RD Sharma Solutions Class 9 Lines And Angles Exercise 8.4

RD Sharma Class 9 Solutions Chapter 8 Ex 8.4 Free Download

RD Sharma Solutions Class 9 Chapter 8 Ex 8.4

Q 1:  In below fig. AB CD and ∠1 and ∠2 are in the ratio 3 : 2. Determine all angles from 1 to 8.

1

Ans :      Let ∠1 = 3θ and ∠2 = 2θ

As you can see, ∠1 and ∠2 are linear pair of angle

Therefore, we can write it as;

=> ∠1 + ∠2 = 180

=>  3θ + 2θ = 180

=> 5θ  = 180

=> θ = 180 / 5

=> θ = 36

∠1 = 3θ = 108°, ∠2 = 2θ = 72°

As we know already from the theorem, the vertically opposite angles are equal, therefore;

∠1 = ∠3 = 108°

∠2 = ∠4 = 72°

∠6 = ∠7 = 108°

∠5 = ∠8 = 72°

We also know, if a transversal intersects any parallel lines, then the corresponding angles are equal

∠1 = ∠5 = 108°

∠2 = ∠6 = 72°

Q 2: In the below fig, I, m and n are parallel lines intersected by transversal p at X. Y and Z respectively. Find ∠1, ∠2 and ∠3.

2

Ans : From the given figure :

∠3+ ∠m YZ = 180°            [From Linear pair angles theorem]

=> ∠3= 180 – 120

=> ∠3= 60°

From the figure, line l is parallel to line m, therefore,

∠1 = ∠3                                         [Both are Corresponding angles]

∠1 = 60°

Also, line m  is parallel to line n;

2 = ∠mYZ                                  [Alternate interior angles are equal]

∴ ∠2 = 120°

Hence, ∠1 = 60°, ∠2 = 120° and ∠3 = 60°.

 

Q.3. In the below fig, AB || CD || EF and GH || KL Find HKL.

Ans :     Extend LK to meet GF at P.

Now, CD and GF are parallel lines, therefore, alternate angles are equal.

∠CHG =∠HGP = 60°

∠HGP =∠KPF = 60°         [Corresponding angles of parallel lines are equal]

Hence, ∠KPG =180 – 60 = 120°

=> ∠GPK = ∠AKL= 120° [Corresponding angles of parallel lines are equal]

∠AKH = ∠KHD = 25°        [alternate angles of parallel lines]

Therefore, HKL = ∠AKH + ∠AKL = 25 + 120 = 145°

 

Q 4 : In the below fig, show that AB || EF.

Ans : Produce EF to intersect AC at N.

Now, DCE + ∠CEF = 35 + 145 = 180°

Therefore, EF || CD        (Since Sum of Co-interior angles of parallel lines is 180)  —–(1)

Now, BAC = 57°

∠ACD = 22°+35° = 57°

Therefore,

=>BA || EF                          [Alternative angles of parallel lines are equal]                   —–(2)

From (1) and (2)

AB || EF                          [Since, Lines parallel to the same line are parallel to each other]

Hence proved.

Q 5 : If below fig.if AB || CD and CD || EF, find ACE.

5

Ans :  Since EF || CD

Therefore,  EFC + ECD = 180°        [sum of co-interior angles are supplementary to each other]

=> ECD = 180° – 130° = 50°

Also, BA || CD

=> BAC = ACD = 70°                           [alternative angles of parallel lines are equal]

But, ACE + ECD =70°

=> ACE = 70° — 50° = 20°

 

Q 6  : In the below fig, PQ || AB and PR || BC. If QPR = 102°, determine ABC. Give reasons.

6

Ans :  Extend line AB to meet line PR at G.

Since PQ || AB,

∴ ∠QPR = ∠BGR =102°                         [corresponding angles of parallel lines are equal]

Since PR || BC,

∴ ∠RGB+ ∠CBG =180°                          [ Since Corresponding angles are supplementary]

∠CBG = 180° – 102° = 78°

Since, ∠CBG = ∠ABC

∴ ∠ABC = 78°

 

Q 7 : In the below fig, state Which lines are parallel and why?

7

Ans : As we know, for parallel lines, vertically opposite angles are equal, therefore;

=> ∠EOC = ∠DOK = 100°

∠DOK =∠ACO = 100°

Here two lines ED and CA are cut by a third line DC and the corresponding angles to it are equal.

Therefore, ED || AC.

 

Q.8. In the below fig. if l||m, n || p and 1 = 85°. find 2.

