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NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles – Kunji

## NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1.

## NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1

Ex 6.1 Class 9 Maths Question 1

In figure, lines AB and CD intersect at 0. If AOC + BOE = 70° and BOD = 40°, find BOE and reflex COE.

Solution:
Since AB is a straight line,
AOC + COE + EOB = 180°
or (AOC + BOE) + COE = 180° or 70° + COE = 180° [ ∵∠AOC + BOE = 70° (Given)]
or COE = 180° – 70° = 110°
Reflex COE = 360° – 110° = 250°
Also, AB and CD intersect at O.
∴∠COA = BOD [Vertically opposite angles]
But BOD = 40° [Given] COA = 40°
Also, AOC + BOE = 70°
40° + BOE = 70° or BOE = 70° -40° = 30°
Thus, BOE = 30° and reflex COE = 250°.

Ex 6.1 Class 9 Maths Question 2.
In figure, lines XY and MN intersect at 0. If POY = 90° , and a : b = 2 : 3. find c.

Solution:
Since XOY is a straight line.
b+a+POY= 180°
But POY = 90° [Given]
b + a = 180° – 90° = 90° …(i)
Also a : b = 2 : 3 b = 3a2 …(ii)
Now from (i) and (ii), we get
3a2 + A = 90°
5a2 = 90°
a = 905×2=36 = 36°
From (ii), we get
b = 32 x 36° = 54°
Since XY and MN interstect at O c = [a + POY] [Vertically opposite angles]
or c = 36° + 90° = 126°
Thus, the required measure of c = 126°.

Ex 6.1 Class 9 Maths Question 3.
In figure, PQR = PRQ, then prove that PQS = PRT.

Solution:
ST is a straight line.
PQR + PQS = 180° …(1) [Linear pair]
Similarly, PRT + PRQ = 180° …(2) [Linear Pair]

From (1) and (2), we have
PQS + PQR = PRT + PRQ
But PQR = PRQ [Given]
PQS = PRT

Ex 6.1 Class 9 Maths Question 4.
In figure, if x + y = w + , then prove that AOB is a line.

Solution:
Sum of all the angles at a point = 360°
x + y + + w = 360° or, (x + y) + ( + w) = 360°
But (x + y) = ( + w) [Given]
(x + y) + (x + y) = 360° or,
2(x + y) = 360°
or, (x + y) = 3602 = 180°
AOB is a straight line.

Ex 6.1 Class 9 Maths Question 5.
In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

Solution:
rara POQ is a straight line. [Given]
POS + ROS + ROQ = 180°
But OR PQ
ROQ = 90°
POS + ROS + 90° = 180°
POS + ROS = 90°
ROS = 90° – POS … (1)
Now, we have ROS + ROQ = QOS
ROS + 90° = QOS
ROS = QOS – 90° ……(2)
Adding (1) and (2), we have
2 ROS = (QOS – POS) ROS = 12(QOSPOS)

Ex 6.1 Class 9 Maths Question 6.
It is given that XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ZYP, find XYQ and reflex QYP.
Solution:
XYP is a straight line.

∴ ∠XYZ + ∠ZYQ + ∠QYP = 180°
⇒ 64° + ∠ZYQ + ∠QYP = 180°
[∵ ∠XYZ = 64° (given)]
⇒ 64° + 2∠QYP = 180°
[YQ bisects ∠ZYP so, ∠QYP = ∠ZYQ]
⇒ 2∠QYP = 180° – 64° = 116°
⇒ ∠QYP = 116∘2 = 58°
∴ Reflex ∠QYP = 360° – 58° = 302°
Since ∠XYQ = ∠XYZ + ∠ZYQ
⇒ ∠XYQ = 64° + ∠QYP [∵∠XYZ = 64°(Given) and ∠ZYQ = ∠QYP]
⇒ ∠XYQ = 64° + 58° = 122° [∠QYP = 58°]
Thus, ∠XYQ = 122° and reflex ∠QYP = 302°.

### NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2

Ex 6.2 Class 9 Maths Question 1.
In figure, find the values of x and y and then show that AB || CD.

Solution:
In the figure, we have CD and PQ intersect at F.

y = 130° …(1)
[Vertically opposite angles]
Again, PQ is a straight line and EA stands on it.
AEP + AEQ = 180° [Linear pair]
or 50° + x = 180°
x = 180° – 50° = 130° …(2)
From (1) and (2), x = y
As they are pair of alternate interior angles.
AB || CD

Ex 6.2 Class 9 Maths Question 2.
In figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Solution:
AB || CD, and CD || EF [Given]
AB || EF
x = z [Alternate interior angles] ….(1)
Again, AB || CD
x + y = 180° [Co-interior angles]
z + y = 180° … (2) [By (1)]
But y : z = 3 : 7
z = 73 y = 73(180°- z) [By (2)] 10z = 7 x 180°
z = 7 x 180° /10 = 126°
From (1) and (3), we have
x = 126°.

Ex 6.2 Class 9 Maths Question 3.
In figure, if AB || CD, EF CD and GED = 126°, find AGE, GEF and FGE.

Solution:
AB || CD and GE is a transversal.
AGE = GED [Alternate interior angles]
But GED = 126° [Given]
∴∠AGE = 126°
Also, GEF + FED = GED
or GEF + 90° = 126° [ EF CD (given)]
x = z [Alternate interior angles]… (1) Again, AB || CD
x + y = 180° [Co-interior angles]
GEF = 126° -90° = 36°
Now, AB || CD and GE is a transversal.
FGE + GED = 180° [Co-interior angles]
or FGE + 126° = 180°
or FGE = 180° – 126° = 54°
Thus, AGE = 126°, GEF=36° and FGE = 54°.

Ex 6.2 Class 9 Maths Question 4.
In figure, if PQ || ST, PQR = 110° and RST = 130°, find QRS.

Solution:
Draw a line EF parallel to ST through R.

Since PQ || ST [Given]
and EF || ST [Construction]
PQ || EF and QR is a transversal
PQR = QRF [Alternate interior angles] But PQR = 110° [Given]
∴∠QRF = QRS + SRF = 110° …(1)
Again ST || EF and RS is a transversal
RST + SRF = 180° [Co-interior angles] or 130° + SRF = 180°
SRF = 180° – 130° = 50°
Now, from (1), we have QRS + 50° = 110°
QRS = 110° – 50° = 60°
Thus, QRS = 60°.

Ex 6.2 Class 9 Maths Question 5.
In figure, if AB || CD, APQ = 50° and PRD = 127°, find x and y.

Solution:
We have AB || CD and PQ is a transversal.
APQ = PQR
[Alternate interior angles]
50° = x [ APQ = 50° (given)]
Again, AB || CD and PR is a transversal.
APR = PRD [Alternate interior angles]
APR = 127° [ PRD = 127° (given)]
APQ + QPR = 127°
50° + y = 127° [ APQ = 50° (given)] y = 127°- 50° = 77°
Thus, x = 50° and y = 77°.

Ex 6.2 Class 9 Maths Question 6.
In 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.

Solution:
Draw ray BL PQ and CM RS

PQ || RS BL || CM
[ BL || PQ and CM || RS]
Now, BL || CM and BC is a transversal.
LBC = MCB …(1) [Alternate interior angles]
Since, angle of incidence = Angle of reflection
ABL = LBC and MCB = MCD
ABL = MCD …(2) [By (1)]
Adding (1) and (2), we get
LBC + ABL = MCB + MCD
ABC = BCD i. e., a pair of alternate interior angles are equal.
AB || CD.

### NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3

Ex 6.3 Class 9 Maths Question 1.
In figure, sides QP and RQ of ∆PQR are produced to points S and T, respectively. If SPR = 135° and PQT = 110°, find PRQ.

