{"id":73468,"date":"2021-02-05T22:29:15","date_gmt":"2021-02-05T16:59:15","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=73468"},"modified":"2021-02-05T22:29:23","modified_gmt":"2021-02-05T16:59:23","slug":"rd-sharma-chapter-12-class-9-maths-exercise-12-2-solutions","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-12-class-9-maths-exercise-12-2-solutions\/","title":{"rendered":"RD Sharma Chapter 12 Class 9 Maths Exercise 12.2 Solutions"},"content":{"rendered":"\n<p>RD Sharma Chapter 12 Class 9 Maths Exercise 12.2 Solutions will help students to learn about the applications of Heron\u2019s Formula. Students will learn about the applications of the Heron\u2019s Formula. For Example- If a landowner wants to find out the land area, which is in a quadrilateral shape. It divides the quadrilateral into triangular parts and uses Heron\u2019s formula for the area of a triangular section. From this chapter, learn about how to apply Heron\u2019s formula.<\/p>\n<p>Although, with the RD Sharma Chapter 12 Class 9 Maths Exercise 12.2 Solutions PDF, students can do practice with various questions. The PDF is prepared by ou experts through RD Sharma, Previous Year\u2019s Question Paper, and CBSE Text Book of Class 9. Go through the article and know more information about Heron\u2019s Formula with stepwise definitions and examples.<\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-12-herons-formula\/\" target=\"_blank\" rel=\"noopener noreferrer\"><strong>Learn about RD Sharma Chapter 12 (Heron\u2019s Formula)<\/strong><\/a><\/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-69dc7213b6b89\" 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-69dc7213b6b89\"><\/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-chapter-12-class-9-maths-exercise-12-2-solutions\/#download-rd-sharma-chapter-12-class-9-maths-exercise-122-solutions-pdf\" title=\"Download RD Sharma Chapter 12 Class 9 Maths Exercise 12.2 Solutions PDF\">Download RD Sharma Chapter 12 Class 9 Maths Exercise 12.2 Solutions 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-chapter-12-class-9-maths-exercise-12-2-solutions\/#important-definitions-rd-sharma-chapter-12-class-9-maths-exercise-122-solutions\" title=\"Important Definitions RD Sharma Chapter 12 Class 9 Maths Exercise 12.2 Solutions\">Important Definitions RD Sharma Chapter 12 Class 9 Maths Exercise 12.2 Solutions<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-12-class-9-maths-exercise-12-2-solutions\/#heron%e2%80%99s-theorem\" title=\"Heron\u2019s Theorem\">Heron\u2019s Theorem<\/a><ul class='ez-toc-list-level-4'><li class='ez-toc-heading-level-4'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-12-class-9-maths-exercise-12-2-solutions\/#examples\" title=\"Examples\">Examples<\/a><\/li><\/ul><\/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-chapter-12-class-9-maths-exercise-12-2-solutions\/#heron%e2%80%99s-formula-for-quadrilateral\" title=\"Heron\u2019s Formula for Quadrilateral\">Heron\u2019s Formula for Quadrilateral<\/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-chapter-12-class-9-maths-exercise-12-2-solutions\/#heron%e2%80%99s-formula-for-equilateral-triangle\" title=\"Heron\u2019s Formula for Equilateral Triangle\">Heron\u2019s Formula for Equilateral Triangle<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-12-class-9-maths-exercise-12-2-solutions\/#frequently-asked-questions-faqs-of-rd-sharma-chapter-12-class-9-maths-exercise-122-solutions\" title=\"Frequently Asked Questions (FAQs) of RD Sharma Chapter 12 Class 9 Maths Exercise 12.2 Solutions\">Frequently Asked Questions (FAQs) of RD Sharma Chapter 12 Class 9 Maths Exercise 12.2 Solutions<\/a><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-chapter-12-class-9-maths-exercise-122-solutions-pdf\"><\/span>Download RD Sharma Chapter 12 Class 9 Maths Exercise 12.2 Solutions PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\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\/02\/Solutions-for-Class-9-Maths-Chapter-12-Herons-Formula-Exercise-12.2.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/02\/Solutions-for-Class-9-Maths-Chapter-12-Herons-Formula-Exercise-12.2.pdf\">Solutions for Class 9 Maths Chapter 12 Heron&#8217;s Formula Exercise 12.2<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"important-definitions-rd-sharma-chapter-12-class-9-maths-exercise-122-solutions\"><\/span>Important Definitions RD Sharma Chapter 12 Class 9 Maths Exercise 12.2 Solutions<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"heron%e2%80%99s-theorem\"><\/span><strong>Heron\u2019s Theorem<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Heron\u2019s Theorem was named after Heron of Alexandria, who has discovered the area of triangles can be measured in terms of the length of sides of a triangle. By obtaining Heron\u2019s formula, we can determine the area of a triangle without measuring the angles or another distance. This formula is known for its easy calculation based on the length of three sides of a triangle.