Students can Download Maths Chapter 3 Axioms, Postulates and Theorems Ex 3.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## 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:

\(\overrightarrow{\mathrm{AB}}\|\overrightarrow{\mathrm{CD}}\| \overrightarrow{\mathrm{EF}} \| \overrightarrow{\mathrm{GH}} \cdot \overline{\mathrm{XY}}\) interrectsthese lines at P, Q, R and S and \(\overrightarrow{\mathrm{XY}} \perp \overline{\mathrm{AB}}\)

Toprove: \(\overline{\mathrm{XY}} \perp \overrightarrow{\mathrm{CD}}, \overline{\mathrm{XY}} \perp \overrightarrow{\mathrm{GH}}\)

Proof :∠XPB = 90° (data)

∠XPB = ∠ PQD = ∠QRF = ∠RSH = 90°

Question 4.

Let \(\overrightarrow{\mathrm{A}} \mathrm{B} \text { and } \overrightarrow{\mathrm{Cb}}\) be two parallel lines and \(\overrightarrow{\mathrm{PQ}}\) 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°

\(\frac { 1 }{ 2 }\) ∠BRS + \(\frac { 1 }{ 2 }\) ∠RSD = \(\frac { 1 }{ 2 }\) × 180°

\(\frac { 1 }{ 2 }\) ∠BRS = ∠TRS =∠TRB

\(\frac { 1 }{ 2 }\) ∠RSQ = ∠TSR =∠TSD

In triangle TRS,

∠TRS +∠TSR + ∠RTS = 180°

90 + ∠RTS = 180°

∠RTS = 90°

∴ RT ⊥ ST