{"id":63793,"date":"2023-03-27T04:47:00","date_gmt":"2023-03-26T23:17:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=63793"},"modified":"2023-11-23T10:10:14","modified_gmt":"2023-11-23T04:40:14","slug":"rd-sharma-solutions-class-9-maths-chapter-16-circles","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-16-circles\/","title":{"rendered":"RD Sharma Solutions Class 9 Maths Chapter 16 &#8211; Circles (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-124557\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-16-Circles.png\" alt=\"RD Sharma Solutions Class 9 Maths Chapter 16\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-16-Circles.png 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Maths-Chapter-16-Circles-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 16 &#8211; Circles: <\/strong>RD Sharma Solutions Class 9 Maths Chapter 16 enables them also to study all topics more effectively &amp; clear their doubts quickly. The students who want to obtain higher marks in the exams must regularly practice mathematics subject exercises with the help of <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths<\/a> Chapter 16 Circles. This also builds the confidence of the students and helps them to revise each important topic during the exam.<\/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-69d763f1793fe\" 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-69d763f1793fe\"><\/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-16-circles\/#download-rd-sharma-solutions-class-9-maths-chapter-16-pdf\" title=\"Download RD Sharma Solutions Class 9 Maths Chapter 16 PDF\">Download RD Sharma Solutions Class 9 Maths Chapter 16 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-16-circles\/#exercise-wise-rd-sharma-solutions-class-9-maths-chapter-16-circles\" title=\"Exercise-wise RD Sharma Solutions Class 9 Maths Chapter 16 Circles\">Exercise-wise RD Sharma Solutions Class 9 Maths Chapter 16 Circles<\/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-16-circles\/#access-answers-of-rd-sharma-solutions-class-9-maths-chapter-16\" title=\"Access answers of RD Sharma Solutions Class 9 Maths Chapter 16\">Access answers of RD Sharma Solutions Class 9 Maths Chapter 16<\/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-16-circles\/#rd-sharma-class-9-chapter-16-circles-ex-161\" title=\"RD Sharma Class 9 Chapter 16 Circles Ex 16.1\">RD Sharma Class 9 Chapter 16 Circles Ex 16.1<\/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-16-circles\/#exercise-163-page-no-1640\" title=\"Exercise 16.3 Page No: 16.40\">Exercise 16.3 Page No: 16.40<\/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-16-circles\/#exercise-164-page-no-1660\" title=\"Exercise 16.4 Page No: 16.60\">Exercise 16.4 Page No: 16.60<\/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-16-circles\/#exercise-165-page-no-1683\" title=\"Exercise 16.5 Page No: 16.83\">Exercise 16.5 Page No: 16.83<\/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-16-circles\/#exercise-vsaqs-page-no-1689\" title=\"Exercise VSAQs Page No: 16.89\">Exercise VSAQs Page No: 16.89<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-16-circles\/#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><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-16-circles\/#important-topics-in-the-exercise\" title=\"Important Topics in the Exercise\">Important Topics in the Exercise<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-16-circles\/#faqs-on-rd-sharma-solutions-class-9-maths-chapter-16-circles\" title=\"FAQs on RD Sharma Solutions Class 9 Maths Chapter 16 Circles\">FAQs on RD Sharma Solutions Class 9 Maths Chapter 16 Circles<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-16-circles\/#from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-9-maths-chapter-16\" title=\"From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 16?