{"id":72551,"date":"2021-02-08T13:27:44","date_gmt":"2021-02-08T07:57:44","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=72551"},"modified":"2021-02-08T13:27:47","modified_gmt":"2021-02-08T07:57:47","slug":"rd-sharma-chapter-17-class-9-maths-exercise-17-3-solutions","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-17-class-9-maths-exercise-17-3-solutions\/","title":{"rendered":"RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 Solutions"},"content":{"rendered":"\n<p><span style=\"font-weight: 400;\">RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 Solutions is based on the concepts of the Constructions of different types of triangles. In this article, we will discuss- Constructions of Triangles, Equilateral Triangles, triangle when its base, total of other two edges, and one base angle are given, etc. The PDF attached below will help students know about several types of questions based on the RD Sharma exercise.<\/span><\/p>\n<p><span style=\"font-weight: 400;\">RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 Solutions clarifies the construction of various kinds of triangles. Going through these solutions and practicing the same, a student will score well in the exam. As the solutions are given in a stepwise manner, any student can understand them easily.<\/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\u00a0 about RD Sharma Class 9 Chapter 17 (Constructions)<\/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-69d4057adfe67\" 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-69d4057adfe67\"><\/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-17-3-solutions\/#download-rd-sharma-chapter-17-class-9-maths-exercise-173-solutions-pdf\" title=\"Download RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 Solutions PDF\">Download RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 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-17-3-solutions\/#what-are-the-topics-covered-within-the-rd-sharma-chapter-17-class-9-maths-exercise-173-solutions\" title=\"What are the topics covered within the RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 Solutions?\">What are the topics covered within the RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 Solutions?<\/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-chapter-17-class-9-maths-exercise-17-3-solutions\/#important-definitions-rd-sharma-chapter-17-class-9-maths-exercise-173-solutions\" title=\"Important Definitions RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 Solutions\">Important Definitions RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 Solutions<\/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-chapter-17-class-9-maths-exercise-17-3-solutions\/#constructions-of-triangles\" title=\"Constructions of Triangles\">Constructions of Triangles<\/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-chapter-17-class-9-maths-exercise-17-3-solutions\/#construction-of-an-equilateral-triangle\" title=\"Construction of an Equilateral Triangle\">Construction of an Equilateral Triangle<\/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-17-class-9-maths-exercise-17-3-solutions\/#construction-of-a-triangle-when-its-base-total-of-the-other-two-sides-and-one-base-angle-are-given\" title=\"Construction of a triangle when its base, total of the other two sides, and one base angle are given\">Construction of a triangle when its base, total of the other two sides, and one base angle are given<\/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-chapter-17-class-9-maths-exercise-17-3-solutions\/#construction-of-a-triangle-when-its-base-difference-of-the-other-two-sides-and-one-base-angle-are-given\" title=\"Construction of a triangle when its base, difference of the other two sides, and one base angle are given\">Construction of a triangle when its base, difference of the other two sides, and one base angle are given<\/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-chapter-17-class-9-maths-exercise-17-3-solutions\/#construction-of-a-triangle-of-given-perimeter-and-two-base-angles\" title=\"Construction of a triangle of given perimeter and two base angles\">Construction of a triangle of given perimeter and two base angles<\/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-chapter-17-class-9-maths-exercise-17-3-solutions\/#examples-of-rd-sharma-chapter-17-class-9-maths-exercise-173-solutions\" title=\"Examples of RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 Solutions\">Examples of RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 Solutions<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-chapter-17-class-9-maths-exercise-173-solutions-pdf\"><\/span><span style=\"font-weight: 400;\">Download RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 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\/01\/Solutions-for-Class-9-Maths-Chapter-17-Construction-Exercise-17.3.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/Solutions-for-Class-9-Maths-Chapter-17-Construction-Exercise-17.3.pdf\">Solutions for Class 9 Maths Chapter 17 Construction Exercise 17.3<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"what-are-the-topics-covered-within-the-rd-sharma-chapter-17-class-9-maths-exercise-173-solutions\"><\/span><span style=\"font-weight: 400;\">What are the topics covered within the RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 Solutions?<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><span style=\"font-weight: 400;\">Basically, it covers the following topics mentioned below-<\/span><\/p>\n<ul>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">First of all, it covers the triangle construction part. Specifically, it covers the construction of the equilateral triangle, i.e., all sides are of equal length.\u00a0\u00a0<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">It covers the construction of a triangle when its base, in which the sum of two different sides of the triangle is provided. At the same time, it also provides one base angle for the students to construct the triangle.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">It covers the construction of a triangle, in which one of its bases is given. Alongside this, here the difference between the lengths of the other two sides is provided. Also, it provides one base angle for the students to construct the triangle.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">It includes the topic meant for constructing a triangle where the perimeter of the triangle is provided. At the same time, the students are provided with two base angles as well.