Class 9 R S AGGARWAL AND V AGGARWAL Solutions Maths Chapter 7 - Lines And Angles
Lines And Angles Exercise MCQ
Solution 1
Correct option: (d)
In a right triangle, one angle is 90° and the sum of acute angles of a right triangle is 90°.
Solution 2
Correct option: (b)
Let each interior opposite angle be x.
Then, x + x = 110° (Exterior angle property of a triangle)
⇒ 2x = 110°
⇒ x = 55°
Solution 3
Solution 4
Correct option: (d)
Let ∠A = 130°
In ΔABC, by angle sum property,
∠B + ∠C + ∠A = 180°
⇒ ∠B + ∠C + 130° = 180°
⇒ ∠B + ∠C = 50°
Solution 5
Correct option: (b)
AOB is a straight line.
⇒ ∠AOB = 180°
⇒ 60° + 5x° + 3x° = 180°
⇒ 60° + 8x° = 180°
⇒ 8x° = 120°
⇒ x = 15°
Solution 6
Correct option: (c)
By angle sum property,
2x + 3x + 4x = 180°
⇒ 9x = 180°
⇒ x = 20°
Hence, largest angle = 4x = 4(20°) = 80°
Solution 7
Correct option: (c)
Through B draw YBZ ∥ OA ∥ CD.
Now, OA ∥ YB and AB is the transversal.
⇒ ∠OAB + ∠YBA = 180° (interior angles are supplementary)
⇒ 110° + ∠YBA = 180°
⇒ ∠YBA = 70°
Also, CD ∥ BZ and BC is the transversal.
⇒ ∠DCB + ∠CBZ = 180° (interior angles are supplementary)
⇒ 130° + ∠CBZ = 180°
⇒ ∠CBZ = 50°
Now, ∠YBZ = 180° (straight angle)
⇒ ∠YBA + ∠ABC + ∠CBZ = 180°
⇒ 70° + x + 50° = 180°
⇒ x = 60°
⇒ ∠ABC = 60°
Solution 8
Correct option: (a)
Two angles are said to be complementary, if the sum of their measures is 90°.
Clearly, the measures of each of the angles have to be less than 90°.
Hence, each angle is an acute angle.
Solution 9
Correct option: (d)
An angle which measures more than 180o but less than 360o is called a reflex angle.
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Correct option: (a)
Option (a) is false, since through a given point we can draw an infinite number of straight lines.
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Correct option: (c)
Let ∠AOC = x°
Draw YOZ ∥ CD ∥ AB.
Now, YO ∥ AB and OA is the transversal.
⇒ ∠YOA = ∠OAB = 60° (alternate angles)
Again, OZ ∥ CD and OC is the transversal.
⇒ ∠COZ + ∠OCD = 180° (interior angles)
⇒ ∠COZ + 110° = 180°
⇒ ∠COZ = 70°
Now, ∠YOZ = 180° (straight angle)
⇒ ∠YOA + ∠AOC + ∠COZ = 180°
⇒ 60° + x + 70° = 180°
⇒ x = 50°
⇒ ∠AOC = 50°
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Lines And Angles Exercise Ex. 7A
Solution 1
(i) Angle: Two rays having a common end point form an angle.
(ii) Interior of an angle: The interior of AOB is the set of all points in its plane, which lie on the same side of OA as B and also on same side of OB as A.
(iii) Obtuse angle: An angle whose measure is more than 90o but less than 180o, is called an obtuse angle.
(iv) Reflex angle: An angle whose measure is more than 180o but less than 360o is called a reflex angle.
(v) Complementary angles: Two angles are said to be complementary, if the sum of their measures is 90o.
(vi) Supplementary angles: Two angles are said to be supplementary, if the sum of their measures is 180o.
Solution 2(ii)
Complement of 16o = 90 - 16o = 74o
Solution 2(iv)
Complement of 46o 30' = 90o - 46o 30' = 43o 30'
Solution 2(i)
Complement of 55° = 90° - 55° = 35°
Solution 2(iii)
Complement of 90° = 90° - 90° = 0°
Solution 3(iv)
Supplement of 75o 36' = 180o - 75o 36' = 104o 24'
Solution 3(i)
Supplement of 42° = 180° - 42° = 138°
Solution 3(ii)
Supplement of 90° = 180° - 90° = 90°
Solution 3(iii)
Supplement of 124° = 180° - 124° = 56°
Solution 4
(i) Let the required angle be xo
Then, its complement = 90o - xo
The measure of an angle which is equal to its complement is 45o.
