{"id":62220,"date":"2023-08-31T14:45:00","date_gmt":"2023-08-31T09:15:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=62220"},"modified":"2023-12-06T11:04:07","modified_gmt":"2023-12-06T05:34:07","slug":"rd-sharma-solutions-class-9-maths-chapter-12-herons-formula","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-12-herons-formula\/","title":{"rendered":"RD Sharma Solutions Class 9 Maths Chapter 12 &#8211; Herons Formula (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-124465\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-12-Herons-Formula.png\" alt=\"RD Sharma Solutions Class 9 Maths Chapter 12\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-12-Herons-Formula.png 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-12-Herons-Formula-768x432.png 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><span style=\"font-weight: 400;\"><strong>RD Sharma Solutions Class 9 Maths Chapter 12 Herons Formula:<\/strong> How do you manage to stay focused for the final Class 9 Maths Exam? What are you doing to stay updated in terms of preparation? There are many chapters that are full of complex problems. So necessary suggestions must be taken into consideration at the time of attempting questions for the Mathematics exam.&nbsp; Hence get ready with <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths<\/a> Chapter 12 Herons Formula.<\/span><\/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-69eaf2163a7dc\" 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-69eaf2163a7dc\"><\/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-solutions-class-9-maths-chapter-12-herons-formula\/#download-rd-sharma-solutions-class-9-maths-chapter-12-pdf\" title=\"Download RD Sharma Solutions Class 9 Maths Chapter 12 PDF\">Download RD Sharma Solutions Class 9 Maths Chapter 12 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-solutions-class-9-maths-chapter-12-herons-formula\/#exercise-wise-rd-sharma-class-9-maths-chapter-12-solutions-pdf\" title=\"Exercise-wise: RD Sharma Class 9 Maths Chapter 12 Solutions PDF\">Exercise-wise: RD Sharma Class 9 Maths Chapter 12 Solutions PDF<\/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-solutions-class-9-maths-chapter-12-herons-formula\/#access-answers-of-rd-sharma-solutions-class-9-maths-chapter-12\" title=\"Access answers of RD Sharma Solutions Class 9 Maths Chapter 12\">Access answers of RD Sharma Solutions Class 9 Maths Chapter 12<\/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-solutions-class-9-maths-chapter-12-herons-formula\/#rd-sharma-class-9-solution-chapter-12-heron%e2%80%99s-formula-ex-121\" title=\"RD Sharma Class 9 Solution Chapter 12 Heron\u2019s Formula Ex 12.1\">RD Sharma Class 9 Solution Chapter 12 Heron\u2019s Formula Ex 12.1<\/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-solutions-class-9-maths-chapter-12-herons-formula\/#rd-sharma-class-9-chapter-12-heron%e2%80%99s-formula-ex-122\" title=\"RD Sharma Class 9 Chapter 12 Heron\u2019s Formula Ex 12.2\">RD Sharma Class 9 Chapter 12 Heron\u2019s Formula Ex 12.2<\/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-solutions-class-9-maths-chapter-12-herons-formula\/#rd-sharma-class-9-solutions-chapter-12-heron%e2%80%99s-formula-ex-123\" title=\"RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula Ex 12.3\">RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula Ex 12.3<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-12-herons-formula\/#rd-sharma-solutions-class-9-chapter-12-heron%e2%80%99s-formula-124\" title=\"RD Sharma Solutions Class 9 Chapter 12 Heron\u2019s Formula 12.4\">RD Sharma Solutions Class 9 Chapter 12 Heron\u2019s Formula 12.4<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-12-herons-formula\/#class-9-rd-sharma-solutions-chapter-12-heron%e2%80%99s-formula-ex-125\" title=\"Class 9 RD Sharma Solutions Chapter 12 Heron\u2019s Formula ex 12.5\">Class 9 RD Sharma Solutions Chapter 12 Heron\u2019s Formula ex 12.5<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-12-herons-formula\/#class-9-rd-sharma-solutions-chapter-12-heron%e2%80%99s-formula-126\" title=\"Class 9 RD Sharma Solutions Chapter 12 Heron\u2019s Formula 12.6\">Class 9 RD Sharma Solutions Chapter 12 Heron\u2019s Formula 12.6<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-12-herons-formula\/#rd-sharma-solutions-class-9-chapter-12-heron%e2%80%99s-formula-vsaqs\" title=\"RD Sharma Solutions Class 9 Chapter 12 Heron\u2019s Formula VSAQS\">RD Sharma Solutions Class 9 Chapter 12 Heron\u2019s Formula VSAQS<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-12-herons-formula\/#class-9-rd-sharma-solutions-chapter-12-heron%e2%80%99s-formula-mcqs\" title=\"Class 9 RD Sharma Solutions Chapter 12 Heron\u2019s Formula MCQS\">Class 9 RD Sharma Solutions Chapter 12 Heron\u2019s Formula MCQS<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-12-herons-formula\/#detailed-exercise-wise-explanation-with-a-listing-of-important-topics-in-the-exercise\" title=\"Detailed Exercise-wise Explanation with a Listing of Important Topics in the Exercise\">Detailed Exercise-wise Explanation with a Listing of Important Topics in the Exercise<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-13\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-12-herons-formula\/#rd-sharma-class-9-solutions-chapter-12-ex-12a\" title=\"RD Sharma&nbsp; Class 9 Solutions Chapter 12 Ex 12A\">RD Sharma&nbsp; Class 9 Solutions Chapter 12 Ex 12A<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-14\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-12-herons-formula\/#rd-sharma-class-9-solutions-chapter-12-ex-12b\" title=\"RD Sharma&nbsp; Class 9 Solutions Chapter 12 Ex 12B\">RD Sharma&nbsp; Class 9 Solutions Chapter 12 Ex 12B<\/a><ul class='ez-toc-list-level-4'><li class='ez-toc-heading-level-4'><a class=\"ez-toc-link ez-toc-heading-15\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-12-herons-formula\/#important-concepts\" title=\"Important concepts\">Important concepts<\/a><\/li><\/ul><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-16\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-12-herons-formula\/#faqs-on-rd-sharma-solutions-class-9-maths-chapter-12-%e2%80%93-herons-formula\" title=\"FAQs on RD Sharma Solutions Class 9 Maths Chapter 12 &#8211; Herons Formula\">FAQs on RD Sharma Solutions Class 9 Maths Chapter 12 &#8211; Herons Formula<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-17\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-12-herons-formula\/#from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-9-maths-chapter-12\" title=\"From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 12?\">From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 12?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-18\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-12-herons-formula\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-12\" title=\"How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 12?\">How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 12?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-19\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-12-herons-formula\/#can-i-access-the-rd-sharma-solutions-for-class-9-maths-chapter-12-pdf-offline\" title=\"Can I access the RD Sharma Solutions for Class 9 Maths Chapter 12\u00a0PDF offline?\">Can I access the RD Sharma Solutions for Class 9 Maths Chapter 12\u00a0PDF offline?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-9-maths-chapter-12-pdf\"><\/span><strong>Download RD Sharma Solutions Class 9 Maths Chapter 12 PDF<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/12-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 12<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/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\/12-1.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"exercise-wise-rd-sharma-class-9-maths-chapter-12-solutions-pdf\"><\/span><strong>Exercise-wise: RD Sharma Class 9 Maths Chapter 12 Solutions PDF<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<table style=\"border-collapse: collapse; width: 100%; height: 102px;\">\n<tbody>\n<tr style=\"height: 51px;\">\n<td style=\"width: 100%; height: 51px;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd%E2%80%8C-%E2%80%8Csharma%E2%80%8C-%E2%80%8Cchapter%E2%80%8C-%E2%80%8C12-%E2%80%8Cclass%E2%80%8C-%E2%80%8C9%E2%80%8C-%E2%80%8Cmaths%E2%80%8C-%E2%80%8Cexercise%E2%80%8C-%E2%80%8C12-1-%E2%80%8Csolutio\/\">RD Sharma Solutions Class 9 Maths Chapter 12 Exercise 12.1<\/a><\/td>\n<\/tr>\n<tr style=\"height: 51px;\">\n<td style=\"width: 100%; height: 51px;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-12-class-9-maths-exercise-12-2-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths Chapter 12 Exercise 12.2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-solutions-class-9-maths-chapter-12\"><\/span><strong>Access answers of RD Sharma Solutions Class 9 Maths Chapter 12<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-solution-chapter-12-heron%e2%80%99s-formula-ex-121\"><\/span>RD Sharma Class 9 Solution Chapter 12 Heron\u2019s Formula Ex 12.1<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>In the figure, the sides BA and CA have been produced such that BA = AD and CA = AE. Prove the segment DE || BC.