8

Ans : Given, 1 = 85°

As we know, when a line cuts the parallel lines, the corresponding angles are equal, therefore,

=> 1 = ∠3 = 85°

Now we know,  co-interior angles are supplementary;

∠2 + ∠3 = 180°

∠2 + 55° =180°

∠2 = 180° – 85°

∠2 = 95°

Q 9 : If two straight lines are perpendicular to the same line, prove that they are parallel to each other.

9

Ans :      Given line m and line l are perpendicular to line n.

∠1= ∠2=90°

Since, lines l and m are cut by a transversal n and the corresponding angles are equal, therefore,

l || m

Hence, proved.

 

Q 10 : Prove that if the two arms of an angle are perpendicular to the two arms of another angle. then the angles are either equal or supplementary.

Ans : Let the angles be  ∠ACB and ∠ABD

Given,  CA perpendicular to AB,  also CD is perpendicular to BD

To prove : ∠ACD = ∠ABD  (or) ∠ACD + ∠ABD =180°

Proof :  In a quadrilateral = ∠A+ ∠C+ ∠D+ ∠B = 360°

[ Sum of angles of quadrilateral is 360 ]

=> 180°  + ∠C + ∠B = 360°

=> ∠C + ∠B = 360° –180°

Hence, ∠ACD + ∠ABD = 180°                                —–(1)

Also, ∠D + ∠ABD = 180°

=> ∠ABD =180° – 90° = ∠ACD = 90°                   —–(2)

From (i) and (ii), ∠ACD = ∠ABD = 90°

Hence, it is proved that, the angles are equal as well as supplementary.

 

Q 11 : In the below fig, lines AB and CD are parallel and P is any point as shown in the figure. Show that ∠ABP + ∠CDP = ∠DPB.

11

Ans :

Given that AB ||CD

Let QR be the parallel line to AB and CD which passes through P

As we know, for parallel lines, alternate interior angles are equal.

Therefore, ∠ABP = ∠BPR …………..(1)

And,

∠CDP = ∠DPR ………………..(2)

Adding equation 1 and 2, we get,

∠ABP + ∠CDP = ∠BPR + ∠DPR

∠ABP + ∠CDP = ∠DPB

Hence proved.

 

Q 12: In the below fig, AB || CD and P is any point shown in the figure. Prove that : angle ABP + ∠BPD + ∠CDP = 360°

Ans:

Given, AB parallel to CD, P is any point.

To prove: ABP+ ∠BPD+ ∠CDP = 360°

Through P, draw a line PM parallel to AB or CD.

Now,

AB || PM => ABP + BPM = 180°

And

CD||PM = MPD + CDP = 180°

Adding (i) and (ii), we get ABP + ( BPM + MPD ) + CDP = 360°

=> ABP + BPD + CDP = 360°

 

Q 13 : Two unequal angles of a parallelogram are in the ratio 2 : 3. Find all its angles in degrees.

Ans :  Let ∠A = 2θ and ∠B = 3θ

Now, ∠A +∠B = 180°                [Co-interior angles are supplementary for parallel lines]

2θ  + 3θ = 180°                      [AD // BC and AB is the transversal)

=> 5θ = 180°

θ = 180° /5

θ = 36°

Therefore, ∠A = 2 × 36 = 72°

∠B = 3 × 36  = 108°

Now, ∠A = ∠C = 72°                  [Opposite angles of a parallelogram are equal]

∠B = ∠D = 108°

Q 14 :  If each of the two lines is perpendicular to the same line, what kind of lines are they to each other?

Ans :

Let lines AB and CD be perpendicular to line mn.

Line mn intersects AB and CD st O and P respectively.

∠AON =  90° [AB perpendicular to mn]                        —– (i)

∠CPN = 90° [CD perpendicular to mn]                        —– (ii)

From (i) and (ii), we get,

∠AON = ∠CPN = 90°

Since the corresponding angles are equal of lines AB and CD.

Therefore, AB // CD.

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

16

Ans :      Given :

 1 = 60° and ∠2 =(2/3)rd of a right angle

To prove : l is parallel to m

Proof:

 1 = 60°

∠2 = (2/3) of 90° = 60°

Since, the corresponding angles ∠ 1 and ∠2, both are equal to 60°. Therefore, l|| m.

 

16. In the below figure, if l||m||n and  ∠ 1 =60°. Find ∠2.

17

Ans :

Given, l||m||n and p is a transversal.

 1 =60°

To find: ∠2 ?

Since ∠1 and ∠3 are corresponding angles,

therefore, ∠1 = ∠3 = 60°

And, ∠3 and ∠4 are linear pair of angles, formed on the transversal intersecting the line m.

Therefore,

∠3 + ∠4  = 180°

60° + ∠4 = 180°

∠4 = 180° — 60°

= 120°

Now, we can see, m||n and p is the transversal, therefore the alternate interior angles will be equal.

i.e. ∠4  = ∠2 = 120°

Hence,  ∠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 are two straight lines perpendicular to the line mn.