(ii) Let the required angle be xo
Then, its supplement = 180o - xo
The measure of an angle which is equal to its supplement is 90o.
Solution 5
Let the required angle be xo
Then its complement is 90o - xo
The measure of an angle which is 36o more than its complement is 63o.
Solution 6
Let the measure of the required angle = x°
Then, measure of its supplement = (180 - x)°
It is given that
x° = (180 - x)° - 30°
⇒ x° = 180° - x° - 30°
⇒ 2x° = 150°
⇒ x° = 75°
Hence, the measure of the required angle is 75°.
Solution 7
Let the required angle be xo
Then, its complement = 90o - xo
The required angle is 72o.
Solution 8
Let the required angle be xo
Then, its supplement is 180o - xo
The required angle is 150o.
Solution 9
Let the required angle be xo
Then, its complement is 90o - xo and its supplement is 180o - xo
That is we have,The required angle is 60o.
Solution 10
Let the required angle be xo
Then, its complement is 90o - xo and its supplement is 180o - xo
The required angle is 45o.
Solution 11
Let the two required angles be xo and 90o - xo.
Then
5x = 4(90 - x)
5x = 360 - 4x
5x + 4x = 360
9x = 360
Thus, the required angles are 40o and 90o - xo = 90 o - 40o = 50o.
Solution 12
(2x - 5)° and (x - 10)° are complementary angles.
∴ (2x - 5)° + (x - 10)° = 90°
⇒ 2x - 5° + x - 10° = 90°
⇒ 3x - 15° = 90°
⇒ 3x = 105°
⇒ x = 35°
Lines And Angles Exercise Ex. 7B
Solution 1
Since BOC and COA form a linear pair of angles, we have
BOC + COA = 180o
xo + 62o = 180o
x = 180 - 62
x = 118o
Solution 2
∠AOB is a straight angle.
⇒ ∠AOB = 180°
⇒ ∠AOC + ∠COD + ∠BOD = 180°
⇒ (3x - 7)° + 55° + (x + 20)° = 180°
⇒ 4x + 68° = 180°
⇒ 4x = 112°
⇒ x = 28°
Thus, ∠AOC = (3x - 7)° = 3(28°) - 7° = 84° - 7° = 77°
And, ∠BOD = (x + 20)° = 28° + 20° = 48°
Solution 3
Since BOD and DOA from a linear pair of angles.
BOD + DOA = 180o
BOD + DOC + COA = 180o
xo + (2x - 19)o + (3x + 7)o = 180o
6x - 12 = 180
6x = 180 + 12 = 192
x = 32
AOC = (3x + 7)o = (3 32 + 7)o = 103o
COD = (2x - 19)o = (2 32 - 19)o = 45o
and BOD = xo = 32o
Solution 4
x: y: z = 5: 4: 6
The sum of their ratios = 5 + 4 + 6 = 15
But x + y + z = 180o
[Since, XOY is a straight line]
So, if the total sum of the measures is 15, then the measure of x is 5.
If the sum of angles is 180o, then, measure of
And, if the total sum of the measures is 15, then the measure of y is 4.
If the sum of the angles is 180o, then, measure of
And z = 180o - x - y
= 180o - 60o - 48o
= 180o - 108o = 72o
x = 60, y = 48 and z = 72.
Solution 5
AOB will be a straight line, if two adjacent angles form a linear pair.
BOC + AOC = 180o
(4x - 36)o + (3x + 20)o = 180o
4x - 36 + 3x + 20 = 180
7x - 16 = 180o
7x = 180 + 16 = 196
The value of x = 28.
Solution 6
Since AOC and AOD form a linear pair.
AOC + AOD = 180o
50o + AOD = 180o
AOD = 180o - 50o = 130o
AOD and BOC are vertically opposite angles.
