{"id":4352,"date":"2023-02-09T06:58:23","date_gmt":"2023-02-09T06:58:23","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4352"},"modified":"2023-02-09T07:00:15","modified_gmt":"2023-02-09T07:00:15","slug":"ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-1","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-1\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 7 Triangles Ex 7.1"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 9 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 7 (Triangles)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 9 Mathematics\u00a0<strong>Chapter &#8211; 7 Triangles <\/strong>Exercise 7.1 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter 7: Triangles<\/strong><\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-2\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 7.2<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-3\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 7.3<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-4\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 7.4<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-5\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 7.5<\/a><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 7.1\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. In quadrilateral ACBD, AC = AD and AB bisect \u2220A (see Fig. 7.16). Show that \u0394ABC\u2245 \u0394ABD. What can you say about BC and BD?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4386\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.1-Q1.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"223\" height=\"246\" \/><br \/>\n<\/strong><strong>Answer &#8211;\u00a0 <\/strong><strong>Given:<\/strong> AC = AD and AB bisects \u2220A<\/span><br \/>\n<span style=\"color: #000000;\"><strong>To Prove:<\/strong> \u0394 ABC \u2245 \u0394 ABD\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">We can show two sides and included angle of\u00a0ABC are equal\u00a0to the corresponding sides and included angle of ABD.<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394 ABC and\u00a0\u0394 ABD,<\/span><br \/>\n<span style=\"color: #000000;\">AC = AD (Given)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220CAB = \u2220DAB (AB bisects \u2220A)<\/span><br \/>\n<span style=\"color: #000000;\">AB = AB (Common)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 <strong>\u0394 ABC \u2245 \u0394 ABD<\/strong> (By\u00a0SAS\u00a0congruence rule)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 BC = BD (By\u00a0CPCT)<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, BC and BD are of equal lengths.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. ABCD is a quadrilateral in which AD = BC and \u2220DAB = \u2220CBA (see Fig. 7.17). Prove that<br \/>\n<\/strong><strong>(i)<\/strong> \u0394ABD\u00a0<strong>\u2245\u00a0<\/strong>\u0394BAC<br \/>\n<strong>(ii)<\/strong> BD = AC<br \/>\n<strong>(iii)\u00a0\u2220<\/strong>ABD =\u00a0<strong>\u2220<\/strong>BAC.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4387\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.1-Q2.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"219\" height=\"243\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; Given:<\/strong> AD = BC and \u2220DAB = \u2220CBA<br \/>\n<strong>To Prove: <\/strong><strong>(i) \u25b3 ABD \u2245 \u25b3 BAC<br \/>\n<\/strong>In \u25b3ABD and \u25b3BAC,<\/span><br \/>\n<span style=\"color: #000000;\">AD = BC (Given)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220DAB = \u2220CBA (Given)<\/span><br \/>\n<span style=\"color: #000000;\">AB = BA (Common)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u25b3ABD \u2245 \u25b3BAC (By\u00a0SAS\u00a0congruence rule)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>To Prove: <\/strong><strong>(ii) BD = AC<br \/>\n<\/strong>Since \u25b3ABD \u2245 \u25b3BAC, <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 BD = AC (By CPCT)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>To Prove: <\/strong><strong>(iii) \u2220ABD = \u2220BAC<br \/>\n<\/strong>Since \u0394ABD <strong>\u2245\u00a0<\/strong>\u0394BAC so,<br \/>\nAngles\u00a0<strong>\u2220<\/strong>ABD =\u00a0<strong>\u2220<\/strong>BAC (by the rule of CPCT).<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4388\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.1-Q3.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"317\" height=\"231\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.1-Q3.png 317w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.1-Q3-300x219.png 300w\" sizes=\"auto, (max-width: 317px) 100vw, 317px\" \/><br \/>\n<\/strong><strong>Answer &#8211;\u00a0 <\/strong><strong>Given:<\/strong> AD \u22a5 AB, BC \u22a5 AB, and AD = BC<\/span><br \/>\n<span style=\"color: #000000;\"><strong>To Prove:<\/strong> CD bisects AB or OA = OB<\/span><br \/>\n<span style=\"color: #000000;\">We can show that the two triangles OBC and OAD are congruent by using\u00a0AAS\u00a0congruency rule and then we can say corresponding parts of\u00a0congruent triangles\u00a0will be equal.