{"id":1794,"date":"2020-09-08T10:53:20","date_gmt":"2020-09-08T05:23:20","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=1794"},"modified":"2021-07-02T15:03:43","modified_gmt":"2021-07-02T09:33:43","slug":"kseeb-solutions-class-9-maths-chapter-5-ex-5-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths-chapter-5-ex-5-3\/","title":{"rendered":"KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3"},"content":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3<\/strong> are part of KSEEB Solutions for Class 9 Maths<\/a>. Here we have given Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.3.<\/p>\n

Karnataka Board Class 9 Maths Chapter 5 Triangles Ex 5.3<\/h2>\n

Question 1.
\n\u2206ABC and \u2206DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that
\n\"KSEEB
\n(i) \u2206ABD \u2245 \u2206ACD
\n(ii) \u2206ABP \u2245 \u2206ACP
\n(iii) AP bisects \u2220A as well as \u2220D.
\n(iv) AP is the perpendicular bisector of BC.
\nSolution:
\nData : \u2206ABC and \u2206DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. AD is extended to intersect BC at P.
\nTo Prove:
\n(i) \u2206ABD \u2245 \u2206ACD
\n(ii) \u2206ABP \u2245 \u2206ACP
\n(iii) AP bisects \u2220A as well as \u2220D.
\n(iv) AP is the perpendicular bisector of BC
\n(v) AD is the angular bisector of \u2220A.
\nProof:
\n(i) In \u2206ABD and \u2206ACD,
\nAB = AC (data)
\nBD = DC (data)
\nAD is common.
\nS.S.S. Congruence rule.
\n\u2234 \u2206ABD \u2245 \u2206ACD<\/p>\n

(ii) In \u2206ABP and \u2206ACP,
\nAB = AC (data)
\n\u2220ABP = \u2220ACP (Opposite angles)
\n\u2220BAP = \u2220CAP (\u2235 \u2206ABD \u2245 \u2206ACD proved)
\nNow ASA postulate.
\n\u2206ABP \u2245 \u2206ACP.<\/p>\n

(iii) \u2206BAD \u2245 \u2206CAD proved.
\nAP bisects \u2220A.
\nIn \u2206BDP and \u2206CDP,
\nBD = DC (data)
\nBP = PC (proved)
\nDP is common.
\n\u2234 \u2206BDP \u2245 \u2206CDP (SSS postulate)
\n\u2234 \u2220BDP = \u2220CDP
\n\u2234 DP bisects \u2220D.
\n\u2234 AP bisects \u2220D.<\/p>\n

(iv) Now, \u2220APB + \u2220APC = 180\u00b0 (Linear pair)
\n\u2220APB + \u2220APB = 180\u00b0
\n2 \u2220APB = 180
\n\u2234 \u2220APB = \\(\\frac{180}{2}\\)
\n\u2234\u2220APB = 90\u00b0
\n\u2220APB = \u2220APC = 90\u00b0
\nBP = PC (proved)
\n\u2234 AP is the perpendicular bisector BC.<\/p>\n

(v) AP is the angular bisector of \u2220A.
\nAngular bisector of \u2220A is aD, because AD, AP is in one line.<\/p>\n

Question 2.
\nAD is an altitude of an isosceles triangle ABC in which AB = AC. Show that A
\n\"KSEEB
\n(i) AD bisects BC
\n(ii) AD bisects \u2220A.
\nSolution:
\nData: AD is an altitude of an isosceles triangle ABC in which AB = AC.
\nTo Prove:
\n(i) AD bisects BC.
\n(ii) AD bisects \u2220A.
\nProof: i) In \u2206ABD and \u2206ACD,
\n\u2220ADB = \u2220ADC (\u2235 AD \u22a5 BC)
\nAB = AC (data)
\nAD is common.
\n\u2234 \u2206ABD \u2245 \u2206ACD
\n\u2234 BD = DC
\n\u2234 AD bisects BC.<\/p>\n

(ii) \u2220BAD = \u2220CAD (\u2235 \u2206ADB \u2245 \u2206ADC)
\n\u2234 AD bisects \u2220A.<\/p>\n

Question 3.
\nTwo sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of \u2206PQR. Show that :
\n\"KSEEB
\n(i) \u2206ABM \u2245 \u2206PQN
\n(ii) \u2206ABC \u2245 \u2206PQR.
\nSolution:
\nData: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of \u2206PQR.
\nTo Prove:
\n(i) \u2206ABM \u2245 \u2206PQN
\n(ii) \u2206ABC \u2245 \u2206PQR.
\nProof: (i) In \u2206ABC,
\nAM is the median drawn to BC.
\n\u2234 BM = \\(\\frac{1}{2} \\) BC
\nSimilarly, in \u2206PQR,
\nQN = \\(\\frac{1}{2}\\) QR
\nBut, BC = QR
\n\\(\\frac{1}{2} \\) BC = \\(\\frac{1}{2}\\) QR
\n\u2234 BM = QN
\nIn \u2206ABM and \u2206PQN,
\nAB = PQ (data)
\nBM = QN (data)
\nAM = PN (proved)
\n\u2234 \u2206ABM \u2245 \u2206PQN (SSS postulate)<\/p>\n

(ii) In \u2206ABC and \u2206PQR,
\nAB = PQ (data)
\n\u2220ABC = \u2220PQR (proved)
\nBC = QR (data)
\n\u2234 \u2206ABC \u2245 \u2206PQR (SSS postulate)<\/p>\n

Question 4.
\nBE and CF are two equal altitudes of a triangle ABC. Using the RHS congruence rule, prove that the triangle ABC is isosceles.
\n\"KSEEB
\nSolution:
\nData: BE and CF are two equal altitudes of a triangle ABC.
\nTo Prove: ABC is an isosceles triangle.
\nProof : BE = CF (data)
\nIn \u2206BCF and \u2206CBE,
\n\u2220BFC = \u2220CEB = 90\u00b0 (data)
\nBC is a common hypotenuse.
\nAs per Right angle, hypotenuse, side postulate,
\n\u2234 \u2206BCF \u2245 \u2206CBE
\n\u2234 \u2220CBF = \u2220BCE
\n\u2234 \u2220CBA = \u2220BCA
\n\u2234 AB = AC
\n\u2234 \u2206ABC is an isosceles triangle.<\/p>\n

Question 5.
\nABC is an isosceles triangle with AB = AC. Draw AP \u22a5 BC to show that \u2220B = \u2220C.
\n\"KSEEB
\nSolution:
\nData: ABC is an isosceles triangle with AB = AC.
\nTo Prove : \u2220B = \u2220C
\nConstruction: Draw AP \u22a5 BC.
\nProof: In \u2206ABC, AP \u22a5 BC and AB = BC.
\n\u2234 In \u2206ABP and \u2206ACP
\n\u2220APB = \u2220APC = 90\u00b0 ( \u2235 AP \u22a5 BC)
\nHypotenuse AB = Hypotenuse AC
\nAP is common.
\nAs per RHS Postulate,
\n\u2206ABP \u2245 \u2206ACP
\n\u2234 \u2220ABP = \u2220ACP
\n\u2234 \u2220ABC = \u2220ACB
\n\u2234\u2220B = \u2220C.<\/p>\n

We hope the KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.3, 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 5.3 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.3. Karnataka Board Class 9 Maths Chapter 5 Triangles Ex 5.3 Question 1. \u2206ABC and \u2206DBC are two isosceles triangles on the …<\/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\/1794"}],"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=1794"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1794\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=1794"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=1794"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=1794"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}