{"id":1715,"date":"2020-09-07T09:43:35","date_gmt":"2020-09-07T04:13:35","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=1715"},"modified":"2021-07-02T15:03:45","modified_gmt":"2021-07-02T09:33:45","slug":"kseeb-solutions-class-9-maths-chapter-5-ex-5-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths-chapter-5-ex-5-1\/","title":{"rendered":"KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1"},"content":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1<\/strong> are part of KSEEB Solutions for Class 9 Maths<\/a>. Here we have given Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.1.<\/p>\n

Karnataka Board Class 9 Maths Chapter 5 Triangles Ex 5.1<\/h2>\n

Question 1.
\nIn quadrilateral ACBD, AC = AD, and AB bisect \u2220A. Show that \u2206ABC \u2245 \u2206ABD. What can you say about BC and BD?
\nSolution:
\nData: In a quadrilateral ABCD,
\nAC = AD and AB bisect \u2220A.
\n\"KSEEB
\nTo Prove: \u2206ABC \u2245 \u2206ABD
\nProof: In \u2206ABC and \u2206ABD
\nAC = AD (Data)
\n\u2220CAB = \u2220DAB .
\n\u2220A bisected
\nAB is common.
\nHere Side, angle, side rule is there.
\n\u2234 \u2206ABC \u2245 \u2206ABD.<\/p>\n

Question 2.
\nABCD is a quadrilateral in which AD = BC and \u2220DAB = \u2220CBA. Prove that
\n(1) \u2206ABD \u2245 \u2206BAC
\n(2) BD = AC
\n(3) \u2220ABD = \u2220BAC A
\nSolution:
\nData: ABCD is a quadrilateral.
\nAD = BC and \u2220DAB = \u2220CBA.
\n\"KSEEB
\nTo Prove:
\n(1) \u2206ABD \u2245 \u2206BAC
\n(2) BD = AC
\n(3) \u2220ABD = \u2220BAC
\nProof: (i) In \u2206ABD and \u2206BAC
\nAD = BC (Data)
\n\u2220DAB = \u2220CBA (Data)
\nAB is common SAS postulate.
\n\u2234 \u2206ABD \u2245 \u2206BAC.<\/p>\n

(ii) \u2206ABD \u2245 \u2206BAC.
\n\u2235 Corresponding sides are equal.
\n\u2234 BD = AC.<\/p>\n

(iii) \u2206ABD \u2245 \u2206BAC. (Proved)
\nEqual sides of Adjacent angles are equal.
\nAs we have AD = BC,
\nThe adjacent angle for AD is ABD
\nThe adjacent angle for BC is BAC
\n\u2234\u2220ABD = \u2220BAC.<\/p>\n

Question 3.
\nAD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.
\nSolution:
\nData: AD and BC are equal perpendiculars to a line segment AB.
\n\"KSEEB
\nTo Prove: CD Bisects AB.
\nProof: In \u2206CBO and \u2206DAO,
\nBC = AD
\n\u2220CBO = \u2220DAO = 90\u00b0
\n\u2220BOC = \u2220AOD (Vertically opposite angles)
\nSAS postulate,
\n\u2234 \u2206CBO = \u2206DAO \u2234 OA = OB
\n\u2234 CD bisects AB at ‘O’.<\/p>\n

Question 4.
\nl and m are two parallel lines intersected by another pair of parallel lines p and q. Show that \u2206ABC \u2245 \u2206CDA.
\n\"KSEEB
\nSolution:
\nData: l and m are two parallel lines intersected by another pair of parallel lines p and q.
\nTo Prove: \u2206ABC \u2245 \u2206CDA
\nProof: In AABC and ACDA,
\n\u2220ACB = \u2220DAC [\u2235 Alternate angles.]
\n\u2220BAC = \u2220ACD
\nAC is common.
\nA.S.A. postulate.
\n\u2206ABC \u2245 \u2206CDA<\/p>\n

Question 5.
\nLine 1 is the bisector of an angle \u2220A and B is any point on l. BP and BQ are perpendiculars from B to the arms of \u2220A. Show that :
\n\"KSEEB
\n(i) \u2206APB \u2245 \u2206AQB
\n(ii) BP = BQ or B is equidistant from the arms of \u2220A.
\nSolution:
\nData: Line l is the bisector of an angle \u2220A and B is any point on l.
\nBP and BQ are perpendiculars from B to the arms of \u2220A.
\nTo Prove:
\n(i) \u2206APB \u2245 \u2206AQB
\n(ii) BP = BQ or B is equidistant from the arms of \u2220A.
\nProof: In \u2206APB and \u2206AQB
\n\u2220APB = \u2220AQB = 90\u00b0
\n\u2220PAB = \u2220QAB (\u2235 l bisects \u2220A).
\nAB is common.
\n\u2234 \u2206APB \u2245 \u2206AQB (ASA postulate)
\n\u2234 BP = BQ.
\nB is equidistant from the arms of \u2220A.<\/p>\n