Now, as per the given question,

∠ABD = 90°    [ AB is perpendicular to mn ]      —–(i)

∠CDn =90°     [CD is perpendicular to mn ]       —–(ii)

On comparing eq. (i) and (ii);

∠ABD = ∠CDN =90° [From (i) and (ii)]

Now, as we know, if a transversal intersects two parallel lines, then their corresponding angles are equal.

Therefore,  AB||CD.

 

Q 18 : The opposite sides of a quadrilateral are parallel. If one angle of the quadrilateral is 60°. Find the other angles.

19

Ans : Let us draw a quadrilateral ABCD, then as per the question,

AB || CD and AC|| BD

Hence, ABCD is a parallelogram, which has opposite angles as equal.

∠A = ∠D and ∠B = ∠C

Since, ∠C = 60°, from the above figure.

Therefore, ∠B = 60°

Since AB || CD and AC is the transversal line intersecting the lines AB and CD.

Therefore, ∠A + ∠C = 180° (Since sum of Co-interior angles equal to 180)

∠A + ∠C = 180°

∠A = 180° – 60°

∠A = 120°

And ∠D = 120°

Hence, ∠C = ∠B = 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.

20.

Ans :

Given, as per the question,

AOC + ∠COB + ∠BOD = 270°

To find : AOC , ∠COB , ∠BOD, ∠DOA = ?

Here, AOC + ∠COB + ∠BOD + ∠AOD = 360° (Sum of all the angles)

=>           270° + AOD = 360°   (given, ∠AOC + ∠COB + ∠BOD = 270°)

=>           AOD = 360° — 270°

=>           AOD  =  90°

Now, AOD + BOD = 180°                 [Linear pair angles]

90° + BOD = 180°

=>  BOD = 180° – 90°

=> BOD  = 90°

AOD  = BOC = 90°               [Vertically opposite angles are equal]

BOD  = AOC = 90°               [Vertically opposite angles are equal]

Hence, all the angles are found.

Q 20. In the below figure, p is a transversal to lines m and n, 2=120°  and 5=60°. Prove that m|| n.

21

Ans :

Given here,

2=120°  and 5=60°

To prove: 2+ ∠1 =180°      [ As Linear pair angles are supplementary ]

120 + ∠1=180°

∠1=180° -120°

∠1=60°

Since 1 = ∠5 =60°

Since, pair of corresponding angles are equal, therefore,

m||n. Proved.

 

Q 21: In the below fig. transversal l intersects two lines m and n, 4 =110° and 7 =65°. Is m|| n?

22

Ans :  Given,

4 =110° and 7 =65°

To find: Is m||n?

Here, 7 = ∠5 = 65°            [Since vertically opposite angles are equal]

Add, ∠4 + ∠5 = 110° + 65° = 175°

4 and ∠5 are the pair of co-interior angles. Then, they should be supplementary to each other. But their sums has resulted in 175° rather than 180°.

Therefore, m is not parallel to n.

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

23

Ans : From the figure;

A+ ∠B = 115 + 65 = 180°

Therefore, AD  || BC [ sum of co interior angles equals 180° or are supplementary]

Similarly,

∠B + ∠C =65+115=180°

Therefore, AB || CD (sum of co interior angles equals 180° or are supplementary]

 

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

Ans :

Given, l||m, n perpendicular to l.

To prove: n is perpendicular to m

Since, l||m and n intersect l.

1 = ∠2                                         [Since the Corresponding angles are equal for parallel lines]

But,  1 = 90°

=> ∠2 =90°

Hence, it is proved, 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.

25

Ans :

Given, as per question,

AB || DE and BC || EF

To prove : ABC= ∠DEF

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

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

ABC = DMP                —–(i)                             [Corresponding angles are equal for parallel lines]

Also, since, BP || EF and DE is the transversal

DMP = DEF             —–(ii)                               [Corresponding angles are equal for parallel lines]

From (i) and (ii), we get,

∠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 °.

 

26

Ans :

Given, as per the question,

AB // DE and BC // EF

To prove: ABC + ∠DEF = 180°

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

27

Proof :

Since AB || EP and BP is the transversal AB and ED.

∠ABC = ∠EPM        —–(i)       [Corresponding angles are equal for parallel lines]

Also,

EF || PM and EP is the transversal.

As we know, co-interior angles for parallel lines are supplementary

∠DEF + ∠EPM = 180°        ………… (ii)

From (i) and (ii) we get,

DEF + ∠ABC = 180°. Hence, proved.

 

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

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

Ans: False

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

Ans: True

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

Ans: False.

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

Ans: True.

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

Ans: 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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