<img src=\"https:\/\/farm2.staticflickr.com\/1903\/44730132835_7b7c8c5382_o.png\" alt=\"RD Sharma Class 9 Chapter 12 Heron's Formula\" width=\"175\" height=\"217\"><br>Solution:<br>Given : Sides BA and CA of \u2206ABC are produced such that BA = AD are CA = AE. ED is joined.<br>To prove : DE || BC<br>Proof: In \u2206ABC and \u2206DAE AB=AD (Given)<br>AC = AE (Given)<br>\u2220BAC = \u2220DAE (Vertically opposite angles)<br>\u2234 \u2206ABC \u2245 \u2206DAE (SAS axiom)<br>\u2234 \u2220ABC = \u2220ADE (c.p.c.t.)<br>But there are alternate angles<br>\u2234 DE || BC<\/p>\n<p>Question 2.<br>In a \u2206PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN.<br>Solution:<br>Given : In \u2206PQR, PQ = QR<br>L, M and N are the mid points of the sides PQ, QR and PR respectively<br><img src=\"https:\/\/farm2.staticflickr.com\/1948\/45644282261_f552938cb9_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 12 Heron's Formula\" width=\"189\" height=\"233\"><br>To prove : LM = MN<br>Proof : In \u2206LPN and \u2206MRH<br>PN = RN (\u2235 M is mid point of PR)<br>LP = MR (Half of equal sides)<br>\u2220P = \u2220R (Angles opposite to equal sides)<br>\u2234 ALPN \u2245 AMRH (SAS axiom)<br>\u2234 LN = MN (c.p.c.t.)<\/p>\n<p>Question 3.<br>Prove that the medians of an equilateral triangle are equal.<br>Solution:<br>Given : In \u2206ABC, AD, BE and CF are the medians of triangle and AB = BC = CA<br><img src=\"https:\/\/farm2.staticflickr.com\/1932\/45644282181_a841575826_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 12 Heron's Formula\" width=\"232\" height=\"221\"><br>To prove : AD = BE = CF<br>Proof : In \u2206BCE and \u2206BCF,<br>BC = BC (Common side)<br>CE = BF (Half of equal sides)<br>\u2220C = \u2220B (Angles opposite to equal sides)<br>\u2234 ABCE \u2245 ABCF (SAS axiom)<br>\u2234 BE = CF (c.p.c.t.) \u2026(i)<br>Similarly, we can prove that<br>\u2234 \u2206CAD \u2245 \u2206CAF<br>\u2234 AD = CF \u2026(ii)<br>From (i) and (ii)<br>BE = CF = AD<br>\u21d2 AD = BE = CF<\/p>\n<p>Question 4.<br>In a \u2206ABC, if \u2220A = 120\u00b0 and AB = AC. Find \u2220B and \u2220C.<br>Solution:<br>In \u2206ABC, \u2220A = 120\u00b0 and AB = AC<br>\u2234 \u2220B = \u2220C (Angles opposite to equal sides)<br>But \u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of angles of a triangle)<br><img src=\"https:\/\/farm2.staticflickr.com\/1917\/44730132395_4e9a28d3b3_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 12 Heron's Formula\" width=\"245\" height=\"196\"><br>\u21d2 120\u00b0 + \u2220B + \u2220B = 180\u00b0<br>\u21d2 2\u2220B = 180\u00b0 \u2013 120\u00b0 = 60\u00b0<br>\u2234 \u2220B =&nbsp;<span id=\"MathJax-Element-1-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-1\" class=\"math\"><span id=\"MathJax-Span-2\" class=\"mrow\"><span id=\"MathJax-Span-3\" class=\"mfrac\"><span id=\"MathJax-Span-4\" class=\"msubsup\"><span id=\"MathJax-Span-5\" class=\"texatom\"><span id=\"MathJax-Span-6\" class=\"mrow\"><span id=\"MathJax-Span-7\" class=\"mn\">60<\/span><\/span><\/span><span id=\"MathJax-Span-8\" class=\"texatom\"><span id=\"MathJax-Span-9\" class=\"mrow\"><span id=\"MathJax-Span-10\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-11\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;= 30\u00b0<br>and \u2220C = \u2220B = 30\u00b0<br>Hence \u2220B = 30\u00b0 and \u2220C = 30\u00b0<\/p>\n<p>Question 5.<br>In a \u2206ABC, if AB = AC and \u2220B = 70\u00b0, find \u2220A.<br>Solution:<br>In \u2206ABC, \u2220B = 70\u00b0<br>AB =AC<br>\u2234 \u2220B = \u2220C (Angles opposite to equal sides)<br>But \u2220B = 70\u00b0<br>\u2234 \u2220C = 70\u00b0<br>But \u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of angles of a triangle)<br>\u21d2 \u2220A + 70\u00b0 + 70\u00b0 = 180\u00b0<br>\u21d2 \u2220A + 140\u00b0= 180\u00b0<br>\u2234\u2220A = 180\u00b0- 140\u00b0 = 40\u00b0<\/p>\n<p>Question 6.<br>The vertical angle of an isosceles triangle is 100\u00b0. Find its base angles.<br>Solution:<br>In \u2206ABC, AB = AC and \u2220A = 100\u00b0<br>But AB = AC (In isosceles triangle)<br><img src=\"https:\/\/farm2.staticflickr.com\/1966\/44730132285_26fb855865_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 12 Heron's Formula\" width=\"217\" height=\"187\"><br>\u2234 \u2220C = \u2220B (Angles opposite to equal sides)<br>\u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of angles of a triangle)<br>\u21d2 100\u00b0 + \u2220B + \u2220B = 180\u00b0 (\u2235 \u2220C = \u2220B)<br>\u21d2 2\u2220B = 180\u00b0 \u2013 100\u00b0 = 80\u00b0<br>\u2234 \u2220C = \u2220B = 40\u00b0<br>Hence \u2220B = 40\u00b0, \u2220C = 40\u00b0<\/p>\n<p>Question 7.<br>In the figure, AB = AC and \u2220ACD = 105\u00b0, find \u2220BAC.<br>Solution:<br>In \u2206ABC, AB = AC<br>\u2234 \u2220B = \u2220C (Angles opposite to equal sides)<br>But \u2220ACB + \u2220ACD = 180\u00b0 (Linear pair)<br>\u21d2 \u2220ACB + 105\u00b0= 180\u00b0<br>\u21d2 \u2220ACB = 180\u00b0-105\u00b0 = 75\u00b0<br>\u2234 \u2220ABC = \u2220ACB = 75\u00b0<br>But \u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of angles of a triangle)<br>\u21d2 \u2220A + 75\u00b0 + 75\u00b0 = 180\u00b0<br>\u21d2 \u2220A + 150\u00b0= 180\u00b0<br>\u21d2 \u2220A= 180\u00b0- 150\u00b0 = 30\u00b0<br>\u2234 \u2220BAC = 30\u00b0<\/p>\n<p>Question 8.<br>Find the measure of each exterior angle of an equilateral triangle.<br>Solution:<br>In an equilateral triangle, each interior angle is 60\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1914\/44730132075_c49c9d71b7_o.png\" alt=\"Class 9 Maths Chapter 12 Heron's Formula RD Sharma Solutions\" width=\"217\" height=\"254\"><br>But interior angle + exterior angle at each vertex = 180\u00b0<br>\u2234 Each exterior angle = 180\u00b0 \u2013 60\u00b0 = 120\u00b0<\/p>\n<p>Question 9.<br>If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.<br>Solution:<br>Given : In an isosceles \u2206ABC, AB = AC<br>and base BC has produced both ways<br><img src=\"https:\/\/farm2.staticflickr.com\/1975\/44730131925_a5a800f44f_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 12 Heron's Formula\" width=\"285\" height=\"194\"><br>To prove: \u2220ACD = \u2220ABE<br>Proof: In \u2206ABC,<br>\u2235 AB = AC<br>\u2234\u2220C = \u2220B (Angles opposite to equal sides)<br>\u21d2 \u2220ACB = \u2220ABC<br>But \u2220ACD + \u2220ACB = 180\u00b0 (Linear pair)<br>and \u2220ABE + \u2220ABC = 180\u00b0<br>\u2234 \u2220ACD + \u2220ACB = \u2220ABE + \u2220ABC<br>But \u2220ACB = \u2220ABC (Proved)<br>\u2234 \u2220ACD = \u2220ABE<br>Hence proved.<\/p>\n<p>Question 10.<br>In the figure, AB = AC and DB = DC, find the ratio \u2220ABD : \u2220ACD.<br><img src=\"https:\/\/farm2.staticflickr.com\/1912\/45644281531_e843baed3b_o.png\" alt=\"RD Sharma Class 9 Book Chapter 12 Heron's Formula\" width=\"204\" height=\"185\"><br>Solution:<br>In the given figure,<br>In \u2206ABC,<br>AB = AC and DB = DC<br>In \u2206ABC,<br>\u2235 AB = AC<br>\u2234 \u2220ACD = \u2220ABE \u2026(i) (Angles opposite to equal sides)<br>Similarly, in \u2206DBC,<br>DB = DC<br>\u2234 \u2220DCB = \u2220DBC .. (ii)<br>Subtracting (ii) from (i)<br>\u2220ACB \u2013 \u2220DCB = \u2220ABC \u2013 \u2220DBC<br>\u21d2 \u2220ACD = \u2220ABD<br>\u2234 Ratio \u2220ABD : \u2220ACD = 1 : 1<\/p>\n<p>Question 11.<br>Determine the measure of each of the equal angles of a right-angled isosceles triangle.<br>OR<br>ABC is a right-angled triangle in which \u2220A = 90\u00b0 and AB = AC. Find \u2220B and \u2220C.<br>Solution:<br>Given: In a right-angled isosceles \u2206ABC, \u2220A = 90\u00b0 and AB = AC<br>To determine, each equal angle of the triangle<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1925\/45644281451_a6a4ba71eb_o.png\" alt=\"Heron's Formula With Solutions PDF RD Sharma Class 9 Solutions\" width=\"260\" height=\"192\"><br>\u2235 \u2220A = 90\u00b0<br>\u2234 \u2220B + \u2220C = 90\u00b0<br>But \u2220B = \u2220C<br>\u2234 \u2220B + \u2220B = 90\u00b0<br>\u21d2 2\u2220B = 90\u00b0<br>90\u00b0<br>\u21d2 \u2220B =&nbsp;<span id=\"MathJax-Element-2-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-12\" class=\"math\"><span id=\"MathJax-Span-13\" class=\"mrow\"><span id=\"MathJax-Span-14\" class=\"mfrac\"><span id=\"MathJax-Span-15\" class=\"msubsup\"><span id=\"MathJax-Span-16\" class=\"texatom\"><span id=\"MathJax-Span-17\" class=\"mrow\"><span id=\"MathJax-Span-18\" class=\"mn\">90<\/span><\/span><\/span><span id=\"MathJax-Span-19\" class=\"texatom\"><span id=\"MathJax-Span-20\" class=\"mrow\"><span id=\"MathJax-Span-21\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-22\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp; = 45\u00b0<br>and \u2220C = \u2220B = 45\u00b0<br>Hence \u2220B = \u2220C = 45\u00b0<\/p>\n<p>Question 12.<br>In the figure, PQRS is square and SRT is an equilateral triangle. Prove that<br>(i) PT = QT<br>(ii) \u2220TQR = 15\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1959\/45644281371_a2b6fa88c2_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 12 Heron's Formula\" width=\"184\" height=\"270\"><br>Solution:<br>Given: PQRS is square and SRT is an equilateral triangle. PT and QT are joined.<br><img src=\"https:\/\/farm2.staticflickr.com\/1929\/44730131405_83471a55ee_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 12 Heron's Formula\" width=\"163\" height=\"274\"><br>To prove : (i) PT = QT; (ii) \u2220TQR = 15\u00b0<br>Proof: In \u2206TSP and \u2206TQR<br>ST = RT (Sides of an equilateral triangle)<br>SP = PQ (Sides of the square)<br>and \u2220TSP = \u2220TRQ (Each = 60\u00b0 + 90\u00b0)<br>\u2234 \u2206TSP \u2245 \u2206TQR (SAS axiom)<br>\u2234 PT = QT (c.p.c.t.)