{"id":25309,"date":"2021-06-22T15:58:18","date_gmt":"2021-06-22T10:28:18","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=25309"},"modified":"2021-06-22T15:58:18","modified_gmt":"2021-06-22T10:28:18","slug":"ncert-solutions-for-class-9-maths-chapter-7-ex-7-3","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-7-ex-7-3\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 7 Triangles Ex 7.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.3<\/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 BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that
\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\"NCERT
\nSolution:
\n(i) In \u2206ABD and \u2206ACD
\nAB = AC (\u2235 ABC is an isosceles \u2206)
\nBD = CD (\u2235 DBC is an isosceles \u2206)
\nAD = AD (Common)
\nBy S-S-S Congruency condition
\n\u2206ABD = \u2206ACD
\n\u2234 \u2220BAD = \u2220CAD ……(A) (By C.P.CT)<\/p>\n

(ii) In \u2206ABP and \u2206ACP
\nAB = AC (\u2235 ABC is an isoscles \u2206)
\n\u2220BAP = \u2220CAP
\nAP = AP
\nBy S-A-S Congruency Condition
\n\u2220ABP = \u2220ACP
\nTherefore, BP = CP …….(i) (By C.P.C.T.)<\/p>\n

\"NCERT<\/p>\n

(iii) We have
\n\u2206ABD = \u2206ACD (From (A))
\n\u2234 \u2220BAD = \u2220CAD (By C.P.C.T.)
\n\u2234 AP bisects \u2220A
\nAgain,
\nIn \u2206BDP and \u2206GDP
\nBD = CD (\u2235 \u2206BDC is an isoscles \u2206)
\nDP = DP (Common)
\nBP = CP (From equation (i))
\nBy S-S-S Congruency Condition
\n\u2206BDP \u2245 \u2206CDP ………(B)
\n\u2234 \u2220BDP = \u2220CDP (By C.P.C.T)
\nTherefore, AP bisects \u2220D …….(iii)
\nFrom equation (ii) and (iii) we can say AP bisects \u2220A as well as \u2220D.<\/p>\n

(iv) we have
\n\u2206BDP = \u2206CDP (from (B))
\n\u2234 BP = CP ……(iv) (By C.P.C.T.)
\nAgain, \u2220BPD = \u2220CPD (By C.P.C.T.)
\nNow, \u2220BPD + \u2220CPD = 180
\nor, \u2220BPD + \u2220BPD = 180
\n2\u2220BPD = 180
\n\u2234 \u2220BPD = 90 ……(v)
\nTherefore, From equations (iv) and (v) we can say AP is the perpendicular bisector of BC.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nAD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
\n(i) AD bisects BC
\n(ii) AD bisects \u2220A
\n\"NCERT
\nSolution:
\n(i) In \u2206ABD and \u2206ACD
\nAB = AC (Given)
\n\u2220ADB = \u2220ADC (each 90\u00b0)
\nAD = AD (Common)
\nBy R.H.S Congruency Condition
\n\u2206ABD = \u2206ACD ……(i)
\n\u2234 BD = CD (By C.P.C.T.)
\nTherefore, AD bisects BC.<\/p>\n

(ii) We have
\n\u2206ABD = \u2206ACD (from equation (i))
\n\u2234 \u2220BAD = \u2220CAD (By C.P.C.T.)
\nTherefore, AD bisects \u2220A.<\/p>\n

\"NCERT<\/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 (see Fig. 7.40). Show that:
\n(i) \u2206ABM \u2245 \u2206PQN
\n(ii) \u2206ABC \u2245 \u2206PQR
\n\"NCERT
\nSolution:
\n(i) In \u2206ABM and \u2206PQN
\nAB = PQ (Given)
\nBM = QN (Given)
\nAM = PN (Given)
\nBy S-S-S Congruency Condition,
\n\u2206ABM \u2245 \u2206PQN …..prove (i)
\n\u2234 \u2220B = \u2220Q (By C.P.C.T)<\/p>\n

(ii) Now, In \u2206ABC and \u2206PQR
\nAB = PQ (Given)
\nBM = QN (Given)
\n\u2234 2BM = 2QN
\nor, BC = QR
\n(Because M and N are respectively the midpoint of side BC and QR)
\n\u2220B = \u2220Q (Prove above)
\nBy S-A-S Congruency Condition
\n\u2206ABC \u2245 \u2206PQR …….prove (ii)<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nBE and CF are two equal altitudes of a triangle ABC. Using the R.H.S congruence rule, prove that the triangle ABC is isosceles.
\n\"NCERT
\nSolution:
\nIn \u2206BFC and \u2206CEB
\nCR = BE (Given)
\n\u2220BFC = \u2220CEB (Each 90\u00b0)
\nBC = BC (Common)
\n\u2234 By R-H-S congruency condition,
\n\u2206BFC \u2245 \u2206CEB
\n\u2234 \u2220B = \u2220C (By C.P.C.T)
\nTherefore, AB = AC (Sides opposite to equal angles of a \u2206)
\nHence, \u2206ABC is isosceles.<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nABC is an isosceles triangle with AB = AC. Draw AP \u22a5 BC to show that \u2220B = \u2220C.
\n\"NCERT
\nSolution:
\nIn \u2206ABP and \u2206ACP
\nAB = AC (Given)
\n\u2220APB = \u2220APC (Each 90\u00b0)
\nAP = AP (Common)
\nBy R-H-S Congruency Condition,
\n\u2206ABP \u2245 \u2206ACP
\nor, \u2220B = \u2220C (By C.P.C.T)<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.3 Question 1. \u2206ABC and \u2206DBC are two isosceles triangles on the same base BC and vertices A and D are on …<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-7-ex-7-3\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3 Questions and Answers are prepared by our highly skilled subject experts. 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