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
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
A. Axiom 1
(ii) Given a point on the plane, one can draw ________
C. finite numberd
D. infinitely many
(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 ________
B. exactly one
C. at most one
D. more than one
(B) Exactly one
(iv) If two angles are supplementary, then their sum is
(v) The measure of an angle which is 5 times its supplement is _________
What is the difference between a pair of supplementary angles and a pair of complementary angles?
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.
What is the least number of non – collinear points required to determine a plane?
When do you say two angles are adjacent?
Two angles are said to be adjacent angles if they have a common vertex and a common side.
Let 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
AD = BC [data]
AD – CD = BC – CD [Axiom 3]
AC = DB [In the fig AC – CD = AC , BC – CD = DB]
Let and be two straight lines intersecting at O. Let be the bisector of ∠BOD. Draw between and such that ⊥ prove that bisects ∠DOA .
Construction: Produce [/latex] such that to A point Z such that [/latex] such that .is between [/latex] such that and [/latex] such that
Proof : ∠AOZ = ∠BOX [vertically opposite angles]
∠BOX = ∠POX [ 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 bisects ∠AOD
Let and be two parallel lines and be a transversal. Let intersect in L. Suppose the bisector of ∠ALP intersect CD in R and the bisector of ∠PLB intersect CDin S prove that ∠LRS + ∠RSL = 90°
∠ALP + ∠BLP = 180° ∠Linear pair]
∠ALP + ∠BLP = 180°
[Multiplying by ]
∠ELP + ∠FLP = 90° [∴ and 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°
In the adjoining figure and are parallel lines. The transversals and intersect at U on the line . Given ∠DWU = 110° and ∠CVP = 70° find the measure of ∠QUS.
∠ AUV – ∠CVP = 70°
∠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°
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]
i. 1.40 hours – 190°
ii. 2.15 hours – 22° 30′
How much would hour’s hand have moved from its position at 12 noon when the time is 4.24 p.m?
Let be a line segment and let C be the midpoint of Extend AB to D such that B lies between A and D. Prove that AD + BD = 2CD
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
Let and be two lines . intersecting at a point O. Let be a ray bisecting [BOD. Prove that the extension of to the left of O bisects ∠AOC.
∠DOX = ∠BOX [ ∵ 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]
∴ bisects ∠AOC
Let be a ray and let and be two rays on the same side of with between and . Let be the bisector of ∠AOB prove that ∠XOA + ∠XOB = 2∠XOC
∠XOA + ∠XOB = 2∠XOC
Let and be two rays and let be a ray between and such ∠AOX > ∠XOB . Let be the bisector of ∠AOB prove that ∠AOX – ∠XQB = 2∠COX
Let , , be three rays such that lies between and . Suppose the bisectors of ∠AOC and ∠COB are perpendicular to each other. Prove that B, O. A are . collinear.
∠XOC + ∠YOC – 90° [data]
∠AOB = 180°
A, O and B are collinear.
In the adjoining figure || . Prove that ∠ABC – [DCB + [CDE = 180°
Construction: Through C draw ||
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
∠ABC – ∠BCD + ∠CDE = 180°
Consider two parallel lines and a transversal among the measures of 8 angles formed how many distinct numbers are there?
There will be two distinct numbers.