{"id":455,"date":"2025-01-09T14:31:14","date_gmt":"2025-01-09T09:01:14","guid":{"rendered":"https:\/\/samacheerkalvi.guide\/?p=455"},"modified":"2025-01-10T15:53:09","modified_gmt":"2025-01-10T10:23:09","slug":"samacheer-kalvi-9th-maths-guide-chapter-4-ex-4-1","status":"publish","type":"post","link":"https:\/\/samacheerkalvi.guide\/samacheer-kalvi-9th-maths-guide-chapter-4-ex-4-1\/","title":{"rendered":"Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.1"},"content":{"rendered":"

Students can download Maths Chapter 4 Geometry Ex 4.1 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide<\/a> Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.<\/p>\n

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 4 Geometry Ex 4.1<\/h2>\n

Question 1.
\nIn the figure, AB is parallel to CD, find x.
\n\"Samacheer
\nSolution:
\n(i) Through T draw TE || AB.
\n\"Samacheer
\n\u2234 \u2220BAT + \u2220ATE = 180\u00b0 (AB || TE)
\n140\u00b0 + \u2220ATE = 180\u00b0
\n\u2220ATE = 180\u00b0- 140\u00b0 = 40\u00b0
\nSimilarly \u2220ETC + \u2220TCD = 180\u00b0 (TE || CD)
\n\u2220ETC+150\u00b0 = 180\u00b0
\n\u2220ETC = 180\u00b0- 150\u00b0 = 30\u00b0
\nx = \u2220ATE + \u2220ETC
\n= 40\u00b0+ 30\u00b0 = 70\u00b0
\nx = 70\u00b0<\/p>\n

\"Samacheer<\/p>\n

(ii) Draw TE || AB.
\n\"Samacheer
\n\u2220ABT + \u2220ETB = 180\u00b0 (AB || TE)
\n48\u00b0 + \u2220ETB = 180\u00b0
\n\u2220ETB = 180\u00b0 – 48\u00b0 = 132\u00b0
\nSimilarly \u2220CDT + \u2220DTE = 180\u00b0
\n24\u00b0 + \u2220DTE = 180\u00b0
\n\u2234 \u2220DTE = 180\u00b0 – 24\u00b0
\n= 156\u00b0
\n\u2234 \u2220BTE + \u2220ETD = 132\u00b0 + 156\u00b0
\n= 288\u00b0
\nx = 288\u00b0<\/p>\n

\"Samacheer<\/p>\n

(iii) In the given figure AB || CD, AD is the transversal.
\n\u2220CDA = \u2220BAD
\n= 53\u00b0 (alternate angles are equal)
\nIn \u0394ECD, \u2220D = \u2220A = 53\u00b0 (Alternate angles are equal)
\n\u2220E + \u2220C + \u2220D = 180\u00b0 (sum of the angles of a triangle)
\nx\u00b0 + 38\u00b0 + 53\u00b0 = 180\u00b0
\nx\u00b0 = 180\u00b0- 91\u00b0
\n= 89\u00b0
\nx = 89\u00b0<\/p>\n

Question 2.
\nThe angles of a triangle are in the ratio 1 : 2 : 3, find the measure of each angle of the triangle.
\nSolution:
\nThe ratio of the angles of a triangle = 1 : 2 : 3.
\nLet the angles of a triangle be x, 2x and 3x.
\nx + 2x + 3x = 180\u00b0 (Total angle of a triangle is 180\u00b0)
\n6x = 180\u00b0
\nx = \\(\\frac{180\u00b0}{6}\\)
\n= 30\u00b0
\nx = 30\u00b0; 2x = 2 \u00d7 30\u00b0 = 60\u00b0; 3x = 3 \u00d7 30\u00b0 = 90\u00b0
\nMeasures of the angles of a triangle = 30\u00b0, 60\u00b0 and 90\u00b0.<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nConsider the given pairs of triangles and say whether each pair is that of congruent triangles. If the triangles are congruent, say \u2018how\u2019; if they are not congruent say \u2018why\u2019 and also say if a small modification would make them congruent:
\n\"Samacheer
\n(i) In \u0394PQR and \u0394ABC
\nPQ = AB (Given)
\nRQ = BC (Given)
\n\u0394ABC is not congruent to \u0394PQR.
\nIf PR = AC then \u0394ABC \u2245 \u0394PQR<\/p>\n

(ii) In \u0394ABD and \u0394CDB
\nAB = CD (Given)
\nAD = BC (Given)
\nBD is common
\nBy SSS congruency
\n\u0394ABD \u2245 \u0394CDB<\/p>\n

