{"id":72438,"date":"2021-02-08T13:27:37","date_gmt":"2021-02-08T07:57:37","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=72438"},"modified":"2021-02-08T13:27:40","modified_gmt":"2021-02-08T07:57:40","slug":"rd-sharma-chapter-17-class-9-maths-exercise-17-2-solutions","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-17-class-9-maths-exercise-17-2-solutions\/","title":{"rendered":"RD Sharma Chapter 17 Class 9 Maths Exercise 17.2 Solutions"},"content":{"rendered":"\n<p><span style=\"font-weight: 400;\">RD Sharma Chapter 17 Class 9 Maths Exercise 17.2 Solutions has been provided here. Primarily, the exercise revolves around constructing a bisector for the provided angle. Moreover, the questions covered fall thor<\/span><span style=\"font-weight: 400;\">oughly as per the syllabus and norms of the Central Board of Secondary Education.\u00a0<\/span><\/p>\n<p><span style=\"font-weight: 400;\">RD Sharma Chapter 17 Class 9 Maths Exercise 17.2 Solutions is useful for teaching students about the modes of constructing angles through only compass and ruler.<\/span><\/p>\n<p><a href=\"https:\/\/byjus.com\/rd-sharma-solutions\/class-9-maths-chapter-17-constructions\/\" target=\"_blank\" rel=\"noopener noreferrer\"><strong>Learn about RD Sharma Chapter 17 (Constructions) Class 9<\/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-69d3d81baa7b4\" 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-69d3d81baa7b4\"><\/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-17-class-9-maths-exercise-17-2-solutions\/#download-rd-sharma-chapter-17-class-9-maths-exercise-172-solutions-pdf\" title=\"Download RD Sharma Chapter 17 Class 9 Maths Exercise 17.2 Solutions PDF\">Download RD Sharma Chapter 17 Class 9 Maths Exercise 17.2 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-17-class-9-maths-exercise-17-2-solutions\/#important-definition-rd-sharma-chapter-17-class-9-maths-exercise-172-solutions\" title=\"Important Definition RD Sharma Chapter 17 Class 9 Maths Exercise 17.2 Solutions\">Important Definition RD Sharma Chapter 17 Class 9 Maths Exercise 17.2 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-17-class-9-maths-exercise-17-2-solutions\/#construction-of-the-bisector-of-an-angle\" title=\"Construction of the Bisector of an Angle\">Construction of the Bisector of an Angle<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-17-class-9-maths-exercise-17-2-solutions\/#construction-of-some-standard-angles\" title=\"Construction of some Standard Angles\">Construction of some Standard Angles<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-17-class-9-maths-exercise-17-2-solutions\/#frequently-asked-questions-faqs-of-rd-sharma-chapter-17-class-9-maths-exercise-172-solutions\" title=\"Frequently Asked Questions (FAQs) of RD Sharma Chapter 17 Class 9 Maths Exercise 17.2 Solutions\">Frequently Asked Questions (FAQs) of RD Sharma Chapter 17 Class 9 Maths Exercise 17.2 Solutions<\/a><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-chapter-17-class-9-maths-exercise-172-solutions-pdf\"><\/span><span style=\"font-weight: 400;\">Download RD Sharma Chapter 17 Class 9 Maths Exercise 17.2 Solutions PDF<\/span><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-17-Construction-Exercise-17.2.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/02\/Solutions-for-Class-9-Maths-Chapter-17-Construction-Exercise-17.2.pdf\">Solutions for Class 9 Maths Chapter 17 Construction Exercise 17.2<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"important-definition-rd-sharma-chapter-17-class-9-maths-exercise-172-solutions\"><\/span><span style=\"font-weight: 400;\">Important Definition RD Sharma Chapter 17 Class 9 Maths Exercise 17.2 Solutions<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><span style=\"font-weight: 400;\">We will learn about RD Sharma Chapter 17 Class 9 Maths Exercise 17.2 Solutions based on the following points-<\/span><\/p>\n<ol>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Construction of the Bisector of an Angle<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Construction of some Standard Angles \u2013 30\u00b0, 60\u00b0, 90\u00b0, 45\u00b0 and 120\u00b0<\/span><\/li>\n<\/ol>\n<p><span style=\"font-weight: 400;\">Students are asked to construct the bisector of an angle, like 30\u00b0, 60\u00b0, 90\u00b0, 45\u00b0, and 120\u00b0.<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"construction-of-the-bisector-of-an-angle\"><\/span><strong>Construction of the Bisector of an Angle<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">Bisection of an angle refers to dividing the angle into two congruent parts, or simply two equal parts, without measuring the angle.\u00a0<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"construction-of-some-standard-angles\"><\/span><strong>Construction of some Standard Angles<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">Construction of Standard angles like 30\u00b0, 60\u00b0, 90\u00b0, 45\u00b0, and 120\u00b0 have to be done using a compass and ruler where the angle measurement is given.<\/span><\/p>\n<p><strong>Ques 1: Draw an angle and name it as \u2220BAC. Construct different angle, equal to \u2220BAC.<\/strong><\/p>\n<p><strong>Solution-<\/strong><\/p>\n<p><span style=\"font-weight: 400;\">Steps of construction-<\/span><\/p>\n<ol>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Draw any angle ABC. (Now will create an angle similar to \u2220BAC).<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Draw a line section QR.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Draw an arc that crosses \u2220BAC at E and D using A as a center and take any radius.