Students can Download Maths Chapter 3 Axioms, Postulates and Theorems Additional 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 State Syllabus Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions

Question 1.

Choose the correct option

(i) If a = 6 and b = a, then b = 60 by _______

A. Axiom 1

B. Axiom 2

C. Axiom 3

D. Axiom 4

Solution:

A. Axiom 1

(ii) Given a point on the plane, one can draw ________

A. unique

B. two

C. finite numberd

D. infinitely many

Solution:

(D) Infinetly many lines through that point.

(iii) Given two points in a plane, the number of lines which can be drawn to pass through these two points is ________

A. Zero

B. exactly one

C. at most one

D. more than one

Solution:

(B) Exactly one

(iv) If two angles are supplementary, then their sum is

A. 90°

B. 180°

C. 270°

D. 360°

Solution:

(B) 180°

(v) The measure of an angle which is 5 times its supplement is _________

A. 30°

B. 60°

C. 120°

D. 150°

Solution:

(D) 150°

Question 2.

What is the difference between a pair of supplementary angles and a pair of complementary angles?

Solution:

If the sum of two angles is 180° then they are supplementary angles. If the sum of two angles is 90° then they are complementary angles.

Question 3.

What is the least number of non – collinear points required to determine a plane?

Solution:

Three (3)

Question 4.

When do you say two angles are adjacent?

Solution:

Two angles are said to be adjacent angles if they have a common vertex and a common side.

Question 5.

Let \(\overline{\mathbf{A B}}\) be a segment with C and D between them such that the order of points on the segment is A, C, D, B suppose AD = BC prove that AC = DB

Solution:

AD = BC [data]

AD – CD = BC – CD [Axiom 3]

AC = DB [In the fig AC – CD = AC , BC – CD = DB]

Question 6.

Let \(\overrightarrow{\mathbf{A B}}\) and \(\overrightarrow{\mathbf{A B}}\) be two straight lines intersecting at O. Let \(\overrightarrow{\mathbf{O X}}\) be the bisector of ∠BOD. Draw \(\overrightarrow{\mathbf{O Y}}\) between \(\overrightarrow{\mathbf{O D}}\) and \(\overrightarrow{\mathbf{O A}}\) such that \(\overrightarrow{\mathbf{O Y}}\) ⊥ \(\overrightarrow{\mathbf{O B}}\) prove that \(\overrightarrow{\mathbf{O Y}}\) bisects ∠DOA .

Solution:

Construction: Produce [/latex] such that \(\overrightarrow{\mathbf{X O}}\) to A point Z such that [/latex] such that \(\overrightarrow{\mathbf{O Z}}\).is between [/latex] such that \(\overrightarrow{\mathbf{O A}}\) and [/latex] such that \(\overrightarrow{\mathbf{O C}}\)

Proof : ∠AOZ = ∠BOX [vertically opposite angles]

∠BOX = ∠POX [\(\overrightarrow{\mathbf{OX}}\) bisects ∠DOB]

∴ ∠AOZ = ∠DOX [Axiom 1] … (i)

[ZOY = [YOX r=90°] … (ii)

(ii) – (i) ∠ZOY – ∠AOZ = ∠YOX – ∠DOX ∠AOY = ∠YOD [Axiom 3]

∴ [/latex] such that \(\overrightarrow{\mathbf{O Y}}\) bisects ∠AOD

Question 7.

Let \(\overrightarrow{\mathbf{A B}}\) and \(\overrightarrow{\mathbf{C D}}\) be two parallel lines and \(\overrightarrow{\mathbf{P Q}}\) be a transversal. Let \(\overrightarrow{\mathbf{P Q}}\) intersect \(\overrightarrow{\mathbf{A B}}\) in L. Suppose the bisector of ∠ALP intersect CD in R and the bisector of ∠PLB intersect CDin S prove that ∠LRS + ∠RSL = 90°

Solution:

∠ALP + ∠BLP = 180° ∠Linear pair]

\(\frac{1}{2}\) ∠ALP + \(\frac{1}{2}\) ∠BLP = \(\frac{1}{2}\)180°

[Multiplying by \(\frac{1}{2}\) ]

∠ELP + ∠FLP = 90° [∴ \(\overrightarrow{\mathbf{E L}}\) and \(\overrightarrow{\mathbf{F L}}\) are bisectors of ∠ALP and ∠BLP ]

∠ELF = 90°

[SLR = ∠ELF = 90°

[vertically opposite angles]

In ∆ SLR, ∠SLR + ∠LRS + ∠RSL = 180° [sum of the angles of the triangle is 180° ]

90 + ∠LRS + ∠RSL = 180°

∠LRS + ∠SL = 180 – 90

∠LRS + [RSL = 90°

Question 8.

In the adjoining figure \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{CD}}\) are parallel lines. The transversals \(\overrightarrow{\mathrm{PQ}}\) and \(\overrightarrow{\mathrm{RS}}\) intersect at U on the line \(\overrightarrow{\mathrm{AB}}\). Given ∠DWU = 110° and ∠CVP = 70° find the measure of ∠QUS.