Question 6.
\nIn AC = AE, AB = AD and \u2220BAD = \u2220EAC. Show that BC = DE.
\n\"KSEEB
\nSolution:
\nData: AC = AE, AB = AD and \u2220BAD = \u2220EAC
\nTo Prove : BC = DE
\nProof: In \u2206ABC and \u2206EAD,
\nAB = AD (Data)
\nAC = AE (Data)
\n\u2206BAC = \u2206EAD
\n(\u2235\u2220BAD + \u2220DAC = AEAC + \u2220DAC and \u2220DAC = \u2220DAC.)
\nSide – Angle – Side (SAS) Postulate
\n\u2234 \u2206ABC = \u2206EAD
\n\u2234 BC = DE.<\/p>\n

Question 7.
\nAB is a line segment and P is its midpoint. D and E are points on the same side of AB such that \u2220BAD = \u2220ABE and \u2220EPA = \u2220DPB. Show that
\n\"KSEEB
\n(i) \u2206DAP \u2245 \u2206EBP
\n(ii) AD = BE.
\nSolution:
\nData: AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that \u2220BAD= \u2220ABE and \u2220EPA = \u2220DPB.
\nTo Prove :
\n(i) \u2206DAP \u2245 \u2206EBP
\n(ii) AD = BE
\nProof : (i) In \u2206DAP and \u2206EBP,
\nAP = BP (\u2235 P is the mid-point of AB)
\n\u2220BAD = \u2220ABE (Data)
\n\u2220APD = \u2220BPE
\n\u2235 (\u2220EPA = \u2220DPB Adding \u2220EPD to both sides)
\n\u2220EPA + \u2220EPD = \u2220DPB + \u2220EPD
\n\u2234 \u2220APD = -BPE.
\nNow, Angle, Side, Angle postulate.
\n\u2234 \u2206DAP \u2245 \u2206EBP.<\/p>\n

(ii) As it is \u2206DAP \u2245 \u2206EBP,
\nThree sides and three angles are equal to each other.
\n\u2234 AD = BE.<\/p>\n

Question 8.
\nIn the right triangle ABC, right-angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B. Show that :
\n\"KSEEB
\n(i) \u2206AMC \u2245 \u2206BMD
\n(ii) \u2220DBC is a right angle.
\n(iii) \u2206DBC \u2245 \u2206ACB
\n(iv) CM = \\(\\frac{1}{2}\\) AB.
\nSolution:
\nData: In a right triangle ABC, right-angled at C, M is the mid-point of hypotenuse AB.
\nC is joined to M and produced to a point D such that DM = CM.
\nPoint D is joined to point B.
\nTo Prove:
\n(i) \u2206AMC \u2245 \u2206BMD
\n(ii) \u2220DBC is a right angle.
\n(iii) \u2206DBC \u2245 \u2206ACB
\n(iv) CM = \\(\\frac{1}{2}\\) AB.
\nProof: (i) In \u2206AMC and \u2206BMD,
\nBM = AM (\u2235 M is the mid-point of AB)
\nDM = CM (Data)
\n\u2220BMD = \u2220AMC (Vertically opposite angles)
\nNow, Side, Angle, Side postulate.
\n\u2234 \u2206AMC \u2245 \u2206BMD<\/p>\n

(ii) \u2206AMC \u2245 \u2206BMD (Proved)
\nAdding AMBC to both sides,
\n\u2206BMD + \u2206MBC = \u2206AMC + \u2206MBC
\n\u2206DBC \u2245 \u2206ACB.
\n\u2234 AB Hypotenuse = DC Hypotenuse
\n\u2220ACB = \u2220DBC = 90\u00b0
\n\u2234 \u2220DBC is a right angle.<\/p>\n

(iii) In \u2206DBC and \u2206ACB,
\nDB = AC
\n\u2235 \u2206DMB \u2245 DMC (Proved)
\n\u2220DBC = \u2220ACB (Proved)
\nBC is Common.
\nSide, angle, side postulate.
\n\u2234 \u2220DBC \u2245 \u2206ACB.<\/p>\n

(iv) \u2206DBC \u2245 \u2206ACB (proved)
\nHypotenuse AB = Hypotenuse DC
\n\\(\\frac{1}{2}\\) AB = \\(\\frac{1}{2}\\) DC
\n\\(\\frac{1}{2}\\) AB = CM
\n\u2234 CM = \\(\\frac{1}{2}\\)AB.<\/p>\n

We hope the KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.1, 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 5 Triangles Ex 5.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.1. Karnataka Board Class 9 Maths Chapter 5 Triangles Ex 5.1 Question 1. In quadrilateral ACBD, AC = AD, and AB bisect …<\/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\/1715"}],"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=1715"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1715\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=1715"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=1715"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=1715"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}