{"id":430,"date":"2020-09-07T12:15:47","date_gmt":"2020-09-07T06:45:47","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=430"},"modified":"2021-07-02T15:03:45","modified_gmt":"2021-07-02T09:33:45","slug":"kseeb-solutions-class-9-maths-chapter-3-ex-3-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths-chapter-3-ex-3-2\/","title":{"rendered":"KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2"},"content":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2<\/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 3.2.<\/p>\n

Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Ex 3.2<\/h2>\n

Question 1.
\nIn Fig. 3.28. find the values of x and y and then shows that AB||CD.
\n\"KSEEB
\nAnswer:
\ni) \u2220x = ?
\nii) \u2220y = ?
\niii) Show tht AB||CD.
\n\"KSEEB
\n(i) PQ Bisects AB at R.
\n\u2220PRA and \u2220ARS are on straight line PQ.
\n\u2234 \u2220PRA + \u2220ARS = 180\u00b0 Adjacent angles
\n50 + \u2220ARS = 180\u00b0
\n\u2220ARS = 180 – 50
\n\u2234 x = \u2220ARS = 130\u00b0 (i)<\/p>\n

(ii) Similarly, PQ straight line intersects straight line CD at S.
\n\u2234 \u2220CSQ = \u2220RSD . Vertically opposite angles
\n\u2234 y = \u2220RSD = 130\u00b0 (ii)
\nFrom (i) and (ii),
\n\u2220ARS = \u2220RYD
\nx = y = 130\u00b0<\/p>\n

(iii) Therse are mutual alternate angles
\n\u2234 x = y = 130\u00b0
\n\u2234 AB || CD.<\/p>\n

Question 2.
\nIn Fig. 3.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.
\n\"KSEEB
\nAnswer:
\nPQ straight line intersects AB||CD]|EF at points R, S, T.
\nIf y : z = 3 : 7, then value of ‘x’ = ?
\nPQ straight line intersects CD||EF at S and T.
\n\u2234 \u2220CSR = \u2220DST = y (\u2235 vertically opposite angle)
\nBut, \u2220DST + \u2220STF = 180\u00b0
\n(\u2235 Sum of interior angles in the same side)
\n\u2234 y + z = 180\u00b0
\nBut, y : z = 3 : 7.
\nSum of ratio 3 + 7 = 10
\nIf ratio 10 means 180
\nIf ratio 9 means ?
\n\\(\\frac{180 \\times 3}{10}=54^{\\circ}\\)
\n\u2234 y = 54\u00b0
\ny + z = 180
\n54 + z = 180
\n\u2234 z = 180 – 54
\n\u2234 z = 126\u00b0
\nBut, x + y = 180
\n(\u2235 Sum of interior angles on one side)
\nx + 54 = 180
\n\u2235 x = 180 – 54
\n\u2235 x = 126\u00b0.<\/p>\n

Question 3.
\nIn Fig. 3.30, If AB||CD, EF \u22a5 CD and \u2220GED = 126\u00b0, find \u2220AGE, \u2220GEF and \u2220FGE.
\n\"KSEEB
\nAnswer:
\nAB||CD and EF\u22a5CD and \u2220GED = 126\u00b0
\n\"KSEEB
\n\u2220AGE =?
\n\u2220GEF =?
\n\u2220FGE =?
\n\u2220GED = 126\u00b0.
\n\u2234 \u2220GEF + \u2220ZFED = 126\u00b0
\n\u2220GEF + 90\u00b0 = 126\u00b0 \u2235 EF \u22a5 CD.
\n\u2234 GEF = 126-90
\n\u2234\u00a0 GEF = 36\u00b0.
\n\u2234 \u2220GFE = 90\u00b0 . EF \u22a5 AB
\n\u2234 In \u2206GEF
\n\u2220FGE + \u2220GFE + \u2220GEF = 180\u00b0
\n\u2220FGE + 90\u00b0 + 36\u00b0 = 180\u00b0
\n\u2220FGE + 126\u00b0 = 180\u00b0
\n\u2234 \u2220FGE = 180-126
\n\u2234 \u2220FGE = 54\u00b0.
\n\u2220AGE + \u2220FGE = 180\u00b0 (Adjacent angles)
\n\u2220AGE + 54\u00b0 = 180\u00b0
\n\u2234 \u2220AGE =180-54
\n\u2220AGE = 126\u00b0
\n\u2234 \u2220AGE = 126\u00b0
\n\u2220GEF = 36\u00b0
\n\u2220FGE = 54\u00b0.<\/p>\n

