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NCERT Class 9 Mathematics Chapter 6 Lines and Angles
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Lines and Angles
Chapter – 6
| Exercise 6.1 |
Q.1. In figure, lines AB and CD intersect at 0. If ∠ AOC + ∠ BOE = 70⁰and ∠ BOD = 40⁰ , find ∠ BOE and reflex ∠COE.
Ans: Lines AB and CD intersect at O
∴ ∠ AOC + ∠ BOD
| Vertically Opposite Angles
But ∠ BOD = 40⁰ , …. (1) | Given
∠ AOC = 40⁰ …(2)
Now, ∠ AOC + ∠BOE = 70⁰
⇒ 40⁰ + ∠ BOE = 70⁰ | Using (2)
⇒ ∠ BOE = 70⁰ – 40⁰
⇒ ∠ BOE = 30⁰
Again, Reflex ∠COE =∠COD + ∠ BOD + ∠BOE
= ∠ COD + 40⁰ + 30⁰
| Using (1) and (2)
= 180⁰ + 40⁰ + 30⁰
∴ Ray OA stands on line CD
= 250⁰ (Linear Pair Axiom)
Q.2. In figure, lines XT and MN intersect at 0. If⇒∠POY=90⁰
and a:b = 2:3, find c.
Ans: Ray OP stands on line XY
∴ ∠POX + ∠POY = 180⁰ | Linear Pair Axioms
⇒ ∠POX +90⁰ = 180⁰ |∵∠POY=90⁰(Given)
⇒ ∠POX = 180⁰-90⁰
⇒ ∠POX = 90⁰
⇒ ∠POM + ∠XOM = 90⁰
⇒ a+b=90 ….(1)
A:b = 2:3
⇒ a = 2k
B = 3k
Putting the values of a and b in (1), we get
2k + 3k = 90⁰
⇒ k=18⁰
…(2)
∴ Ray OX stands on line MN
∴ ∠XOM + ∠ XON = 180⁰ | Linear Pair Axioms
⇒b+c=180⁰ ⇒54⁰ +c=180⁰ | Using (2)
⇒c = 180⁰-54⁰⇒c = 126⁰
Q.3. In figure,∠PQR=∠PRQ then prove that ∠PQS = ZPRT.
Ans: Ray QP stands on line ST
∴ ∠ PQS+ ∠ PQR = 180⁰ …(1)| Linear Pair Axiom
∴ Ray RP stands on line ST
∴ ∠ PRQ + ∠ PRT = 180⁰
…(2) Linear Pair Axiom
From (1) and (2), we obtain
∠ PQS + ∠ PQR=∠PRQ+ ∠ PRT
⇒ ∠ PQS = ∠PRT
|∵ ∠ PQR = ∠ PRQ(Given)
Q.4. In figure, if x + y = w + z then prove that AOB is a line.
Ans: x + y = w + z …(1)| Given
∴ The sum of all the angles around a point is equal to 360⁰
∴ x + y + w + z = 360⁰
⇒ x + y + x + y = 360⁰ | Using (1)
⇒ 2(x + y) = 360⁰
⇒ x + y = 180⁰
∴ AOB is a line If the sum of two adjacent angles is 180⁰, then the non-common arms of the angles form a line
Q.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
Ans: Ray OR is perpendicular to the line PQ
From (2) and (3),
∴ ∠ QOS – ∠ POS =(∠QOR- ∠POR) +2 ∠ROS=2∠ROS
| Using (1)
Q.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
Ans: Ray YZ stands on line PX
∴ ∠ XYZ+ ∠ ZYP = 180⁰ l Linear Pair Axiom
⇒ 64^ 0 + ∠ ZYP = 180⁰
| ∵ ∠ XYZ = 64⁰ Given)
⇒ ∠ ZYP = 180⁰- 64⁰
⇒∠ZYP = 116⁰ …(1)
∵ Ray YQ bisects ∠ ZYP
=58⁰ …(2)
∴ Reflex ∠ QYP = 360⁰ – 58⁰=302⁰
∵ The sum of all the angles round a point is equal to 360⁰
Again, ∠ XYQ = ∠ XYZ+ ∠ ZYQ
= 64⁰ + 58⁰
= 122⁰
| Exercise 6.2 |
Q.1. In figure, find the values of x and y and then show that AB||CD
Ans: Let transversal l intersects AB and CD at P and Q respectively.