\">From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 16?<\/a><\/li><li class='ez-toc-page-1 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-16-circles\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-9-maths-chapter-16\" title=\"How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 16?\">How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 16?<\/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-16-circles\/#can-i-access-the-rd-sharma-solutions-for-class-9-maths-chapter-16-pdf-offline\" title=\"Can I access the RD Sharma Solutions for Class 9 Maths Chapter 16\u00a0PDF offline?\">Can I access the RD Sharma Solutions for Class 9 Maths Chapter 16\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-16-pdf\"><\/span><strong>Download RD Sharma Solutions Class 9 Maths Chapter 16 PDF<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/16-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 16<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/16-1.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"exercise-wise-rd-sharma-solutions-class-9-maths-chapter-16-circles\"><\/span><strong>Exercise-wise RD Sharma Solutions Class 9 Maths Chapter 16 Circles<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-16-class-9-maths-exercise-16-1-solutions\/\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">RD Sharma class 9 chapter 16 exercise 16a<\/span><\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-16-class-9-maths-exercise-16-2-solutions\/\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">RD Sharma class 9 chapter 16 exercise 16b<\/span><\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-16-class-9-maths-exercise-16-3-solutions\/\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">RD Sharma class 9 chapter 16 exercise 16c<\/span><\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-16-class-9-maths-exercise-16-4-solutions\/\" target=\"_blank\" rel=\"noopener\"><span style=\"font-weight: 400;\">RD Sharma class 9 chapter 16 exercise 16d<\/span><\/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-16\"><\/span><strong>Access answers of RD Sharma Solutions Class 9 Maths Chapter 16<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-9-chapter-16-circles-ex-161\"><\/span>RD Sharma Class 9 Chapter 16 Circles Ex 16.1<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Question 1: The radius of a circle is 8 cm and the length of one of its chords is 12 cm. Find the distance of the chord from the centre.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.2 solution 1\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-2-solutio.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.2 solution 1\" width=\"386\" height=\"277\" \/><\/strong><\/p>\n<p>Radius of circle (OA) = 8 cm (Given)<\/p>\n<p>Chord (AB) = 12cm (Given)<\/p>\n<p>Draw a perpendicular OC on AB.<\/p>\n<p>We know, perpendicular from the centre to a chord bisects the chord<\/p>\n<p>Which implies, AC = BC = 12\/2 = 6 cm<\/p>\n<p>In right \u0394OCA:<\/p>\n<p>Using Pythagoras&#8217; theorem,<\/p>\n<p>OA<sup>2<\/sup>\u00a0= AC<sup>2<\/sup>\u00a0+ OC<sup>2<\/sup><\/p>\n<p>64 = 36 + OC<sup>2<\/sup><\/p>\n<p>OC<sup>2<\/sup>\u00a0= 64 \u2013 36 = 28<\/p>\n<p>or OC = \u221a28 = 5.291 (approx.)<\/p>\n<p><em>The distance of the chord from the centre is 5.291 cm.<\/em><\/p>\n<p><strong>Question 2: Find the length of a chord which is at a distance of 5 cm from the centre of a circle of radius 10 cm.