{"id":391,"date":"2020-09-07T11:04:04","date_gmt":"2020-09-07T05:34:04","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=391"},"modified":"2021-07-02T15:03:45","modified_gmt":"2021-07-02T09:33:45","slug":"kseeb-solutions-class-9-maths-chapter-3-ex-3-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths-chapter-3-ex-3-1\/","title":{"rendered":"KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1"},"content":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1<\/strong> are part of KSEEB Solutions for Class 9 Maths<\/a>. Here we have given Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Exercise 3.1.<\/p>\n

Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Ex 3.1<\/h2>\n

Question 1.
\nIn Fig 3.13, lines AB and CD intersect at O. If \u2220AOC + \u2220BOE = 70\u00b0 and \u2220BOD = 40\u00b0, find \u2220BOE and reflex \u2220COE.
\n\"KSEEB
\nAnswer:
\n\"KSEEB
\n\u2220AOC = \u2220BOD = 40\u00b0
\n\u2220BOD = 40\u00b0 (Date)
\n\u2220AOC = \u2220BOD = 40\u00b0 because vertifically opposite angles.
\n\u2234 \u2220AOC + \u2220BOE = 70\u00b0
\n40\u00b0 + \u2220BOE = 70\u00b0
\n\u2234 \u2220BOE = 70\u00b0 – 40\u00b0 = 30\u00b0, and
\n\u2220AOD = 180\u00b0 – \u2220BOD = 180\u00b0 – 40\u00b0 = 140\u00b0
\nReflex angle COE = \u2220COA + \u2220AOD + \u2220BOD + \u2220BOE
\n= 40\u00b0 + 140\u00b0 + 40\u00b0 + 30\u00b0
\n\u2234 Reflex angle, \u2220COE = 250\u00b0.<\/p>\n

Question 2.
\nIn Fig. 3.14, lines XY and MN intersect at O. If \u2220POY = 90\u00b0 and a : b = 2 : 3, find c.
\n\"KSEEB
\nAnswer:
\n\u2220XOP + \u2220POY = 180\u00b0
\n\u2235 straight supplementary
\n\u2220XOP + 90\u00b0 =180\u00b0
\n\u2234 \u2220XOP = 180\u00b0 – 90\u00b0 = 90\u00b0
\nBut, \u2220XOP = a + b = 90\u00b0
\na : b = 2 : 3
\n2 + 3 = 5 Ratio
\nRatio 5 means 90\u00b0
\n\\(=\\frac{2 \\times 90}{5}=2 \\times 18=36^{\\circ}\\)
\n\u2234 If a = 36\u00b0 then, \u2220b = 54\u00b0.
\n\u2220XOM = \u2220YON = b = 54\u00b0 (\u2235 Vertically opposite angles)
\n\u2220XON + \u2220YON = 180\u00b0 (\u2235 Straight angle)
\n\u2234 c + 54\u00b0 = 180\u00b0
\nc = 180 – 54
\n\u2234 c = 126\u00b0.<\/p>\n

Question 3.
\nIn Fig. 3.15, \u2220PQR = \u2220PRQ, then prove that \u2220PQS = \u2220PRT.
\n\"KSEEB
\nAnswer:
\n\"KSEEB
\nData: In this figure, \u2220PQR = \u2220PRQ.
\nTo prove: \u2220PQS = \u2220PRT
\nProof: In \u2206PQR, \u2220Q = \u2220R.
\n\u2234 This is an isosceles triangle.
\nLet \u2220Q = 70\u00b0, then \u2220R = 70\u00b0 .
\n\u2220PQS + \u2220PQR = 180\u00b0
\n\u2220PQS + 70\u00b0 = 180\u00b0
\n\u2234 \u2220PQS =180 – 70
\n\u2220PQS = 110\u00b0 … (i)
\nSimilarly,
\n\u2220PRT + \u2220PRQ = 180\u00b0
\n\u2220PRT + 70\u00b0 = 180\u00b0
\n\u2234 \u2220PRT = 180 – 70
\n\u2220PRT =100 … (ii)
\nfrom (i) and (ii)
\n\u2220PQS = \u2220PRT =110\u00b0
\n\u2234 \u2220PQS = \u2220PRS proved.<\/p>\n