<br>In \u2206TQR,<br>\u2235 RT = RQ (Square sides)<br>\u2220RTQ = \u2220RQT<br>But \u2220TRQ = 60\u00b0 + 90\u00b0 = 150\u00b0<br>\u2234 \u2220RTQ + \u2220RQT = 180\u00b0 \u2013 150\u00b0 = 30\u00b0<br>\u2235 \u2220PTQ = \u2220RQT (Proved)<br>\u2220RQT =&nbsp;<span id=\"MathJax-Element-3-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-23\" class=\"math\"><span id=\"MathJax-Span-24\" class=\"mrow\"><span id=\"MathJax-Span-25\" class=\"mfrac\"><span id=\"MathJax-Span-26\" class=\"msubsup\"><span id=\"MathJax-Span-27\" class=\"texatom\"><span id=\"MathJax-Span-28\" class=\"mrow\"><span id=\"MathJax-Span-29\" class=\"mn\">30<\/span><\/span><\/span><span id=\"MathJax-Span-30\" class=\"texatom\"><span id=\"MathJax-Span-31\" class=\"mrow\"><span id=\"MathJax-Span-32\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-33\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp; = 15\u00b0<br>\u21d2 \u2220TQR = 15\u00b0<\/p>\n<p>Question 13.<br>AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from points A and B (see figure). Show that the line PQ is the perpendicular bisector of AB.<br><img src=\"https:\/\/farm2.staticflickr.com\/1906\/44730131275_501ac831c5_o.png\" alt=\"Solution Of Rd Sharma Class 9 Chapter 12 Heron's Formula\" width=\"153\" height=\"248\"><br>Solution:<br>Given: AB is a line segment.<br>P and Q are points such that they are equidistant from A and B<br>i.e. PA = PB and QA = QB AP, PB, QA, QB, PQ are joined<br>To prove : PQ is perpendicular bisector of AB<br><img src=\"https:\/\/farm2.staticflickr.com\/1911\/43826225520_1584eee6f8_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 12 Heron's Formula\" width=\"154\" height=\"258\"><br>Proof: In \u2206PAQ and \u2206PBQ,<br>PA = PB (Given)<br>QA = QB (Given)<br>PQ = PQ (Common)<br>\u2234 \u2206PAQ \u2245 \u2206PBQ (SSS axiom)<br>\u2234 \u2220APQ = \u2220BPQ (c.p.c.t.)<br>Now in \u2206APC = \u2206BPC<br>PA = PB (Given)<br>\u2206APC \u2245 \u2206BPC (Proved)<br>PC = PC (Common)<br>\u2234 \u2206APC = \u2206BPC (SAS axiom)<br>\u2234 AC = BC (c.p.c.t.)<br>and \u2220PCA = \u2220PCB (c.p.c.t.)<br>But \u2220PCA + \u2220PCB = 180\u00b0 (Linear pair)<br>\u2234 \u2220PCA = \u2220PCB = 90\u00b0<br>\u2234 PC or PQ is perpendicular bisector of AB<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-chapter-12-heron%e2%80%99s-formula-ex-122\"><\/span>RD Sharma Class 9 Chapter 12 Heron\u2019s Formula Ex 12.2<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>BD and CE are bisectors of \u2220B and \u2220C of an isosceles \u2220ABC with AB = AC. Prove that BD = CE.<br>Solution:<br>Given: In \u2206ABC, AB = AC<br>BD and CE are the bisectors of \u2220B and \u2220C respectively<br>To prove: BD = CE<br>Proof: In \u2206ABC, AB = AC<br>\u2234 \u2220B = \u2220C (Angles opposite to equal sides)<br>\u2234&nbsp;<span id=\"MathJax-Element-4-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-34\" class=\"math\"><span id=\"MathJax-Span-35\" class=\"mrow\"><span id=\"MathJax-Span-36\" class=\"mfrac\"><span id=\"MathJax-Span-37\" class=\"mn\">1<\/span><span id=\"MathJax-Span-38\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;\u2220B =&nbsp;<span id=\"MathJax-Element-5-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-39\" class=\"math\"><span id=\"MathJax-Span-40\" class=\"mrow\"><span id=\"MathJax-Span-41\" class=\"mfrac\"><span id=\"MathJax-Span-42\" class=\"mn\">1<\/span><span id=\"MathJax-Span-43\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;\u2220C<br>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula Ex 12.2 \u2013 1<br>\u2220DBC = \u2220ECB<br>Now, in \u2206DBC and \u2206EBC,<br>BC = BC (Common)<br>\u2220C = \u2220B (Equal angles)<br>\u2220DBC = \u2220ECB (Proved)<br>\u2234 \u2206DBC \u2245 \u2206EBC (ASA axiom)<br>\u2234 BD = CE<\/p>\n<p>Question 2.<br>In the figure, it is given that RT = TS, \u22201 = 2\u22202 and \u22204 = 2\u22203. Prove that: \u2206RBT = \u2206SAT.<br>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula Ex 12.2 \u2013 2<br>Solution:<br>Given: In the figure, RT = TS<br>\u22201 = 2\u22202 and \u22204 = 2\u22203<br>To prove: \u2206RBT \u2245 \u2206SAT<br>Proof: \u2235 \u22201 = \u22204 (Vertically opposite angles)<br>But \u22201 = 2\u22202 and 4 = 2\u22203<br>\u2234 2\u22202 = 2\u22203 \u21d2 \u22202 = \u22203<br>\u2235 RT = ST (Given)<br>\u2234\u2220R = \u2220S (Angles opposite to equal sides)<br>\u2234 \u2220R \u2013 \u22202 = \u2220S \u2013 \u22203<br>\u21d2 \u2220TRB = \u2220AST<br>Now in \u2206RBT and \u2206SAT<br>\u2220TRB = \u2220SAT (prove)<br>RT = ST (Given)<br>\u2220T = \u2220T (Common)<br>\u2234 \u2206RBT \u2245 \u2206SAT (SAS axiom)<\/p>\n<p>Question 3.<br>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.<br>Solution:<br>Given: Two lines AB and CD intersect each other at O such that AD = BC and AD <span id=\"MathJax-Element-6-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-44\" class=\"math\"><span id=\"MathJax-Span-45\" class=\"mrow\"><span id=\"MathJax-Span-46\" class=\"mo\">\u2225<\/span><\/span><\/span><\/span><br>BC<br>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula Ex 12.2 \u2013 3<br>To prove: AB and CD bisect each other<br>i. e. AO = OB and CO = OD<br>Proof: In \u2206AOD and \u2206BOC,<br>AD = BC (Given)<br>\u2220A = \u2220B (Alternate angles)<br>\u2220D = \u2220C (Alternate angles)<br>\u2234 \u2206AOD \u2245 \u2206BOC (ASA axiom)<br>AO = OB and AO = OC (c.p.c.t.)<br>Hence AB and CD bisect each other.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-solutions-chapter-12-heron%e2%80%99s-formula-ex-123\"><\/span>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula Ex 12.3<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>In two right triangles, one side has an acute angle of one are 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 <span id=\"MathJax-Element-7-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-47\" class=\"math\"><span id=\"MathJax-Span-48\" class=\"mrow\"><span id=\"MathJax-Span-49\" class=\"mo\">\u2225<\/span><\/span><\/span><\/span>&nbsp;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&nbsp;<span id=\"MathJax-Element-8-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-50\" class=\"math\"><span id=\"MathJax-Span-51\" class=\"mrow\"><span id=\"MathJax-Span-52\" class=\"mo\">\u2225<\/span><\/span><\/span><\/span>&nbsp;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=&nbsp;<span id=\"MathJax-Element-9-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-53\" class=\"math\"><span id=\"MathJax-Span-54\" class=\"mrow\"><span id=\"MathJax-Span-55\" class=\"mfrac\"><span id=\"MathJax-Span-56\" class=\"msubsup\"><span id=\"MathJax-Span-57\" class=\"texatom\"><span id=\"MathJax-Span-58\" class=\"mrow\"><span id=\"MathJax-Span-59\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-60\" class=\"texatom\"><span id=\"MathJax-Span-61\" class=\"mrow\"><span id=\"MathJax-Span-62\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-63\" class=\"mn\">6<\/span><\/span><\/span><\/span><\/span>&nbsp;= 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 =&nbsp;<span id=\"MathJax-Element-10-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-64\" class=\"math\"><span id=\"MathJax-Span-65\" class=\"mrow\"><span id=\"MathJax-Span-66\" class=\"mfrac\"><span id=\"MathJax-Span-67\" class=\"msubsup\"><span id=\"MathJax-Span-68\" class=\"texatom\"><span id=\"MathJax-Span-69\" class=\"mrow\"><span id=\"MathJax-Span-70\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-71\" class=\"texatom\"><span id=\"MathJax-Span-72\" class=\"mrow\"><span id=\"MathJax-Span-73\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-74\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>= 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 =&nbsp;<span id=\"MathJax-Element-11-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-75\" class=\"math\"><span id=\"MathJax-Span-76\" class=\"mrow\"><span id=\"MathJax-Span-77\" class=\"mfrac\"><span id=\"MathJax-Span-78\" class=\"msubsup\"><span id=\"MathJax-Span-79\" class=\"texatom\"><span id=\"MathJax-Span-80\" class=\"mrow\"><span id=\"MathJax-Span-81\" class=\"mn\">90<\/span><\/span><\/span><span id=\"MathJax-Span-82\" class=\"texatom\"><span id=\"MathJax-Span-83\" class=\"mrow\"><span id=\"MathJax-Span-84\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-85\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;= 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 a \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 =&nbsp;<span id=\"MathJax-Element-12-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-86\" class=\"math\"><span id=\"MathJax-Span-87\" class=\"mrow\"><span id=\"MathJax-Span-88\" class=\"mfrac\"><span id=\"MathJax-Span-89\" class=\"mn\">1<\/span><span id=\"MathJax-Span-90\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\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 draw which 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 bisectors 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, 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<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-9-chapter-12-heron%e2%80%99s-formula-124\"><\/span>RD Sharma Solutions Class 9 Chapter 12 Heron\u2019s Formula 12.4<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>In the figure, it is given that AB = CD and AD = BC. Prove that \u2206ADC \u2245 \u2206CBA.