{"id":126731,"date":"2023-09-12T11:49:00","date_gmt":"2023-09-12T06:19:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126731"},"modified":"2023-11-23T12:00:36","modified_gmt":"2023-11-23T06:30:36","slug":"rd-sharma-class-9-solutions-chapter-12-exercise-12-3","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-12-exercise-12-3\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-126732\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-12-Exercise-12.3.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-12-Exercise-12.3.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-12-Exercise-12.3-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3:<\/strong> No internet for accessing the Class 9 Maths PDF every time? No need for that more than once. You can access the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-for-maths\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths<\/a> PDF even without the internet, once you have downloaded it online. Prepare for your Maths exam with the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-12-herons-formula\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 12<\/a> Exercise 12.3 and ace it. To know more about the RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3, read the whole blog.\u00a0<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69d0725f3d77e\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg 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Chapter 12 Exercise 12.3 PDF<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-12-exercise-12-3\/#access-answers-of-rd-sharma-class-9-solutions-chapter-12-exercise-123\" title=\"Access answers of RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3\">Access answers of RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-12-exercise-12-3\/#faqs-on-rd-sharma-class-9-solutions-chapter-12-exercise-123\" title=\"FAQs on RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3.\">FAQs on RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3.<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-12-exercise-12-3\/#from-where-can-i-download-the-pdf-of-rd-sharma-class-9-solutions-chapter-12-exercise-123\" title=\"From where can I download the PDF of RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3?\">From where can I download the PDF of RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-12-exercise-12-3\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-class-9-solutions-chapter-12-exercise-123\" title=\"How much does it cost to download the PDF of RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3?\">How much does it cost to download the PDF of RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-12-exercise-12-3\/#can-i-access-the-rd-sharma-class-9-solutions-chapter-12-exercise-123-pdf-offline\" title=\"Can I access the RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3\u00a0PDF offline?\">Can I access the RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3\u00a0PDF offline?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-9-solutions-chapter-12-exercise-123-pdf\"><\/span><strong>Download RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3 PDF<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-12-Exercise-12.3-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-12-Exercise-12.3-1.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-class-9-solutions-chapter-12-exercise-123\"><\/span><strong>Access answers of RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Question 1.<br \/>In two right triangles, one side has an acute angle of one is equal to the corresponding side and angle of the other. Prove that the triangles are congruent.<br \/>Solution:<br \/>Given: In \u2206ABC and \u2206DEF,<br \/>\u2220B = \u2220E = 90\u00b0<br \/>\u2220C = \u2220F<br \/>AB = DE<br \/>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula Ex 12.3 \u2013 1<br \/>To prove: \u2206ABC = \u2206DEF<br \/>Proof: In \u2206ABC and \u2206DEF,<br \/>\u2220B = \u2220E (Each = 90\u00b0)<br \/>\u2220C = \u2220F (Given)<br \/>AB = DE (Given)<br \/>\u2206ABC = \u2206DEF (AAS axiom)<\/p>\n<p>Question 2.