii) If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are _____________**

**(iii) Two lines perpendicular to the same line are _______ to each other**

**(Iv) Two lines parallel to the same line are __________ to each other.**

**(v) If a transversal intersects a pair of lines in such a way that a pair of alternate angles we equal. then the lines are ___________**

**(vi) If a transversal intersects a pair of lines in such a way that the sum of interior angles on the seine side of transversal is 180′. then the lines are _____________**

**Ans :**

(i) Equal

(ii) Parallel

(iii) Supplementary

(iv) Parallel

(v) Parallel

(vi) Parallel

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