According to given fig.,
50⁰ + x = 180⁰ (linear pair)
⇒ x = 180⁰ – 50⁰
⇒ x= 130⁰ …(i)
y = 130⁰
(vertically opposite angles) … (ii)
From (i) and (ii), we observe that: x = y
It shows that interior, alternate interior angles are equal.
As we know that if a transversal (say l) intersects two lines in such a way that a pair of alternate interior angles are equal, then the two lines are parallel.
Therefore, ABIICD
Q.2. In figure, if AB|| CD, CD || EF and y:z = 3:7 find x
Ans:
∴ AB|| CD and CD ED
∴ AB|| EF | Lines parallel to the same line are parallel to each other
∴ x = z …(1)
x+y=180⁰ …(2)
Alternate Interior Angles
Consecutive interior angles on the same side of the transversal GH to parallel lines AB and CD
From (1) and (2),
z + y = 180⁰
y:z = 3:7
Sum of the ratios =3+7=10
∴ x = z = 126⁰
Q.3. In figure, if AB II CD, EF⊥CD and ∠ GED = 126 deg , find∠AGE, GEF and ∠ FGE.
Ans: (i) ∠ AGE= ∠ GED = 126⁰
| Alternate Interior Angles
(ii) ∠ GED = 126⁰
⇒ ∠GEF+ ∠ FED = 126⁰
⇒ ∠GEF + 90⁰ = 126⁰
| ∵ EF⊥CD ∠ FED = 90⁰
⇒ ∠GEF = 126⁰- 90⁰= 36⁰
(iii) ∠ GEC+ ∠GEF+ ∠ FED = 180⁰ |∵ CD is a line
Now, ∠FGE = ∠ GEC = 54⁰ | Alternate Interior Angles
Q.4. In figure, if PQ || ST, ∠PQR = 110⁰ and ∠RST=130⁰, find ∠QRS
[Hint. Draw a line parallel to ST through point R ]
Ans: Construction: Draw a line RU parallel to ST through Point R
∠RST + SRU=180⁰ Sum of the consecutive
interior angles on the
⇒130⁰+SRU=180⁰ same side of the
transversal is 180⁰
⇒ ∠SRU=180⁰-130⁰ 50⁰ ….(1)
∠QRU = ∠ PQR = 110⁰ | Alternate Interior Angles
⇒ ∠QRS +∠SRU=110⁰
⇒ ∠QRS+50⁰ 110⁰ | Using (1)
⇒∠QRS=110⁰-50⁰ = 60⁰
Q.5. In figure, If AB|| CD, ∠APQ = 50⁰ and PRD=127⁰, find x and y.
Ans: x = APQ = 50⁰
| Alternate Interior Angles
x=∠PRD=127⁰
| Alternate Interior Angles
Q.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 againreflects back along CD. Prove that AB|| CD.
Or
In figure, m and n are two plane mirrors parallel to each other. Prove that incident ray CA is parallel to reflected ray BD.
[Hint. Draw perpendiculars (normals) at A and B to the plane mirrors. Recall that angle of incidence is equal to angle of reflection.]
Ans: Construction: Draw ray BL ⊥ PQ and ray CM⊥RS
Proof: ∴ BL⊥PQ, CM⊥RS and PQ II RS
∴ BL | CM
= Angle of reflection
From (1), (2) and (3), we get
∠ ABL = ∠ MCD …(4)
Adding (1) and (4), we get
∠ LBC + ∠ ABL=∠MCB+ ∠ MCD
⇒ ∠ ABC =∠BCD
But these are alternate interior angles and they are equal.
So, AB || CD
