Class 9 - Maths - Lines and Angles

                                                                      Exercise 6.1

Question 1:

In the figure lines AB and CD intersect at O.  If ∠ AOC + ∠ BOE = 700 and ∠ BOD = 400,find ∠ BOE and reflex ∠ COE.

Class_9_Lines_Angles_Figure1

Answer:

Lines AB and CD intersect at O.

Class_9_Lines_Angles_Figure2

∠AOC + ∠BOE = 700            (Given) ………………1

∠BOD = 400                          (Given) ………………2

Since, ∠AOC = ∠BOD           (Vertically opposite angles)

Therefore, ∠AOC = 400      [From equation 2]

and 400 + ∠BOE = 700        [From equation 1]

=> ∠BOE = 700 – 400 = 300

Also, ∠AOC + ∠BOE + ∠COE = 1800         [Since AOB is a straight line]

=> 700 + ∠COE = 1800        [Form equation 1]

=> ∠COE = 1800 – 700 = 1100

Now, reflex ∠COE = 3600 – 1100 = 2500

Hence, ∠BOE = 300 and reflex ∠COE = 2500

 

Question 2:

In the figure, lines XY and MN intersect at O. If ∠POY = 900 and a : b = 2 : 3, find c.

          Class_9_Lines_Angles_Figure3                                            

Answer:

In the figure, lines XY and MN intersect at O and ∠ POY = 900.

Also, given a : b = 2 : 3

Let a = 2x and b = 3x.

Since, ∠XOM + ∠POM + ∠POY = 1800      [Linear pair axiom]

=> 3x + 2x + 900 = 1800

=> 5x = 1800 – 900

=> x = 900/50 = 180

So, ∠XOM = b = 3x = 3 * 180 = 540

and ∠POM = a = 2x = 2 * 180 = 360

Now, ∠XON = c = ∠MOY = ∠POM + ∠POY  [Vertically opposite angles]

=> c = 360 + 900 = 1260

Hence, c = 1260

 Class_9_Lines_Angles_Figure4

Question 3:

In the figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠ PRT.

Class_9_Lines_Angles_Triangle1

Answer:

∠PQS + ∠PQR = 180° ……………1      (Linear pair axiom)

∠PRQ + ∠PRT = 180° ……………2       (Linear pair axiom)

But, ∠PQR = ∠PRQ               (Given)   

From equation 1 and 2, we get  

∠PQS = ∠PRT

Class_9_Lines_Angles_Triangle2

Question 4:

In the figure, if x + y = w + z, then prove that AOB is a line.

             Class_9_Lines_Angles_Figure5                                    

Answer:

           Class_9_Lines_Angles_Figure6                                       

Assume AOB is a line.

Therefore, x + y = 180°  …………….1         [Linear pair axiom]

and w + z = 180° .......................2            [Linear pair axiom]

Now, from equation 1 and 2, we get

x + y = w + z

Question 5:

In the figure, POQ is a line. Ray OR is perpendicular to line PQ.OS is another ray lying between rays OP and OR.

Prove that ∠ROS = (∠QOS – ∠ POS)/2

Class_9_Lines_Angles_Figure7

Answer:

From the figure,

Class_9_Lines_Angles_Figure7

∠ROS = ∠ROP – ∠POS   ……………..1

and ∠ROS = ∠QOS – ∠QOR   .......2

Adding equation 1 and 2, we get

∠ROS + ∠ROS = ∠QOS – ∠QOR + ∠ROP – ∠POS

=> 2∠ROS = ∠QOS – ∠POS            [Since ∠QOR = ∠ROP = 900]

=> ∠ROS = (∠QOS – ∠POS)/2

Question 6:

It is given that ∠ XYZ = 640 and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ ZYP, find ∠ XYQ and reflex ∠QYP.

Class_9_Lines_Angles_Figure8

Answer:

From the figure,

∠XYZ = 640             (Given)

Now, ∠ZYP + ∠XYZ = 1800             [Linear pair axiom]

=> ∠ZYP + 640 = 1800

=> ∠ZYP = 1800 – 640 – 1160

Also, given that ray YQ bisects ∠ZYP.

But, ∠ZYP = ∠QYP = ∠QYZ = 1160

Therefore, ∠QYP = 580 and ∠QYZ = 580

Also, ∠XYQ = ∠XYZ + ∠QYZ

=> ∠XYQ = 640 + 580 = 1220

and reflex ∠QYP = 3600 – ∠QYP = 3600 – 580 = 3020          [Since ∠QYP = 580]

Hence, ∠XYQ = 1220 and reflex ∠QYP = 3020

 

                                                                       Exercise 6.2

Question 1:

In the figure, find the values of x and y and then show that AB || CD.

