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>\nNCERT 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 \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
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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 \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 \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
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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 \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 \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
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Question 6. \nIn Fig. 7.21, AC = AE, AB = AD and \u2220BAD = \u2220EAC. Show that BC = DE. \n \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 \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
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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 \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
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(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 \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n