<br>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula Ex 12.4 \u2013 1<br>Solution:<br>Given : In the figure, AB = CD, AD = BC<br>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula Ex 12.4 \u2013 2<br>To prove : \u2206ADC = \u2206CBA<br>Proof : In \u2206ADC and \u2206CBA<br>CD = AB (Given)<br>AD = BC (Given)<br>CA = CA (Common)<br>\u2234 \u2206ADC \u2245 \u2206CBA (SSS axiom)<\/p>\n<p>Question 2.<br>In a APQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN.<br>Solution:<br>Given : In \u2206PQR, PQ = QR<br>L, M and N are the mid-points of sides PQ, QR and RP respectively. Join LM, MN and LN<br>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula Ex 12.4 \u2013 2A<br>To prove : \u2220PNM = \u2220PLM<br>Proof : In \u2206PQR,<br>\u2235 M and N are the mid points of sides PR and QR respectively<br>\u2234 MN || PQ and MN =&nbsp;<span id=\"MathJax-Element-13-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-91\" class=\"math\"><span id=\"MathJax-Span-92\" class=\"mrow\"><span id=\"MathJax-Span-93\" class=\"mfrac\"><span id=\"MathJax-Span-94\" class=\"mn\">1<\/span><span id=\"MathJax-Span-95\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;PQ \u2026(i)<br>\u2234 MN = PL<br>Similarly, we can prove that<br>LM = PN<br>Now in \u2206NML and \u2206LPN<br>MN = PL (Proved)<br>LM = PN (Proved)<br>LN = LN (Common)<br>\u2234 \u2206NML = \u2206LPN (SSS axiom)<br>\u2234 \u2220MNL = \u2220PLN (c.p.c.t.)<br>and \u2220MLN = \u2220LNP (c.p.c.t.)<br>\u21d2 \u2220MNL = \u2220LNP = \u2220PLM = \u2220MLN<br>\u21d2 \u2220PNM = \u2220PLM<\/p>\n<h3><span class=\"ez-toc-section\" id=\"class-9-rd-sharma-solutions-chapter-12-heron%e2%80%99s-formula-ex-125\"><\/span>Class 9 RD Sharma Solutions Chapter 12 Heron\u2019s Formula ex 12.5<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>ABC is a triangle and D is the mid-point of BC. The perpendiculars from D to AB and AC are equal. Prove that the triangle is isosceles.<br>Solution:<br>Given : In \u2206ABC, D is mid-point of BC and DE \u22a5 AB, DF \u22a5 AC and DE = DF<br><img src=\"https:\/\/farm2.staticflickr.com\/1934\/44730127325_804fda0f77_o.png\" alt=\"RD Sharma Class 9 Chapter 12 Heron's Formula\" width=\"268\" height=\"184\"><br>To Prove : \u2206ABC is an isosceles triangle<br>Proof : In right \u2206BDE and \u2206CDF,<br>Side DE = DF<br>Hyp. BD = CD<br>\u2234 \u2206BDE \u2245 \u2206CDF (RHS axiom)<br>\u2234 \u2220B = \u2220C (c.p.c.t.)<br>Now in \u2206ABC,<br>\u2220B = \u2220C (Prove)<br>\u2234 AC = AB (Sides opposite to equal angles)<br>\u2234 AABC is an isosceles triangle<\/p>\n<p>Question 2.<br>ABC is a triangle in which BE and CF are, respectively, the perpendiculars to the sides AC and AB. If BE = CF, prove that \u2206ABC is an isosceles.<br>Solution:<br>Given : In \u2206ABC,<br>BE \u22a5 AC and CF \u22a5 AB<br>BE = CF<br><img src=\"https:\/\/farm2.staticflickr.com\/1921\/44730127195_55c1a313e8_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 12 Heron's Formula\" width=\"283\" height=\"191\"><br>To prove : AABC is an isosceles triangle<br>Proof : In right ABCE and ABCF Side<br>BE = CF (Given)<br>Hyp. BC = BC (Common)<br>\u2234 \u2206BCE \u2245 \u2206BCF (RHS axiom)<br>\u2234 \u2220BCE = \u2220CBF (c.p.c.t.)<br>\u2234 AB = AC (Sides opposite to equal angles)<br>\u2234 \u2206ABC is an isosceles triangle<\/p>\n<p>Question 3.<br>If perpendiculars from any point within an angle on its arms are congruent, prove that it lies on the bisector of that angle.<br>Solution:<br>Given : A point P lies in the angle ABC and PL \u22a5 BA and PM \u22a5 BC and PL = PM. PB is joined<br><img src=\"https:\/\/farm2.staticflickr.com\/1966\/43826222590_859785f919_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 12 Heron's Formula\" width=\"213\" height=\"224\"><br>To prove : PB is the bisector \u2220ABC,<br>Proof : In right \u2206PLB and \u2206PMB<br>Side PL = PM (Given)<br>Hyp. PB = PB (Common)<br>\u2234 \u2206PLB \u2245 \u2206PMB (RHS axiom)<br>\u2234 \u2206PBL = \u2206PBM (c.p.c.t.)<br>\u2234 PB is the bisector of \u2220ABC<\/p>\n<p>Question 4.<br>In the figure, AD \u22a5 CD and CB \u22a5 CD. If AQ = BP and DP = CQ, prove that \u2220DAQ = \u2220CBP.<br><img src=\"https:\/\/farm2.staticflickr.com\/1916\/44730127005_5984cd92a8_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 12 Heron's Formula\" width=\"195\" height=\"176\"><br>Solution:<br>Given : In the figure,<br>AD \u22a5 CD and CB \u22a5 CD, AQ = BP and DP = CQ<br><img src=\"https:\/\/farm2.staticflickr.com\/1924\/44730126825_a4baef6812_o.png\" alt=\"Heron's Formula Class 9 RD Sharma Solutions\" width=\"211\" height=\"170\"><br>To prove : \u2220DAQ = \u2220CBP<br>Proof : \u2235 DP = CQ<br>\u2234 DP + PQ = PQ + QC<br>\u21d2 DQ = PC<br>Now in right \u2206ADQ and \u2206BCP<br>Side DQ = PC (Proved)<br>Hyp. AQ = BP<br>\u2234 \u2206ADQ \u2245 \u2206BCP (RHS axiom)<br>\u2234 \u2220DAQ = \u2220CBP (c.p.c.t.)<\/p>\n<p>Question 5.<br>Which of the following statements are true (T) and which are false (F):<br>(i) Sides opposite to equal angles of a triangle may be unequal.<br>(ii) Angles opposite to equal sides of a triangle are equal.<br>(iii) The measure of each angle of an equilateral triangle is 60\u00b0.<br>(iv) If the altitude from one vertex of a triangle bisects the opposite side, then the triangle may be isosceles.<br>(v) The bisectors of two equal angles of a triangle are equal.<br>(vi) If the bisector of the vertical angle of a triangle bisects the base, then the triangle may be isosceles.<br>(vii) The two altitudes corresponding to two equal sides of a triangle need not be equal.<br>(viii)If any two sides of a right triangle are respectively equal to two sides of other right triangle, then the two triangles are congruent.<br>(ix) Two right triangles are congruent if hypotenuse and a side of one triangle are respectively equal to the hypotenuse and a side of the other triangle.<br>Solution:<br>(i) False : Sides opposite to equal angles of a triangle are equal.<br>(ii) True.<br>(iii) True.<br>(iv) False : The triangle is an isosceles triangle.<br>(v) True.<br>(vi) False : The triangle is an isosceles.<br>(vii) False : The altitude an equal.<br>(viii) False : If one side and hypotenuse of one right triangle on one side and hypotenuse of the other right triangle are equal, then triangles are congruent.<br>(ix) True.<\/p>\n<p>Question 6.<br>Fill in the blanks in the following so that each of the following statements is true.<br>(i) Sides opposite to equal angles of a triangle are \u2026\u2026.<br>(ii) Angle opposite to equal sides of a triangle are \u2026\u2026.<br>(iii) In an equilateral triangle all angles are \u2026\u2026.<br>(iv) In a \u2206ABC if \u2220A = \u2220C, then AB = \u2026\u2026.<br>(v) If altitudes CE and BF of a triangle ABC are equal, then AB = \u2026\u2026..<br>(vi) In an isosceles triangle ABC with AB = AC, if BD and CE are its altitudes, then BD is \u2026\u2026\u2026 CE.<br>(vii) In right triangles ABC and DEF, if hypotenuse AB = EF and side AC = DE, then \u2206ABC \u2245 \u2206\u2026\u2026<br>Solution:<br>(i) Sides opposite to equal angles of a triangle are equal.<br>(ii) Angles opposite to equal sides of a triangle are equal.<br>(iii) In an equilateral triangle all angles are equal.<br>(iv) In a \u2206ABC, if \u2220A = \u2220C, then AB = BC.<br>(v) If altitudes CE and BF of a triangle ABC are equal, then AB = AC.<br>(vi) In an isosceles triangle ABC with AB = AC, if BD and CE are its altitudes, then BD is equal to CE.<br>(vii) In right triangles ABC and DEF, it hypotenuses AB = EF and side AC = DE, then \u2206ABC \u2245 \u2206EFD.<\/p>\n<p>Question 7.<br>ABCD is a square, X and Y are points on sides AD and BC respectively such that AY = BX. Prove that BY = AX and \u2220BAY = \u2220ABX.<br>Solution:<br>Given : In square ABCD, X and Y are points on side AD and BC respectively and AY = BX<br>To prove : BY = AX<br>\u2220BAY = \u2220ABX<br>Proof: In right \u2206BAX and \u2206ABY<br>AB =AB (Common)<br>Hyp. BX = AY (Given)<br>\u2234 \u2206BAX \u2245 \u2206ABY (RHS axiom)<br>\u2234 AX = BY (c.p.c.t.)<br>\u2220ABX = \u2220BAY (c.p.c.t.)<br>Hence, BY = AX and \u2220BAY = \u2220ABX.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"class-9-rd-sharma-solutions-chapter-12-heron%e2%80%99s-formula-126\"><\/span>Class 9 RD Sharma Solutions Chapter 12 Heron\u2019s Formula 12.6<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>In \u2206ABC, if \u2220A = 40\u00b0 and \u2220B = 60\u00b0. Determine the longest and shortest sides of the triangle.<br>Solution:<br>In \u2206ABC, \u2220A = 40\u00b0, \u2220B = 60\u00b0<br>But \u2220A + \u2220B + \u2220C = 180\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1963\/44730126345_9616d37d35_o.png\" alt=\"Class 9 Maths Chapter 12 Heron's Formula RD Sharma Solutions\" width=\"213\" height=\"144\"><br>\u21d2 40\u00b0 + 60\u00b0 + \u2220C = 180\u00b0<br>\u21d2 \u2220C = 180\u00b0 = (40\u00b0 + 60\u00b0)<br>= 180\u00b0 \u2013 100\u00b0 = 80\u00b0<br>\u2235 \u2220C = 80\u00b0, which is the greatest angle and<br>\u2220A = 40\u00b0 is the smallest angle<br>\u2234 Side AB which is opposite to the greatest angle is the longest and side BC which is opposite to the smallest angle is the shortest.<\/p>\n<p>Question 2.