{"id":126564,"date":"2023-09-03T12:56:00","date_gmt":"2023-09-03T07:26:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126564"},"modified":"2023-10-31T10:08:50","modified_gmt":"2023-10-31T04:38:50","slug":"rd-sharma-class-9-solutions-chapter-10-exercise-10-3","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-10-exercise-10-3\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-126566\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-10-Exercise-10.3.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-10-Exercise-10.3.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-10-Exercise-10.3-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3:\u00a0<\/strong>Want to have the perfect score in the Class 9 Maths exam? Start studying 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>. It is a much need guide that will clear your doubts and make you score good marks. All the solutions of <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-10-congruent-triangles\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 10<\/a> Exercise 10.3 are designed by subject matter experts.\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-69e77be08dd9d\" 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 style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input  type=\"checkbox\" id=\"item-69e77be08dd9d\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-10-exercise-10-3\/#download-rd-sharma-class-9-solutions-chapter-10-exercise-103-pdf\" title=\"Download RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3 PDF\">Download RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3 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-10-exercise-10-3\/#access-answers-of-rd-sharma-class-9-solutions-chapter-10-exercise-103\" title=\"Access answers of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3\">Access answers of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3<\/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-10-exercise-10-3\/#faqs-on-rd-sharma-class-9-solutions-chapter-10-exercise-103\" title=\"FAQs on RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3\">FAQs on RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3<\/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-10-exercise-10-3\/#can-i-access-the-rd-sharma-class-9-solutions-chapter-10-exercise-103-pdf-offline\" title=\"Can I access the RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3\u00a0PDF offline?\">Can I access the RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3\u00a0PDF offline?<\/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-10-exercise-10-3\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-class-9-solutions-chapter-10-exercise-103\" title=\"How much does it cost to download the PDF of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3?\">How much does it cost to download the PDF of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-9-solutions-chapter-10-exercise-103-pdf\"><\/span><strong>Download RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3 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-Solutions-Class-9-Maths-Chapter-10-Ex-10.3.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3<\/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-Solutions-Class-9-Maths-Chapter-10-Ex-10.3.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-class-9-solutions-chapter-10-exercise-103\"><\/span><strong>Access answers of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Question 1: In two right triangles, one side and an acute angle of one triangle are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.<\/strong><\/p>\n<p><strong>Solution<\/strong>:<\/p>\n<p>In two right triangles, one side and an acute angle of one triangle are equal to the corresponding side and angles of the other. (Given)<\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 10 ex 10.3 question 1\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-10-ex-10-3-questio-3.png\" alt=\"RD sharma class 9 maths chapter 10 ex 10.3 question 1\" width=\"264\" height=\"164\" \/><\/p>\n<p>To prove: Both triangles are congruent.<\/p>\n<p>Consider two right triangles such that<\/p>\n<p>\u2220 B = \u2220 E = 90<sup>o<\/sup>\u00a0\u2026\u2026.(i)<\/p>\n<p>AB = DE \u2026\u2026.(ii)<\/p>\n<p>\u2220 C = \u2220 F \u2026\u2026(iii)<\/p>\n<p>Here we have two right triangles, \u25b3 ABC and \u25b3 DEF<\/p>\n<p>From (i), (ii) and (iii),<\/p>\n<p>By the AAS congruence criterion, we have \u0394 ABC \u2245 \u0394 DEF<\/p>\n<p>Both triangles are congruent. Hence proved.