\"Samacheer<\/p>\n

(iii) In \u0394PXY and \u0394PXZ
\nPX is common.
\nXY = XZ (Given)
\nPY = PZ (Given)
\nBy SSS congruency
\n\u0394PXY \u2245 \u0394PXZ<\/p>\n

(iv) In the given figure BD bisect AC
\nIn \u0394AOB and \u0394OCD
\nOA = OC (Given)
\n\u2220AOB = \u2220DOC (vertically opposite angles)
\n\u2220B = \u2220D (Given)
\nBy ASA congruency \u0394AOB \u2245 \u0394OCD<\/p>\n

(v) In the given figure AC and BD bisect each other at O.
\n\u2234 OA = OC (Given); OB = OD (Given)
\n\u2220AOB = \u2220COD (vertically opposite angles)
\nBy SAS congruency
\n\u0394AOB \u2245 \u0394OCD<\/p>\n

\"Samacheer<\/p>\n

(vi) In the given figure
\nAB = AC (Given)
\nBM = MC (AM is the median of the \u0394ABC)
\nAM is common (By SSS congruency)
\n\u2234 \u0394ABM \u2245 \u0394ACM<\/p>\n

Question 4.
\n\u0394ABC and \u0394DEF are two triangles in which AB = DF, \u2220ACB = 70\u00b0, \u2220ABC = 60\u00b0; \u2220DEF = 70\u00b0 and \u2220EDF = 60\u00b0. Prove that the triangles are congruent.
\nSolution:
\n\"Samacheer
\nIn \u0394ABC \u2220B = 60\u00b0 and \u2220C = 70\u00b0
\n\u2234 \u2220A = 180\u00b0 – (60\u00b0 + 70\u00b0)
\n= 180\u00b0 – 130\u00b0
\n= 50\u00b0
\nIn \u0394DEF \u2220E = 70\u00b0 and \u2220D = 60\u00b0
\n\u2220F = 180\u00b0 – (70\u00b0 + 60\u00b0)
\n= 180\u00b0 – 130\u00b0
\n= 50\u00b0
\n\u2220A = \u2220F = 50\u00b0
\n\u2220B = \u2220D = 60\u00b0
\n\u2220C = \u2220E = 70\u00b0
\nBy AAA congruency
\n\u0394ABC \u2245 \u0394FDE
\n(or)
\n\u2220B = \u2220D = 60\u00b0
\n\u2220C = \u2220E = 70\u00b0
\nAB = FE
\nBy ASA congruency
\n\u0394ABC \u2245 \u0394FDE<\/p>\n

\"Samacheer<\/p>\n

Question 5.
\nFind all the three angles of the \u0394ABC.
\n\"Samacheer
\nSolution:
\n\u2220A + \u2220B = \u2220ACD (An exterior angle of a triangle is sum of its interior opposite angles)
\nx + 35 + 2x – 5 = 4x – 15
\n3x + 30 = 4x – 15
\n30 + 15 = 4x – 3x
\n45\u00b0 = x
\n\u2220A = x + 35\u00b0
\n= 45\u00b0 + 35\u00b0
\n= 80\u00b0
\n\u2220B = 2x – 5
\n= 2(45\u00b0) – 5\u00b0
\n= 90\u00b0 – 5\u00b0
\n= 85\u00b0
\n\u2220ACD = 4x – 15
\n= 4 (45\u00b0) – 15\u00b0
\n= 180\u00b0 – 15\u00b0
\n= 165\u00b0
\n\u2220ACB = 180\u00b0 – \u2220ACD
\n= 180\u00b0 – 165\u00b0
\n= 15\u00b0
\n\u2220A = 80\u00b0, \u2220B = 85\u00b0 and \u2220C = 15\u00b0.<\/p>\n

\"Samacheer<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can download Maths Chapter 4 Geometry Ex 4.1 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams. Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 4 Geometry Ex 4.1 Question …<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":false,"jetpack_social_options":{"image_generator_settings":{"template":"highway","enabled":false},"version":2}},"categories":[3],"tags":[],"class_list":["post-455","post","type-post","status-publish","format-standard","hentry","category-class-9"],"jetpack_publicize_connections":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/455"}],"collection":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/comments?post=455"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/455\/revisions"}],"predecessor-version":[{"id":39830,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/455\/revisions\/39830"}],"wp:attachment":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/media?parent=455"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/categories?post=455"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/tags?post=455"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}