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. Answer:

Lines AB and CD intersect at O. ∠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. 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 Question 3:

In the figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠ PRT. 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 Question 4:

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

From the figure, ∠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. 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. Answer:

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

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

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

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

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

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

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