{"id":126517,"date":"2023-03-28T00:42:00","date_gmt":"2023-03-27T19:12:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126517"},"modified":"2023-11-22T10:18:03","modified_gmt":"2023-11-22T04:48:03","slug":"rd-sharma-class-9-solutions-chapter-10-exercise-10-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-10-exercise-10-2\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-126543\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-10-Exercise-10.2.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-10-Exercise-10.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-10-Exercise-10.2-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2:\u00a0<\/strong>You can easily download the PDF 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 10.2 from the link mentioned in this blog. All the solutions of <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> are easy to understand. Clear your doubts with this ultimate help book and to more about the RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2 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-69d7ee26f0243\" 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-69d7ee26f0243\"><\/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-10-2\/#download-rd-sharma-class-9-solutions-chapter-10-exercise-102-pdf\" title=\"Download RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2 PDF\">Download RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2 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-10-2\/#access-answers-of-rd-sharma-class-9-solutions-chapter-10-exercise-102\" title=\"Access answers of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2\">Access answers of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2<\/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-10-2\/#faqs-on-rd-sharma-class-9-maths-chapter-10-exercise-102\" title=\"FAQs on RD Sharma Class 9 Maths Chapter 10 Exercise 10.2\">FAQs on RD Sharma Class 9 Maths Chapter 10 Exercise 10.2<\/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-10-2\/#from-where-can-i-download-the-pdf-of-rd-sharma-class-9-solutions-chapter-10-exercise-102\" title=\"From where can I download the PDF of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2?\">From where can I download the PDF of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2?<\/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-10-2\/#can-i-access-the-rd-sharma-class-9-solutions-chapter-10-exercise-102-pdf-offline\" title=\"Can I access the RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2\u00a0PDF offline?\">Can I access the RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2\u00a0PDF offline?<\/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-10-exercise-10-2\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-class-9-solutions-chapter-10-exercise-102\" title=\"How much does it cost to download the PDF of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2?\">How much does it cost to download the PDF of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-9-solutions-chapter-10-exercise-102-pdf\"><\/span><strong>Download RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2 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-10.2.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2<\/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-10.2.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-class-9-solutions-chapter-10-exercise-102\"><\/span><strong>Access answers of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Question 1: In the figure, it is given that RT = TS, \u2220 1 = 2 \u2220 2, and \u22204 = 2(\u22203). Prove that \u0394RBT \u2245 \u0394SAT.<\/strong><\/p>\n<p><strong><img title=\"RD sharma class 9 maths chapter 10 ex 10.2 question 1\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-10-ex-10-2-questio-3.png\" alt=\"RD sharma class 9 maths chapter 10 ex 10.2 question 1\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>In the figure,<\/p>\n<p>RT = TS \u2026\u2026(i)<\/p>\n<p>\u2220 1 = 2 \u2220 2 \u2026\u2026(ii)<\/p>\n<p>And \u2220 4 = 2 \u2220 3 \u2026\u2026(iii)<\/p>\n<p>To prove: \u0394RBT \u2245 \u0394SAT<\/p>\n<p>Let the point of intersection RB and SA be denoted by O<\/p>\n<p>\u2220 AOR = \u2220 BOS [Vertically opposite angles]<\/p>\n<p>or \u2220 1 = \u2220 4<\/p>\n<p>2 \u2220 2 = 2 \u2220 3 [From (ii) and (iii)]<\/p>\n<p>or \u2220 2 = \u2220 3 \u2026\u2026(iv)<\/p>\n<p>Now in \u0394 TRS, we have RT = TS<\/p>\n<p>\u21d2 \u0394 TRS is an isosceles triangle<\/p>\n<p>\u2220 TRS = \u2220 TSR \u2026\u2026(v)<\/p>\n<p>But, \u2220 TRS = \u2220 TRB + \u2220 2 \u2026\u2026(vi)<\/p>\n<p>\u2220 TSR = \u2220 TSA + \u2220 3 \u2026\u2026(vii)<\/p>\n<p>Putting (vi) and (vii) in (v), we get<\/p>\n<p>\u2220 TRB + \u2220 2 = \u2220 TSA + \u2220 3<\/p>\n<p>\u21d2 \u2220 TRB = \u2220 TSA [From (iv)]<\/p>\n<p>Consider \u0394 RBT and \u0394 SAT<\/p>\n<p>RT = ST [From (i)]<\/p>\n<p>\u2220 TRB = \u2220 TSA [From (iv)]<\/p>\n<p>By the ASA criterion of congruence, we have<\/p>\n<p>\u0394 RBT\u00a0<strong>\u2245<\/strong>\u00a0\u0394 SAT<\/p>\n<p><strong>Question 2: Two lines AB and CD intersect at O such that BC is equal and parallel to AD. Prove that the lines AB and CD bisect at O.