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.2 solution 2\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-2-solutio-1.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.2 solution 2\" width=\"376\" height=\"270\" \/><\/p>\n<p>Distance of the chord from the centre = OC = 5 cm (Given)<\/p>\n<p>Radius of the circle = OA = 10 cm (Given)<\/p>\n<p>In \u0394OCA:<\/p>\n<p>Using Pythagoras&#8217; theorem,<\/p>\n<p>OA<sup>2<\/sup>\u00a0= AC<sup>2<\/sup>\u00a0+ OC<sup>2<\/sup><\/p>\n<p>100 = AC<sup>2\u00a0<\/sup>+ 25<\/p>\n<p>AC<sup>2<\/sup>\u00a0= 100 \u2013 25 = 75<\/p>\n<p>AC = \u221a75 = 8.66<\/p>\n<p>As perpendicular from the centre to the chord bisects the chord.<\/p>\n<p>Therefore, AC = BC = 8.66 cm<\/p>\n<p>=&gt; AB = AC + BC = 8.66 + 8.66 = 17.32<\/p>\n<p><em>Answer: AB = 17.32 cm<\/em><\/p>\n<p><strong>Question 3: Find the length of a chord which is at a distance of 4 cm from the centre of a circle of radius 6 cm.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-2-solutio-2.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.2 solution 3\" width=\"360\" height=\"258\" \/><\/p>\n<p>Distance of the chord from the centre = OC = 4 cm (Given)<\/p>\n<p>Radius of the circle = OA = 6 cm (Given)<\/p>\n<p>In \u0394OCA:<\/p>\n<p>Using Pythagoras&#8217; theorem,<\/p>\n<p>OA<sup>2<\/sup>\u00a0= AC<sup>2<\/sup>\u00a0+ OC<sup>2<\/sup><\/p>\n<p>36 = AC<sup>2\u00a0<\/sup>+ 16<\/p>\n<p>AC<sup>2<\/sup>\u00a0= 36 \u2013 16 = 20<\/p>\n<p>AC = \u221a20 = 4.47<\/p>\n<p>Or AC = 4.47cm<\/p>\n<p>As perpendicular from the centre to the chord bisects the chord.<\/p>\n<p>Therefore, AC = BC = 4.47 cm<\/p>\n<p>=&gt; AB = AC + BC = 4.47 + 4.47 = 8.94<\/p>\n<p><em>Answer: AB = 8.94 cm<\/em><\/p>\n<p><strong>Question 4: Two chords AB, and CD of lengths 5 cm, and 11 cm respectively of a circle are parallel. If the distance between AB and CD is 3 cm, find the radius of the circle.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.2 solution 4\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-2-solutio-3.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.2 solution 4\" width=\"407\" height=\"292\" \/><\/p>\n<p>Given: AB = 5 cm, CD = 11 cm, PQ = 3 cm<\/p>\n<p>Draw perpendiculars OP on CD and OQ on AB<\/p>\n<p>Let OP = x cm and OC = OA = r cm<\/p>\n<p>We know perpendicular from the centre to the chord bisects it.<\/p>\n<p>Since OP\u22a5CD, we have<\/p>\n<p>CP = PD = 11\/2 cm<\/p>\n<p>And OQ\u22a5AB<\/p>\n<p>AQ = BQ = 5\/2 cm<\/p>\n<p>In \u0394OCP:<\/p>\n<p>By Pythagoras&#8217; theorem,<\/p>\n<p>OC<sup>2<\/sup>\u00a0= OP<sup>2<\/sup>\u00a0+ CP<sup>2<\/sup><\/p>\n<p>r<sup>2\u00a0<\/sup>= x<sup>2\u00a0<\/sup>+ (11\/2)<sup>\u00a02\u00a0<\/sup>\u2026..(1)<\/p>\n<p>In \u0394OQA:<\/p>\n<p>By Pythagoras&#8217; theorem,<\/p>\n<p>OA<sup>2<\/sup>=OQ<sup>2<\/sup>+AQ<sup>2<\/sup><\/p>\n<p>r<sup>2<\/sup>= (x+3)<sup>\u00a02\u00a0<\/sup>+ (5\/2)<sup>\u00a02<\/sup>\u00a0\u2026..(2)<\/p>\n<p>From equations (1) and (2), we get<\/p>\n<p>(x+3)<sup>\u00a02\u00a0<\/sup>+ (5\/2)<sup>\u00a02<\/sup>\u00a0= x<sup>2\u00a0<\/sup>+ (11\/2)<sup>\u00a02<\/sup><\/p>\n<p>Solve the above equation and find the value of x.<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 6x + 9 + 25\/4 = x<sup>2<\/sup>\u00a0+ 121\/4<\/p>\n<p>(using identity, (a+b)<sup>\u00a02<\/sup>\u00a0= a<sup>2<\/sup>\u00a0+ b<sup>2<\/sup>\u00a0+ 2ab )<\/p>\n<p>6x = 121\/4 \u2013 25\/4 \u2212 9<\/p>\n<p>6x = 15<\/p>\n<p>or x = 15\/6 = 5\/2<\/p>\n<p>Substitute the value of x in equation (1), and find the length of the radius,<\/p>\n<p>r<sup>2\u00a0<\/sup>= (5\/2)<sup>2\u00a0<\/sup>+ (11\/2)<sup>\u00a02<\/sup><\/p>\n<p>= 25\/4 + 121\/4<\/p>\n<p>= 146\/4<\/p>\n<p><em>or r = \u221a146\/4 cm<\/em><\/p>\n<p><strong>Question 5: Give a method to find the centre of a given circle.