KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2

Students can Download Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 9 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

Karnataka State Syllabus Class 9 Maths Chapter 3 Lines and Angles Ex 3.2

Question 1.
In Figure, find the values of x and y and then shows that AB || CD.
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 1
Answer:
i) ∠x = ?
ii) ∠y = ?
iii) Show tht AB || CD.
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 2

(i) PQ Bisects AB at R.
∠PRA and ∠ARS are on straight line PQ.
∴ ∠PRA + ∠ARS = 180° Adjacent angles
50 + ∠ARS = 180°
∠ARS = 180 – 50
∴ x = ∠ARS = 130° (i)

(ii) Similarly, PQ straight line intersects straight line CD at S.
∴∠CSQ = ∠RSD . Vertically opposite angles
∴y = ∠RSD = 130° (ii)
From (i) and (ii),
∠ARS = ∠RYD
x = y = 130°

(iii) Therse are mutual alternate angles
∴x = y = 130°
∴AB || CD.

KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2

Question 2.
In Fig. 3.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 3
Answer:
PQ straight line intersects AB||CD]|EF at points R, S, T.
If y : z = 3 : 7, then value of ‘x’ = ?
PQ straight line intersects CD||EF at S and T.
∴ ∠CSR = ∠DST = y (∵ vertically opposite angle)
But, ∠DST + ∠STF = 180°
(∵ Sum of interior angles in the same side)
∴ y + z = 180°
But, y : z = 3 : 7.
Sum of ratio 3 + 7 = 10
If ratio 10 means 180
If ratio 9 means ?
\(\frac{180 \times 3}{10}=54^{\circ}\)
∴ y = 54°
y + z = 180
54 + z = 180
∴ z = 180 – 54
∴ z = 126°
But, x + y = 180
(∵ Sum of interior angles on one side)
x + 54 = 180
∵ x = 180 – 54
∵ x = 126°.

Question 3.
In Fig. 3.30, If AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 4
Answer:
AB || CD and EF ⊥ CD and ∠GED = 126°
∠AGE =?
∠GEF =?
∠FGE =?
∠GED = 126°.
∴ ∠GEF + ∠FED = 126°
∠GEF + 90° = 126° ∵ EF ⊥ CD.
∴ GEF = 126-90
∴  GEF = 36°.
∴ ∠GFE = 90° . EF ⊥ AB
∴ In ∆GEF
∠FGE + ∠GFE + ∠GEF = 180°
∠FGE + 90° + 36° = 180°
∠FGE + 126° = 180°
∴ ∠FGE = 180 – 126
∴ ∠FGE = 54°.
∠AGE + ∠FGE = 180° (Adjacent angles)
∠AGE + 54° = 180°
∴ ∠AGE =180 – 54
∠AGE = 126°
∴ ∠AGE = 126°
∠GEF = 36°
∠FGE = 54°.

KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2

Question 4.
In Fig. 3.31, If PQ||ST, ∠PQR =110°, and ∠RST = 130°, find ∠QRS.
(Hint: Draw a line parallel to ST through point R).
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 5
Answer:
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 6
Data: PQ || ST and ∠PQR = 110° and ∠RST= 130°.
To Prove: ∠QRS =?
Construction: Draw ST || UV through ‘R’
Proof: PQ || ST (Data)
ST || UV (Construction)
∴ PQ || ST || UV
PQ || UV.
∴ ∠PQR + ∠URQ = 180° (Sum of interior angles)
110 + ∠URQ = 180°
∠URQ = 180 – 110
∴ ∠URQ = 70° (i)
Similarly, ST || UV.
∴ ∠RST + ∠SRV = 180°
130 + ∠SRV = 180
∴ ∠SRV = 180 – 130
∴ ∠SRV = 50° (ii)
But, URV is a straight line.
∴ ∠URV = 180°.
∠URQ + ∠QRS + ∠RSV = 180°
70 + ∠QRS + 50 = 180
∠QRS + 120 = 180°
∴ ∠QRS = 180 – 120
∴ ∠QRS = 60°.

KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2

Question 5.
In Figure if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 7
Answer:
If AB || CD and ∠APQ= 50° and ∠PRD = 127°, then ∠x = ? ∠y = ?
AB || CD PQ straight line intersects these at P and Q.
∴ ∠APQ = ∠PQR . alternate angles.
50 = ∠PQR = x
∴ x = 50°
Similarly, AB||CD PR straight line intersects these at P and R.
∴ ∠APR = ∠PRD
∠APQ + ∠QPR = ∠PRD
50 + ∠QPR = 127
∠QPR = 127 – 50
∴∠QPR = 77 °
∠QPR = y = 77 °
∴ x = 50 °,
y = 77 °.

Question 6.
In Fig. 3.33, 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 || CD.
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 8
Answer:
Data: 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.
To Prove AB||CD.
Proof: PQ is a mirror.
AB is an incident ray, BC is a reflected ray. ∠ABC is the angle of incidence.
∠BCD is the angle of reflection.
The incident ray formed in the PQ mirror is equal to the reflected ray formed in the RS mirror.
∴ ∠ABC = ∠BCD.
These are pair of alternate interior angles, then AB || CD.

KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2