  Class_9_Lines_Angles_Figure9                                                     

Answer:

In the given figure, a transversal intersects two lines AB and CD such that

Class_9_Lines_Angles_Figure9

      x + 500 = 1800                [Linear pair axiom]

=> x = 1800 – 500 = 1300

and y = 1300                       [Vertically opposite angles]

Therefore, ∠x = ∠y = 1300    [Alternate angles]

Hence, AB || CD                    [Converse of alternate angles axiom]

Question 2:

In the figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.

       Class_9_Lines_Angles_Figure11                                             

 

Answer:

In the given figure, AB || CD, CD || EF and y : z = 3 : 7.

Class_9_Lines_Angles_Figure10

Let y = 3a and z = 7a

∠DHI = y                         [vertically opposite angles]

∠DHI + ∠FIH = 1800      [Interior angles on the same side of the transversal]

=> y + z = 1800

=> 3a + 7a = 1800

=> 10a = 1800

=> a = 180

So, y = 3 * 18° = 540

and z = 180 * 7 = 1260

Also, x + y = 1800

=> x + 540 = 1800

=> x = 1800 – 540 = 1260

Hence, x = 1260

Question 3:

In the figure, if AB || CD, EF ⊥ CD and ∠GED = 126°. Find ∠ AGE, ∠GEF and ∠FGE.

Class_9_Lines_Angles_Figure12

Answer:

In the given figure, AB || CD, EF ⊥ CD and ∠GED = 1260

Class_9_Lines_Angles_Figure12

∠AGE = ∠LGE                   [Alternate angle]

So, ∠AGE = 1260

Now, ∠GEF = ∠GED – ∠DEF

                      = 1260 – 900 = 360    [Since ∠DEF = 900]

Also, ∠AGE + ∠FGE = 1800           [Linear pair axiom]

=> 1260 + FGE = 1800

=> ∠FGE = 1800 – 1260 = 540

Question 4:

In the figure, if PQ || ST, ∠PQR = 110and ∠ RST = 1300, find ∠QRS.

Class_9_Lines_Angles_Figure13

Answer:

Extend PQ to Y and draw LM || ST through R.

Class_9_Lines_Angles_Figure14

∠TSX = ∠QXS                      [Alternate angles]

=> ∠QXS = 1300

∠QXS + ∠RXQ = 1800       [Linear pair axiom]

=> ∠RXQ = 1800 – 1300 = 500    …………..1

∠PQR = ∠QRM                 [Alternate angles]

=> ∠QRM = 1100   ………………2

∠RXQ = ∠XRM                 [Alternate angles]

=> ∠XRM = 500                [from equation 1]

∠QRS = ∠QRM – ∠XRM

           = 1100 – 500 = 600

 

Question 5:

In the figure, if AB || CD, ∠APQ = 500 and ∠PRD = 127°, find x and y.

 Class_9_Lines_Angles_Figure15

Answer:

In the given figure, AB || CD, ∠APQ = 500 and ∠PRD = 1270

Class_9_Lines_Angles_Figure15

∠APQ + ∠PQC = 1800                     [Pair of consecutive interior angles are supplementary]

=> 500 + ∠PQC = 1800

=> ∠PQC = 1800 – 500 = 1300

Now, ∠PQC + ∠PQR = 1800          [Linear pair axiom]

=> 1300 + x = 1800

=> x = 1800 – 1300 = 500

Also, x + y = 1270   [Exterior angle of a triangle is equal to the sum of the two interior opposite                         

                                  angles]

=> 500 + y = 1270

=> y = 1270 – 500 = 770

Hence, x = 500 and y = 770

Question 6:

In the figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along

the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.

Class_9_Lines_Angles_Figure16

Answer:

At point B, draw BE ⊥ PQ and at point C, draw CF ⊥ RS.

Class_9_Lines_Angles_Figure17

∠1 = ∠2 …………..1          [Angle of incidence is equal to angle of reflection]

∠3 = ∠4 …………..2          [Angle of incidence is equal to angle of reflection]

Also, ∠2 = ∠3 ......3        [Alternate angles]

=> ∠1 = ∠4                      [From equation 1, 2, and 3]

=> 2∠1 = 2∠4

=> ∠1 + ∠1 = ∠4 + ∠4

=> ∠1 + ∠2 = ∠3 + ∠4       [From (i) and (ii)]

=> ∠BCD = ∠ABC

Hence, AB || CD.          [Alternate angles are equal]    

Hence, Proved.

 

                                                                   Exercise 6.3

Question 1:

In the figure, sides QP and RQ of ∆ PQR are produced to points S and T respectively. If ∠SPR = 1350 and ∠PQT = 1100, find ∠PRQ.

Class_9_Lines_Angles_Figure18

Answer:

In the given figure, ∠SPR = 1350 and ∠PQT = 1100.