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Steps of Construction:<\/p>\n<p>Step 1: Consider three points A, B and C on a circle.<\/p>\n<p>Step 2: Join AB and BC.<\/p>\n<p>Step 3: Draw perpendicular bisectors of chord AB and BC which intersect each other at a point, say O.<\/p>\n<p>Step 4: This point O is the centre of the circle because we know that, the Perpendicular bisectors of a chord always pass through the centre.<\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.2 solution 5\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-2-solutio-4.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.2 solution 5\" width=\"354\" height=\"254\" \/><\/p>\n<p><strong>Question 6: Prove that the line joining the mid-point of a chord to the centre of the circle passes through the mid-point of the corresponding minor arc.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.2 solution 6\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-2-solutio-5.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.2 solution 6\" width=\"342\" height=\"245\" \/><\/strong><\/p>\n<p>From the figure, Let C is the mid-point of chord AB.<\/p>\n<p>To prove: D is the mid-point of arc AB.<\/p>\n<p>Now, In \u0394OAC and \u0394OBC<\/p>\n<p>OA = OB [Radius of the circle]<\/p>\n<p>OC = OC [Common]<\/p>\n<p>AC = BC [C is the mid-point of chord AB (given)]<\/p>\n<p>So, by SSS condition: \u0394OAC \u2245 \u0394OBC<\/p>\n<p>So, \u2220AOC = \u2220BOC (BY CPCT)<\/p>\n<p><img title=\"RD sharma class 9 maths chapter 16 ex 16.2 mid-point theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-2-mid-poi.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.2 mid-point theorem\" \/><\/p>\n<p><em>Therefore, D is the mid-point of arc AB. Hence Proved.<\/em><\/p>\n<p><strong>Question 7: Prove that a diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.2 solution 7\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-2-solutio-6.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.2 solution 7\" width=\"323\" height=\"232\" \/><\/p>\n<p>Form figure: PQ is the diameter of the circle which bisects the chord AB at C. (Given)<\/p>\n<p>To Prove: PQ bisects \u2220AOB<\/p>\n<p>Now,<\/p>\n<p>In \u0394BOC and \u0394AOC<\/p>\n<p>OA = OB [Radius]<\/p>\n<p>OC = OC [Common side]<\/p>\n<p>AC = BC [Given]<\/p>\n<p>Then, by SSS condition: \u0394AOC \u2245 \u0394BOC<\/p>\n<p>So, \u2220AOC = \u2220BOC [By c.p.c.t.]<\/p>\n<p><em>Therefore, PQ bisects \u2220AOB. Hence proved.<\/em><\/p>\n<hr \/>\n<h3><span class=\"ez-toc-section\" id=\"exercise-163-page-no-1640\"><\/span>Exercise 16.3 Page No: 16.40<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Question 1: Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20 m drawn in a park. Ishita throws a ball to Isha, Isha to Nisha and Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24 m each, what is the distance between Ishita and Nisha?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let R, S and M be the position of Ishita, Isha and Nisha respectively.<\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.3 solution 1\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-3-solutio.