<br>In a \u2206ABC, if \u2220B = \u2220C = 45\u00b0. which is the longest side?<br>Solution:<br>In \u2206ABC, \u2220B = \u2220C = 45\u00b0<br>But \u2220A + \u2220B + \u2220C = 180\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1950\/44730126105_13e0b07e87_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 12 Heron's Formula\" width=\"288\" height=\"214\"><br>\u21d2 \u2220A + 45\u00b0 + 45\u00b0 = 180\u00b0<br>\u21d2 \u2220A + 90\u00b0 = 180\u00b0<br>\u2234 \u2220A = 180\u00b0-90\u00b0 = 90\u00b0<br>\u2234\u2220A is the greatest<br>\u2234 Side BC opposite to it is the longest<\/p>\n<p>Question 3.<br>In \u2206ABC, side AB is produced to D so that BD = BC. If \u2220B = 60\u00b0 and \u2220A = 70\u00b0, prove that :<br>(i) AD &gt; CD<br>(ii) AD &gt; AC<br>Solution:<br>Given: In AABC, side BC is produced to D such that BD = BC<br>\u2220A = 70\u00b0 and \u2220B = 60\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1937\/44730125985_e0cee8cebc_o.png\" alt=\"RD Sharma Class 9 Book Chapter 12 Heron's Formula\" width=\"308\" height=\"204\"><br>To prove :<br>(i) AD &gt; CD (ii) AD &gt; AC<br>Proof: In \u2206ABC,<br>\u2220A = 70\u00b0, \u2220B = 60\u00b0<br>But Ext. \u2220CBD + \u2220CBA = 180\u00b0 (Linear pair)<br>\u2220CBD + 60\u00b0 = 180\u00b0 3<br>\u21d2 \u2220CBD = 180\u00b0 \u2013 60\u00b0 = 120\u00b0<br>But in \u2206BCD,<br>BD = BC<br>\u2234 \u2220D = \u2220BCD<br>But \u2220D + \u2220BCD = 180\u00b0 \u2013 120\u00b0 = 60\u00b0<br>\u2234\u2220D = \u2220BCD =&nbsp;<span id=\"MathJax-Element-14-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-96\" class=\"math\"><span id=\"MathJax-Span-97\" class=\"mrow\"><span id=\"MathJax-Span-98\" class=\"mfrac\"><span id=\"MathJax-Span-99\" class=\"msubsup\"><span id=\"MathJax-Span-100\" class=\"texatom\"><span id=\"MathJax-Span-101\" class=\"mrow\"><span id=\"MathJax-Span-102\" class=\"mn\">60<\/span><\/span><\/span><span id=\"MathJax-Span-103\" class=\"texatom\"><span id=\"MathJax-Span-104\" class=\"mrow\"><span id=\"MathJax-Span-105\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-106\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp; = 30\u00b0<br>and in \u2206ABC,<br>\u2220A + \u2220B + \u2220C = 180\u00b0<br>\u21d2 70\u00b0 + 60\u00b0 + \u2220C = 180\u00b0<br>\u21d2 130\u00b0 + \u2220C = 180\u00b0<br>\u2234 \u2220C =180\u00b0- 130\u00b0 = 50\u00b0<br>Now \u2220ACD = \u2220ACB + \u2220BCD = 50\u00b0 + 30\u00b0 = 80\u00b0<br>(i) Now in \u2206ACB,<br>\u2220ACD = 80\u00b0 and \u2220A = 70\u00b0<br>\u2234 Side AD &gt; CD<br>(Greater angle has greatest side opposite to it)<br>(ii) \u2235 \u2220ACD = 80\u00b0 and \u2220D = 30\u00b0<br>\u2234 AD &gt; AC<\/p>\n<p>Question 4.<br>Is it possible to draw a triangle with sides of length 2 cm, 3 cm and 7 cm?<br>Solution:<br>We know that in a triangle, sum of any two sides is greater than the third side and 2 cm + 3 cm = 5 cm and 5 cm &lt; 7 cm<br>\u2234 This triangle is not possible to draw<\/p>\n<p>Question 5.<br>In \u2206ABC, \u2220B = 35\u00b0, \u2220C = 65\u00b0 and the bisector of \u2220BAC meets BC in P. Arrange AP, BP and CP in descending order.<br>Solution:<br>In \u2206ABC, \u2220B = 35\u00b0, \u2220C = 65\u00b0 and AP is the bisector of \u2220BAC which meets BC in P.<br>Arrange PA, PB and PC in descending order In \u2206ABC,<br>\u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of angles of a triangle)<br><img src=\"https:\/\/farm2.staticflickr.com\/1979\/43826221430_0c2b9e9e28_o.png\" alt=\"Heron's Formula With Solutions PDF RD Sharma Class 9 Solutions\" width=\"248\" height=\"169\"><br>\u21d2 \u2220A + 35\u00b0 + 65\u00b0 = 180\u00b0<br>\u2220A + 100\u00b0= 180\u00b0<br>\u2234 \u2220A =180\u00b0- 100\u00b0 = 80\u00b0<br>\u2235 PA is a bisector of \u2220BAC<br>\u2234 \u22201 = \u22202 =&nbsp;<span id=\"MathJax-Element-15-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-107\" class=\"math\"><span id=\"MathJax-Span-108\" class=\"mrow\"><span id=\"MathJax-Span-109\" class=\"mfrac\"><span id=\"MathJax-Span-110\" class=\"msubsup\"><span id=\"MathJax-Span-111\" class=\"texatom\"><span id=\"MathJax-Span-112\" class=\"mrow\"><span id=\"MathJax-Span-113\" class=\"mn\">80<\/span><\/span><\/span><span id=\"MathJax-Span-114\" class=\"texatom\"><span id=\"MathJax-Span-115\" class=\"mrow\"><span id=\"MathJax-Span-116\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-117\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp; = 40\u00b0<br>Now in \u2206ACP, \u2220ACP &gt; \u2220CAP<br>\u21d2 \u2220C &gt; \u22202<br>\u2234 AP &gt; CP \u2026(i)<br>Similarly, in \u2206ABP,<br>\u2220BAP &gt; \u2220ABP \u21d2 \u22201 &gt; \u2220B<br>\u2234 BP &gt; AP \u2026(ii)<br>From (i) and (ii)<br>BP &gt; AP &gt; CP<\/p>\n<p>Question 6.<br>Prove that the perimeter of a triangle is greater than the sum of its altitudes<br>Solution:<br>Given: In \u2206ABC,<br>AD, BE and CF are altitudes<br><img src=\"https:\/\/farm2.staticflickr.com\/1965\/43826221340_823dce5e3a_o.png\" alt=\"RD Sharma Class 9 Chapter 12 Heron's Formula\" width=\"250\" height=\"219\"><br>To prove : AB + BC + CA &gt; AD + BC + CF<br>Proof : We know that side opposite to the greater angle is greater.<br>In \u2206ABD, \u2220D = 90\u00b0<br>\u2234 \u2220D &gt; \u2220B<br>\u2234 AB &gt;AD \u2026(i)<br>Similarly, we can prove that<br>BC &gt; BE and<br>CA &gt; CF<br>Adding we get,<br>AB + BC + CA &gt; AD + BE + CF<\/p>\n<p>Question 7.<br>In the figure, prove that:<br>(i) CD + DA + AB + BC &gt; 2AC<br>(ii) CD + DA + AB &gt; BC<br><img src=\"https:\/\/farm2.staticflickr.com\/1955\/43826221210_9ed0297410_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 12 Heron's Formula\" width=\"136\" height=\"239\"><br>Solution:<br>Given : In the figure, ABCD is a quadrilateral and AC is joined<br>To prove :<br>(i) CD + DA + AB + BC &gt; 2AC<br>(ii) CD + DA + AB &gt; BC<br>Proof:<br>(i) In \u2206ABC,<br>AB + BC &gt; AC \u2026(i)<br>(Sum of two sides of a triangle is greater than its third side)<br>Similarly in \u2206ADC,<br>CD + DA &gt; AC \u2026(ii)<br>Adding (i) and (ii)<br>CD + DA + AB + BC &gt; AC + AC<br>\u21d2 CD + DA + AB + BC &gt; 2AC<br>(ii) In \u2206ACD,<br>CD + DA &gt; CA<br>(Sum of two sides of a triangle is greater than its third side)<br>Adding AB to both sides,<br>CD + DA + AB &gt; CA + AB<br>But CA + AB &gt; BC (in \u2206ABC)<br>\u2234 CD + DA + AD &gt; BC<\/p>\n<p>Question 8.<br>Which of the following statements are true (T) and which are false (F)?<br>(i) Sum of the three sides of a triangle is less than the sum of its three altitudes.<br>(ii) Sum of any two sides of a triangle is greater than twice the median drawn to the third side.<br>(iii) Sum of any two sides of a triangle is greater than the third side.<br>(iv) Difference of any two sides of a triangle is equal to the third side.<br>(v) If two angles of a triangle are unequal, then the greater angle has the larger side opposite to it.<br>(vi) Of all the line segments that can be drawn from a point to a line not containing it, the perpendicular line segment is the shortest one.<br>Solution:<br>(i) False. The sum of the three sides of a triangle is greater than the sum of its altitudes.<br>(ii) True.<br>(iii) True.<br>(iv) False. The difference of the two sides is less than the third side.<br>(v) True.<br>(vi) True.<\/p>\n<p>Question 9.<br>Fill in the blanks to make the following statements true.<br>(i) In a right triangle, the hypotenuse is the \u2026\u2026. side.<br>(ii) The sum of three altitudes of a triangle is \u2026\u2026. than its perimeter.<br>(iii) The sum of any two sides of a triangle is \u2026\u2026.. than the third side.<br>(iv) If two angles of a triangle are unequal, then the smaller angle has the \u2026.. side opposite to it.<br>(v) Difference of any two sides of a triangle is\u2026\u2026. than the third side.<br>(vi) If two sides of a triangle are unequal, then the larger side has \u2026\u2026\u2026 angle opposite to it.<br>Solution:<br>(i) In a right triangle, the hypotenuse is the longest side.<br>(ii) The sum of three altitudes of a triangle is less than its perimeter.<br>(iii) The sum of any two sides of a triangle is greater than the third side.<br>(iv) If two angles of a triangle are unequal, then the smaller angle has the smaller side opposite to it.<br>(v) Difference of any two sides of a triangle is less than the third side.<br>(vi) If two sides of a triangle are unequal, then the larger side has a greater angle opposite to it.<\/p>\n<p>Question 10.<br>O is any point in the interior of \u2206ABC. Prove that<br>(i) AB + AC &gt; OB + OC<br>(ii) AB + BC + CA &gt; OA + OB + OC<br>(iii) OA + OB + OC &gt;&nbsp;<span id=\"MathJax-Element-16-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-118\" class=\"math\"><span id=\"MathJax-Span-119\" class=\"mrow\"><span id=\"MathJax-Span-120\" class=\"mfrac\"><span id=\"MathJax-Span-121\" class=\"mn\">1<\/span><span id=\"MathJax-Span-122\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;(AB + BC + CA)<br>Solution:<br>Given: In \u2206ABC, O is any point in the interior of the \u2206ABC, OA, OB and OC are joined<br>To prove :<br>(i) AB + AC &gt; OB + OC<br>(ii) AB + BC + CA &gt; OA + OB + OC<br>(iii) OA + OB + OC &gt;&nbsp;<span id=\"MathJax-Element-17-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-123\" class=\"math\"><span id=\"MathJax-Span-124\" class=\"mrow\"><span id=\"MathJax-Span-125\" class=\"mfrac\"><span id=\"MathJax-Span-126\" class=\"mn\">1<\/span><span id=\"MathJax-Span-127\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;(AB + BC + CA)<br>Construction: Produce BO to meet AC in D.