<\/span><\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"important-definitions-rd-sharma-chapter-17-class-9-maths-exercise-173-solutions\"><\/span><span style=\"font-weight: 400;\">Important Definitions RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 Solutions<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><span style=\"font-weight: 400;\">The concepts of the Constructions of different types of triangles are mentioned in the following points-<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"constructions-of-triangles\"><\/span><strong>Constructions of Triangles<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">Triangle is basically a polygon having three sides and thus three angles. Triangles can be constructed of equal\/unequal sides and angles. One can construct a triangle using rulers, compasses, and protractors.\u00a0\u00a0\u00a0\u00a0<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"construction-of-an-equilateral-triangle\"><\/span><strong>Construction of an Equilateral Triangle<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">The equilateral triangle is one having equal sides and angles. It can be constructed using compass, scale, and rulers.\u00a0\u00a0<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"construction-of-a-triangle-when-its-base-total-of-the-other-two-sides-and-one-base-angle-are-given\"><\/span><strong>Construction of a triangle when its base, total of the other two sides, and one base angle are given<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">These triangles can be constructed using scale, compass, protractor, and ruler. However, one has to determine the other two angles upon calculating concerning given angle, and 180 degrees.\u00a0<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"construction-of-a-triangle-when-its-base-difference-of-the-other-two-sides-and-one-base-angle-are-given\"><\/span><strong>Construction of a triangle when its base, difference of the other two sides, and one base angle are given<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">These triangles can also be constructed using scale, compass, protractor, and ruler. Again, one needs to first determine the other two angles upon calculating concerning given angle, and 180 degrees.\u00a0<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"construction-of-a-triangle-of-given-perimeter-and-two-base-angles\"><\/span><strong>Construction of a triangle of given perimeter and two base angles<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">Here one needs to draw a line of length as of given perimeter and then draw two base angles from both its terminal points, using compass. Then one needs to bisect both the angles and draw a perpendicular bisector of both. The point where these perpendicular bisectors meet the perimeter length becomes the base of the desired triangle, and the point where the bisectors meet become the other point.<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"examples-of-rd-sharma-chapter-17-class-9-maths-exercise-173-solutions\"><\/span><strong>Examples of RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 Solutions<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Ques- Construct a \u25b3ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and \u2220B = 60\u00b0.<\/strong><\/p>\n<p><strong>Solution- <\/strong><span style=\"font-weight: 400;\">Steps of Construction of triangle is mentioned below-<\/span><\/p>\n<ol>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Draw a line section BC = 3.6 cm.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">At spot B, draw \u2220XBC = 60\u00b0.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Draw an arc that crosses XB at point \u2018D\u2019 from point B\u2019\u2019 and with a radius of 4.8 cm.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Join DC.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Draw a perpendicular bisector of DC, which crosses DB at \u2018A.\u2019<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Join AC.<\/span><\/li>\n<\/ol>\n<p><span style=\"font-weight: 400;\">Hence, \u25b3ABC is the required triangle.<\/span><\/p>\n<p><strong>Ques- Using rulers and compasses (protractor) only, construct a \u25b3ABC, given base BC = 7cm, \u2220ABC = 60\u00b0, and AB + AC = 12cm.<\/strong><\/p>\n<p><strong>Solution- <\/strong><span style=\"font-weight: 400;\">Steps of Construction of a triangle is mentioned below-<\/span><\/p>\n<ol>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Draw a line segment BC = 7cm.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Draw an arc from point \u2018B\u2019 cutting BC at \u2018N.\u2019 (Choose any radius.)<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Keep compass (protractor) at point \u2018N\u2019 with the same radius (selected in step 2), make the previous arc at \u2018M.\u2019<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Join the line segment BM.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Produce BM to any point \u2018P.\u2019<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Mark BR = 12cm, from BP.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Join CR.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Draw a perpendicular bisector of RC which crosses BR at \u2018A.\u2019<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Join the line segment AC.<\/span><\/li>\n<\/ol>\n<p><span style=\"font-weight: 400;\">Hence, \u25b3ABC is the required triangle.<\/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 17.3 Solutions is based on the concepts of the Constructions of different types of triangles. In this article, we will discuss- Constructions of Triangles, Equilateral Triangles, triangle when its base, total of other two edges, and one base angle are given, etc. The PDF attached below will &#8230; <a title=\"RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 Solutions\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-17-class-9-maths-exercise-17-3-solutions\/\" aria-label=\"More on RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 Solutions\">Read more<\/a><\/p>\n","protected":false},"author":241,"featured_media":72552,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[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\/72551"}],"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=72551"}],"version-history":[{"count":3,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/72551\/revisions"}],"predecessor-version":[{"id":73958,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/72551\/revisions\/73958"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/72552"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=72551"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=72551"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=72551"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}