<\/p>\n<p><strong>Question 2: If the bisector of the exterior vertical angle of a triangle is parallel to the base. Show that the triangle is isosceles.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let ABC be a triangle such that AD is the angular bisector of the exterior vertical angle, \u2220EAC and AD \u2225 BC.<\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 10 ex 10.3 question 2\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-10-ex-10-3-questio-4.png\" alt=\"RD sharma class 9 maths chapter 10 ex 10.3 question 2\" width=\"340\" height=\"211\" \/><\/p>\n<p>From figure,<\/p>\n<p>\u22201 = \u22202 [AD is a bisector of \u2220 EAC]<\/p>\n<p>\u22201 = \u22203 [Corresponding angles]<\/p>\n<p>and \u22202 = \u22204 [alternative angle]<\/p>\n<p>From above, we have \u22203 = \u22204<\/p>\n<p>This implies, AB = AC<\/p>\n<p>Two sides, AB and AC, are equal.<\/p>\n<p>\u21d2 \u0394 ABC is an isosceles triangle.<\/p>\n<p><strong>Question 3: In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let \u0394 ABC be isosceles where AB = AC and \u2220 B = \u2220 C<\/p>\n<p>Given: Vertex angle A is twice the sum of the base angles B and C. i.e., \u2220 A = 2(\u2220 B + \u2220 C)<\/p>\n<p>\u2220 A = 2(\u2220 B + \u2220 B)<\/p>\n<p>\u2220 A = 2(2 \u2220 B)<\/p>\n<p>\u2220 A = 4(\u2220 B)<\/p>\n<p>Now, We know that the sum of angles in a triangle =180\u00b0<\/p>\n<p>\u2220 A + \u2220 B + \u2220 C =180\u00b0<\/p>\n<p>4 \u2220 B + \u2220 B + \u2220 B = 180\u00b0<\/p>\n<p>6 \u2220 B =180\u00b0<\/p>\n<p>\u2220 B = 30\u00b0<\/p>\n<p>Since, \u2220 B = \u2220 C<\/p>\n<p>\u2220 B = \u2220 C = 30\u00b0<\/p>\n<p>And \u2220 A = 4 \u2220 B<\/p>\n<p>\u2220 A = 4 x 30\u00b0 = 120\u00b0<\/p>\n<p>Therefore, the angles of the given triangle are 30\u00b0 and 30\u00b0 and 120\u00b0.<\/p>\n<p><strong>Question 4: PQR is a triangle in which PQ = PR and 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.<\/strong><\/p>\n<p><strong>Solution:<\/strong>\u00a0Given that PQR is a triangle such that PQ = PR and S is any point on the side PQ and ST \u2225 QR.<\/p>\n<p>To prove: PS = PT<\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 10 ex 10.3 question 4\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-10-ex-10-3-questio-5.png\" alt=\"RD sharma class 9 maths chapter 10 ex 10.3 question 4\" width=\"297\" height=\"189\" \/><\/p>\n<p>Since, PQ= PR, so \u25b3PQR is an isosceles triangle.<\/p>\n<p>\u2220 PQR = \u2220 PRQ<\/p>\n<p>Now, \u2220 PST = \u2220 PQR and \u2220 PTS = \u2220 PRQ<\/p>\n<p>[Corresponding angles as ST parallel to QR]<\/p>\n<p>Since, \u2220 PQR = \u2220 PRQ<\/p>\n<p>\u2220 PST = \u2220 PTS<\/p>\n<p>In \u0394 PST,<\/p>\n<p>\u2220 PST = \u2220 PTS<\/p>\n<p>\u0394 PST is an isosceles triangle.<\/p>\n<p>Therefore, PS = PT.<\/p>\n<p>Hence proved.<\/p>\n<p>This is the complete blog on RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3. 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-10-exercise-103\"><\/span><strong>FAQs on RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3<\/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-1631269489547\" 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-10-exercise-103-pdf-offline\"><\/span>Can I access the RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3\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 id=\"faq-question-1631269526937\" 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-10-exercise-103\"><\/span>How much does it cost to download the PDF of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3?<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>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3:\u00a0Want to have the perfect score in the Class 9 Maths exam? Start studying the RD Sharma Solutions Class 9 Maths. It is a much need guide that will clear your doubts and make you score good marks. All the solutions of RD Sharma Class 9 Solutions &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-10-exercise-10-3\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":126566,"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\/126564"}],"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=126564"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126564\/revisions"}],"predecessor-version":[{"id":499585,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126564\/revisions\/499585"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126566"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126564"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126564"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126564"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}