<\/span><br \/>\n<span style=\"color: #000000;\">Consider two triangles \u25b3 BOC and \u25b3 AOD,<\/span><br \/>\n<span style=\"color: #000000;\">In \u25b3 BOC and \u25b3 AOD,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BOC = \u2220AOD (Vertically opposite angles)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220CBO = \u2220DAO (Each 90\u00ba, since AD and BC are \u22a5 to AB)<\/span><br \/>\n<span style=\"color: #000000;\">BC = AD (Given) <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u25b3BOC \u2245\u00a0\u25b3AOD (AAS congruence rule)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 BO = AO (By\u00a0CPCT)<\/span><br \/>\n<span style=\"color: #000000;\">Thus, CD bisects AB and O is the mid-point of AB.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4.\u00a0<em>l\u00a0<\/em>and\u00a0<em>m<\/em>\u00a0are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that \u0394ABC \u2245 \u0394CDA.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4389\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.1-Q4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"234\" height=\"188\" \/><br \/>\n<\/strong><strong>Answer &#8211;\u00a0 <\/strong><strong>Given:<\/strong> <em>l<\/em> || <em>m<\/em> and <em>p<\/em> || <em>q<br \/>\n<\/em><strong>To Prove:<\/strong> \u0394ABC \u2245 \u0394CDA.<\/span><br \/>\n<span style=\"color: #000000;\">We can show both the triangles are\u00a0congruent\u00a0by using ASA congruency criterion<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394ABC and \u0394CDA,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAC = \u2220DCA (Alternate interior angles, as p and\u00a0q are\u00a0parallel lines)<\/span><br \/>\n<span style=\"color: #000000;\">AC = CA (Common)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BCA = \u2220DAC (Alternate interior angles, as l || m)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394ABC \u2245 \u0394CDA (By\u00a0ASA\u00a0congruence rule)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. Line l is the bisector of an angle \u2220A and B is any point on<em>\u00a0l<\/em>. BP and BQ are perpendiculars from B to the arms of \u2220A (see Fig. 7.20). Show that:<br \/>\n<\/strong><strong>(i)<\/strong> \u0394APB \u2245 \u0394AQB<br \/>\n<strong>(ii)<\/strong> BP = BQ or B is equidistant from the arms of \u2220A.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4391\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.1-Q5.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"223\" height=\"194\" \/><br \/>\n<strong>Answer &#8211;\u00a0 <\/strong><strong>Given:<\/strong> <em>l<\/em> is the bisector of an angle \u2220A and BP \u22a5 AP and BQ \u22a5 AQ<\/span><br \/>\n<span style=\"color: #000000;\"><strong>To Prove:<\/strong> \u0394APB \u2245 \u0394AQB and BP = BQ<\/span><br \/>\n<span style=\"color: #000000;\"><strong>(i) <\/strong>We can show two triangles APB and AQB are\u00a0congruent\u00a0by using AAS congruency rule and then show that the\u00a0corresponding parts of congruent triangles\u00a0will be equal.<br \/>\nIn \u0394APB and \u0394AQB,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAP = \u2220BAQ (l is the\u00a0angle bisector\u00a0of \u2220A)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220APB = \u2220AQB (Each 90\u00b0)<\/span><br \/>\n<span style=\"color: #000000;\">AB = AB (Common)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394APB \u2245 \u0394AQB (By\u00a0AAS\u00a0congruence rule)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) <\/strong>Since, \u0394APB \u2245 \u0394AQB<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 BP = BQ (By CPCT)<\/span><br \/>\n<span style=\"color: #000000;\">Or, it can be said that point B is equidistant from the arms of \u2220A.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. In Fig. 7.21, AC = AE, AB = AD and \u2220BAD = \u2220EAC. Show that BC = DE.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4392\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.1-Q6.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"258\" height=\"219\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong><strong>Given:<\/strong> AC = AE, AB = AD and \u2220BAD = \u2220EAC<\/span><br \/>\n<span style=\"color: #000000;\"><strong>To Prove:<\/strong> BC = DE<\/span><br \/>\n<span style=\"color: #000000;\">We can show two triangles BAC and DAE are\u00a0congruent triangles\u00a0by using SAS congruency rule and then we can say corresponding parts of congruent triangles will be equal.\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">It is given that \u2220BAD = \u2220EAC<\/span><br \/>\n<span style=\"color: #000000;\">Thus, by adding\u00a0\u2220DAC to both sides of this equation, we get\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAD + \u2220DAC = \u2220EAC + \u2220DAC (\u2220DAC is common)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAC = \u2220DAE<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394BAC and \u0394DAE,<\/span><br \/>\n<span style=\"color: #000000;\">AB = AD (Given)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAC = \u2220DAE (Proven above)<\/span><br \/>\n<span style=\"color: #000000;\">AC = AE (Given)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394BAC \u2245 \u0394DAE (By\u00a0SAS\u00a0congruence rule)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 BC = DE (By\u00a0CPCT)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that \u2220BAD = \u2220ABE and \u2220EPA = \u2220DPB (see Fig. 7.22). Show that<br \/>\n<\/strong><strong>(i) <\/strong>\u0394DAP \u2245 \u0394EBP<strong><br \/>\n<\/strong><strong>(ii) <\/strong>AD = BE<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4393\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.1-Q7.