{"id":4743,"date":"2020-09-09T10:43:34","date_gmt":"2020-09-09T05:13:34","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=4743"},"modified":"2021-07-02T15:03:35","modified_gmt":"2021-07-02T09:33:35","slug":"kseeb-solutions-class-8-maths-chapter-3-ex-3-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-8-maths-chapter-3-ex-3-3\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Ex 3.3"},"content":{"rendered":"

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

Karnataka Board Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Ex 3.3<\/h2>\n

Question 1.
\nFind all the angles in the following
\n\"KSEEB
\nAnswer:
\n\u2220CMP + \u2220PMD = 180\u00b0 [Linear pair]
\n\u2220CMP+ 135\u00b0 = 180\u00b0
\n\u2220CMP = 180\u00b0 – 135
\n\u2220CMP = 45\u00b0
\n\u2220LMD = \u2220CMP = 45\u00b0 [Vertical opposite angles]
\n\u2220LMC =\u2220PMD = 135\u00b0 [Vertical opposite angles]
\n\u2220ALM = \u2220 LMC = 135\u00b0 [Alternate angles]
\n\u2220BLM = \u2220 LMC = 135\u00b0 [Alternate angle]
\n\u2220QLB = \u2220LMD = 45\u00b0 [Corresponding angles]
\n\u2220QLA = \u2220 LMC = 135\u00b0 [Corresponding angles<\/p>\n

Question 2.
\nFind the value of x in the diagram below.
\n\"KSEEB
\nAnswer:
\n\u2220PQS = \u2220EPT = 130\u00b0 [Corresponding angle]
\n\u2220PQS + \u2220SQR = 180\u00b0 [Linear pair]
\nSQR = 180 – 130
\n\u2220SQR = 50\u00b0
\n\u2220QRS + \u2220FRS = 180\u00b0 [Linear pair]
\n\u2220QRS + 90\u00b0 =180\u00b0
\n\u2220QRS = 180 – 90
\n\u2220QRS = 90\u00b0
\n\u2220SQR+ \u2220QRS +\u2220QSR = 180\u00b0
\n[Sum of the angles of triangle is 180\u00b0 ]
\n50 + 90 + \u2220QSR = 180\u00b0
\n140 + \u2220QSR = 180\u00b0
\n\u2220QSR = 180 – 140
\n\u2220QSR =40\u00b0
\n\u2220TSD = \u2220QSR [Vertically opposite angles]
\nx = 40\u00b0.<\/p>\n

Question 3.
\nShow 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.
\nAnswer:
\n\"KSEEB
\n\\(\\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}}\\)
\nToprove: \\(\\overline{\\mathrm{XY}} \\perp \\overrightarrow{\\mathrm{CD}}, \\overline{\\mathrm{XY}} \\perp \\overrightarrow{\\mathrm{GH}}\\)
\nProof :\u2220XPB = 90\u00b0 (data)
\n\u2220XPB = \u2220 PQD = \u2220QRF = \u2220RSH = 90\u00b0<\/p>\n

Question 4.
\nLet \\(\\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.
\nAnswer:
\n\"KSEEB
\nLet the bisectors of \u2220BRS and\u2220RSD intersect at T
\nTo prove: RT\u22a5ST i.e., \u2220RTS = 90\u00b0
\nProof: \u2220BRS +\u2220RSD = 180\u00b0
\n\\(\\frac { 1 }{ 2 }\\) \u2220BRS + \\(\\frac { 1 }{ 2 }\\) \u2220RSD = \\(\\frac { 1 }{ 2 }\\) \u00d7 180\u00b0
\n\\(\\frac { 1 }{ 2 }\\) \u2220BRS = \u2220TRS =\u2220TRB
\n\\(\\frac { 1 }{ 2 }\\) \u2220RSQ = \u2220TSR =\u2220TSD
\nIn triangle TRS,
\n\u2220TRS +\u2220TSR + \u2220RTS = 180\u00b0
\n90 + \u2220RTS = 180\u00b0
\n\u2220RTS = 90\u00b0
\n\u2234 RT \u22a5 ST<\/p>\n","protected":false},"excerpt":{"rendered":"

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. …<\/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\/4743"}],"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=4743"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4743\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=4743"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=4743"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=4743"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}