Question 4.
\nIn Fig. 3.31, If PQ||ST, \u2220PQR =110\u00b0, and \u2220RST = 130\u00b0, find \u2220QRS.
\n(Hint: Draw a line parallel to ST through point R).
\n\"KSEEB
\nAnswer:
\n\"KSEEB
\nData: PQ||ST and \u2220PQR= 110\u00b0 and \u2220RST= 130\u00b0.
\nTo Prove: \u2220QRS =?
\nConstruction: Draw ST||UV through ‘R’
\nProof: PQ||ST (Data)
\nST || UV (Construction)
\n\u2234 PQ||ST||UV
\nPQ||UV.
\n\u2234 \u2220PQR + \u2220URQ = 180\u00b0 (Sum of interior angles)
\n110 + \u2220URQ = 180\u00b0
\n\u2220URQ = 180- 110
\n\u2234 \u2220URQ = 70\u00b0 (i)
\nSimilarly, ST||UV.
\n\u2234 \u2220RST + \u2220SRV = 180\u00b0
\n130 + \u2220SRV = 180
\n\u2234 \u2220SRV = 180 – 130
\n\u2234 \u2220SRV = 50\u00b0 (ii)
\nBut, URV is a straight line.
\n\u2234 \u2220URV = 180\u00b0.
\n\u2220URQ + \u2220QRS + \u2220RSV = 180\u00b0
\n70 + \u2220QRS + 50 = 180
\n\u2220QRS + 120 = 180\u00b0
\n\u2234 \u2220QRS = 180 – 120
\n\u2234 \u2220QRS = 60\u00b0.<\/p>\n

Question 5.
\nIn Fig. 3.32. if AB||CD, \u2220APQ = 50\u00b0 and \u2220PRD = 127\u00b0, find x and y.
\n\"KSEEB
\nAnswer:
\nIf AB || CD and \u2220APQ= 50\u00b0 and \u2220PRD = 127\u00b0, then \u2220x = ? \u2220y = ?
\nAB || CD PQ straight line intersects these at P and Q.
\n\u2234 \u2220APQ = \u2220PQR . alternate angles.
\n50 = \u2220PQR = x
\n\u2234 x = 50\u00b0
\nSimilarly, AB || CD PR straight line intersects these at P and R.
\n\u2234 \u2220APR = \u2220PRD
\n\u2220APQ + \u2220QPR = \u2220PRD
\n50 + \u2220QPR = 127
\n\u2220QPR = 127 – 50
\n\u2234\u2220QPR = 77 \u00b0
\n\u2220QPR = y = 77 \u00b0
\n\u2234 x = 50 \u00b0,
\ny = 77 \u00b0.<\/p>\n

Question 6.
\nIn Fig. 3.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD.
\nProve that AB||CD.
\n\"KSEEB
\nAnswer:
\nData: PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B. the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD.
\nTo Prove: AB || CD.
\nProof: PQ is a mirror.
\nAB is an incident ray, BC is a reflected ray. \u2220ABC is the angle of incidence.
\n\u2220BCD is an angle of reflection.
\nThe incident ray formed in the PQ mirror is equal to the reflected ray formed in the RS mirror.
\n\u2234 \u2220ABC = \u2220BCD.
\nThese are pair of alternate interior angles, then AB || CD.<\/p>\n

We hope the KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Exercise 3.2, 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 3.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Exercise 3.2. Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 Question 1. In Fig. 3.28. …<\/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\/430"}],"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=430"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/430\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=430"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=430"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=430"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}