{"id":1851,"date":"2020-09-08T11:20:57","date_gmt":"2020-09-08T05:50:57","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=1851"},"modified":"2021-07-02T15:03:43","modified_gmt":"2021-07-02T09:33:43","slug":"kseeb-solutions-class-9-maths-chapter-5-ex-5-4","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths-chapter-5-ex-5-4\/","title":{"rendered":"KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4"},"content":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4<\/strong> are part of KSEEB Solutions for Class 9 Maths<\/a>. Here we have given Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.4.<\/p>\n

Karnataka Board Class 9 Maths Chapter 5 Triangles Ex 5.4<\/h2>\n

Question 1.
\nShow that in a right angled triangle, the hypotenuse is the longest side.
\n\"KSEEB
\nSolution:
\nIn \u2206ABC, \u2220ABC = 90\u00b0
\n\u2220A + \u2220C = 90\u00b0
\n\u2220ABC > \u2220BAC and \u2220ABC < \u2220BCA
\n\u2234 D is the largest side of \u2220ABC.
\n\u2234 AC is opposite side of larger angle.
\n\u2234 AC hypoternuse is largest side of \u2206ABC.<\/p>\n

Question 2.
\nIn sides AB and AC of \u2206ABC are extended to points P and Q respectively. Also, \u2220PBC < \u2220QCB. Show that AC > AB.
\n\"KSEEB
\nSolution:
\nData: AB and AC are the sides of \u2206ABC and AB and AC are produced to P and Q respectively.
\n\u2220PBC < \u2220QCB.
\nTo Prove: AC > AB
\nProof: \u2220PBC < \u2220QCB
\nNow, \u2220PBC + \u2220ABC = 180\u00b0
\n\u2220ABC = 180 – \u2220PBC ………. (i)
\nSimilarly, \u2220QCB + \u2220ACB = 180\u00b0
\n\u2220ACB = 180 – \u2220QCB …………. (ii)
\nBut, \u2220PBC < \u2220QCB (Data)
\n\u2234 \u2220ABC > \u2220ACB Comparing (i) and (ii), AC > AB
\n(\u2235 Angle opposite to larger side is larger)<\/p>\n

Question 3.
\nIn fig \u2220B < \u2220A and \u2220C < \u2220D. Show that AD < BC.
\n\"KSEEB
\nSolution:
\nData : \u2220B < \u2220A and \u2220C < \u2220D.
\nTo Prove: AD < BC
\nProof: \u2220B < \u2220A
\n\u2234 OA < OB …………. (i)
\nSimilarly, \u2220C < \u2220D
\n\u2234 OD < OC …………. (ii)
\nAdding (i) and (ii), we have
\nOA + OD < OB + OC
\n\u2234 AD < BC.<\/p>\n

Question 4.
\nAB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. Show that \u2220A > \u2220C and \u2220B > \u2220D.
\n\"KSEEB
\nSolution:
\nData : AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD.
\nTo Prove :
\n(i) \u2220A > \u2220C
\n(ii) \u2220B > \u2220D
\nConstruction: Join AC and BD
\n\"KSEEB
\nProof: In \u2206ABC, AB < BC (data)
\n\u2234 \u22203 < \u22204 ………. (i)
\nIn \u2206ADC, AD < CD (data)
\n\u2234 \u22202 <\u22201 …………. (ii)
\nAdding (i) and (ii),
\n\u22203 + \u22202 < \u22204 + \u22201
\n\u2220C < \u2220A
\n\u2220A > \u2220C
\nBy adding B,
\nIn \u2206ABD, AB < AD
\n\u22205 < \u22208
\nIn \u2206BCD, BC < CD
\n\u22206 < \u22207
\n\u2234 \u22205 + \u22206 < \u22208 + \u22207
\n\u2220D < \u2220B
\nOR \u2220B > \u2220D.<\/p>\n

Question 5.
\nIn Fig. PR > PQ and PS bisect \u2220QPR. Prove that \u2220PSR > \u2220PSQ.
\n\"KSEEB
\nSolution:
\nData: PR > PQ and PS bisect \u2220QPR.
\nTo Prove: \u2220PSR > \u2220PSQ
\nProof: In \u2206PQR, PR > PQ
\n\u2234 \u2220PQR > \u2220PRQ ………. (i)
\nPS bisects \u2220QPR.
\n\u2234 \u2220QPS = \u2220RPS ………… (ii)
\nBy adding (i) and (ii),
\n\u2220PQR + \u2220QPS > \u2220PRQ + \u2220RPS ………. (iii)
\nBut, in \u2206PQS, \u2220PSR is an exterior angle.
\n\u2234 \u2220PSR = \u2220PQR + \u2220QPS ………… (iv)
\nSimilarly, in \u2206PRS, \u2220PSQ is the exterior angle.
\n\u2234 \u2220PSQ = \u2220PRS + \u2220RPS ……… (v)
\nIn (iii), subtracting (iv) and (v),
\n\u2220PSR > \u2220PSQ.<\/p>\n

Question 6.
\nShow that of all line segments drawn from a given point, not on it, the perpendicular line segment is the shortest.
\n\"KSEEB
\nSolution:
\nP is a point on straight line l.
\nPR is the segment for l drawn such that PQ \u22a5 l.
\nNow, in \u2206PQR,
\n\u2220PQR = 90\u00b0 (Construction)
\n\u2234 \u2220QPR + \u2220QRP = 90\u00b0
\n\u2234 \u2220QRP < \u2220PQR
\n\u2234 PQ < PR
\nAny line segment is drawn from P to l they form a small angle.
\n\u2234 PQ \u22a5 l is smaller.<\/p>\n

We hope the KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.4, 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 5 Triangles Ex 5.4 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.4. Karnataka Board Class 9 Maths Chapter 5 Triangles Ex 5.4 Question 1. Show that in a right angled triangle, the hypotenuse …<\/p>\n","protected":false},"author":6,"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\/1851"}],"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\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=1851"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1851\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=1851"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=1851"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=1851"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}