AOD = BOC
BOC = 130o
BOD and AOC are vertically opposite angles.
BOD = AOC
BOD = 50o
Solution 7
Since COE and DOF are vertically opposite angles, we have,
COE = DOF
z = 50o
Also BOD and COA are vertically opposite angles.
So, BOD = COA
t = 90o
As COA and AOD form a linear pair,
COA + AOD = 180o
COA + AOF + FOD = 180o [t = 90o]
t + x + 50o = 180o
90o + xo + 50o = 180o
x + 140 = 180
x = 180 - 140 = 40
Since EOB and AOF are vertically opposite angles
So, EOB = AOF
y = x = 40
Thus, x = 40 = y = 40, z = 50 and t = 90
Solution 8
Since COE and EOD form a linear pair of angles.
COE + EOD = 180o
COE + EOA + AOD = 180o
5x + EOA + 2x = 180
5x + BOF + 2x = 180
[EOA and BOF are vertically opposite angles so, EOA = BOF]
5x + 3x + 2x = 180
10x = 180
x = 18
Now AOD = 2xo = 2 18o = 36o
COE = 5xo = 5 18o = 90o
and, EOA = BOF = 3xo = 3 18o = 54o
Solution 9
Let the two adjacent angles be 5x and 4x.
Now, since these angles form a linear pair.
So, 5x + 4x = 180o
9x = 180o
The required angles are 5x = 5x = 5 20o = 100o
and 4x = 4 20o = 80o
Solution 10
Let two straight lines AB and CD intersect at O and let AOC = 90o.
Now, AOC = BOD [Vertically opposite angles]
BOD = 90o
Also, as AOC and AOD form a linear pair.
90o + AOD = 180o
AOD = 180o - 90o = 90o
Since, BOC = AOD [Verticallty opposite angles]
BOC = 90o
Thus, each of the remaining angles is 90o.
Solution 11
Since, AOD and BOC are vertically opposite angles.
AOD = BOC
Now, AOD + BOC = 280o [Given]
AOD + AOD = 280o
2AOD = 280o
AOD =
BOC = AOD = 140o
As, AOC and AOD form a linear pair.
So, AOC + AOD = 180o
AOC + 140o = 180o
AOC = 180o - 140o = 40o
Since, AOC and BOD are vertically opposite angles.
AOC = BOD
BOD = 40o
BOC = 140o, AOC = 40o , AOD = 140o and BOD = 40o.
Solution 12
Let ∠AOC = 5x and ∠AOD = 7x
Now, ∠AOC + ∠AOD = 180° (linear pair of angles)
⇒ 5x + 7x = 180°
⇒ 12x = 180°
⇒ x = 15°
⇒ ∠AOC = 5x = 5(15°) = 75° and ∠AOD = 7x = 7(15°) = 105°
Now, ∠AOC = ∠BOD (vertically opposite angles)
⇒ ∠BOD = 75°
Also, ∠AOD = ∠BOC (vertically opposite angles)
⇒ ∠BOC = 105°
Solution 13
∠BOD = 40°
⇒ AOC = ∠BOD = 40° (vertically opposite angles)
∠AOE = 35°
⇒ ∠BOF = ∠AOE = 35° (vertically opposite angles)
∠AOB is a straight angle.
⇒ ∠AOB = 180°
⇒ ∠AOE + ∠EOD + ∠BOD = 180°
⇒ 35° + ∠EOD + 40° = 180°
⇒ ∠EOD + 75° = 180°
⇒ ∠EOD = 105°
Now, ∠COF = ∠EOD = 105° (vertically opposite angles)
Solution 14
∠AOC + ∠BOC = 180° (linear pair of angles)
⇒ x + 125 = 180°
⇒ x = 55°
Now, ∠AOD = ∠BOC (vertically opposite angles)
⇒ y = 125°
Also, ∠BOD = ∠AOC (vertically opposite angles)
⇒ z = 55°
Solution 15
Given : AB and CD are two lines which are intersecting at O. OE is a ray bisecting the BOD. OF is a ray opposite to ray OE.
To Prove: AOF = COF
Proof : Since are two opposite rays, is a straight line passing through O.
