{"id":4652,"date":"2020-09-09T10:07:43","date_gmt":"2020-09-09T04:37:43","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=4652"},"modified":"2021-07-02T15:03:35","modified_gmt":"2021-07-02T09:33:35","slug":"kseeb-solutions-class-8-maths-chapter-15-ex-15-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-8-maths-chapter-15-ex-15-3\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 15 Quadrilaterals Ex 15.3"},"content":{"rendered":"

Students can Download Maths Chapter 15 Quadrilaterals Ex 15.3 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 15 Quadrilaterals Ex 15.3<\/h2>\n

Question 1.
\nThe adjacent angles of a parallelogram are in the ratio 2 : 1. Find the measures of all the angles.
\nAnswer:
\nThe adjacent angles are in the ratio 2 : 1.
\nLet the angles be 2x and x
\n2x + x = 180\u00b0
\n[adjacent angles of a parallelogram are supplementary]
\n3x = 180\u00b0 .
\nx =\u00a0\"\\frac\u00a0= 60\u00b0
\n2x = 2 \u00d7 60\u00b0 = 120\u00b0
\n\u2234The angles of the parallelogram are 60\u00b0, 120\u00b0. 60\u00b0 and 120\u00b0<\/p>\n

Question 2.
\nA field is in the form of a parallelogram whose perimeter is 450m and one of its sides is larger than the other by 75m. Find the lengths of all the sides.
\nAnswer:
\nLet one side be x, then the other side is x + 75
\n\"KSEEB
\nPerimeter = 450
\nx + (x + 75) + x + (x + 75) = 450
\n4x + 150 = 450
\n4x = 450 – 150
\n4x = 300
\nx = \\(\\frac { 300 }{ 4 }\\) = 75
\nx = 75m
\nx + 75 = 75 + 75 = 150 m
\n\u2234 The four sides are 75m, 150m, 75m, 150m.<\/p>\n

Question 3.
\nIn the figure ABCD is a parallelogram. The diagonals AC and BD intersect at 0; and \u2220DAC = 40\u00b0, \u2220CAB = 35\u00b0 and \u2220DOC = 110\u00b0 calculate the measure \u2220ABO, \u2220ADC, \u2220ACB and \u2220CBD
\nAnswer:
\n\"KSEEB
\n\u2220AOB = \u2220DOC = 110\u00b0
\n[Vertically opposite angles]
\nIn \u0394AOB,
\n\u2220AOB + \u2220OAB + \u2220OBA = 180\u00b0
\n110 + 55 +\u2220OBA = 180\u00b0
\n\u2220OBA = 180 \u2013 165
\n\u2220ABO = 15\u00b0
\n\u2220ADC + \u2220DAB = 180\u00b0
\n[adjacent angles of a parallelogram , are supplementary]
\n\u2220ADC + 95 = 180\u00b0
\n\u2220ADC = 180 \u2013 95
\n\u2220 ADC = 85\u00b0
\n\u2220ACB = \u2220DAC = 40\u00b0
\n[AD 11BC alternate angle]
\n\u2220ACB = 40\u00b0
\n\u2220ABC = \u2220 ADC = 85\u00b0
\n[Opposite angles pf a parallelogram]
\n[\u2220ABC = \u2220ABO + \u2220CBO]
\n15 + \u2220CBO = 85\u00b0
\n\u2220CBO = 85 \u2013 15
\n\u2220CBO = 70
\n\u2234 \u2220CBO = \u2220CBD = 70\u00b0
\n\u2234 \u2220CBD = 70\u00b0<\/p>\n

Question 4.
\nIn a parallelogram ABCD, the side DC is produced to E and \u2220BCE = 105\u00b0 calculate \u2220A, \u2220E, \u2220C, and\u2220D.
\n\"KSEEB
\nAnswer:
\n\u2220DCB + \u2220DCE = 180\u00b0 [Linear pair] \u2220DCB + 1050 = 180\u00b0
\n\u2220DCB = 180 \u2013 105
\n\u2220DCB = 75\u00b0
\n\u2220A = \u2220DCB = 75\u00b0
\n\u2220A = \u2220C = 75\u00b0
\n[Opposite angles of a parallelogram]
\n\u2220A +\u2220B = 180\u00b0
\n[adjacent angles of a parallelogram]
\n75\u00b0 + \u2220B = 180\u00b0
\n\u2220B = 180 + 75
\n\u2220B = 105\u00b0
\n\u2220B = \u2220D = 105\u00b0
\n[Opposite angles of a parallelogram]<\/p>\n

Question 5.
\nIn a parallelogram KLMN, \u2220K = 60 \u00b0. Find the measures of all the angles.
\nAnswer:
\n\"KSEEB
\n\u2220K = \u2220M = 60\u00b0 [Opposite angles of parallelogram]
\n\u2220K + \u2220N = 180\u00b0 [adjacent angles of a parallelogram]
\n60 + \u2220N = 180\u00b0
\n\u2220N = 180 \u2013 60
\n\u2220N = 120\u00b0
\n\u2220N = \u2220L = 120\u00b0
\n[Opposite angles of a parallelograom]
\n\u2234 Angles of a paralelogram are 60\u00b0, 120\u00b0, 60\u00b0 & 120\u00b0<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 15 Quadrilaterals Ex 15.3 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 15 Quadrilaterals Ex 15.3 Question 1. The adjacent angles of a parallelogram …<\/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\/4652"}],"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=4652"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4652\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=4652"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=4652"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=4652"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}