Solution:
We have, TQP + PQR = 180°
[Linear pair]
110° + PQR = 180°
PQR = 180° – 110° = 70°
Since, the side QP of ∆PQR is produced to S.
PQR + PRQ = 135°
[Exterior angle property of a triangle]
70° + PRQ = 135° [PQR = 70°]
PRQ = 135° – 70° PRQ = 65°

Ex 6.3 Class 9 Maths Question 2.
In figure, X = 62°, XYZ = 54°, if YO and ZO are the bisectors of XYZ and XZY respectively of ∆XYZ, find OZY and YOZ.

Solution:
In ∆XYZ, we have XYZ + YZX + ZXY = 180°
[Angle sum property of a triangle]
But XYZ = 54° and ZXY = 62°
54° + YZX + 62° = 180°
YZX = 180° – 54° – 62° = 64°
YO and ZO are the bisectors of XYZ and XZY respectively.
OYZ = 12XYZ = 12(54°) = 27°
and OZY = 12YZX = 12(64°) = 32°
Now, in ∆OYZ, we have
YOZ + OYZ + OZY = 180°
[Angle sum property of a triangle]
YOZ + 27° + 32° = 180°
YOZ = 180° -27° – 32° = 121°
Thus, OZY = 32° and YOZ = 121°

Ex 6.3 Class 9 Maths Question 3.
In figure, if AB || DE, BAC = 35° and CDE = 539 , find DCE.

Solution:
AB || DE and AE is a transversal.
So, BAC = AED
[Alternate interior angles]
and BAC = 35° [Given]
AED = 35°
Now, in ∆CDE, we have CDE + DEC + DCE = 180°
{Angle sum property of a triangle]
53° + 35° + DCE =180°
[ DEC = AED = 35° andCDE = 53° (Given)]
DCE = 180° – 53° – 35° = 92° Thus, DCE = 92°

Ex 6.3 Class 9 Maths Question 4.
In figure, if lines PQ and RS intersect at point T, such that PRT = 40°, RPT = 95° and TSQ = 75°, find SQT.

Solution:
In ∆PRT, we have P + R + PTR = 180°
[Angle sum property of a triangle]
95° + 40° + PTR =180°
[ P = 95°, R = 40° (given)]
PTR = 180° – 95° – 40° = 45°
But PQ and RS intersect at T.
PTR = QTS
[Vertically opposite angles]
QTS = 45° [ PTR = 45°]
Now, in ∆ TQS, we have TSQ + STQ + SQT = 180°
[Angle sum property of a triangle]
75° + 45° + SQT = 180° [ TSQ = 75° and STQ = 45°]
SQT = 180° – 75° – 45° = 60°
Thus, SQT = 60°

Ex 6.3 Class 9 Maths Question 5.
In figure, if PQ PS, PQ||SR, SQR = 2S° and QRT = 65°, then find the values of x and y.

Solution:
In ∆ QRS, the side SR is produced to T.
QRT = RQS + RSQ
[Exterior angle property of a triangle]
But RQS = 28° and QRT = 65°
So, 28° + RSQ = 65°
RSQ = 65° – 28° = 37°
Since, PQ || SR and QS is a transversal.
PQS = RSQ = 37°
[Alternate interior angles]
x = 37°
Again, PQ PS AP = 90°
Now, in ∆PQS,
we have P + PQS + PSQ = 180°
[Angle sum property of a triangle] 90° + 37° + y = 180°
y = 180° – 90° – 37° = 53°
Thus, x = 37° and y = 53°

Ex 6.3 Class 9 Maths Question 6.
In 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

Solution:
In ∆PQR, side QR is produced to S, so by exterior angle property,
PRS = P + PQR
12PRS = 12P + 12PQR
TRS = 12P + TQR …(1)
[ QT and RT are bisectors of PQR and PRS respectively.]
Now, in ∆QRT, we have
TRS = TQR + T …(2)
[Exterior angle property of a triangle]
From (1) and (2),
we have TQR + 12P = TQR + T
12P = T
12QPR = QTR or QTR = 12QPR

### Kunji Team

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