{"id":27382,"date":"2020-06-30T15:27:12","date_gmt":"2020-06-30T09:57:12","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=27382"},"modified":"2021-07-02T15:07:10","modified_gmt":"2021-07-02T09:37:10","slug":"kseeb-solutions-for-class-8-maths-chapter-3-additional-questions","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-8-maths-chapter-3-additional-questions\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions"},"content":{"rendered":"

Students can Download Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions<\/h2>\n

Question 1.
\nChoose the correct option
\n(i) If a = 6 and b = a, then b = 60 by _______
\nA. Axiom 1
\nB. Axiom 2
\nC. Axiom 3
\nD. Axiom 4
\nSolution:
\nA. Axiom 1<\/p>\n

(ii) Given a point on the plane, one can draw ________
\nA. unique
\nB. two
\nC. finite numberd
\nD. infinitely many
\nSolution:
\n(D) Infinetly many lines through that point.<\/p>\n

(iii) Given two points in a plane, the number of lines which can be drawn to pass through these two points is ________
\nA. Zero
\nB. exactly one
\nC. at most one
\nD. more than one
\nSolution:
\n(B) Exactly one<\/p>\n

(iv) If two angles are supplementary, then their sum is
\nA. 90\u00b0
\nB. 180\u00b0
\nC. 270\u00b0
\nD. 360\u00b0
\nSolution:
\n(B) 180\u00b0<\/p>\n

(v) The measure of an angle which is 5 times its supplement is _________
\nA. 30\u00b0
\nB. 60\u00b0
\nC. 120\u00b0
\nD. 150\u00b0
\nSolution:
\n(D) 150\u00b0<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nWhat is the difference between a pair of supplementary angles and a pair of complementary angles?
\nSolution:
\nIf the sum of two angles is 180\u00b0 then they are supplementary angles. If the sum of two angles is 90\u00b0 then they are complementary angles.<\/p>\n

Question 3.
\nWhat is the least number of non – collinear points required to determine a plane?
\nSolution:
\nThree (3)<\/p>\n

Question 4.
\nWhen do you say two angles are adjacent?
\nSolution:
\nTwo angles are said to be adjacent angles if they have a common vertex and a common side.<\/p>\n

Question 5.
\nLet \\(\\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
\nSolution:
\n\"KSEEB
\nAD = BC [data]
\nAD – CD = BC – CD [Axiom 3]
\nAC = DB [In the fig AC – CD = AC , BC – CD = DB]<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nLet \\(\\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 \u2220BOD. Draw \\(\\overrightarrow{\\mathbf{O Y}}\\) between \\(\\overrightarrow{\\mathbf{O D}}\\) and \\(\\overrightarrow{\\mathbf{O A}}\\) such that \\(\\overrightarrow{\\mathbf{O Y}}\\) \u22a5 \\(\\overrightarrow{\\mathbf{O B}}\\) prove that \\(\\overrightarrow{\\mathbf{O Y}}\\) bisects \u2220DOA .
\nSolution:
\n\"KSEEB
\nConstruction: 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}}\\)
\nProof : \u2220AOZ = \u2220BOX [vertically opposite angles]
\n\u2220BOX = \u2220POX [\\(\\overrightarrow{\\mathbf{OX}}\\) bisects \u2220DOB]
\n\u2234 \u2220AOZ = \u2220DOX\u00a0 [Axiom 1] … (i)
\n[ZOY = [YOX r=90\u00b0] … (ii)
\n(ii) – (i) \u2220ZOY – \u2220AOZ = \u2220YOX – \u2220DOX \u2220AOY = \u2220YOD [Axiom 3]
\n\u2234 [\/latex] such that \\(\\overrightarrow{\\mathbf{O Y}}\\) bisects \u2220AOD<\/p>\n

Question 7.
\nLet \\(\\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 \u2220ALP intersect CD in R and the bisector of \u2220PLB intersect CDin S prove that \u2220LRS + \u2220RSL = 90\u00b0
\n\"KSEEB
\nSolution:
\n\u2220ALP + \u2220BLP = 180\u00b0 \u2220Linear pair]
\n\\(\\frac{1}{2}\\) \u2220ALP + \\(\\frac{1}{2}\\) \u2220BLP = \\(\\frac{1}{2}\\)180\u00b0
\n[Multiplying by \\(\\frac{1}{2}\\) ]
\n\u2220ELP + \u2220FLP = 90\u00b0 [\u2234 \\(\\overrightarrow{\\mathbf{E L}}\\) and \\(\\overrightarrow{\\mathbf{F L}}\\) are bisectors of \u2220ALP and \u2220BLP ]
\n\u2220ELF = 90\u00b0
\n[SLR = \u2220ELF = 90\u00b0
\n[vertically opposite angles]
\nIn \u2206 SLR, \u2220SLR + \u2220LRS + \u2220RSL = 180\u00b0 [sum of the angles of the triangle is 180\u00b0 ]
\n90 + \u2220LRS + \u2220RSL = 180\u00b0
\n\u2220LRS + \u2220SL = 180 – 90
\n\u2220LRS + [RSL = 90\u00b0<\/p>\n

\"KSEEB<\/strong><\/p>\n

Question 8.
\nIn 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 \u2220DWU = 110\u00b0 and \u2220CVP = 70\u00b0 find the measure of \u2220QUS.
\nSolution:
\n\"KSEEB<\/p>\n