Question 4.
\nIf Fig. 3.16, if x + y = w + z, then prove that AOB is a line.
\n\"KSEEB
\nAnswer:
\nData: In this fiure, \u2220BOC = x\u00b0
\n\u2220AOC = y\u00b0
\n\u2220BOD = w\u00b0
\n\u2220AOD = z\u00b0 and
\nx + y = w + z.
\nTo Prove: AOB is a straight line.
\nProof: x + y + w + z = 360\u00b0 (\u2235 one completre angle)
\nBut, x + y = w + z.
\n\u2234 x + y = w + z= \\(\\frac{360}{2}\\) = 180\u00b0
\n\u2234 x + y = 180\u00b0
\n\u2234 w + z = 180\u00b0 proved.
\nBut, \u2220AOC and \u2220BOC are adjacent angles,
\n\u2220AOC + \u2220BOC = 180\u00b0
\nx = y = 180\u00b0
\n\u2234 \u2220AOB = 180\u00b0.
\n\u2234 AOB is a straight line.<\/p>\n

Question 5.
\nIn Fig. 3.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that
\n\u2220ROS = \\(\\frac{1}{2}\\) (\u2220QOS – \u2220POS).
\n\"KSEEB
\nAnswer:
\nData: POQ is a straight line. Ray OR is perpendicular on straight line PQ. OS Ray is in between Rays OP and OR.
\nTo Prove: \u2220ROS = \\(\\frac{1}{2}\\) (\u2220QOS – \u2220POS)
\nProof : \u2220QOR = \u2220POR = 90\u00b0 (Data)
\n\u2220POR = \u2220POS + \u2220ROS
\n\u2234 \u2220ROS = \u2220POR – \u2220POS
\n\u2220ROS = 90\u00b0- \u2220POS (i)
\nNow, QOS = \u2220QOR + \u2220ROS
\n\u2220QOS = 90\u00b0 + \u2220ROS
\n\u2234 \u2220ROS = \u2220QOS – 90\u00b0 (ii)
\nBy adding equation (i) and (ii)
\n\"KSEEB
\n\u2234 \u2220ROS = \\(\\frac{1}{2}\\) (\u2220QOS – \u2220POS)<\/p>\n

Question 6.
\nIt is given that \u2220XYZ = 64\u00b0 and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects \u2220ZYP, find \u2220XYO and reflex \u2220QYP.
\nAnswer:
\n\"KSEEB<\/p>\n

Data: \u2220XYZ = 64\u00b0 and XY are produced up to P. \u2220ZYP is bisected.
\nTo Prove: \u2220XYQ = ? Reflex \u2220QYP = ?
\nProof: YQ bisects \u2220ZYP
\n\u2234 Let \u2220PYQ = \u2220ZYQ = x\u00b0.
\n\u2220XYZ + \u2220ZYQ + \u2220QYP = 180\u00b0 (\u2235 \u2220XYP is straight angle)
\n64 + x + x = 180
\n\u2234 64 + 2x = 180
\n\u2234 2x = 180 – 64
\n2x = 116
\n\u2234 \\(x=\\frac{116}{2}=58^{\\circ}\\)
\n\u2234 \u2220PYQ = \u2220ZYQ = 58\u00b0<\/p>\n

(i) \u2234 \u2220XYQ = \u2220XYZ + \u2220ZYQ
\n= 64 + 58
\n\u2220XYQ = 122\u00b0<\/p>\n

(ii) Reflex \u2220QYP = \u2220PYX + \u2220XYZ + \u2220ZYQ
\n= 180 \u00b0 + 64\u00b0 + 58\u00b0
\n= 180 + 122
\n\u2234 Reflex \u2220QYP = 302\u00b0.<\/p>\n

We hope the KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Exercise 3.1, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Exercise 3.1. Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Ex 3.1 Question 1. In Fig 3.13, …<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/391"}],"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\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=391"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/391\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=391"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=391"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=391"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}