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.3 solution 1\" width=\"343\" height=\"214\" \/><\/p>\n<p>Since OA is a perpendicular bisector on RS, so AR = AS = 24\/2 = 12 cm<\/p>\n<p>Radii of circle = OR = OS = OM = 20 cm (Given)<\/p>\n<p>In \u0394OAR:<\/p>\n<p>By Pythagoras&#8217; theorem,<\/p>\n<p>OA<sup>2<\/sup>+AR<sup>2<\/sup>=OR<sup>2<\/sup><\/p>\n<p>OA<sup>2<\/sup>+12<sup>2<\/sup>=20<sup>2<\/sup><\/p>\n<p>OA<sup>2\u00a0<\/sup>= 400 \u2013 144 = 256<\/p>\n<p>Or OA = 16 m \u2026(1)<\/p>\n<p>From the figure, OABC is a kite since OA = OC and AB = BC. We know that the diagonals of a kite are perpendicular and the diagonal common to both the isosceles triangles is bisected by another diagonal.<\/p>\n<p>So in \u0394RSM, \u2220RCS = 90<sup>0<\/sup>\u00a0and RC = CM \u2026(2)<\/p>\n<p>Now, Area of \u0394ORS = Area of \u0394ORS<\/p>\n<p>=&gt;1\/2\u00d7OA\u00d7RS = 1\/2 x RC x OS<\/p>\n<p>=&gt; OA \u00d7RS = RC x OS<\/p>\n<p>=&gt; 16 x 24 = RC x 20<\/p>\n<p>=&gt; RC = 19.2<\/p>\n<p>Since RC = CM (from (2), we have<\/p>\n<p>RM = 2(19.2) = 38.4<\/p>\n<p><em>So, the distance between Ishita and Nisha is 38.4 m.<\/em><\/p>\n<p><strong>Question 2: A circular park of a radius of 40 m is situated in a colony. Three boys Ankur, Amit and Anand are sitting at equal distances on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"RD sharma class 9 maths chapter 16 ex 16.3 solution 2\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-3-solutio-1.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.3 solution 2\" \/><\/p>\n<p>Since, AB = BC = CA. So, ABC is an equilateral triangle<\/p>\n<p>Radius = OA = 40 m (Given)<\/p>\n<p>We know, medians of equilateral triangles pass through the circumcentre and intersect each other at the ratio of 2:1.<\/p>\n<p>Here AD is the median of an equilateral triangle ABC, we can write:<\/p>\n<p>OA\/OD = 2\/1<\/p>\n<p>or 40\/OD = 2\/1<\/p>\n<p>or OD = 20 m<\/p>\n<p>Therefore, AD = OA + OD = (40 + 20) m = 60 m<\/p>\n<p>Now, In \u0394ADC:<\/p>\n<p>By Pythagoras&#8217; theorem,<\/p>\n<p>AC<sup>2<\/sup>\u00a0= AD<sup>2<\/sup>\u00a0+ DC<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>\u00a0= 60<sup>2<\/sup>\u00a0+ (AC\/2)<sup>\u00a02<\/sup><\/p>\n<p>AC<sup>2\u00a0<\/sup>= 3600 + AC<sup>2<\/sup>\u00a0\/ 4<\/p>\n<p>3\/4 AC<sup>2\u00a0<\/sup>= 3600<\/p>\n<p>AC<sup>2\u00a0<\/sup>= 4800<\/p>\n<p>or AC = 40\u221a3 m<\/p>\n<p><em>Therefore, the length of the string of each phone will be 40\u221a3 m.<\/em><\/p>\n<hr \/>\n<h3><span class=\"ez-toc-section\" id=\"exercise-164-page-no-1660\"><\/span>Exercise 16.4 Page No: 16.60<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Question 1: In the figure, O is the centre of the circle. If \u2220APB = 50<sup>0<\/sup>, find \u2220AOB and \u2220OAB.<\/strong><\/p>\n<p><img title=\"RD sharma class 9 maths chapter 16 ex 16.4 question 1\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-4-questio.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.4 question 1\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u2220APB = 50<sup>0<\/sup>\u00a0(Given)<\/p>\n<p>By degree measure theorem: \u2220AOB = 2\u2220APB<\/p>\n<p>\u2220AOB = 2 \u00d7 50<sup>0<\/sup>\u00a0= 100<sup>0<\/sup><\/p>\n<p>Again, OA = OB [Radius of the circle]<\/p>\n<p>Then \u2220OAB = \u2220OBA [Angles opposite to equal sides]<\/p>\n<p>Let \u2220OAB = m<\/p>\n<p>In \u0394OAB,<\/p>\n<p>By angle sum property: \u2220OAB+\u2220OBA+\u2220AOB=180<sup>0<\/sup><\/p>\n<p>=&gt; m + m + 100<sup>0<\/sup>\u00a0= 180<sup>0<\/sup><\/p>\n<p>=&gt;2m = 180<sup>0<\/sup>\u00a0\u2013 100<sup>0<\/sup>\u00a0= 80<sup>0<\/sup><\/p>\n<p>=&gt;m = 80<sup>0<\/sup>\/2 = 40<sup>0<\/sup><\/p>\n<p><em>\u2220OAB = \u2220OBA = 40<sup>0<\/sup><\/em><\/p>\n<p><strong>Question 2: In the figure, it is given that O is the centre of the circle and \u2220AOC = 150<sup>0<\/sup>. Find \u2220ABC.