\u2220 AUV – \u2220CVP = 70\u00b0
\n[corresponding angles]
\n\u2220QUB – \u2220AUV = 70\u00b0 [vertically opposite angles]
\n\u2220BUW + \u2220DWU = 180\u00b0 [Interior angles on the same side of transversal]
\n\u2220BUW + 110\u00b0= 180\u00b0
\n\u2220BUW = 180-110\u00b0 = 70\u00b0
\n\u2220BUW = \u2220AUS= 70\u00b0 [vertically opposite angles]
\n\u2220AUS + \u2220SUQ + \u2220QUB = 180\u00b0 \u2220AUB is a straight angle]
\n70 + \u2220SUQ + 70 = 180\u00b0
\n\u2220SUQ + 140\u00b0 = 180\u00b0
\n\u2220SUQ = 180\u00b0 – 140\u00b0
\n\u2220SUQ = 40\u00b0<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nWhat is the angle between the hours hand and minutes hand of a clock at (i) 1.40 hours (ii) 2.15 hours [use 1\u00b0 = 60 minutes]
\nSolution:
\ni. 1.40 hours – 190\u00b0
\nii. 2.15 hours – 22\u00b0 30′<\/p>\n

Question 10.
\nHow much would hour\u2019s hand have moved from its position at 12 noon when the time is 4.24 p.m?
\nSolution:
\n144\u00b0<\/p>\n

Question 11.
\nLet \\(\\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
\nSolution:
\n\"KSEEB
\nAD + BD = 2CD L.H.S – AD + BD
\n= AC + CB + BD + BD [\u2235 AD = AC + CB + BD]
\n= CB + CB + BD + BD
\n[\u2234 AC = CB]
\n= CB + BD + CB + BD
\n= CD + CD
\n= 2CD = RHS
\n\u2234 AD + B D = 2CD<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nLet \\(\\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 \u2220AOC.
\nSolution:
\n\"KSEEB<\/p>\n

\u2220DOX = \u2220BOX [ \u2235\\(\\overrightarrow{\\mathbf{O X}}\\) bisects \u2220BOD]
\n\u2220DOX =\u2220COY [vertically opposite angles] ….(i)
\n\u2220BOX = \u2220AOY
\n[vertically opposite anglesj … (ii)
\nFrom (i) and (ii)
\n[vertically opposite] …(ii)
\n\u2220AQY – \u2220COY [Axiom 1]
\n\u2234 \\(\\overrightarrow{\\mathbf{O X}}\\) bisects \u2220AOC<\/p>\n

Question 13.
\nLet \\(\\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 \u2220AOB prove that \u2220XOA + \u2220XOB = 2\u2220XOC
\nSolution:
\n\u2220XOA + \u2220XOB = 2\u2220XOC
\n\"KSEEB<\/p>\n

\"KSEEB<\/strong><\/p>\n

Question 14.
\nLet \\(\\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 \u2220AOX > \u2220XOB . Let \\(\\overrightarrow{\\mathbf{O C}}\\) be the bisector of \u2220AOB prove that \u2220AOX – \u2220XQB = 2\u2220COX
\nSolution:
\n\"KSEEB
\n\"KSEEB<\/p>\n

Question 15.
\nLet \\(\\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 \u2220AOC and \u2220COB are perpendicular to each other. Prove that B, O. A are . collinear.
\nSolution:
\n\u2220XOC + \u2220YOC – 90\u00b0 [data]
\n\"KSEEB
\n\u2220AOB = 180\u00b0
\nA, O and B are collinear.<\/p>\n

\"KSEEB<\/p>\n

Question 16.
\nIn the adjoining figure \\(\\overrightarrow{\\mathrm{AB}}\\) || \\(\\overrightarrow{\\mathrm{DE}}\\) . Prove that \u2220ABC – [DCB + [CDE = 180\u00b0
\nSolution:
\nConstruction: Through C draw \\(\\overrightarrow{\\mathrm{XY}}\\) || \\(\\overrightarrow{\\mathrm{AB}}\\)
\nProof : \u2220ABC + \u2220BCX = 180\u00b0 [sum of interior angles on the same side of trnasversal]
\n\u2234 \u2220BCX = 180 – \u2220ABC
\n\u2220CDE + \u2220DCY – 180\u00b0 [sum of interior angles on the same side of transversal]
\n\"KSEEB<\/p>\n

\u2234 \u2220DCY = 180\u00b0 – \u2220CDE
\n\u2220BCX + \u2220BCD + \u2220DCY = 180\u00b0 [\u2235 XCY is straight angle]
\n180 – \u2220ABC + \u2220BCD + 180 – \u2220CDE = 180\u00b0 [By substituting]
\n360 – \u2220ABC + \u2220BCD – \u2220CDE – 180\u00b0
\n360 – 180 – \u2220ABC – \u2220BCD + CDE
\n180\u00b0 = \u2220ABC – \u2220BCD + \u2220CDE
\nor
\n\u2220ABC – \u2220BCD + \u2220CDE = 180\u00b0<\/p>\n

Question 17.
\nConsider two parallel lines and a transversal among the measures of 8 angles formed how many distinct numbers are there?
\nSolution:
\nThere will be two distinct numbers.<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths\u00a0helps 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 …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27382"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=27382"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27382\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=27382"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=27382"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=27382"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}