Class_9_Lines_Angles_Figure18

∠PQT + ∠PQR = 1800                     [Linear pair axiom]

=> 1100 + ∠PQR = 1800

=> ∠PQR = 1800 – 1100 = 700

Also, ∠SPR + ∠QPR = 1800           [Linear pair axiom]

=> 1350 + ∠QPR = 1800

=> ∠QPS = 1800 – 1350 = 450

Now, in the triangle PQR

∠PQR + ∠PRQ + ∠QPR = 1800    [Angle sum property of a triangle]

=> 700 + ∠PRQ + 450 = 1800

=> ∠PRQ + 1150 = 1800

=> ∠PRQ = 1800 – 1150 = 650

Hence, ∠ PRQ = 650

 

Question 2:

In the figure, ∠ X = 620, ∠ XYZ = 540. If YO and ZO are the bisectors of ∠ XYZ and ∠ XZY respectively of

∆ XYZ, find ∠OZY and ∠YOZ.

Answer:

In the given figure,

Class_9_Lines_Angles_Figure19

∠X = 620 and ∠XYZ = 540.

∠XYZ + ∠XZY + ∠YXZ = 1800 ………1    [Angle sum property of a triangle]

=> 540 + ∠XZY + 620 = 1800

=> ∠XZY + 1160 = 1800

=> ∠XZY = 1800 – 1160 = 640

Now, ∠OZY = 1/2 * ∠XZY                   [Since ZO is bisector of ∠XZY]

                      = 1/2 * 640 = 320

Similarly, ∠OYZ = 1/2 * 540 = 270

Now, in ∆OYZ, we have

∠OYZ + ∠OZY + ∠YOZ = 1800           [Angle sum property of a triangle]

=> 270 + 320 + ∠YOZ = 1800

=> ∠YOZ = 1800 – 590 = 1210

Hence, ∠OZY = 320 and ∠YOZ = 1210

 

Question 3:

In the figure, if AB || DE, ∠ BAC = 35and ∠CDE = 530, find ∠ DCE.

Answer:

In the given figure,

Class_9_Lines_Angles_Figure20

∠BAC = ∠CED                            [Alternate angles] 

=> ∠CED = 350

In ∆CDE,

∠CDE + ∠DCE + ∠CED = 1800     [Angle sum property of a triangle]

=> 530 + ∠DCE + 350 = 1800

=> ∠DCE + 880 = 1800

=> ∠DCE = 1800 – 880 = 920

Hence, ∠DCE = 920

Question 4:

In the figure, if lines PQ and RS intersect at point T, such that ∠ PRT = 400, ∠ RPT = 950

and ∠TSQ = 750, find ∠SQT.

Class_9_Lines_Angles_Figure21

Answer:

In the given figure, lines PQ and RS

intersect at point T, such that ∠PRT = 400,

∠RPT = 950 and ∠TSQ = 750.

In ∆PRT

∠PRT + ∠RPT + ∠PTR = 1800      [Angle sum property of a triangle]

=> 400 + 950 + ∠PTR = 1800

=> 1350 + ∠PTR = 1800

=> ∠PTR = 1800 – 1350 = 450

Also, ∠PTR = ∠STQ                     [Vertical opposite angles]

So, ∠STQ = 450

Now, in ∆STQ,

∠STQ + ∠TSQ + ∠SQT = 180°    [Angle sum property of a triangle]

=> 450 + 750 + ∠SQT = 1800

=> 1200 + ∠SQT = 1800

=> ∠SQT = 1800 – 1200 = 600

Hence, ∠SQT = 600

Question 5:

In the figure, if PT ⊥ PS, PQ || SR, ∠SQR = 28and ∠QRT = 650, then find the values of x and y.

Answer:

In the given figure, lines PQ ⊥ PS, PQ || SR,  

Class_9_Lines_Angles_Figure22

∠SQR = 280 and ∠QRT = 650

∠PQR = ∠QRT                    [Alternate angles]

=> x + 280 = 650

=> x = 650 – 280 = 370

In ∆PQS,

∠SPQ + ∠PQS + ∠QSP = 1800        [Angle sum property of a triangle]

=> 900 + 370 + y = 1800     [Since PQ ⊥ PS, ∠PQS = x = 370 and ∠QSP = y)

=> 1270 + y = 1800

=> y = 1800 – 1270 = 530

Hence, x = 370 and y = 530

Question 6:

In the figure, the side QR of ∆PQR is produced to a point S. If the bisectors of ∠ PQR and ∠ PRS meet at point T, then prove that ∠QTR = ∠QPR/2.

 Class_9_Lines_Angles_Figure23

Answer:

Exterior ∠PRS = ∠PQR + ∠QPR                     [Exterior angle property]

Therefore, ∠PRS/2 = ∠PQR/2 + ∠QPR/2

=> ∠TRS = ∠TQR + ∠QPR/2   ……………1

But in ∆QTR,

Exterior ∠TRS = ∠TQR + ∠QTR   ……….2      [Exterior angles property]

Therefore, from equation 1 and 2, we get

∠TQR + ∠QTR = ∠TQR + ∠QPR/2

=> ∠QTR = ∠QPR/2

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