AOF = BOE
and COF = DOE
[Vertically opposite angles]
But BOE = DOE (Given)
AOF = COF
Hence, proved.
Solution 16
Given: is the bisector of BCD and is the bisector of ACD.
To Prove: ECF = 90o
Proof: Since ACD and BCD forms a linear pair.
ACD + BCD = 180o
ACE + ECD + DCF + FCB = 180o
ECD + ECD + DCF + DCF = 180o
because ACE = ECD
and DCF = FCB
2(ECD) + 2 (CDF) = 180o
2(ECD + DCF) = 180o
ECD + DCF =
ECF = 90o (Proved)
Lines And Angles Exercise Ex. 7C
Solution 1
Given, ∠1 = 120°
Now, ∠1 + ∠2 = 180° (linear pair)
⇒ 120° + ∠2 = 180°
⇒ ∠2 = 60°
∠1 = ∠3 (vertically opposite angles)
⇒ ∠3 = 120°
Also, ∠2 = ∠4 (vertically opposite angles)
⇒ ∠4 = 60°
Line l ∥ line m and line t is a transversal.
⇒ ∠5 = ∠1 = 120° (corresponding angles)
∠6 = ∠2 = 60° (corresponding angles)
∠7 = ∠3 = 120° (corresponding angles)
∠8 = ∠4 = 60° (corresponding angles)
Solution 2
Given, ∠7 = 80°
Now, ∠7 + ∠8 = 180° (linear pair)
⇒ 80° + ∠8 = 180°
⇒ ∠8 = 100°
∠7 = ∠5 (vertically opposite angles)
⇒ ∠5 = 80°
Also, ∠6 = ∠8 (vertically opposite angles)
⇒ ∠6 = 100°
Line l ∥ line m and line t is a transversal.
⇒ ∠1 = ∠5 = 80° (corresponding angles)
∠2 = ∠6 = 100° (corresponding angles)
∠3 = ∠7 = 80° (corresponding angles)
∠4 = ∠8 = 100° (corresponding angles)
Solution 3
Given, ∠1 : ∠2 = 2 : 3
Now, ∠1 + ∠2 = 180° (linear pair)
⇒ 2x + 3x = 180°
⇒ 5x = 180°
⇒ x = 36°
⇒ ∠1 = 2x = 72° and ∠2 = 3x = 108°
∠1 = ∠3 (vertically opposite angles)
⇒ ∠3 = 72°
Also, ∠2 = ∠4 (vertically opposite angles)
⇒ ∠4 = 108°
Line l ∥ line m and line t is a transversal.
⇒ ∠5 = ∠1 = 72° (corresponding angles)
∠6 = ∠2 = 108° (corresponding angles)
∠7 = ∠3 = 72° (corresponding angles)
∠8 = ∠4 = 108° (corresponding angles)
Solution 4
Lines l and m will be parallel if 3x - 20 = 2x + 10
[Since, if corresponding angles are equal, lines are parallel]
3x - 2x = 10 + 20
x = 30
Solution 5
For lines l and m to be parallel to each other, the corresponding angles (3x + 5)° and (4x)° should be equal.
⇒ (3x + 5)° = 4x°
⇒ x = 5°
Solution 6
Since AB || CD and BC is a transversal.
So, BCD = ABC = xo [Alternate angles]
As BC || ED and CD is a transversal.
BCD + EDC = 180o
BCD + 75o =180o
BCD = 180o - 75o = 105o
ABC = 105o [since BCD = ABC]
xo = ABC = 105o
Hence, x = 105.
Solution 7
Since AB || CD and BC is a transversal.
So, ABC = BCD [atternate interior angles]
70o = xo + ECD(i)
Now, CD || EF and CE is transversal.
So,ECD + CEF = 180o [sum of consecutive interior angles is 180o]
ECD + 130o = 180o
ECD = 180o - 130o = 50o
Putting ECD = 50o in (i) we get,
70o = xo + 50o
x = 70 - 50 = 20
Solution 8
AB ∥ CD and EF is transversal.