<br \/>If the bisector of the exterior vertical angle of a triangle is parallel to the base. Show that the triangle is isosceles.<br \/>Solution:<br \/>Given: In \u2206ABC, AE is the bisector of vertical exterior \u2220A and AE \\(\\parallel\\) BC<br \/>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula Ex 12.3 \u2013 2<br \/>To prove : \u2206ABC is an isosceles<br \/>Proof: \u2235 AE \\(\\parallel\\) BC<br \/>\u2234 \u22201 = \u2220B (Corresponding angles)<br \/>\u22202 = \u2220C (Alternate angle)<br \/>But \u22201 = \u22202 (\u2235 AE is the bisector of \u2220CAD)<br \/>\u2234 \u2220B = \u2220C<br \/>\u2234 AB = AC (Sides opposite to equal angles)<br \/>\u2234 \u2206ABC is an isosceles triangle<\/p>\n<p>Question 3.<br \/>In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.<br \/>Solution:<br \/>Given: In \u2206ABC, AB = AC<br \/>\u2220A = 2(\u2220B + \u2220C)<br \/>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula Ex 12.3 \u2013 3<br \/>To calculate Base angles,<br \/>Let \u2220B = \u2220C = x<br \/>Then \u2220A = 2(\u2220B + \u2220C)<br \/>= 2(x + x) = 2 x 2x = 4x<br \/>\u2235 The sum of angles of a triangle = 180\u00b0<br \/>\u2234 4x + x + x \u2013 180\u00b0 \u21d2 6x = 180\u00b0<br \/>\u21d2 x= \\(\\frac { { 180 }^{ \\circ } }{ 6 }\\) = 30\u00b0 o<br \/>\u2234 \u2220B = \u2220C = 30 and \u2220A = 4 x 30\u00b0 = 120<\/p>\n<p>Question 4.<br \/>Prove that each angle of an equilateral triangle is 60\u00b0. [NCERT]<br \/>Solution:<br \/>Given: \u2206ABC is an equilateral triangle<br \/>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula Ex 12.3 \u2013 4<br \/>Proof: In \u2206ABC,<br \/>AB = AC (Sides of an equilateral triangle)<br \/>\u2234 \u2220C = \u2220B \u2026(i)<br \/>(Angles opposite to equal angles)<br \/>Similarly, AB = BC<br \/>\u2234 \u2220C = \u2220A \u2026(ii)<br \/>From (i) and (ii),<br \/>\u2220A = \u2220B = \u2220C<br \/>But \u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of angles of a triangle)<br \/>\u2234 \u2220A = \u2220B = \u2220C = \\(\\frac { { 180 }^{ \\circ } }{ 3 }\\)= 60\u00b0<\/p>\n<p>Question 5.<br \/>Angles A, B, and C of a triangle ABC are equal to each other. Prove that \u2206ABC is equilateral.<br \/>Solution:<br \/>Given: In \u2206ABC, \u2220A = \u2220B = \u2220C<br \/>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula Ex 12.3 \u2013 5<br \/>To prove : \u2206ABC is an equilateral<br \/>Proof: In \u2206ABC,<br \/>\u2234 \u2220B = \u2220C (Given)<br \/>\u2234 AC = AB \u2026(i) (Sides opposite to equal angles)<br \/>Similarly, \u2220C = \u2220A<br \/>\u2234 BC =AB \u2026(ii)<br \/>From (i) and (ii)<br \/>AB = BC = CA<br \/>Hence \u2206ABC is an equilateral triangle<\/p>\n<p>Question 6.<br \/>ABC is a right-angled triangle in which \u2220A = 90\u00b0 and AB = AC. Find \u2220B and \u2220C.<br \/>Solution:<br \/>In \u2206ABC, \u2220A = 90\u00b0<br \/>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula Ex 12.3 \u2013 6<br \/>AB =AC (Given)<br \/>\u2234 \u2220C = \u2220B (Angles opposite to equal sides)<br \/>But \u2220B + \u2220C = 90\u00b0 (\u2235 \u2220B = 90\u00b0)<br \/>\u2234 \u2220B = \u2220C = \\(\\frac { { 90 }^{ \\circ } }{ 2 }\\) = 45\u00b0<br \/>Hence \u2220B = \u2220C = 45\u00b0<\/p>\n<p>Question 7.<br \/>PQR is a triangle in which PQ = PR and S is any point on the side PQ. Through S, a line is drawn parallel to QR and intersecting PR at T. Prove that PS = PT.<br \/>Solution:<br \/>Given: In \u2206PQR, PQ = PR<br \/>S is a point on PQ and PT || QR<br \/>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula Ex 12.3 \u2013 7<br \/>To prove: PS = PT<br \/>Proof: \u2235ST || QR<br \/>\u2234 \u2220S = \u2220Q and \u2220T = \u2220R (Corresponding angles)<br \/>But \u2220Q = \u2220R (\u2235 PQ = PR)<br \/>\u2234 PS = PT (Sides opposite to equal angles)<\/p>\n<p>Question 8.<br \/>In an \u2206ABC, it is given that AB = AC and the bisectors of \u2220B and \u2220C intersect at O. If M is a point on BO produced, prove that \u2220MOC = \u2220ABC.<br \/>Solution:<br \/>Given: In \u2206ABC, AB = AC the bisectors of \u2220B and \u2220C intersect at O. M is any point on BO produced.