{"id":4355,"date":"2023-02-09T06:58:37","date_gmt":"2023-02-09T06:58:37","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4355"},"modified":"2023-02-09T07:00:32","modified_gmt":"2023-02-09T07:00:32","slug":"ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-3","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-7-triangles-ex-7-3\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 7 Triangles Ex 7.3"},"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 7.3 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-7-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 7.1<\/a><\/li>\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-7-4\/\" 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; 7.3<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. \u0394ABC and \u0394DBC are two isosceles triangles on the same base BC, and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4406\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.3-Q1.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"248\" height=\"204\" \/><\/strong><br \/>\n<strong>(i)<\/strong> \u0394ABD \u2245 \u0394ACD<br \/>\n<strong>(ii)<\/strong> \u0394ABP \u2245 \u0394ACP<br \/>\n<strong>(iii)<\/strong> AP bisects \u2220A as well as \u2220D<br \/>\n<strong>(iv)<\/strong> AP is the perpendicular bisector of BC<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211;\u00a0Given:\u00a0<\/strong>\u0394 ABC and \u0394 DBC are\u00a0isosceles triangles.<br \/>\n<strong>(i) \u0394ABD \u2245 \u0394ACD<br \/>\n<\/strong>In \u0394ABD and \u0394ACD,<\/span><br \/>\n<span style=\"color: #000000;\">AB = AC (Equal sides of\u00a0isosceles\u00a0\u0394ABC)<\/span><br \/>\n<span style=\"color: #000000;\">BD = CD (Equal sides of isosceles \u0394DBC)<\/span><br \/>\n<span style=\"color: #000000;\">AD = AD (Common)<\/span><br \/>\n<span style=\"color: #000000;\"><strong>\u0394ABD \u2245 \u0394ACD<\/strong> (By\u00a0SSS\u00a0congruence rule)<\/span><\/p>\n<p><span style=\"color: #000000;\">By CPCT, we get <\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAD = \u2220CAD<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAP = \u2220CAP\u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;-\u00a0 (1)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ADB = \u2220ADC\u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (2)<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(ii)<\/strong> \u0394ABP \u2245 \u0394ACP<br \/>\nAB = AC (Given)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BAP = \u2220CAP\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[From equation (1)]<\/span><br \/>\n<span style=\"color: #000000;\">AP = AP (Common)<\/span><br \/>\n<span style=\"color: #000000;\"><strong>\u2234 \u0394ABP \u2245 \u0394ACP<\/strong>\u00a0 \u00a0 \u00a0 (By SAS congruence rule)<\/span><\/p>\n<p><span style=\"color: #000000;\">\u2234 BP = CP (By CPCT)\u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (3)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) AP bisects \u2220A as well as \u2220D<br \/>\n<\/strong>From Equation (1) we know that \u2220BAP = \u2220CAP<\/span><br \/>\n<span style=\"color: #000000;\">Hence, AP is the\u00a0angle\u00a0bisector\u00a0of\u00a0\u2220A.<\/span><br \/>\n<span style=\"color: #000000;\">From equation (2), we know that \u2220ADB\u00a0= \u2220ADC<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2\u00a0180\u00b0 &#8211; \u2220ADB\u00a0= 180\u00b0\u00a0&#8211; \u2220ADC<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 \u2220BDP = \u2220CDP\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (4)<\/span><br \/>\n<span style=\"color: #000000;\">Hence, AP is the bisector of\u00a0\u2220D.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) AP is the perpendicular bisector of BC<br \/>\n<\/strong>\u2220BPD = \u2220CPD (by CPCT as \u0394BPD \u0394CPD)<br \/>\nBP = CP\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;-\u2014 (5)<br \/>\nalso,<br \/>\n\u2220BPD + \u2220CPD = 180\u00b0 (Since BC is a straight line)<br \/>\n\u21d2 2 \u2220BPD = 180\u00b0<br \/>\n\u21d2 \u2220BPD = 90\u00b0\u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;-\u2014 (6)<br \/>\nNow, from equations (5) and (6), it can be said that<br \/>\nAP is the perpendicular bisector of BC.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that <\/strong><br \/>\n<strong>(i)<\/strong> AD bisects BC<br \/>\n<strong>(ii)<\/strong> AD bisects \u2220A.<br \/>\n<strong>Answer &#8211; Given:<\/strong> AB = AC<br \/>\nLet&#8217;s construct an isosceles triangle ABC in which AB = AC as shown below.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4407\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.3-Ans2.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"205\" height=\"177\" \/><br \/>\n<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(i) AD bisects BC<\/strong><\/span><br \/>\n<span style=\"color: #000000;\">In\u00a0\u0394BAD and \u0394CAD,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ADB = \u2220ADC (Each 90\u00b0 as AD is an\u00a0altitude)<\/span><br \/>\n<span style=\"color: #000000;\">AB = AC (Given)<\/span><br \/>\n<span style=\"color: #000000;\">AD = AD (Common)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394BAD \u2245 \u0394CAD (By\u00a0RHS\u00a0Congruence rule)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 BD = CD (By\u00a0CPCT)<\/span><br \/>\n<span style=\"color: #000000;\">Hence, AD bisects BC.