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">With the equal measurements (set in step 2), Draw an arc from point Q.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">With &#8216;S&#8217; as center and radius equal to DE, draw an arc that crosses the previous arc at T.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Join Q and T.<\/span><\/li>\n<\/ol>\n<p><span style=\"font-weight: 400;\">Therefore \u2220PQR= \u2220BAC<\/span><\/p>\n<p><strong>Ques- Draw an obtuse angle. Divide it. Measure each of the angles so formed.<\/strong><\/p>\n<p><strong>Solution-<\/strong><\/p>\n<p><span style=\"font-weight: 400;\">Steps of construction:<\/span><\/p>\n<ol>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Draw an obtuse angle. We take \u2220ABC = 120\u00b0.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Draw an arc that crosses AB at P and BC at Q, from center B, and take any radius.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Draw an arc from spot P by setting a radius of more than half of PQ.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Copy step 3 using Q as a center and cut the previous arc at R.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Join BR.<\/span><\/li>\n<\/ol>\n<p><span style=\"font-weight: 400;\">Therefore \u2220ABR= \u2220RBC = 60\u00b0<\/span><\/p>\n<p><strong>Ques- Utilizing your compass (protractor), draw an angle of 108\u00b0. With this given detail of an angle, draw an angle of 54\u00b0.<\/strong><\/p>\n<p><strong>Solution-<\/strong><\/p>\n<p><span style=\"font-weight: 400;\">Steps of construction:<\/span><\/p>\n<ol>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Draw \u2220ABC = 108\u00b0.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Draw an arc that crosses AB at P and BC at Q from point B. (Take any radius)<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Draw an arc from spot P by setting a radius of more than half of PQ.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Copy Step 3 using Q as the center and touch the previous arc at R.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Join BR.<\/span><\/li>\n<\/ol>\n<p><span style=\"font-weight: 400;\">Therefore \u2220RBC = 54\u00b0<\/span><\/p>\n<p><strong>Ques- Utilizing the protractor, draw a right angle. Divide it to make an angle of measure 45\u00b0.<\/strong><\/p>\n<p><strong>Solution- <\/strong><span style=\"font-weight: 400;\">Steps of construction:<\/span><\/p>\n<ol>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Draw \u2220ABC = 90\u00b0.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Draw an arc that crosses AB at P and BC at Q from point B. (Take any radius)<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Draw an arc from spot P by setting a radius of more than half of PQ.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Copy step 3 using Q as a center and touch the previous arc at R.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Join RB.<\/span><\/li>\n<\/ol>\n<p><span style=\"font-weight: 400;\">Therefore \u2220RBC= 45\u00b0<\/span><\/p>\n<h2><span class=\"ez-toc-section\" id=\"frequently-asked-questions-faqs-of-rd-sharma-chapter-17-class-9-maths-exercise-172-solutions\"><\/span><span style=\"font-weight: 400;\">Frequently Asked Questions (FAQs) of RD Sharma Chapter 17 Class 9 Maths Exercise 17.2 Solutions<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><span style=\"font-weight: 400;\">Ques 1- Why does angle bisector construction work?<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Ans- To bisect an angle means that we have to divide the angle into two equal (congruent) parts without measuring the angle. This Euclidean construction works by making two congruent (equal) triangles.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Ques 2- Which tool is used to construct an angle bisector?<\/span><\/p>\n<p><span style=\"font-weight: 400;\">Ans- It concerns the construction of the angle equal to one-third of the given arbitrary angle, using only two tools- an unmarked straightedge and a protractor (compass).<\/span><\/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 17 Class 9 Maths Exercise 17.2 Solutions has been provided here. Primarily, the exercise revolves around constructing a bisector for the provided angle. Moreover, the questions covered fall thoroughly as per the syllabus and norms of the Central Board of Secondary Education.\u00a0 RD Sharma Chapter 17 Class 9 Maths Exercise 17.2 Solutions &#8230; <a title=\"RD Sharma Chapter 17 Class 9 Maths Exercise 17.2 Solutions\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-17-class-9-maths-exercise-17-2-solutions\/\" aria-label=\"More on RD Sharma Chapter 17 Class 9 Maths Exercise 17.2 Solutions\">Read more<\/a><\/p>\n","protected":false},"author":241,"featured_media":72440,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[1,2917,73719,73411,2985,73410],"tags":[3081,3086,3085],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/72438"}],"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=72438"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/72438\/revisions"}],"predecessor-version":[{"id":73957,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/72438\/revisions\/73957"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/72440"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=72438"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=72438"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=72438"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}