NCERT solutions class 9 maths exercise 6.2-Lines and Angles
NCERT solutions class 9 maths exercise 6.2 of chapter 6- Lines and Angles are presented here for helping the class 9 maths students of CBSE in doing their homework and preparation for class tests and the CBSE board exam.NCERT solutions for class 9 maths NCERT of chapter 6 are the most important input material for clearing the doubts on the chapter Lines and Angles which is the basic maths chapter of class 9 maths in the steps of understanding the geometry of class 9 and class 10 level maths. All NCERT solutions of the class 9 maths chapter 6 are created by an expert of maths.
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Q1.In fig.6.28, find the values of x and y and then show that AB ll CD.
Ans. It can be observed from the figure that 50° and x forms linear pair
So, x + 50° =180°
x = 180° – 50° = 130°
130° and y forms vertically opposite angles
So, y = 130°
x = y = 130°, implies that interior alternative angles are equal
Since alternative angles are equal, therefore AB ll CD, Hence proved
Q2.In fig. 6.29,if AB ll CD, CD ll EF and y: z = 3 : 7, find x.
Ans.If AB ll CD and CD ll EF
Then AB ll EF
Therefore x = z (alternate angle)
We are given
y: z = 3 : 7
Putting z = x
y : x = 3 : 7
Let y = 3a and x = 7a
y and x are co-interior angles ,so the sum of both x and y is 180°
y + x = 180°
3a + 7a = 180°
10 a = 180°
a = 18
Therefore y = 3×18 = 54° and y =7× 18 = 126°
Hence the value of x is 126°
Q3.In fig.6.30 if AB ll CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE,∠GEF and ∠FGE.
Ans. We are given that ∠GED = 126°
EF ⊥ CD, so ∠DEF = 90°
∠AGE = ∠GED (alternate angle)
∠AGE =126°
From fig.we have
∠GEF = ∠GED -∠DEF
Putting the value of ∠GED =126° and ∠DEF= 90°
∠GEF = 126° – 90°= 36°
∠AGE and ∠FGE are linear pair
So,∠AGE + ∠FGE = 180°
Putting the value ∠AGE = 126°
126°+ ∠FGE = 180°
∠FGE = 180° – 126° = 54°
Hence the value of ∠AGE= 126°,∠GEF=36° and ∠FGE= 54°
Q4. In fig. 6.31 ,if PQ ll ST,∠PQR = 110° and ∠RST = 130°, find ∠QRS.(Hint: draw a line parallel to ST through point R).
Ans.In the fig. we are given
∠PQR = 110° and ∠RST = 130°
PQ ll ST
Drawing a line XY ll PQ ll ST
Since XY ll PQ and QR is the transversal
∠PQR and ∠QRX are co-interior angles
So, ∠PQR + ∠QRX = 180°
Putting the value of ∠PQR = 110°
110° +∠QRX = 180°
∠QRX = 180°- 110° = 70°
From the fig. we have
XY ll ST and SR is the transversal
So,∠RST and ∠SRY are the co-interior angles
∠RST + ∠SRY = 180°
Putting the value of ∠RST = 130°
130° + ∠SRY = 180°
∠SRY = 180° – 130° = 50°
It can be observed in the figure that ∠SRX and ∠SRY are the linear pair
∴ ∠SRX + ∠SRY = 180°
∠QRX + ∠QRS + ∠SRY = 180°
Putting the value of ∠QRX = 70° and ∠SRY = 50°
∠QRS + 70° + 50° = 180°
∠QRS + 120° = 180°
∠QRS = 180° – 120° = 60°
Therefore the value of ∠QRS is 60°
Q5.In fig.6.32 , if AB ll CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
Ans.We are given in the figure
AB ll CD , ∠APQ = 50° and ∠PRD = 127°
AB ll CD and SR is the transversal, so ∠APR and ∠PRD are the alternates angle
∠APR = ∠PRD
∠APQ + y = 127°
50° + y = 127°
y = 127°- 50° = 77°
AB ll CD and PQ is the transversal, so ∠APQ and ∠PQR are the alternates angle
∠APQ = ∠PQR
x = 50°
Hence the value of x is 50° and of y is 77°
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Q6. In fig. 6.33, PQ and RS are the 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 ll CD.
Ans. We are given the plane mirrors PQ ll RS, the ray AB is incident on PQ and ray CD incident on RS.
Let BM is normal to the plane PQ and CN is normal to the plane RS
Let ∠ABM = i_{1}, is the incident angle on PQ, r_{1} is the reflected angle on PQ, i_{2} is the incident angle on RS and r_{2} is the reflected angle on RS
PQ and RS can be supposed as parallel lines and BC as transversal then ∠ABC and ∠BCD are the alternate angles
∴ ∠ABC = ∠BCD
i_{1}= r_{1}and i_{2}= r_{2 }(incident angle = reflected angle)
BM ⊥PQ and CN ⊥ RS
So, BM ll CN (Perpendiculars on parallel lines are parallel) and BC transversal
Therefore r_{1}= i_{2 }(alternate angles)
2r_{1}= 2i_{2}
_{i1}+r_{1} = i_{2} + r_{2}
∠ABC = ∠BCD (alternate angles)
∴AB ll CD
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