NCERT Solutions For Class 9 Maths Chapter 6 Lines and Angles
NCERT Solutions For Class 9 Maths Chapter 6 Lines and Angles on this step-by-step answer guide. In some of State Boards and CBSE schools, college students are taught thru NCERT books. As the chapter involves an end, students are requested few questions in an exercise to evaluate their knowledge of the chapter. Students regularly want guidance managing those NCERT Solutions For Class 9 Maths Chapter 6 Lines and Angles.
It’s most effective natural to get stuck withinside the exercises while solving them so that you could assist students score higher marks, we have provided step by step NCERT answers for all exercises of Class 9 Mathematics Lines And Angles so you can search for assist from them. Students should solve those exercises carefully as questions withinside the final exams are asked from those so these exercises without delay have an effect on students’ final score. Find all NCERT Solutions for Class 9 Mathematics Lines And Angles below and prepare for your exams easily.
NCERT Solutions For Class 9 Maths Chapter 6 Lines and Angles
Exercise: 6.1
(Page No: 96)
1. In Fig. 6.13, lines AB and CD intersect at O. If AOC +BOE = 70° and BOD = 40°, find BOE and reflex COE.
Solution:
From the diagram, we have
(∠AOC +∠BOE +∠COE) and (∠COE +∠BOD +∠BOE) forms a straight line.
So, ∠AOC+∠BOE +∠COE = ∠COE +∠BOD+∠BOE = 180°
Now, by putting the values of ∠AOC + ∠BOE = 70° and ∠BOD = 40° we get
∠COE = 110°, ∠BOE = 30° and reflex ∠COE = 360o – 110o = 250o
2. In Fig. 6.14, lines XY and MN intersect at O. If POY = 90° and a : b = 2 : 3, find c.
Solution:
We know that the sum of linear pair are always equal to 180°
So,
∠POY +a +b = 180°
Putting the value of ∠POY = 90° (as given in the question) we get,
a+b = 90°
Now, it is given that a : b = 2 : 3 so,
Let a be 2x and b be 3x
∴ 2x+3x = 90°
Solving this we get
5x = 90°
So, x = 18°
∴ a = 2×18° = 36°
Similarly, b can be calculated and the value will be
b = 3×18° = 54°
From the diagram, b+c also forms a straight angle so,
b+c = 180°
c+54° = 180°
∴ c = 126°
3. In Fig. 6.15, PQR = PRQ, then prove that PQS = PRT.
Solution:
Since ST is a straight line so,
∠PQS+∠PQR = 180° (linear pair) and
∠PRT+∠PRQ = 180° (linear pair)
Now, ∠PQS + ∠PQR = ∠PRT+∠PRQ = 180°
Since ∠PQR =∠PRQ (as given in the question)
∠PQS = ∠PRT. (Hence proved).
4. In Fig. 6.16, if x+y = w+z, then prove that AOB is a line.
Solution:
For proving AOB is a straight line, we will have to prove x+y is a linear pair
i.e. x+y = 180°
We know that the angles around a point are 360° so,
x+y+w+z = 360°
In the question, it is given that,
x+y = w+z
So, (x+y)+(x+y) = 360°
2(x+y) = 360°
∴ (x+y) = 180° (Hence proved).
5. In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = ½ (QOS – POS).
Solution:
In the question, it is given that (OR ⊥ PQ) and ∠POQ = 180°
So, ∠POS+∠ROS+∠ROQ = 180°
Now, ∠POS+∠ROS = 180°- 90° (Since ∠POR = ∠ROQ = 90°)
∴ ∠POS + ∠ROS = 90°
Now, ∠QOS = ∠ROQ+∠ROS
It is given that ∠ROQ = 90°,
∴ ∠QOS = 90° +∠ROS
Or, ∠QOS – ∠ROS = 90°
As ∠POS + ∠ROS = 90° and ∠QOS – ∠ROS = 90°, we get
∠POS + ∠ROS = ∠QOS – ∠ROS
Þ2 ∠ROS + ∠POS = ∠QOS
Or, ∠ROS = ½ (∠QOS – ∠POS) (Hence proved).