<br><img src=\"https:\/\/farm2.staticflickr.com\/1936\/31771970738_5c37b467db_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 12 Heron's Formula\" width=\"250\" height=\"176\"><br>Proof: In \u2206ABD,<br>(i) AB + AD &gt; BD (Sum of any two sides of a triangle is greater than third)<br>\u21d2 AB + AD &gt; BO + OD \u2026(i)<br>Similarly, in \u2206ODC,<br>OD + DC &gt; OC \u2026(ii)<br>Adding (i) and (ii)<br>AB + AD + OD + DC &gt; OB + OD + OC<br>\u21d2 AB + AD + DC &gt; OB + OC<br>\u21d2 AB + AC &gt; OB + OC<br>(ii) Similarly, we can prove that<br>BC + AB &gt; OA + OC<br>and CA + BC &gt; OA + OB<br>(iii) In \u2206OAB, AOBC and \u2206OCA,<br>OA + OB &gt; AB<br>OB + OC &gt; BC<br>and OC + OA &gt; CA<br>Adding, we get<br>2(OA + OB + OC) &gt; AB + BC + CA<br>\u2234 OA + OB + OO &gt;&nbsp;<span id=\"MathJax-Element-18-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-128\" class=\"math\"><span id=\"MathJax-Span-129\" class=\"mrow\"><span id=\"MathJax-Span-130\" class=\"mfrac\"><span id=\"MathJax-Span-131\" class=\"mn\">1<\/span><span id=\"MathJax-Span-132\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;(AB + BC + CA)<\/p>\n<p>Question 11.<br>Prove that in a quadrilateral the sum of all the sides is greater than the sum of its diagonals.<br>Solution:<br>Given: In quadrilateral ABCD, AC, and BD are its diagonals,<br><img src=\"https:\/\/farm2.staticflickr.com\/1952\/44919305094_fd163b5e85_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 12 Heron's Formula\" width=\"177\" height=\"190\"><br>To prove : AB + BC + CD + DA &gt; AC + BD<br>Proof: In \u2206ABC,<br>AB + BC &gt; AC \u2026(i)<br>(Sum of any two sides of a triangle is greater than its third side)<br>Similarly, in \u2206ADC,<br>DA + CD &gt; AC \u2026(ii)<br>In \u2206ABD,<br>AB + DA &gt; BD \u2026(iii)<br>In \u2206BCD,<br>BC + CD &gt; BD \u2026(iv)<br>Adding (i), (ii), (iii), and (iv)<br>2(AB + BC + CD + DA) &gt; 2AC + 2BD<br>\u21d2 2(AB + BC + CD + DA) &gt; 2(AC + BD)<br>\u2234 AB + BC + CD + DA &gt; AC + BD<\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-9-chapter-12-heron%e2%80%99s-formula-vsaqs\"><\/span>RD Sharma Solutions Class 9 Chapter 12 Heron\u2019s Formula VSAQS<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Question 1.<br>Solution:<br>In two congruent triangles ABC and DEF, if AB = DE and BC = EF. Name the pairs of equal angles.<\/p>\n<p>In AABC and ADEF,<br>\u2206ABC \u2245 \u2206DEF<br>and AB = DE, BC = EF<br>\u2234 \u2220A = \u2220D, \u2220B = \u2220E and \u2220C = \u2220F<br>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula VSAQS \u2013 1<\/p>\n<p>Question 2.Solution:<br>In two triangles ABC and DEF, it is given that \u2220A = \u2220D, \u2220B = \u2220E and \u2220C = \u2220F. Are the two triangles necessarily congruent?<\/p>\n<p>No, as the triangles are equiangular, so similar.<\/p>\n<p>Question 3. Solution:<br>If ABC and DEF are two triangles such that AC = 2.5 cm, BC = 5 cm, \u2220C = 75\u00b0, DE = 2.5 cm, DF = 5 cm, and \u2220D = 75\u00b0. Are two triangles congruent?<\/p>\n<p>Yes, triangles are congruent (SAS axiom)<br>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula VSAQS \u2013 3<\/p>\n<p>Question 4. Solution:<br>In two triangles ABC and ADC, if AB = AD and BC = CD. Are they congruent?<\/p>\n<p>Yes, these are congruent<br>In two triangles ABC are ADC,<br>AB = AD (Given)<br>BC = CD (Given)<br>and AC = AC (Common)<br>\u2234 \u2206sABC \u2245 AADC (SSS axiom)<br>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula VSAQS \u2013 4<\/p>\n<p>Question 5.Solution:<br>In triangles ABC and CDE, if AC = CE, BC = CD, \u2220A = 60\u00b0, \u2220C \u2013 30\u00b0, and \u2220D = 90\u00b0. Are two triangles congruent?<\/p>\n<p>Yes, triangles are congruent because,<br>In \u2206ABC, and \u2206CDE,<br>AC = CE<br>BC = CD \u2220C = 30\u00b0<br>\u2234 \u2206ABC \u2245 \u2206CDE (SAS axiom)<br>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula VSAQS \u2013 5<\/p>\n<p>Question 6.Solution:<br>ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.<\/p>\n<p>Given: In \u2206ABC, AB = AC<br>BE and CF are two medians<br>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula VSAQS \u2013 6<br>To prove : BE = CF<br>Proof: In \u2206ABE and \u2206ACF.<br>AB = AC (Given)<br>\u2220A = \u2220A (Common)<br>AE = AF (Half of equal sides)<br>\u2234 \u2206ABE \u2245 \u2206ACF (SAS axiom)<br>\u2234 BE = CF (c.p.c.t.)<\/p>\n<p>Question 7.Solution:<br>Find the measure of each angle of an equilateral triangle.<\/p>\n<p>In \u2206ABC,<br>AB = AC = BC<br>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula VSAQS \u2013 7<br>\u2235 AB = AC<br>\u2234 \u2220C = \u2220B \u2026(i)<br>(Angles opposite to equal sides)<br>Similarly,<br>AC = BC<br>\u2234 \u2220B = \u2220A \u2026(ii)<br>From (i) and (ii),<br>\u2220A = \u2220B = \u2220C<br>But \u2220A + \u2220B + \u2220C = 180\u00b0<br>(Sum of angles of a triangle)<br>\u2234 \u2220A + \u2220B + \u2220C =&nbsp;<span id=\"MathJax-Element-19-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-133\" class=\"math\"><span id=\"MathJax-Span-134\" class=\"mrow\"><span id=\"MathJax-Span-135\" class=\"mfrac\"><span id=\"MathJax-Span-136\" class=\"msubsup\"><span id=\"MathJax-Span-137\" class=\"texatom\"><span id=\"MathJax-Span-138\" class=\"mrow\"><span id=\"MathJax-Span-139\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-140\" class=\"texatom\"><span id=\"MathJax-Span-141\" class=\"mrow\"><span id=\"MathJax-Span-142\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-143\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp;= 60\u00b0<\/p>\n<p>Question 8.Solution:<br>CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that \u2206ADE \u2245 \u2206BCE.<\/p>\n<p>Given : An equilateral ACDE is formed on the side of square ABCD. AE and BE are joined<br>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula VSAQS \u2013 8<br>To prove : \u2206ADE \u2245 \u2206BCE<br>Proof : In \u2206ADE and \u2206BCE,<br>AD = BC (Sides of a square)<br>DE = CE (Sides of an equilateral triangle)<br>\u2220ADE = \u2220BCE(Each = 90\u00b0 + 60\u00b0 = 150\u00b0)<br>\u2234 AADE \u2245 ABCE (SAS axiom)<\/p>\n<p>Question 9.Solution:<br>Prove that the sum of three altitude of a triangle is less than the sum of its sides.<\/p>\n<p>Given : In \u2206ABC, AD, BE and CF are the altitude of \u2206ABC<br>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula VSAQS \u2013 9<br>To prove : AD + BE + CF &lt; AB + BC + CA<br>Proof : In right \u2206ABD, \u2220D = 90\u00b0<br>Then other two angles are acute<br>\u2235 \u2220B &lt; \u2220D<br>\u2234 AD &lt; AB \u2026(i)<br>Similarly, in \u2206BEC and \u2206ABE we can prove thatBE and CF &lt; CA \u2026(iii)<br>Adding (i), (ii), (iii)<br>AD + BE -t CF &lt; AB + BC + CA<\/p>\n<p>Question 10.Solution:<br>In the figure, if AB = AC and \u2220B = \u2220C. Prove that BQ = CP.<br>RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula VSAQS \u2013 10<\/p>\n<p>Given : In the figure, AB = AC, \u2220B = \u2220C<br>To prove : BQ = CP<br>Proof : In \u2206ABQ and \u2206ACP<br>AB = AC (Given)<br>\u2220A = \u2220A (Common)<br>\u2220B = \u2220C (Given)<br>\u2234 \u2206ABQ \u2245 \u2206ACP (ASA axiom)<br>\u2234 BQ = CP (c.p.c.t.)<\/p>\n<h3><span class=\"ez-toc-section\" id=\"class-9-rd-sharma-solutions-chapter-12-heron%e2%80%99s-formula-mcqs\"><\/span>Class 9 RD Sharma Solutions Chapter 12 Heron\u2019s Formula MCQS<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Mark the correct alternative in each of the following:<br>Question 1.<br>In \u2206ABC \u2245 \u2206LKM, then side of \u2206LKM equal to side AC of \u2206ABC is<br>(a) LX<br>(b) KM<br>(c) LM<br>(d) None of these<br>Solution:<br>Side AC of \u2206ABC = LM of \u2206LKM (c)<br><img src=\"https:\/\/farm2.staticflickr.com\/1932\/44919314804_79a50068c6_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 12 Heron's Formula\" width=\"332\" height=\"129\"><\/p>\n<p>Question 2.<br>In \u2206ABC \u2245 \u2206ACB, then \u2206ABC is isosceles with<br>(a) AB=AC<br>(b) AB = BC<br>(c) AC = BC<br>(d) None of these<br>Solution:<br>\u2235 \u2206ABC \u2245 \u2206ACB<br>\u2234 AB = AC (a)<br><img src=\"https:\/\/farm2.staticflickr.com\/1975\/44919314624_9e40ac59ee_o.png\" alt=\"Heron's Formula Class 9 RD Sharma Solutions\" width=\"150\" height=\"130\"><\/p>\n<p>Question 3.<br>In \u2206ABC \u2245 \u2206PQR, then \u2206ABC is congruent to \u2206RPQ, then which of the following is not true:<br>(a) BC = PQ<br>(b) AC = PR<br>(c) AB = PQ<br>(d) QR = BC<br>Solution:<br>\u2235 \u2206ABC = \u2206PQR<br>\u2234 AB = PQ, BC = QR and AC = PR<br>\u2234 BC = PQ is not true (a)<\/p>\n<p>Question 4.<br>In triangles ABC and PQR three equality relations between some parts are as follows: AB = QP, \u2220B = \u2220P and BC = PR State which of the congruence conditions applies:<br>(a) SAS<br>(b) ASA<br>(c) SSS<br>(d) RHS<br>Solution:<br>In two triangles \u2206ABC and \u2206PQR,<br>AB = QP, \u2220B = \u2220P and BC = PR<br>The condition apply : SAS (a)<br><img src=\"https:\/\/farm2.staticflickr.com\/1930\/44919314404_cf5f72cf06_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 12 Heron's Formula\" width=\"345\" height=\"163\"><\/p>\n<p>Question 5.