⇒ ∠AEF = ∠EFG (alternate angles)
Given, ∠AEF = 75°
⇒ ∠EFG = y = 75°
Now, ∠EFC + ∠EFG = 180° (linear pair)
⇒ x + y = 180°
⇒ x + 75° = 180°
⇒ x = 105°
∠EGD = ∠EFG + ∠FEG (Exterior angle property)
⇒ 125° = y + z
⇒ 125° = 75° + z
⇒ z = 50°
Thus, x = 105°, y = 75° and z = 50°
Solution 9(i)
Through E draw EG || CD. Now since EG||CD and ED is a transversal.
So,GED = EDC = 65o[Alternate interior angles]
Since EG || CD and AB || CD,
EG||AB and EB is transversal.
So,BEG = ABE = 35o[Alternate interior angles]
So,DEB = xo
BEG + GED = 35o + 65o = 100o.
Hence, x = 100.
Solution 9(ii)
Through O draw OF||CD.
Now since OF || CD and OD is transversal.
CDO + FOD = 180o
[sum of consecutive interior angles is 180o]
25o + FOD = 180o
FOD = 180o - 25o = 155o
As OF || CD and AB || CD [Given]
Thus, OF || AB and OB is a transversal.
So,ABO + FOB = 180o [sum of consecutive interior angles is 180o]
55o + FOB = 180o
FOB = 180o - 55o = 125o
Now, xo = FOB + FOD = 125o + 155o = 280o.
Hence, x = 280.
Solution 9(iii)
Through E, draw EF || CD.
Now since EF || CD and EC is transversal.
FEC + ECD = 180o
[sum of consecutive interior angles is 180o]
FEC + 124o = 180o
FEC = 180o - 124o = 56o
Since EF || CD and AB ||CD
So, EF || AB and AE is a trasveral.
So,BAE + FEA = 180o
[sum of consecutive interior angles is 180o]
116o + FEA = 180o
FEA = 180o - 116o = 64o
Thus,xo = FEA + FEC
= 64o + 56o = 120o.
Hence, x = 120.
Solution 10
Through C draw FG || AE
Now, since CG || BE and CE is a transversal.
So, GCE = CEA = 20o [Alternate angles]
DCG = 130o - GCE
= 130o - 20o = 110o
Also, we have AB || CD and FG is a transversal.
So, BFC = DCG = 110o [Corresponding angles]
As, FG || AE, AF is a transversal.
BFG = FAE [Corresponding angles]
xo = FAE = 110o.
Hence, x = 110
Solution 11
Since AB || PQ and EF is a transversal.
So, CEB = EFQ [Corresponding angles]
EFQ = 75o
EFG + GFQ = 75o
25o + yo = 75o
y = 75 - 25 = 50
Also, BEF + EFQ = 180o [sum of consecutive interior angles is 180o]
BEF = 180o - EFQ= 180o - 75o
BEF = 105o
FEG + GEB = BEF = 105o
FEG = 105o - GEB = 105o - 20o = 85o
In EFG we have,
xo + 25o + FEG = 180o
Hence, x = 70.
Solution 12
Since AB || CD and AC is a transversal.
So, BAC + ACD = 180o [sum of consecutive interior angles is 180o]
ACD = 180o - BAC
= 180o - 75o = 105o
ECF = ACD [Vertically opposite angles]
ECF = 105o
Now in CEF,
ECF + CEF + EFC =180o
x = 180 - 30 - 105 = 45
Hence, x = 45.
Solution 13
Since AB || CD and PQ a transversal.
So, PEF = EGH [Corresponding angles]
EGH = 85o
EGH and QGH form a linear pair.
So, EGH + QGH = 180o
QGH = 180o - 85o = 95o
Similarly, GHQ + 115o = 180o
GHQ = 180o - 115o = 65o
In GHQ, we have,
xo + 65o + 95o = 180o
x = 180 - 65 - 95 = 180 - 160
x = 20
Solution 14
Since AB || CD and BC is a transversal.
So, ABC = BCD
x = 35
Also, AB || CD and AD is a transversal.
So, BAD = ADC
z = 75
In ABO, we have,
xo + 75o + yo = 180o
35 + 75 + y = 180
y = 180 - 110 = 70
x = 35, y = 70 and z = 75.