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:
Here, XP is a straight line
So, ∠XYZ +∠ZYP = 180°
Putting the value of ∠XYZ = 64° we get,
64° +∠ZYP = 180°
∴ ∠ZYP = 116°
From the diagram, we also know that ∠ZYP = ∠ZYQ + ∠QYP
Now, as YQ bisects ∠ZYP,
∠ZYQ = ∠QYP
Or, ∠ZYP = 2∠ZYQ
∴ ∠ZYQ = ∠QYP = 58°
Again, ∠XYQ = ∠XYZ + ∠ZYQ
By putting the value of ∠XYZ = 64° and ∠ZYQ = 58° we get.
∠XYQ = 64°+58°
Or, ∠XYQ = 122°
Now, reflex ∠QYP = 180°+∠XYQ
We computed that the value of ∠XYQ = 122°.
So,
∠QYP = 180°+122°
∴ ∠QYP = 302°
Exercise: 6.2
(Page No: 103)
1. In Fig. 6.28, find the values of x and y and then show that AB CD.
Solution:
We know that a linear pair is equal to 180°.
So, x+50° = 180°
∴ x = 130°
We also know that vertically opposite angles are equal.
So, y = 130°
In two parallel lines, the alternate interior angles are equal. In this,
x = y = 130°
This proves that alternate interior angles are equal and so, AB II CD.
2. In Fig. 6.29, if AB II CD, CD II EF and y : z = 3 : 7, find x.
Solution:
It is known that AB II CD and CD II EF
As the angles on the same side of a transversal line sums up to 180°,
x + y = 180° —–(i)
Also,
∠O = z (Since they are corresponding angles)
and, y +∠O = 180° (Since they are a linear pair)
So, y+z = 180°
Now, let y = 3w and hence, z = 7w (As y : z = 3 : 7)
∴ 3w+7w = 180°
Or, 10 w = 180°
So, w = 18°
Now, y = 3×18° = 54°
and, z = 7×18° = 126°
Now, angle x can be calculated from equation (i)
x+y = 180°
Or, x+54° = 180°
∴ x = 126°
3. In Fig. 6.30, if AB II CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
Solution:
Since AB II CD, GE is a transversal.
It is given that ∠GED = 126°
So, ∠GED = ∠AGE = 126° (As they are alternate interior angles)
Also,
∠GED = ∠GEF +∠FED
As EF⊥ CD, ∠FED = 90°
∴ ∠GED = ∠GEF+90°
Or, ∠GEF = 126° – 90° = 36°
Again, ∠FGE +∠GED = 180° (Transversal)
Putting the value of ∠GED = 126° we get,
∠FGE = 54°
So,
∠AGE = 126°
∠GEF = 36° and
∠FGE = 54°
4. In Fig. 6.31, if PQ ST, PQR = 110° and RST = 130°, find QRS.
[Hint : Draw a line parallel to ST through point R.]
Solution:
First, construct a line XY parallel to PQ.
We know that the angles on the same side of transversal is equal to 180°.
So, ∠PQR+∠QRX = 180°
Or, ∠QRX = 180°-110°
∴ ∠QRX = 70°
Similarly,
∠RST +∠SRY = 180°
Or, ∠SRY = 180°- 130°
∴ ∠SRY = 50°
Now, for the linear pairs on the line XY-
∠QRX+∠QRS+∠SRY = 180°
Putting their respective values, we get,
∠QRS = 180° – 70° – 50°
Hence, ∠QRS = 60°
5. In Fig. 6.32, if AB CD, APQ = 50° and PRD = 127°, find x and y.
Solution:
From the diagram,
∠APQ = ∠PQR (Alternate interior angles)
Now, putting the value of ∠APQ = 50° and ∠PQR = x we get,
x = 50°
Also,
∠APR = ∠PRD (Alternate interior angles)
Or, ∠APR = 127° (As it is given that ∠PRD = 127°)
We know that
∠APR = ∠APQ+∠QPR
Now, putting values of ∠QPR = y and ∠APR = 127° we get,
127° = 50°+ y
Or, y = 77°
Thus, the values of x and y are calculated as:
x = 50° and y = 77°
6. In Fig. 6.33, 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 II CD.
Solution:
First, draw two lines BE and CF such that BE ⊥ PQ and CF ⊥ RS.