Solution:

∠ AUV – ∠CVP = 70°

[corresponding angles]

∠QUB – ∠AUV = 70° [vertically opposite angles]

∠BUW + ∠DWU = 180° [Interior angles on the same side of transversal]

∠BUW + 110°= 180°

∠BUW = 180-110° = 70°

∠BUW = ∠AUS= 70° [vertically opposite angles]

∠AUS + ∠SUQ + ∠QUB = 180° ∠AUB is a straight angle]

70 + ∠SUQ + 70 = 180°

∠SUQ + 140° = 180°

∠SUQ = 180° – 140°

∠SUQ = 40°

Question 9.

What is the angle between the hours hand and minutes hand of a clock at (i) 1.40 hours (ii) 2.15 hours [use 1° = 60 minutes]

Solution:

i. 1.40 hours – 190°

ii. 2.15 hours – 22° 30′

Question 10.

How much would hour’s hand have moved from its position at 12 noon when the time is 4.24 p.m?

Solution:

144°

Question 11.

Let \(\overrightarrow{\mathrm{AB}}\) be a line segment and let C be the midpoint of \(\overrightarrow{\mathrm{AB}}\) Extend AB to D such that B lies between A and D. Prove that AD + BD = 2CD

Solution:

AD + BD = 2CD L.H.S – AD + BD

= AC + CB + BD + BD [∵ AD = AC + CB + BD]

= CB + CB + BD + BD

[∴ AC = CB]

= CB + BD + CB + BD

= CD + CD

= 2CD = RHS

∴ AD + B D = 2CD

Question 12.

Let \(\overrightarrow{\mathbf{A B}}\) and \(\overrightarrow{\mathbf{C D}}\) be two lines . intersecting at a point O. Let \(\overrightarrow{\mathbf{O X}}\) be a ray bisecting [BOD. Prove that the extension of \(\overrightarrow{\mathbf{O X}}\) to the left of O bisects ∠AOC.

Solution:

∠DOX = ∠BOX [ ∵\(\overrightarrow{\mathbf{O X}}\) bisects ∠BOD]

∠DOX =∠COY [vertically opposite angles] ….(i)

∠BOX = ∠AOY

[vertically opposite anglesj … (ii)

From (i) and (ii)

[vertically opposite] …(ii)

∠AQY – ∠COY [Axiom 1]

∴ \(\overrightarrow{\mathbf{O X}}\) bisects ∠AOC

Question 13.

Let \(\overrightarrow{\mathbf{O X}}\) be a ray and let \(\overrightarrow{\mathbf{O A}}\) and \(\overrightarrow{\mathbf{O B}}\) be two rays on the same side of \(\overrightarrow{\mathbf{O X}}\) with \(\overrightarrow{\mathbf{O A}}\) between \(\overrightarrow{\mathbf{O X}}\) and \(\overrightarrow{\mathbf{O B}}\) . Let \(\overrightarrow{\mathbf{O C}}\) be the bisector of ∠AOB prove that ∠XOA + ∠XOB = 2∠XOC

Solution:

∠XOA + ∠XOB = 2∠XOC

Question 14.

Let \(\overrightarrow{\mathbf{O A}}\) and \(\overrightarrow{\mathbf{O B}}\) be two rays and let \(\overrightarrow{\mathbf{O X}}\) be a ray between \(\overrightarrow{\mathbf{O A}}\) and \(\overrightarrow{\mathbf{O B}}\) such ∠AOX > ∠XOB . Let \(\overrightarrow{\mathbf{O C}}\) be the bisector of ∠AOB prove that ∠AOX – ∠XQB = 2∠COX

Solution:

Question 15.

Let \(\overrightarrow{\mathbf{OA}}\), \(\overrightarrow{\mathbf{O B}}\), \(\overrightarrow{\mathbf{OC}}\) be three rays such that \(\overrightarrow{\mathbf{OC}}\) lies between \(\overrightarrow{\mathbf{OA}}\) and \(\overrightarrow{\mathbf{O A}}\). Suppose the bisectors of ∠AOC and ∠COB are perpendicular to each other. Prove that B, O. A are . collinear.

Solution:

∠XOC + ∠YOC – 90° [data]

∠AOB = 180°

A, O and B are collinear.

Question 16.

In the adjoining figure \(\overrightarrow{\mathrm{AB}}\) || \(\overrightarrow{\mathrm{DE}}\) . Prove that ∠ABC – [DCB + [CDE = 180°

Solution:

Construction: Through C draw \(\overrightarrow{\mathrm{XY}}\) || \(\overrightarrow{\mathrm{AB}}\)

Proof : ∠ABC + ∠BCX = 180° [sum of interior angles on the same side of trnasversal]

∴ ∠BCX = 180 – ∠ABC

∠CDE + ∠DCY – 180° [sum of interior angles on the same side of transversal]

∴ ∠DCY = 180° – ∠CDE

∠BCX + ∠BCD + ∠DCY = 180° [∵ XCY is straight angle]

180 – ∠ABC + ∠BCD + 180 – ∠CDE = 180° [By substituting]

360 – ∠ABC + ∠BCD – ∠CDE – 180°

360 – 180 – ∠ABC – ∠BCD + CDE

180° = ∠ABC – ∠BCD + ∠CDE

or

∠ABC – ∠BCD + ∠CDE = 180°

Question 17.

Consider two parallel lines and a transversal among the measures of 8 angles formed how many distinct numbers are there?

Solution:

There will be two distinct numbers.