In this chapter, we provide KSEEB SSLC Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Ex 3.3 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Ex 3.3 pdf, free KSEEB SSLC Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Ex 3.3 pdf download. Now you will get step by step solution to each question.

**Karnataka State Syllabus Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Ex 3.3**

## Karnataka Board Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Ex 3.3

Question 1.

Find all the angles in the following

Answer:

∠CMP + ∠PMD = 180° [Linear pair]

∠CMP+ 135° = 180°

∠CMP = 180° – 135

∠CMP = 45°

∠LMD = ∠CMP = 45° [Vertical opposite angles]

∠LMC =∠PMD = 135° [Vertical opposite angles]

∠ALM = ∠ LMC = 135° [Alternate angles]

∠BLM = ∠ LMC = 135° [Alternate angle]

∠QLB = ∠LMD = 45° [Corresponding angles]

∠QLA = ∠ LMC = 135° [Corresponding angles

Question 2.

Find the value of x in the diagram below.

Answer:

∠PQS = ∠EPT = 130° [Corresponding angle]

∠PQS + ∠SQR = 180° [Linear pair]

SQR = 180 – 130

∠SQR = 50°

∠QRS + ∠FRS = 180° [Linear pair]

∠QRS + 90° =180°

∠QRS = 180 – 90

∠QRS = 90°

∠SQR+ ∠QRS +∠QSR = 180°

[Sum of the angles of triangle is 180° ]

50 + 90 + ∠QSR = 180°

140 + ∠QSR = 180°

∠QSR = 180 – 140

∠QSR =40°

∠TSD = ∠QSR [Vertically opposite angles]

x = 40°.

Question 3.

Show that if a straight line is perpendicular to one of the two or more parallel lines, then it is also perpendicular to the remaining lines.

Answer:

AB−→−∥CD−→−∥EF−→∥GH−→−⋅XY¯¯¯¯¯¯¯¯ interrectsthese lines at P, Q, R and S and XY−→−⊥AB¯¯¯¯¯¯¯

Toprove: XY¯¯¯¯¯¯¯¯⊥CD−→−,XY¯¯¯¯¯¯¯¯⊥GH−→−

Proof :∠XPB = 90° (data)

∠XPB = ∠ PQD = ∠QRF = ∠RSH = 90°

Question 4.

Let A→B and Cb−→ be two parallel lines and PQ−→− be a transversal. Show that the angle bisectors of a pair of two internal angles on the same side of the transversal are perpendicular to each other.

Answer:

Let the bisectors of ∠BRS and∠RSD intersect at T

To prove: RT⊥ST i.e., ∠RTS = 90°

Proof: ∠BRS +∠RSD = 180°

12 ∠BRS + 12 ∠RSD = 12 × 180°

12 ∠BRS = ∠TRS =∠TRB

12 ∠RSQ = ∠TSR =∠TSD

In triangle TRS,

∠TRS +∠TSR + ∠RTS = 180°

90 + ∠RTS = 180°

∠RTS = 90°

∴ RT ⊥ ST

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