<\/p>\n<p>Let\u2019s say, if the length of the three sides are P, Q, and R, then its semi-perimeter is-<\/p>\n<p>S= (P + Q + R)\/ 2<\/p>\n<p>Thus, the area will be-<\/p>\n<p>Area = \u221a[S (S &#8211; P) (S &#8211; Q) (S &#8211; R)]<\/p>\n<h4><span class=\"ez-toc-section\" id=\"examples\"><\/span><strong>Examples<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h4>\n<p><strong>Ques:<\/strong> A triangle XYZ has sides 4cm, 13cm, and 15cm. Find the area of the triangle.<\/p>\n<p>Solution- Semiperimeter of triangle XYZ, s = (4+ 13+ 15)\/ 2 = 32\/ 2 = 16<\/p>\n<p>By Heron\u2019s Formula, we know-<\/p>\n<p>Area (A) = \u221a[s (s &#8211; a) (s &#8211; b) (s &#8211; c)]<\/p>\n<p>Hence, A = \u221a[16 (16 &#8211; 4) (16 &#8211; 13) (16 &#8211; 15)] = \u221a(16 x 12 x 3 x 1) = \u221a576 = 24 square cm<\/p>\n<h3><span class=\"ez-toc-section\" id=\"heron%e2%80%99s-formula-for-quadrilateral\"><\/span><strong>Heron\u2019s Formula for Quadrilateral<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>In the following points, we will learn how to find the area of quadrilateral using Heron\u2019s formula\u2019-<\/p>\n<p>If PQRS is a quadrilateral, where PQ||RS and PR &amp; QS are the diagonals. (ABCD)<\/p>\n<p>PR divides the quadrilateral PQRS into two triangles PSR and PQR.<\/p>\n<p>Now we have two triangles here.<\/p>\n<p>Areas of quadrilateral PQRS = Area of \u2206PSR + Area of \u2206PQR<\/p>\n<p>If we know the lengths of all sides of the quadrilateral and the length of diagonal PR, we can apply Heron\u2019s formula to get the total area.<\/p>\n<p>Hence, we will first find the area of \u2206PSR and area of \u2206PQR applying Heron\u2019s formula and will add them to obtain the final value.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"heron%e2%80%99s-formula-for-equilateral-triangle\"><\/span><strong>Heron\u2019s Formula for Equilateral Triangle<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>We know that the equilateral triangle has all equal sides. To find the area of the equilateral triangle, first, we have to find the semi perimeter of the equilateral triangle-<\/p>\n<p>s = (a+ a+ a)\/ 2<\/p>\n<p>s= 3a\/ 2<\/p>\n<p>where \u2018a\u2019 is the length of the side.<\/p>\n<p>Now, apply Heron\u2019s formula, we know-<\/p>\n<p>Area (A) = \u221as(s \u2212 a) (s \u2212 b) (s \u2212 c)<\/p>\n<p>As, a = b = c<\/p>\n<p>Therefore,<\/p>\n<p>Area (A) = \u221a[s(s-a)3] (required formula).<\/p>\n<h2><span class=\"ez-toc-section\" id=\"frequently-asked-questions-faqs-of-rd-sharma-chapter-12-class-9-maths-exercise-122-solutions\"><\/span>Frequently Asked Questions (FAQs) of RD Sharma Chapter 12 Class 9 Maths Exercise 12.2 Solutions<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Ques 1- When we apply Heron\u2019s formula?<\/p>\n<p>Ans- Heron\u2019s formula is applied to find the area of a triangle when all the three side-lengths of a triangle are known to us.<\/p>\n<p>Ques 2- What is Heron\u2019s formula for the equilateral triangle?<\/p>\n<p>Ans- An equilateral triangle has all three sides similar, so Heron\u2019s formula to obtain the area of the equilateral triangle is-<\/p>\n<p>Area (A) = \u221a[s(s-a)3]<\/p>\n<p>Ques 3- Is Heron\u2019s Formula accurate?<\/p>\n<p>Ans- Heron\u2019s formula calculates the area of a triangle if the length of each side is given. If it is a very thin triangle, one where two sides are approx equal and the third side is much smaller, a direct implementation of Heron\u2019s formula may not be accurate.<\/p>\n<p><a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener noreferrer\"><strong>Know about CBSE Board<\/strong><\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>RD Sharma Chapter 12 Class 9 Maths Exercise 12.2 Solutions will help students to learn about the applications of Heron\u2019s Formula. Students will learn about the applications of the Heron\u2019s Formula. For Example- If a landowner wants to find out the land area, which is in a quadrilateral shape. It divides the quadrilateral into triangular &#8230; <a title=\"RD Sharma Chapter 12 Class 9 Maths Exercise 12.2 Solutions\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-12-class-9-maths-exercise-12-2-solutions\/\" aria-label=\"More on RD Sharma Chapter 12 Class 9 Maths Exercise 12.2 Solutions\">Read more<\/a><\/p>\n","protected":false},"author":241,"featured_media":73474,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[2985,2917,73411,73410],"tags":[3081,3086,3085],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/73468"}],"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\/241"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=73468"}],"version-history":[{"count":2,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/73468\/revisions"}],"predecessor-version":[{"id":73798,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/73468\/revisions\/73798"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/73474"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=73468"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=73468"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=73468"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}