<br>In triangles ABC and PQR, if \u2220A = \u2220R, \u2220B = \u2220P and AB = RP, then which one of the following congruence conditions applies:<br>(a) SAS<br>(b) ASA<br>(c) SSS<br>(d) RHS<br>Solution:<br>In \u2206ABC and \u2206PQR,<br>\u2220A = \u2220R<br>\u2220B = \u2220P<br>AB = RP<br>\u2234 \u2206ABC \u2245 \u2206PQR (ASA axiom) (b)<br><img src=\"https:\/\/farm2.staticflickr.com\/1979\/44919314174_5539faa607_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 12 Heron's Formula\" width=\"321\" height=\"153\"><\/p>\n<p>Question 6.<br>If \u2206PQR \u2245 \u2206EFD, then ED =<br>(a) PQ<br>(b) QR<br>(c) PR<br>(d) None of these<br>Solution:<br>\u2235 \u2206PQR = \u2206EFD<br>\u2234 ED = PR (c)<br><img src=\"https:\/\/farm2.staticflickr.com\/1943\/44919313894_f5cd67b696_o.png\" alt=\"Class 9 Maths Chapter 12 Heron's Formula RD Sharma Solutions MCQS\" width=\"319\" height=\"131\"><\/p>\n<p>Question 7.<br>If \u2206PQR \u2245 \u2206EFD, then \u2220E =<br>(a) \u2220P<br>(b) \u2220Q<br>(c) \u2220R<br>(d) None of these<br>Solution:<br>\u2235 \u2206PQR \u2245 \u2206EFD<br>\u2234 \u2220E = \u2220P (a)<\/p>\n<p>Question 8.<br>In a \u2206ABC, if AB = AC and BC is produced to D such that \u2220ACD = 100\u00b0, then \u2220A =<br>(a) 20\u00b0<br>(b) 40\u00b0<br>(c) 60\u00b0<br>(d) 80\u00b0<br>Solution:<br>In \u2206ABC, AB = AC<br>\u2234 \u2220B = \u2220C<br>But Ext. \u2220ACD = \u2220A + \u2220B<br><img src=\"https:\/\/farm2.staticflickr.com\/1978\/44919313424_7b21477a97_o.png\" alt=\"RD Sharma Class 9 Book Chapter 12 Heron's Formula\" width=\"257\" height=\"161\"><br>\u2220ACB + \u2220ACD = 180\u00b0 (Linear pair)<br>\u2234 \u2220ACB + 100\u00b0 = 180\u00b0<br>\u21d2 \u2220ACB = 180\u00b0-100\u00b0 = 80\u00b0<br>\u2234 \u2220B = \u2220ACD = 80\u00b0<br>But \u2220A + \u2220B 4- \u2220C = 180\u00b0<br>\u2234 \u2220A + 80\u00b0 + 80\u00b0 = 180\u00b0<br>\u21d2\u2220A+ 160\u00b0= 180\u00b0<br>\u2234 \u2220A= 180\u00b0- 160\u00b0 = 20\u00b0 (a)<\/p>\n<p>Question 9.<br>In an isosceles triangle, if the vertex angle is twice the sum of the base angles, then the measure of vertex angle of the triangle is<br>(a) 100\u00b0<br>(b) 120\u00b0<br>(c) 110\u00b0<br>(d) 130\u00b0<br>Solution:<br>In \u2206ABC,<br>\u2220A = 2(\u2220B + \u2220C)<br>= 2\u2220B + 2\u2220C<br><img src=\"https:\/\/farm2.staticflickr.com\/1966\/44919313174_72b4a0c8d5_o.png\" alt=\"Heron's Formula With Solutions PDF RD Sharma Class 9 Solutions\" width=\"175\" height=\"166\"><br>Adding 2\u2220A to both sides,<br>\u2220A + 2\u2220A = 2\u2220A + 2\u2220B + 2\u2220C<br>\u21d2 3\u2220A = 2(\u2220A + \u2220B + \u2220C)<br>\u21d2 3\u2220A = 2 x 180\u00b0 (\u2235\u2220A + \u2220B + \u2220C = 180\u00b0 )<br>\u21d2 3\u2220A = 360\u00b0<br>\u21d2\u2220A =&nbsp;<span id=\"MathJax-Element-20-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-144\" class=\"math\"><span id=\"MathJax-Span-145\" class=\"mrow\"><span id=\"MathJax-Span-146\" class=\"mfrac\"><span id=\"MathJax-Span-147\" class=\"msubsup\"><span id=\"MathJax-Span-148\" class=\"texatom\"><span id=\"MathJax-Span-149\" class=\"mrow\"><span id=\"MathJax-Span-150\" class=\"mn\">360<\/span><\/span><\/span><span id=\"MathJax-Span-151\" class=\"texatom\"><span id=\"MathJax-Span-152\" class=\"mrow\"><span id=\"MathJax-Span-153\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-154\" class=\"mn\">3<\/span><\/span><\/span><\/span><\/span>&nbsp; = 120\u00b0<br>\u2234 \u2220A = 120\u00b0 (b)<\/p>\n<p>Question 10.<br>Which of the following is not a criterion for the congruence of triangles?<br>(a) SAS<br>(b) SSA<br>(c) ASA<br>(d) SSS<br>Solution:<br>SSA is not the criterion of congruence of triangles. (b)<\/p>\n<p>Question 11.<br>In the figure, the measure of \u2220B\u2019A\u2019C\u2019 is<br>(a) 50\u00b0<br>(b) 60\u00b0<br>(c) 70\u00b0<br>(d) 80\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1947\/44919313094_3524f9bfe1_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 12 Heron's Formula\" width=\"296\" height=\"188\"><br>Solution:<br>In the figure,<br>\u2206ABC \u2245 \u2206A\u2019B\u2019C\u2019<br>\u2234 \u2220A = \u2220A<br>\u21d23x = 2x- + 20<br>\u21d2 3x \u2013 2x = 20<br>\u21d2 x = 20<br>\u2220B\u2019A\u2019C\u2019 = 2x + 20 = 2 x 20 + 20<br>= 40 + 20 = 60\u00b0 (b)<\/p>\n<p>Question 12.<br>If ABC and DEF are two triangles such that \u2206ABC \u2245 \u2206FDE and AB = 5 cm, \u2220B = 40\u00b0 and \u2220A = 80\u00b0. Then, which of the following is true?<br>(a) DF = 5 cm, \u2220F = 60\u00b0<br>(b) DE = 5 cm, \u2220E = 60\u00b0<br>(c) DF = 5 cm, \u2220E = 60\u00b0<br>(d) DE = 5 cm, \u2220D = 40\u00b0<br>Solution:<br>\u2235 \u2206ABC \u2245 \u2206FDE,<br>AB = 5 cm, \u2220A = 80\u00b0, \u2220B = 40\u00b0<br>\u2234 DF = 5 cm, \u2220F = 80\u00b0, \u2220D = 40\u00b0<br>\u2234 \u2220C =180\u00b0- (80\u00b0 + 40\u00b0) = 180\u00b0 \u2013 120\u00b0 = 60\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1965\/44919312834_765b32c112_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 12 Heron's Formula\" width=\"350\" height=\"150\"><br>\u2234 \u2220E = \u2220C = 60\u00b0<br>\u2234 DF = 5 cm, \u2220E = 60\u00b0 (c)<\/p>\n<p>Question 13.<br>In the figure, AB \u22a5 BE and FE \u22a5 BE. If BC = DE and AB = EF, then \u2206ABD is congruent to<br>(a) \u2206EFC<br>(b) \u2206ECF<br>(c) \u2206CEF<br>(d) \u2206FEC<br><img src=\"https:\/\/farm2.staticflickr.com\/1911\/44919312604_dbcc508b35_o.png\" alt=\"Solution Of Rd Sharma Class 9 Chapter 12 Heron's Formula\" width=\"227\" height=\"146\"><br>Solution:<br>In the figure, AB \u22a5 BE, FE \u22a5 BE<br>BC = DE, AB = EF,<br>then CD + BC = CD + DE BD = CE<br>In \u2206ABD and \u2206CEF,<br>BD = CE (Prove)<br>AB = FE (Given)<br>\u2220B = \u2220E (Each 90\u00b0)<br>\u2234 \u2206ABD \u2245 \u2206FCE (b)<\/p>\n<p>Question 14.<br>In the figure, if AE || DC and AB = AC, the value of \u2220ABD is<br>(a) 70\u00b0<br>(b) 110\u00b0<br>(c) 120\u00b0<br>(d) 130\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1928\/44919312374_616030b1bc_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 12 Heron's Formula\" width=\"271\" height=\"210\"><br>Solution:<br>In the figure, AE || DC<br>\u2234 \u22201 = 70\u00b0 (Vertically opposite angles)<br>\u2234 \u22201 = \u22202 (Alternate angles)<br>\u22202 = \u2220ABC (Base angles of isosceles triangle)<br><img src=\"https:\/\/farm2.staticflickr.com\/1964\/44919312144_28d4106b1a_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 12 Heron's Formula\" width=\"270\" height=\"226\"><br>\u2234 ABC = 90\u00b0<br>But \u2220ABC + \u2220ABD = 180\u00b0 (Linear pair)<br>\u21d2 70\u00b0 +\u2220ABD = 180\u00b0<br>\u21d2\u2220ABD = 180\u00b0-70\u00b0= 110\u00b0<br>\u2234 \u2220ABD =110\u00b0 (b)<\/p>\n<p>Question 15.<br>In the figure, ABC is an isosceles triangle whose side AC is produced to E. Through C, CD is drawn parallel to BA. The value of x is<br>(a) 52\u00b0<br>(b) 76\u00b0<br>(c) 156\u00b0<br>(d) 104\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1976\/44919311914_318a3b4d8e_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 12 Heron's Formula\" width=\"255\" height=\"231\"><br>Solution:<br>In \u2206ABC, AB = AC<br>AC is produced to E<br><img src=\"https:\/\/farm2.staticflickr.com\/1970\/44919311674_7e65bd86ea_o.png\" alt=\"RD Sharma Class 9 Chapter 12 Heron's Formula\" width=\"247\" height=\"240\"><br>CD || BA is drawn<br>\u2220ABC = 52\u00b0<br>\u2234 \u2220ACB = 52\u00b0 (\u2235 AB = AC)<br>\u2234 \u2220BAC = 180\u00b0-(52\u00b0 +52\u00b0)<br>= 180\u00b0-104\u00b0 = 76\u00b0<br>\u2235 AB || CD<br>\u2234 \u2220ACD = \u2220BAC (Alternate angles)<br>= 76\u00b0<br>and \u2220BCE + \u2220DCB = 180\u00b0 (Linear pair)<br>\u2220BCE + 52\u00b0 = 180\u00b0<br>\u21d2\u2220BCE = 180\u00b0-52\u00b0= 128\u00b0<br>\u2220x + \u2220ACD = 380\u00b0<br>\u21d2 x + 76\u00b0 = 180\u00b0<br>\u2234 x= 180\u00b0-76\u00b0= 104\u00b0 (d)<\/p>\n<p>Question 16.<br>In the figure, if AC is the bisector of \u2220BAD such that AB = 3 cm and AC = 5 cm, then CD =<br>(a) 2 cm<br>(b) 3 cm<br>(c) 4 cm<br>(d) 5 cm<br><img src=\"https:\/\/farm2.staticflickr.com\/1942\/44919311484_27c7e28fc3_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 12 Heron's Formula\" width=\"200\" height=\"160\"><br>Solution:<br>In the figure, AC is the bisector of \u2220BAD, AB = 3 cm, AC = 5 cm<br><img src=\"https:\/\/farm2.staticflickr.com\/1912\/44919311314_c68203585d_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 12 Heron's Formula\" width=\"193\" height=\"159\"><br>In \u2206ABC and \u2206ADC,<br>AC = AC (Common)<br>\u2220B = \u2220D (Each 90\u00b0)<br>\u2220BAC = \u2220DAC (\u2235 AC is the bisector of \u2220A)<br>\u2234 \u2206ABC \u2245 \u2206ADC (AAS axiom)<br>\u2234 BC = CD and AB = AD (c.p.c.t.)<br>Now in the right \u2206ABC,<br>AC<sup>2<\/sup>&nbsp;= AB<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><br>\u21d2 (5)<sup>2<\/sup>&nbsp;= (3)<sup>2<\/sup>&nbsp;+ BC<sup>2<\/sup><br>\u21d225 = 9 + BC<sup>2<\/sup><br>\u21d2 BC<sup>2<\/sup>&nbsp;= 25 \u2013 9 = 16 = (4)<sup>2<\/sup><br>\u2234 BC = 4 cm<br>But CD = BC<br>\u2234 CD = 4 cm (c)<\/p>\n<p>Question 17.