{"id":4387,"date":"2020-09-09T11:31:25","date_gmt":"2020-09-09T06:01:25","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=4387"},"modified":"2021-07-02T15:03:35","modified_gmt":"2021-07-02T09:33:35","slug":"kseeb-solutions-class-8-maths-chapter-11-ex-11-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-8-maths-chapter-11-ex-11-3\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 11 Congruency of Triangles Ex 11.3"},"content":{"rendered":"

Students can Download Maths Chapter 11 Congruency of Triangles Ex 11.3 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 11 Congruency of Triangles Ex 11.3<\/h2>\n

Question 1.
\nIn a \u0394 ABC, AB = AC andlA.= 50\u00b0 find \u2220B and \u2220C.
\nAnswer:
\n\u2220A+ \u2220B + \u2220C = 180\u00b0
\n(Sum of the angles of a triangle is 180\u00b0 )
\n50 +\u2220[B + \u2220B = 180\u00b0
\n\u2220B = \u2220C Base angles of an isosceles triangle
\n50 + 2\u2220B= 180\u00b0
\n2\u2220B = 180 – 50
\n2\u2220B = 130\u00b0
\n\u2220B = \\(\\frac{130^{\\circ}}{2}\\)
\n\u2220B = 65\u00b0
\n\u2220B = \u2220C = 65\u00b0<\/p>\n

Question 2.
\nIn AABC,AB = BCand|B = 64\u00b0find|\u00a3,
\nAnswer:
\nAB = BC [data]
\n\u2234 \u2220C = \u2220A [Theorem 1]
\n\u2220A + \u2220B + \u2220C = 180\u00b0
\n(Sum of the angles of a triangle is 180\u00b0)
\n\u2220C + 64 + \u2220C = 180\u00b0 [\u2220A = \u2220C]
\n64 + 2\u2220C = 180\u00b0
\n2\u2220C = 180 – 64
\n2\u2220C = 116
\n\u2220C = \\(\\frac { 116 }{ 2 }\\) = 58\u00b0<\/p>\n

Question 3.
\nIn each of the following figure find the value of x :
\ni.
\n\"KSEEB
\nAnswer:
\nIn \u2206ABC, AB = AC
\n\u2234 \u2220ABC = \u2220ACB
\n\u2234 \u2220BAC + \u2220ABC + \u2220ACB = 180\u00b0
\n(Sum of the angles of a triangle is 180\u00b0)
\n40 + \u2220ABC + \u2220ABC = 180\u00b0
\n(\u2220ABC =\u2220ACB)
\n40 + 2\u2220ABC = 180\u00b0
\n2\u2220ABC = 180 \u2013 40
\n2\u2220ABC = 140\u00b0
\n\u2220ABC = 70\u00b0
\n\u2220ACB = \u2220 ABC = 70\u00b0
\n\u2220ACB + \u2220ACD = 180\u00b0
\n70 + x = 180\u00b0
\nx = 180 \u2013 70
\nx = 110\u00b0<\/p>\n

ii
\n\"KSEEB
\nAC = CD
\n\u2220CAD = \u2220CDA
\n\u2220CAD = \u2220CAD = 30\u00b0
\n\u2220ACD + \u2220CAD + \u2220CDA = 180\u00b0
\n(Sum of the angles of a triangle is 180\u00b0)
\n\u2220ACD + 30 + 30=180\u00b0
\n\u2220ACD + 60=180\u00b0
\n\u2220ACD = 180-60
\n\u2220ACD = 120\u00b0
\n\u2220ACD = \u2220BAC + \u2220ABC
\n120 = 65\u00b0 + x
\n120 – 65 = x
\n55 = x
\nx=55\u00b0<\/p>\n

iiii
\n\"KSEEB
\nAnswer:
\nAB = AC
\n\u2220ABC = \u2220ACB = 55\u00b0 [Theorem l]
\nExterior \u2220APB = \u2220DAC +\u2220ACD
\n75 = x + 55
\n75 – 55 = x
\n20 = x
\nx = 20\u00b0<\/p>\n