<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(ii) AD bisects \u2220A<br \/>\n<\/strong>Since,\u00a0\u0394BAD \u2245 \u0394CAD<\/span><br \/>\n<span style=\"color: #000000;\">By CPCT,\u00a0\u2220BAD = \u2220CAD<\/span><br \/>\n<span style=\"color: #000000;\">Hence, AD bisects \u2220A.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of \u0394PQR (see Fig. 7.40). Show that:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4408\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.3-Q3.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"368\" height=\"190\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.3-Q3.png 368w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.3-Q3-300x155.png 300w\" sizes=\"auto, (max-width: 368px) 100vw, 368px\" \/><br \/>\n<\/strong><strong>(i)<\/strong> \u0394ABM \u2245 \u0394PQN<br \/>\n<strong>(ii)<\/strong> \u0394ABC \u2245 \u0394PQR <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Answer &#8211; Given:<\/strong> AB = PQ, AM = PN, BM = QN<br \/>\n<strong>(i) \u0394ABM \u2245 \u0394PQN<br \/>\n<\/strong>In \u0394ABC, AM is the\u00a0median\u00a0to BC.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 BM = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> BC<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394PQR, PN is the median to QR. <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 QN = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> QR<\/span><br \/>\n<span style=\"color: #000000;\">It is given that BC = QR <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> BC = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> QR<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 BM = QN\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (1)<\/span><br \/>\n<span style=\"color: #000000;\">In \u0394ABM and \u0394PQN,<\/span><br \/>\n<span style=\"color: #000000;\">AB = PQ (Given)<\/span><br \/>\n<span style=\"color: #000000;\">BM = QN [From equation (1)]<\/span><br \/>\n<span style=\"color: #000000;\">AM = PN (Given)<\/span><br \/>\n<span style=\"color: #000000;\"><strong>\u2234\u00a0\u0394ABM \u2245 \u0394PQN<\/strong> (Using\u00a0SSS congruence criterion)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">\u21d2\u00a0\u2220ABM\u00a0= \u2220PQN\u00a0(By\u00a0CPCT)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 \u2220ABC = \u2220PQR\u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (2)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) \u0394ABC \u2245 \u0394PQR<br \/>\n<\/strong>In \u0394 ABC and \u0394 PQR,<\/span><br \/>\n<span style=\"color: #000000;\">AB = PQ (Given)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220ABC = \u2220PQR [From Equation (2)]<\/span><br \/>\n<span style=\"color: #000000;\">BC = QR (Given)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394ABC \u2245 \u0394PQR (By\u00a0SAS\u00a0congruence rule)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. BE and CF are two equal altitudes of a triangle ABC. Using the R.H.S. congruence rule, prove that the triangle ABC is isosceles.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0 <\/strong>Let&#8217;s construct a diagram according to the given question as shown below.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4409\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.3-Ans4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"334\" height=\"237\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.3-Ans4.png 334w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.3-Ans4-300x213.png 300w\" sizes=\"auto, (max-width: 334px) 100vw, 334px\" \/><br \/>\nIn \u0394BEC and \u0394CFB,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220BEC = \u2220CFB (Each 90\u00b0)<\/span><br \/>\n<span style=\"color: #000000;\">BC = CB (Common)<\/span><br \/>\n<span style=\"color: #000000;\">BE = CF (altitudes\u00a0are equal given)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u0394BEC \u2245 \u0394CFB (By\u00a0RHS\u00a0congruency)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220BCE = \u2220CBF (By\u00a0CPCT)<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 AB = AC (Sides opposite to equal angles of a triangle are equal)<\/span><br \/>\n<span style=\"color: #000000;\">Hence, \u0394ABC is isosceles.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. ABC is an isosceles triangle with AB = AC. Draw AP \u22a5 BC to show that \u2220B = \u2220C.<br \/>\n<\/strong><strong>Answer &#8211;\u00a0<\/strong>Let&#8217;s construct a diagram according to the given question.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4410\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-7.3-Ans5.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"224\" height=\"218\" \/><br \/>\nIn triangles APB and APC,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220APB = \u2220APC (Each 90\u00b0)<\/span><br \/>\n<span style=\"color: #000000;\">AB = AC (Since ABC is an\u00a0isosceles triangle)<\/span><br \/>\n<span style=\"color: #000000;\">AP = AP (Common)<\/span><br \/>\n<span style=\"color: #000000;\">\u0394APB \u2245 \u0394APC (Using\u00a0RHS\u00a0congruence rule)<\/span><br \/>\n<span style=\"color: #000000;\">Thus, \u2220B = \u2220C (CPCT)<\/span><\/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 7.3 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; 7.1 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-4355","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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