<br>D, E, F are the mid-point of the sides BC, CA, and AB respectively of \u2206ABC. Then \u2206DEF is congruent to the triangle<br>(a) ABC<br>(b) AEF<br>(c) BFD, CDE<br>(d) AFE, BFD, CDE<br>Solution:<br>In \u2206ABC, D, E, F are the mid-points of the sides BC, CA, AB respectively<br>DE, EF and FD are joined<br><img src=\"https:\/\/farm2.staticflickr.com\/1934\/44919311044_8284d37e41_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 12 Heron's Formula\" width=\"259\" height=\"185\"><br>\u2235 E and F are the mid-points<br>AC and AB,<br>\u2234 EF =&nbsp;<span id=\"MathJax-Element-21-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-155\" class=\"math\"><span id=\"MathJax-Span-156\" class=\"mrow\"><span id=\"MathJax-Span-157\" class=\"mfrac\"><span id=\"MathJax-Span-158\" class=\"mn\">1<\/span><span id=\"MathJax-Span-159\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;BC and EF || BC<br>Similarly,<br>DE =&nbsp;<span id=\"MathJax-Element-22-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-160\" class=\"math\"><span id=\"MathJax-Span-161\" class=\"mrow\"><span id=\"MathJax-Span-162\" class=\"mfrac\"><span id=\"MathJax-Span-163\" class=\"mn\">1<\/span><span id=\"MathJax-Span-164\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;AB and DE || AB<br>DF =&nbsp;<span id=\"MathJax-Element-23-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-165\" class=\"math\"><span id=\"MathJax-Span-166\" class=\"mrow\"><span id=\"MathJax-Span-167\" class=\"mfrac\"><span id=\"MathJax-Span-168\" class=\"mn\">1<\/span><span id=\"MathJax-Span-169\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;AC and DF || AC<br>\u2234 \u2206DEF is congruent to each of the triangles so formed<br>\u2234 \u2206DEF is congruent to triangle AFE, BFD, CDE (d)<\/p>\n<p>Question 18.<br>ABC is an isosceles triangle such that AB = AC and AD is the median to base BC. Then, \u2220BAD =<br>(a) 55\u00b0<br>(b) 70\u00b0<br>(c) 35\u00b0<br>(d) 110\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1919\/44919310844_8ed1b382db_o.png\" alt=\"Heron's Formula Class 9 RD Sharma Solutions\" width=\"266\" height=\"154\"><br>Solution:<br>In \u2206ABC, AB = AC<br>AD is median to BC<br><img src=\"https:\/\/farm2.staticflickr.com\/1928\/44919310614_5e6b534f45_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 12 Heron's Formula\" width=\"265\" height=\"154\"><br>\u2234 BD = DC<br>In \u2206ADB, \u2220D = 90\u00b0, \u2220B = 35\u00b0<br>But \u2220B + BAD + \u2220D = 180\u00b0 (Sum of angles of a triangle)<br>\u21d2 35\u00b0 + \u2220BAD + 90\u00b0 = 180\u00b0<br>\u21d2\u2220BAD + 125\u00b0= 180\u00b0<br>\u21d2 \u2220BAD = 180\u00b0- 125\u00b0<br>\u21d2\u2220BAD = 55\u00b0 (a)<\/p>\n<p>Question 19.<br>In the figure, X is a point in the interior of square ABCD. AXYZ is also a square. If DY = 3 cm and AZ = 2 cm, then BY =<br>(a) 5 cm<br>(b) 6 cm<br>(c) 7 cm<br>(d) 8 cm<br><img src=\"https:\/\/farm2.staticflickr.com\/1906\/44919310454_cf39e76d03_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 12 Heron's Formula\" width=\"278\" height=\"238\"><br>Solution:<br>In the figure, ABCD and AXYZ are squares<br>DY = 3 cm, AZ = 2 cm<br><img src=\"https:\/\/farm2.staticflickr.com\/1967\/44919310264_e100c590fc_o.png\" alt=\"Class 9 Maths Chapter 12 Heron's Formula RD Sharma Solutions\" width=\"288\" height=\"246\"><br>DZ = DY + YZ<br>= DY + Z = 3 + 2 = 5 cm<br>In \u2206ADZ, \u22202 = 90\u00b0<br>AD<sup>2<\/sup>&nbsp;+ AZ<sup>2<\/sup>&nbsp;+ DZ<sup>2<\/sup>&nbsp;= 2<sup>2<\/sup>&nbsp;+ 5<sup>2<\/sup>&nbsp;cm<br>= 4 + 25 = 29<br>In \u2220ABX, \u2220X = 90\u00b0<br>AB<sup>2<\/sup>&nbsp;= AX<sup>2<\/sup>&nbsp;+ BX<sup>2<\/sup><br>AD<sup>2<\/sup>&nbsp;= AZ<sup>2<\/sup>&nbsp;+ BX<sup>2<\/sup><br>(\u2235 AB = AD, AX = AZ sides of square)<br>29 = 2<sup>2<\/sup>&nbsp;+ BX<sup>2<\/sup><br>\u21d2 29 = 4 + BX<sup>2<\/sup><br>\u21d2 BX<sup>2<\/sup>&nbsp;= 29 \u2013 4 = 25 = (5)<sup>2<\/sup><br>\u2234 BX = 5 cm (a)<\/p>\n<p>Question 20.<br>In the figure, ABC is a triangle in which \u2220B = 2\u2220C. D is a point on side BC such that AD bisects \u2220BAC and AB = CD. BE is the bisector of \u2220B. The measure of \u2220BAC is<br>(a) 72\u00b0<br>(b) 73\u00b0<br>(c) 74\u00b0<br>(d) 95\u00b0<br>Solution:<br>In the figure, \u2220B = 2\u2220C, AD and BE are the bisectors of \u2220A and \u2220B respectively,<br>AB = CD<br><img src=\"https:\/\/farm2.staticflickr.com\/1963\/45644284461_f20fb36311_o.png\" alt=\"RD Sharma Class 9 Book Chapter 12 Heron's Formula\" width=\"312\" height=\"390\"><\/p>\n<h2><span class=\"ez-toc-section\" id=\"detailed-exercise-wise-explanation-with-a-listing-of-important-topics-in-the-exercise\"><\/span><strong>Detailed Exercise-wise Explanation with a Listing of Important Topics in the Exercise<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-solutions-chapter-12-ex-12a\"><\/span>RD Sharma&nbsp; Class 9 Solutions Chapter 12 Ex 12A<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">In the <\/span>RD Sharma Class 9 Chapter 12 Exercise 12A Solutions <span style=\"font-weight: 400;\">you will be exposed to all kinds of tricky problems from Herons Formula. Try to attempt all the questions from <\/span>RD Sharma Class 9 Chapter 12A <span style=\"font-weight: 400;\">without seeing the answers from Chapter 12 solutions. Assess your performance and make a note of it. <\/span><\/p>\n<p><span style=\"font-weight: 400;\">After some time you check all the answers from <\/span>RD Sharma Solutions Class 9 Chapter 10A. <span style=\"font-weight: 400;\">If you make a habit you will get the maximum benefit in your final Maths exam. While focusing on the problems based on <\/span><span style=\"font-weight: 400;\">Herons Formula <\/span><span style=\"font-weight: 400;\">from <\/span>RD Sharma Class 9 Solutions Chapter 10 Exercise 10A <span style=\"font-weight: 400;\">look for the ways the concepts have been presented. <\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-solutions-chapter-12-ex-12b\"><\/span>RD Sharma&nbsp; Class 9 Solutions Chapter 12 Ex 12B<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">In <\/span>Chapter 12B RD Sharma Class 9 <span style=\"font-weight: 400;\">the<\/span> <span style=\"font-weight: 400;\">students will be able to examine their performances by attempting <\/span><span style=\"font-weight: 400;\">objective-type questions that are associated with the Herons Formula. After going through the solutions of RD Sharma Class 9 Chapter 12B the students will be able to take risks fearlessly. <\/span><\/p>\n<p><span style=\"font-weight: 400;\">The preparation style of the students will undergo a lot of changes which will be beneficial for their performances in the final Maths Exam. Hence, you must buy a copy for yourself right now.&nbsp;<\/span><span style=\"font-weight: 400;\">In the final exercise of RD Sharma Solutions, students will get to know about the techniques which can be applied in the case of Herons Formula. <\/span><\/p>\n<h4><span class=\"ez-toc-section\" id=\"important-concepts\"><\/span><strong>Important concepts<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h4>\n<ul>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Heron\u2019s Formula Introduction<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Applications of Heron\u2019s Formula<\/span><\/li>\n<\/ul>\n<p>This is the complete blog on RD Sharma Solutions Class 9 Maths Chapter 12. 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.&nbsp;<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-9-maths-chapter-12-%e2%80%93-herons-formula\"><\/span><strong>FAQs on RD Sharma Solutions Class 9 Maths Chapter 12 &#8211; Herons Formula<\/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-1630748446745\" 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-solutions-class-9-maths-chapter-12\"><\/span>From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 12?<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-1630748481443\" 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-solutions-for-class-9-maths-chapter-12\"><\/span>How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 12?<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-1630748513337\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rd-sharma-solutions-for-class-9-maths-chapter-12-pdf-offline\"><\/span>Can I access the RD Sharma Solutions for Class 9 Maths Chapter 12\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 Solutions Class 9 Maths Chapter 12 Herons Formula: How do you manage to stay focused for the final Class 9 Maths Exam? What are you doing to stay updated in terms of preparation? There are many chapters that are full of complex problems. So necessary suggestions must be taken into consideration at the &#8230; <a title=\"RD Sharma Solutions Class 9 Maths Chapter 12 &#8211; Herons Formula (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-12-herons-formula\/\" aria-label=\"More on RD Sharma Solutions Class 9 Maths Chapter 12 &#8211; Herons Formula (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":124465,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3081,3037,3086],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/62220"}],"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=62220"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/62220\/revisions"}],"predecessor-version":[{"id":514536,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/62220\/revisions\/514536"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/124465"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=62220"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=62220"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=62220"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}