iv.
\n\"KSEEB
\nBD = DC = Ad
\nBD = DC = AD & \u2220ABD = 50 \u00b0
\n\u2220ABD = \u2220BAD(Th. l)
\n\u2220BAD = 50\u00b0
\n\u2220ABD + \u2220BAD + \u2220ADB = 180\u00b0
\n50 + 50 +\u2220ADB = 180\u00b0
\n\u2220ADB = 180 – 100 = 80\u00b0
\n\u2220APB = 80\u00b0
\n\u2234 \u2220APB +\u2220ADC = 180\u00b0
\n80 +\u2220ADC = 180\u00b0
\n\u2220ADC = 100\u00b0
\nNow AD = DC
\n\u2234\u2220DAC = \u2220DCA = x\u00b0
\n\u2234 x + x +\u2220ADC = 180\u00b0
\n2x + 100 = 180\u00b0
\n2x = 180-100=80\u00b0
\nx=40\u00b0<\/p>\n

Question 4.
\nSuppose ABC is an equilateral triangle. Its base BC is produced to D such that BC = CD.
\nCalculate: l.\u2220ACD
\n2.\u2220ADC
\nAnswer:
\n\u2220ABC = \u2220ACB = \u2220BAC = 60\u00b0
\n(ABC is an equilateral triangle)
\n\u2220ACB +\u2220ACD = 180\u00b0 (Linearpoint)
\n60 + \u2220ACD = 180\u00b0
\n\u2220ACD = 180 – 60
\n\u2220ACD = 120\u00b0
\n\"KSEEB
\nIn \u2206ACD,AC=CD
\n\u2220 CAD = \u2220CPA (Theorem l)
\n\u2220ACB +\u2220ACD = 180\u00b0 [linear pair]
\n60\u00b0 +\u2220ACD = 180\u00b0
\n\u2220ACD = 180\u00b0 – 60\u00b0 = 120\u00b0
\n\u2220ACA +\u2220CAD + \u2220CDA = 180\u00b0
\n2\u2220CDA = 180 – 120\u00b0
\n2\u2220CDA = 60\u00b0
\n\u2220CDA = \\(\\frac { 60 }{ 2 }\\) = 30\u00b0
\n\u2220CDA = 30<\/p>\n

Question 5.
\nShow that the perpendicular drawn from the vertices of the base of an isosceles triangle to the opposite sides are equal. ,
\nAnswer:
\nData : In \u2206ABC, AB = AC,
\nBD \u22a5AC & CE\u22a5 AB
\nTo prove : BD = CE
\n\"KSEEB
\nProof:
\nIn \u2206ABC, AB = AC [data]
\n\u2220ABC = \u2220ACB [Theorem l]
\nIn \u2206EBC and \u2206DCB
\n\u2220EBC =\u2220DCB( Base angles )
\n\u2220BEC = \u2220CDB [= 90\u00b0 ]
\nBC = BC (Common side)
\n\u2206EBC = \u2206DCB [ASA postulate]
\nBD = CE [Corresponding sides]<\/p>\n

Question 6.
\nProve that an \u2206ABC is an isosceles triangle if the altitude AD from A on BC bisects BC.
\nAnswer:
\nIn \u2206ADB and \u2206ADC
\nAD = AD [Common side]
\n\u2220APB =\u2220 ADC [90\u00b0 ]
\n\"KSEEB
\nBD = DC [AD bisects BC]
\n\u2234\u2206ADB \u2245 \u2206ADC [SAS postulate]
\n\u2234AB = AC [Correspondingsides]
\n\u2234\u2206 ABC is an isosceles triangle<\/p>\n

Question 7.
\nSuppose a triangle is equilateral, prove that it is equiangular.
\nAnswer:
\n\"KSEEB
\nTo prove: \u2220A = \u2220B = \u2220C
\nProof: In \u2206ABC, AB = BC
\n\u2220C = \u2220B [Theorem l]\u2026.(i)
\nBC = AC
\n\u2220A =\u2220B [Theorem l]\u2026(ii)
\nFrom (i) and (ii)
\n\u2220A =\u2220B = \u2220C
\n\u2206ABC is equiangular<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 11 Congruency of Triangles Ex 11.3 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 11 Congruency of Triangles Ex 11.3 Question 1. In a …<\/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\/4387"}],"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=4387"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4387\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=4387"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=4387"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=4387"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}