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"238\" height=\"176\" \/><br \/>\n<strong>Answer &#8211;\u00a0<\/strong><strong>Given:<\/strong> P is its mid-point of AB, \u2220BAD = \u2220ABE and \u2220EPA = \u2220DPB<\/span><br \/>\n<span style=\"color: #000000;\">We can show two triangles DAP and EBP congruent\u00a0by using ASA congruency rule and then we can say corresponding parts of congruent triangles will be equal.\u00a0<\/span><br \/>\n<span style=\"color: #000000;\"><strong>To Prove:<\/strong> <strong>(i)<\/strong> \u0394DAP \u2245 \u0394EBP<\/span><br \/>\n<span style=\"color: #000000;\">It is given that \u2220EPA = \u2220DPB<\/span><br \/>\n<span style=\"color: #000000;\">\u2220EPA + \u2220DPE = \u2220DPB + \u2220DPE (\u2220DPE is common)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220DPA = \u2220EPB<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394DAP and \u0394EBP,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220DAP = \u2220EBP (Given)<\/span><br \/>\n<span style=\"color: #000000;\">AP = BP (P is mid &#8211; point of AB)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220DPA = \u2220EPB (Proven above)<\/span><br \/>\n<span style=\"color: #000000;\"><strong>\u2234 \u0394DAP \u2245 \u0394EBP<\/strong> (ASA\u00a0congruence rule)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>To Prove:<\/strong> <strong>(ii)<\/strong> AD = BE.<\/span><br \/>\n<span style=\"color: #000000;\">Since,\u00a0\u0394DAP \u2245 \u0394EBP\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AD = BE (By\u00a0CPCT)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see Fig. 7.23). Show that:<\/strong><br \/>\n<strong>(i)<\/strong> \u0394AMC \u2245 \u0394BMD<br \/>\n<strong>(ii)<\/strong> \u2220DBC is a right angle.<br \/>\n<strong>(iii)<\/strong> \u0394DBC \u2245 \u0394ACB<br \/>\n<strong>(iv)<\/strong> CM = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> AB<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4394\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.1-Q8.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"209\" height=\"199\" \/><br \/>\n<strong>Answer &#8211;\u00a0 <\/strong><strong>Given:<\/strong> M is the mid-point of hypotenuse AB, \u2220C = 90\u00b0 and DM = CM <\/span><br \/>\n<span style=\"color: #000000;\"><strong>To Prove: (i)<\/strong> \u0394AMC \u2245 \u0394BMD <\/span><br \/>\n<span style=\"color: #000000;\">In \u0394AMC and \u0394BMD,<\/span><br \/>\n<span style=\"color: #000000;\">AM = BM (M is the mid &#8211; point of AB)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220AMC = \u2220BMD (Vertically opposite angles)<\/span><br \/>\n<span style=\"color: #000000;\">CM = DM (Given)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394AMC \u2245 \u0394BMD (By\u00a0SAS\u00a0congruence rule)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AC = BD (By\u00a0CPCT)<\/span><br \/>\n<span style=\"color: #000000;\">Also, \u2220ACM = \u2220BDM (By CPCT)<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>To Prove: (ii) \u2220DBC is a right angle.<br \/>\n<\/strong>\u2220DBC is a right angle.<\/span><br \/>\n<span style=\"color: #000000;\">We know that, \u2220ACM = \u2220BDM (proved above)<\/span><br \/>\n<span style=\"color: #000000;\">But, \u2220ACM and \u2220BDM are\u00a0alternate interior angles. Since alternate angles are equal, it can be said that DB || AC.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220DBC + \u2220ACB = 180\u00b0 (Co-interior angles)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220DBC + 90\u00b0 = 180\u00b0 [Since,\u00a0\u0394ACB is a right angled triangle]<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220DBC = 90\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Thus,\u00a0\u2220DBC is a right angle.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>To Prove: (iii) \u0394DBC \u2245 \u0394ACB<br \/>\n<\/strong>In \u0394DBC and \u0394ACB,<\/span><br \/>\n<span style=\"color: #000000;\">DB = AC (Already proved)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220DBC = \u2220ACB = 90\u00b0 (Proved above)<\/span><br \/>\n<span style=\"color: #000000;\">BC = CB(Common)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394 DBC \u2245 \u0394 ACB (SAS congruence rule)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>To Prove: (iv) CM = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{2}}\" alt=\"\\mathbf{\\frac{1}{2}}\" align=\"absmiddle\" \/>AB<br \/>\n<\/strong>DC = AB (Since \u0394DBC \u2245 \u0394ACB)<br \/>\n\u21d2 DM = CM = AM = BM (Since M the is mid-point)<br \/>\nSo, DM + CM = BM + AM<br \/>\nHence, CM + CM = AB<br \/>\n\u21d2 CM = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> AB<\/span><\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 9 Maths\u00a0 Chapter &#8211; 7 (Triangles)\u00a0 The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter &#8211; 7 Triangles Exercise 7.1 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 7: Triangles NCERT Solution Class 9 Maths Ex &#8211; 7.2 NCERT Solution Class 9 Maths [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[698],"tags":[713,702,186,714,701,5],"class_list":["post-4352","post","type-post","status-publish","format-standard","hentry","category-class-9-maths","tag-class-9th-chapter-7-triangles-in-english","tag-ncert-solution-class-9","tag-ncert-solutions","tag-ncert-solutions-class-9-maths-chapter-7-in-english","tag-ncert-solutions-class-9-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.4) - 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