<br \/>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula Ex 12.3 \u2013 8<br \/>To prove: \u2220MOC = \u2220ABC<br \/>Proof: In \u2206ABC, AB = BC<br \/>\u2234 \u2220C = \u2220B<br \/>\u2235 OB and OC are the bisectors of \u2220B and \u2220C<br \/>\u2234 \u22201 =\u22202 = \\(\\frac { 1 }{ 2 }\\)\u2220B<br \/>Now in \u2220OBC,<br \/>Ext. \u2220MOC = Interior opposite angles \u22201 + \u22202<br \/>= \u22201 + \u22201 = 2\u22201 = \u2220B<br \/>Hence \u2220MOC = \u2220ABC<\/p>\n<p>Question 9.<br \/>P is a point on the bisector of an angle \u2220ABC. If the line through P parallel to AB meets BC at Q, prove that triangle BPQ is isosceles.<br \/>Solution:<br \/>Given: In \u2206ABC, P is a point on the bisector of \u2220B and from P, RPQ || AB is a draw that meets BC in Q<br \/>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula Ex 12.3 \u2013 9<br \/>To prove : \u2206BPQ is an isosceles<br \/>Proof: \u2235 BD is the bisector of CB<br \/>\u2234 \u22201 = \u22202<br \/>\u2235 RPQ || AB<br \/>\u2234 \u22201 = \u22203 (Alternate angles)<br \/>But \u22201 == \u22202 (Proved)<br \/>\u2234 \u22202 = \u22203<br \/>\u2234 PQ = BQ (sides opposite to equal angles)<br \/>\u2234 \u2206BPQ is an isosceles<\/p>\n<p>Question 10.<br \/>ABC is a triangle in which \u2220B = 2\u2220C, and D is a point on BC such that AD bisects \u2220BAC = 72\u00b0.<br \/>Solution:<br \/>Given: In \u2206ABC,<br \/>\u2220B = 2\u2220C, AD is the bisector of \u2220BAC AB = CD<br \/>To prove : \u2220BAC = 72\u00b0<br \/>Construction: Draw bisector of \u2220B which meets AD at O.<\/p>\n<p>This is the complete blog of RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3. To know more about the <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> Class 9 Maths exam, ask in the comments.\u00a0<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-9-solutions-chapter-12-exercise-123\"><\/span><strong>FAQs on RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631297249350\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-pdf-of-rd-sharma-class-9-solutions-chapter-12-exercise-123\"><\/span>From where can I download the PDF of RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link from the above blog.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631297309311\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-class-9-solutions-chapter-12-exercise-123\"><\/span>How much does it cost to download the PDF of RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631297345782\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rd-sharma-class-9-solutions-chapter-12-exercise-123-pdf-offline\"><\/span>Can I access the RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3\u00a0PDF offline?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Once you have downloaded the PDF online, you can access it offline as well.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3: No internet for accessing the Class 9 Maths PDF every time? No need for that more than once. You can access the RD Sharma Solutions Class 9 Maths PDF even without the internet, once you have downloaded it online. Prepare for your Maths exam with the &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-12-exercise-12-3\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 12 Exercise 12.3 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":126732,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411],"tags":[3086,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126731"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/243"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=126731"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126731\/revisions"}],"predecessor-version":[{"id":504795,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126731\/revisions\/504795"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126732"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126731"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126731"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126731"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}