{"id":25230,"date":"2021-06-22T14:25:55","date_gmt":"2021-06-22T08:55:55","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=25230"},"modified":"2022-03-02T10:36:32","modified_gmt":"2022-03-02T05:06:32","slug":"ncert-solutions-for-class-9-maths-chapter-7-ex-7-1","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-9-maths-chapter-7-ex-7-1\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 7 Triangles Ex 7.1 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.1<\/h2>\n

Question 1.
\nIn quadrilateral ACBD (See Fig. 7.16), AC = AD, and AB bisects \u2220A. Show that \u0394ABC \u2245 \u0394ABD. What can you say about BC and BD?
\n\"NCERT
\nSolution:
\nIn \u0394ABC and \u0394ABD
\nAC = AD (Given)
\n\u2220CAB = \u2220DAB (Given)
\nAB = AB (Common)
\nTherefore, By SAS congruency condition
\n\u0394ABC \u2245 \u0394ABD
\nSo, BC = BD (By C.P.C.T)<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nABCD is a quadrilateral in which AD = BC and \u2220DAB = \u2220CBA (see Fig. 7.17)
\nProve that:
\n(i) \u0394ABD = \u0394BAC
\n(ii) BD = AC
\n(iii) \u2220ABD = \u2220BAC
\n\"NCERT
\nSolution:
\n(i) In \u0394ABD and \u0394BAC
\nAD = BC
\n\u2220DAB = \u2220CBA (Given)
\nand AB = BA (Common)
\nBy SAS Congruency Condition
\n\u0394ABD = \u0394BAC
\n(ii) BD= AC (By C.P.C.T)
\n(iii) \u2220ABD = \u2220BAC (Again by C.P.C.T)<\/p>\n

Question 3.
\nAD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.
\n\"NCERT
\nSolution:
\nIn \u0394AOD and \u0394BOC,
\nAD = BC (Given)
\n\u2220OAD = \u2220OBC (each 90\u00b0)
\n\u2220AOD = \u2220BOC (Vertically opposite angles)
\nTherefore, by ASA congruency condition.
\n\u0394AOD = \u0394BOC
\nSo, OA = OB (by C.P.C.T)
\nHence, CD bisects line segment AB.<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nl and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that \u0394ABC = \u0394CDA.
\n\"NCERT
\nSolution:
\nWe have given that l || m and p || q
\nTherefore, In \u0394ABC and \u0394CDA
\n\u2220BAC = \u2220DCA
\n(Alternate interior angles as AB || DC)
\n\u2220ACB = \u2220CAD
\n(Alternate interior angles as BC || DA)
\nAC = CA (common)
\nSo, By A-S-A congruency condition,
\n\u0394ABC = \u0394CDA<\/p>\n

Question 5.
\nLine l 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 (see Fig. 7.20) show that:
\n(i) \u0394APB \u2245 \u0394AQB
\n(ii) BP = BQ or B is equidistance from the arms of \u2220A.
\n\"NCERT
\nSolution:
\nIn \u0394ABP and \u0394ABQ,
\n\u2220BAP = \u2220BAQ (Given)
\n\u2220APB = \u2220AQB (Each 90\u00b0)
\nAB = AB (Common)
\nBy A-A-S congruency condition.
\nSo, \u0394ABP \u2245 \u0394ABQ
\n(ii) BP = BQ (By C.P.C.T)<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nIn Fig. 7.21, AC = AE, AB = AD and \u2220BAD = \u2220EAC. Show that BC = DE.
\n\"NCERT
\nSolution:
\nIn \u0394BAC and \u0394DAE
\nAB = AD (Given)
\nAC = AE (Given)
\n\u2220BAD = \u2220EAC ……(i) (Given)
\nAdding \u2220DAC both side in equation (i)
\n\u2234 \u2220BAD + \u2220DAC = \u2220EAC + \u2220DAC
\n\u2220BAC = \u2220DAE
\nTherefore by S-A-S Congruency Condition
\n\u0394BAC = \u0394DAE
\nSo, BO = DE (By C.P.C.T)<\/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 (see Fig. 7.22) show that
\n(i) \u0394DAP \u2245 \u0394EBP
\n(ii) AD = BE
\n\"NCERT
\nSolution:
\n(i) In \u0394DAP and \u0394EBP
\n\u2220DAP= \u2220EBP (Given)
\n\u2220APE = \u2220DPB (Given)
\n\u2234 \u2220APE + \u2220EPD = \u2220DPB + \u2220EPD (Add \u2220EPD both side)
\n\u2220APD = \u2220BPE
\nAP = BP (Given P is the mid point of AB)
\n\u2234 By A-S-A Congruency Condition,
\n\u0394DAP \u2245 \u0394EBP
\n(ii) AD = BE (By C.P.C.T.)<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nIn the right triangle ABC right angled at C, M is the midpoint of hypotenuse AB. C is joined to M and produces to a point D such that DM = CM. Point D is joined to point B (see Fig. 7.23). Show that
\n(i) \u0394AMC \u2245 \u0394BMD
\n(ii) \u2220DBC is right angle.
\n(iii) \u0394DBC \u2245 \u0394ACB
\n(iv) CM = \\(\\frac {1}{2}\\) AB
\n\"NCERT
\nSolution:
\n(i) In \u0394AMC and \u0394BMD,
\nAM = BM (Given)
\nCM = DM (Given)
\n\u2220AMC = \u2220BMD (Vertically opposite angles)
\n\u2234 By S-A-S Congruency Condition.
\n\u0394AMC = \u0394BMD<\/p>\n

(ii) \u2220CAM = \u2220DBM (by C.P.C.T)
\nAlso, \u2220CAM + \u2220MBC = 90\u00b0 (Since \u2220C = 90\u00b0)
\n\u2234 \u2220DBM + \u2220MBC = 90\u00b0 (\u2235 \u2220CAM = \u2220DBM)
\nor, \u2220DBC = 90\u00b0<\/p>\n

(iii) In \u0394DBC and \u0394ACB,
\nBC = BC (Common)
\nDB = AC (\u2235 \u0394BMD \u2245 \u0394AMC by C.PC.T)
\nand \u2220DBC = \u2220ACB (each 90\u00b0 proved above)
\nTherefore, by S-A-S Congruency Condition,
\n\u2234 \u0394DBC \u2245 \u0394ACB<\/p>\n

\"NCERT<\/p>\n

(iv) Since, \u0394DBC \u2245 \u0394ACB,
\nDC = AB
\n\u2234 \\(\\frac {1}{2}\\) DC = \\(\\frac {1}{2}\\) AB
\nCM = AM
\nor CM = \\(\\frac {1}{2}\\) AB<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.1 Question 1. In quadrilateral ACBD (See Fig. 7.16), AC = AD, and AB bisects \u2220A. Show that \u0394ABC \u2245 \u0394ABD. What …<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1<\/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":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 - MCQ Questions<\/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:\/\/mcq-questions.com\/ncert-solutions-for-class-9-maths-chapter-7-ex-7-1\/\" \/>\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 7.1 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 Questions and Answers are prepared by our highly skilled subject experts. 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