Solution 16
Through F, draw KH || AB || CD
Now, KF || CD and FG is a transversal.
KFG = FGD = ro (i)
[alternate angles]
Again AE || KF, and EF is a transversal.
So,AEF + KFE = 180o
KFE = 180o - po (ii)
Adding (i) and (ii) we get,
KFG + KFE = 180 - p + r
EFG = 180 - p + r
q = 180 - p + r
i.e.,p + q - r = 180
Solution 17
PRQ = xo = 60o [vertically opposite angles]
Since EF || GH, and RQ is a transversal.
So, x = y [Alternate angles]
y = 60
AB || CD and PR is a transversal.
So, [Alternate angles]
[since ]
x + QRD = 110o
QRD = 110o - 60o = 50o
In QRS, we have,
QRD + to + yo = 180o
50 + t + 60 = 180
t = 180 - 110 = 70
Since, AB || CD and GH is a transversal
So, zo = to = 70o [Alternate angles]
x = 60 , y = 60, z = 70 and t = 70
Solution 18
AB ∥ CD and a transversal t cuts them at E and F respectively.
⇒ ∠BEF + ∠DFE = 180° (interior angles)
⇒ ∠GEF + ∠GFE = 90° ….(i)
Now, in ΔGEF, by angle sum property
∠GEF + ∠GFE + ∠EGF = 180°
⇒ 90° + ∠EGF = 180° ….[From (i)]
⇒ ∠EGF = 90°
Solution 19
Since AB ∥ CD and t is a transversal, we have
∠AEF = ∠EFD (alternate angles)
⇒ ∠PEF = ∠EFQ
But, these are alternate interior angles formed when the transversal EF cuts EP and FQ.
∴ EP ∥ FQ
Solution 20
Construction: Produce DE to meet BC at Z.
Now, AB ∥ DZ and BC is the transversal.
⇒ ∠ABC = ∠DZC (corresponding angles) ….(i)
Also, EF ∥ BC and DZ is the transversal.
⇒ ∠DZC = ∠DEF (corresponding angles) ….(ii)
From (i) and (ii), we have
∠ABC = ∠DEF
Solution 21
Construction: Produce ED to meet BC at Z.
Now, AB ∥ EZ and BC is the transversal.
⇒ ∠ABZ + ∠EZB = 180° (interior angles)
⇒ ∠ABC + ∠EZB = 180° ….(i)
Also, EF ∥ BC and EZ is the transversal.
⇒ ∠BZE = ∠ZEF (alternate angles)
⇒ ∠BZE = ∠DEF ….(ii)
From (i) and (ii), we have
∠ABC + ∠DEF = 180°
Solution 22
Let the normal to mirrors m and n intersect at P.
Now, OB ⊥ m, OC ⊥ n and m ⊥ n.
⇒ OB ⊥ OC
⇒ ∠APB = 90°
⇒ ∠2 + ∠3 = 90° (sum of acute angles of a right triangle is 90°)
By the laws of reflection, we have
∠1 = ∠2 and ∠4 = ∠3 (angle of incidence = angle of reflection)
⇒ ∠1 + ∠4 = ∠2 + ∠3 = 90°
⇒ ∠1 + ∠2 + ∠3 + ∠4 = 180°
⇒ ∠CAB + ∠ABD = 180°
But, ∠CAB and ∠ABD are consecutive interior angles formed, when the transversal AB cuts CA and BD.
∴ CA ∥ BD
Solution 23
In the given figure,
∠BAC = ∠ACD = 110°
But, these are alternate angles when transversal AC cuts AB and CD.
Hence, AB ∥ CD.
Solution 24
Let the two parallel lines be m and n.
Let p ⊥ m.
⇒ ∠1 = 90°
Let q ⊥ n.
⇒ ∠2 = 90°
Now, m ∥ n and p is a transversal.
⇒ ∠1 = ∠3 (corresponding angles)
⇒ ∠3 = 90°
⇒ ∠3 = ∠2 (each 90°)
But, these are corresponding angles, when transversal n cuts lines p and q.
∴ p ∥ q.
Hence, two lines which are perpendicular to two parallel lines, are parallel to each other.