Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions Ex-8.4, (Part -2), Lines And Angles, Class 9, Maths

RD Sharma Solutions Ex-8.4, (Part -2), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

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

RD Sharma Solutions Ex-8.4, (Part -2), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

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.

RD Sharma Solutions Ex-8.4, (Part -2), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

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.

RD Sharma Solutions Ex-8.4, (Part -2), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

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.

RD Sharma Solutions Ex-8.4, (Part -2), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

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

RD Sharma Solutions Ex-8.4, (Part -2), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics  

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.

RD Sharma Solutions Ex-8.4, (Part -2), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

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 ?

RD Sharma Solutions Ex-8.4, (Part -2), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

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.

RD Sharma Solutions Ex-8.4, (Part -2), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

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.

RD Sharma Solutions Ex-8.4, (Part -2), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

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.

RD Sharma Solutions Ex-8.4, (Part -2), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-8.4, (Part -2), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

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

RD Sharma Solutions Ex-8.4, (Part -2), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-8.4, (Part -2), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans :

Given:

AB II DE, BC II EF

To prove: ∠ABC+∠DEF = 180

Construction: Produce BC to intersect DE at M

RD Sharma Solutions Ex-8.4, (Part -2), Lines And Angles, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

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

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

1. What are the different types of angles?
Ans. There are several types of angles, including acute angles, right angles, obtuse angles, straight angles, and reflex angles. An acute angle measures less than 90 degrees, a right angle measures exactly 90 degrees, an obtuse angle measures more than 90 degrees but less than 180 degrees, a straight angle measures exactly 180 degrees, and a reflex angle measures more than 180 degrees.
2. How can we determine if two lines are parallel?
Ans. Two lines are parallel if they never intersect, regardless of how far they are extended. One way to determine if two lines are parallel is by comparing their slopes. If the slopes of the lines are equal, then the lines are parallel. Another method is to check if the corresponding angles formed by a transversal cutting the lines are equal. If the corresponding angles are equal, then the lines are parallel.
3. What is the sum of the angles of a triangle?
Ans. The sum of the angles of a triangle is always 180 degrees. This property is known as the Triangle Sum Theorem. It means that the three interior angles of any triangle, when added together, will always equal 180 degrees. This property holds true for all types of triangles, whether they are equilateral, isosceles, or scalene.
4. How can we find the measure of an angle using a protractor?
Ans. To find the measure of an angle using a protractor, follow these steps: 1. Place the center of the protractor on the vertex (common endpoint) of the angle. 2. Align one side of the angle with the baseline of the protractor. 3. Read the degree measurement where the other side of the angle intersects the protractor scale. 4. The degree measurement is the measure of the angle.
5. What is the difference between a line segment and a ray?
Ans. A line segment is a portion of a line that has two distinct endpoints. It has a definite length and can be measured. On the other hand, a ray is a portion of a line that starts at a single point (the endpoint) and extends infinitely in one direction. A ray does not have a specific length and cannot be measured.
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