<\/strong><\/p>\n<p><strong>Solution:<\/strong>\u00a0Lines AB and CD Intersect at O<\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 10 ex 10.2 question 2\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-10-ex-10-2-questio-4.png\" alt=\"RD sharma class 9 maths chapter 10 ex 10.2 question 2\" width=\"302\" height=\"199\" \/><\/p>\n<p>Such that BC \u2225 AD and<\/p>\n<p>BC = AD \u2026\u2026.(i)<\/p>\n<p>To prove AB and CD bisect at O.<\/p>\n<p>First, we have to prove that \u0394 AOD \u2245 \u0394 BOC<\/p>\n<p>\u2220OCB =\u2220ODA [AD||BC and CD is transversal]<\/p>\n<p>AD = BC [from (i)]<\/p>\n<p>\u2220OBC = \u2220OAD [AD||BC and AB is transversal]<\/p>\n<p>By ASA Criterion:<\/p>\n<p>\u0394 AOD \u2245 \u0394 BOC<\/p>\n<p>OA = OB and OD = OC (By c.p.c.t.)<\/p>\n<p>Therefore, AB and CD bisect each other at O.<\/p>\n<p>Hence Proved.<\/p>\n<p><strong>Question 3: BD and CE are bisectors of \u2220 B and \u2220 C of an isosceles \u0394 ABC with AB = AC. Prove that BD = CE.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u0394 ABC is isosceles with AB = AC, and BD and CE are bisectors of \u2220 B and \u2220 C We have to prove BD = CE. (Given)<\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 10 ex 10.2 question 3\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-10-ex-10-2-questio-5.png\" alt=\"RD sharma class 9 maths chapter 10 ex 10.2 question 3\" width=\"323\" height=\"251\" \/><\/p>\n<p>Since AB = AC<\/p>\n<p>\u21d2 \u2220ABC = \u2220ACB \u2026\u2026(i)<\/p>\n<p>[Angles opposite to equal sides are equal]<\/p>\n<p>Since BD and CE are bisectors of \u2220 B and \u2220 C<\/p>\n<p>\u2220 ABD = \u2220 DBC = \u2220 BCE = ECA = \u2220B\/2 = \u2220C\/2 \u2026(ii)<\/p>\n<p>Now, Consider \u0394 EBC = \u0394 DCB<\/p>\n<p>\u2220 EBC = \u2220 DCB [From (i)]<\/p>\n<p>BC = BC [Common side]<\/p>\n<p>\u2220 BCE = \u2220 CBD [From (ii)]<\/p>\n<p>By ASA congruence criterion, \u0394 EBC \u2245 \u0394 DCB<\/p>\n<p>Since corresponding parts of congruent triangles are equal.<\/p>\n<p>\u21d2 CE = BD<\/p>\n<p>or, BD = CE<\/p>\n<p>Hence proved.<\/p>\n<p><span style=\"font-size: inherit; background-color: initial;\">This is the complete blog on RD Sharma Class 9 Maths Chapter 10 Exercise 10.2. 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.<\/span><\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-9-maths-chapter-10-exercise-102\"><\/span><strong>FAQs on <span style=\"font-size: inherit; background-color: initial;\">RD Sharma Class 9 Maths Chapter 10 Exercise 10.2<\/span><\/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-1631267678629\" 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-10-exercise-102\"><\/span>From where can I download the PDF of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2?<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-1631267813496\" 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-102-pdf-offline\"><\/span>Can I access the RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2\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-1631267835092\" 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-102\"><\/span>How much does it cost to download the PDF of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2?<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 10.2:\u00a0You can easily download the PDF of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2 from the link mentioned in this blog. All the solutions of RD Sharma Solutions Class 9 Maths are easy to understand. Clear your doubts with this ultimate help book and to &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-10-exercise-10-2\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 10 Exercise 10.2 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":126543,"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\/126517"}],"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=126517"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126517\/revisions"}],"predecessor-version":[{"id":499691,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126517\/revisions\/499691"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126543"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126517"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126517"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126517"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}