Now, since PQ II RS,
So, BE II CF
We know that,
Angle of incidence = Angle of reflection (By the law of reflection)
So,
∠1 = ∠2 and
∠3 = ∠4
We also know that alternate interior angles are equal. Here, BE ⊥ CF and the transversal line BC cuts them at B and C
So, ∠2 = ∠3 (As they are alternate interior angles)
Now, ∠1 +∠2 = ∠3 +∠4
Or, ∠ABC = ∠DCB
So, AB II CD alternate interior angles are equal)
Exercise: 6.3
(Page No: 107)
1. In Fig. 6.39, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.
Solution:
It is given the TQR is a straight line and so, the linear pairs (i.e. ∠TQP and ∠PQR) will add up to 180°
So, ∠TQP +∠PQR = 180°
Now, putting the value of ∠TQP = 110° we get,
∠PQR = 70°
Consider the ΔPQR,
Here, the side QP is extended to S and so, ∠SPR forms the exterior angle.
Thus, ∠SPR (∠SPR = 135°) is equal to the sum of interior opposite angles. (Triangle property)
Or, ∠PQR +∠PRQ = 135°
Now, putting the value of ∠PQR = 70° we get,
∠PRQ = 135°-70°
Hence, ∠PRQ = 65°
2. In Fig. 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of Δ XYZ, find ∠OZY and ∠YOZ.
Solution:
We know that the sum of the interior angles of the triangle.
So, ∠X +∠XYZ +∠XZY = 180°
Putting the values as given in the question we get,
62°+54° +∠XZY = 180°
Or, ∠XZY = 64°
Now, we know that ZO is the bisector so,
∠OZY = ½ ∠XZY
∴ ∠OZY = 32°
Similarly, YO is a bisector and so,
∠OYZ = ½ ∠XYZ
Or, ∠OYZ = 27° (As ∠XYZ = 54°)
Now, as the sum of the interior angles of the triangle,
∠OZY +∠OYZ +∠O = 180°
Putting their respective values, we get,
∠O = 180°-32°-27°
Hence, ∠O = 121°
3. In Fig. 6.41, if AB II DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.
Solution:
We know that
AE is a transversal since AB II DE
Here ∠BAC and ∠AED are alternate interior angles.
Hence, ∠BAC = ∠AED
It is given that ∠BAC = 35°
Þ∠AED = 35°
Now consider the triangle CDE. We know that the sum of the interior angles of a triangle is 180°.
∴ ∠DCE+∠CED+∠CDE = 180°
Putting the values, we get
∠DCE+35°+53° = 180°
Hence, ∠DCE = 92°
4. In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
Solution:
Consider triangle PRT.
∠PRT +∠RPT + ∠PTR = 180°
So, ∠PTR = 45°
Now ∠PTR will be equal to ∠STQ as they are vertically opposite angles.
So, ∠PTR = ∠STQ = 45°
Again, in triangle STQ,
∠TSQ +∠PTR + ∠SQT = 180°
Solving this we get,
∠SQT = 60°
5. In Fig. 6.43, if PQ ⊥ PS, PQ II SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.
Solution:
x +∠SQR = ∠QRT (As they are alternate angles since QR is transversal)
So, x+28° = 65°
∴ x = 37°
It is also known that alternate interior angles are same and so,
∠QSR = x = 37°
Also, Now,
∠QRS +∠QRT = 180° (As they are a Linear pair)
Or, ∠QRS+65° = 180°
So, ∠QRS = 115°
Now, we know that the sum of the angles in a quadrilateral is 360°. So,
∠P +∠Q+∠R+∠S = 360°
Putting their respective values, we get,
∠S = 360°-90°-65°-115°
In Δ SPQ
∠SPQ + x + y = 1800
900 + 370 + y = 1800
Þy = 1800 – 1270 = 530
Hence, y = 53°
6. In Fig. 6.44, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = ½ ∠QPR.
Solution:
Consider the ΔPQR. ∠PRS is the exterior angle and ∠QPR and ∠PQR are interior angles.
So, ∠PRS = ∠QPR+∠PQR (According to triangle property)
Or, ∠PRS -∠PQR = ∠QPR ———–(i)
Now, consider the ΔQRT,
∠TRS = ∠TQR+∠QTR
Or, ∠QTR = ∠TRS-∠TQR
We know that QT and RT bisect ∠PQR and ∠PRS respectively.
So, ∠PRS = 2 ∠TRS and ∠PQR = 2∠TQR
Now, ∠QTR = ½ ∠PRS – ½∠PQR
Or, ∠QTR = ½ (∠PRS -∠PQR)
From (i) we know that ∠PRS -∠PQR = ∠QPR
So, ∠QTR = ½ ∠QPR (hence proved).