<\/strong><\/p>\n<p><strong><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.4 question 2\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-4-questio-1.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.4 question 2\" width=\"304\" height=\"218\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u2220AOC = 150<sup>0<\/sup>\u00a0(Given)<\/p>\n<p>By degree measure theorem: \u2220ABC = (reflex\u2220AOC)\/2 \u2026(1)<\/p>\n<p>We know, \u2220AOC + reflex(\u2220AOC) = 360<sup>0<\/sup>\u00a0[Complex angle]<\/p>\n<p>150<sup>0<\/sup>\u00a0+ reflex\u2220AOC = 360<sup>0<\/sup><\/p>\n<p>or reflex \u2220AOC = 360<sup>0<\/sup>\u2212150<sup>0<\/sup>\u00a0= 210<sup>0<\/sup><\/p>\n<p>From (1)\u00a0<em>=&gt; \u2220ABC = 210<sup>\u00a0o<\/sup>\u00a0\/2 = 105<sup>o<\/sup><\/em><\/p>\n<p><strong>Question 3: In the figure, O is the centre of the circle. Find \u2220BAC.<\/strong><\/p>\n<p><strong><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.4 question 3\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-4-questio-2.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.4 question 3\" width=\"353\" height=\"253\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: \u2220AOB = 80<sup>0<\/sup>\u00a0and \u2220AOC = 110<sup>0<\/sup><\/p>\n<p>Therefore, \u2220AOB+\u2220AOC+\u2220BOC=360<sup>0<\/sup>\u00a0[Completeangle]<\/p>\n<p>Substitute given values,<\/p>\n<p>80<sup>0<\/sup>\u00a0+ 100<sup>0<\/sup>\u00a0+ \u2220BOC = 360<sup>0<\/sup><\/p>\n<p>\u2220BOC = 360<sup>0<\/sup>\u00a0\u2013 80<sup>0<\/sup>\u00a0\u2013 110<sup>0<\/sup>\u00a0= 170<sup>0<\/sup><\/p>\n<p>or \u2220BOC = 170<sup>0<\/sup><\/p>\n<p>Now, by degree measure theorem<\/p>\n<p>\u2220BOC = 2\u2220BAC<\/p>\n<p>170<sup>0<\/sup>\u00a0= 2\u2220BAC<\/p>\n<p>Or\u00a0<em>\u2220BAC = 170<sup>0<\/sup>\/2 = 85<sup>0<\/sup><\/em><\/p>\n<p><strong>Question 4: If O is the centre of the circle, find the value of x in each of the following figures.<\/strong><\/p>\n<p><strong>(i)<\/strong><\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 1\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-4-questio-3.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 1\" width=\"358\" height=\"257\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u2220AOC = 135<sup>0<\/sup>\u00a0(Given)<\/p>\n<p>From the figure, \u2220AOC + \u2220BOC = 180<sup>0<\/sup>\u00a0[Linear pair of angles]<\/p>\n<p>135<sup>0<\/sup>\u00a0+\u2220BOC = 180<sup>0<\/sup><\/p>\n<p>or \u2220BOC=180<sup>0<\/sup>\u2212135<sup>0<\/sup><\/p>\n<p>or \u2220BOC=45<sup>0<\/sup><\/p>\n<p>Again, by degree measure theorem<\/p>\n<p>\u2220BOC = 2\u2220CPB<\/p>\n<p>45<sup>0<\/sup>\u00a0= 2x<\/p>\n<p><em>x = 45<sup>0<\/sup>\/2<\/em><\/p>\n<p><strong>(ii)<\/strong><\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 2\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-4-questio-4.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 2\" width=\"314\" height=\"231\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u2220ABC=40<sup>0<\/sup>\u00a0(given)<\/p>\n<p>\u2220ACB = 90<sup>0<\/sup>\u00a0[Angle in semicircle]<\/p>\n<p>In \u0394ABC,<\/p>\n<p>\u2220CAB+\u2220ACB+\u2220ABC=180<sup>0<\/sup>\u00a0[angle sum property]<\/p>\n<p>\u2220CAB+90<sup>0<\/sup>+40<sup>0<\/sup>=180<sup>0<\/sup><\/p>\n<p>\u2220CAB=180<sup>0<\/sup>\u221290<sup>0<\/sup>\u221240<sup>0<\/sup><\/p>\n<p>\u2220CAB=50<sup>0<\/sup><\/p>\n<p>Now, \u2220CDB = \u2220CAB [Angle is on the same segment]<\/p>\n<p><em>This implies, x = 50<sup>0<\/sup><\/em><\/p>\n<p><strong>(iii)<\/strong><\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 3\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-4-questio-5.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 3\" width=\"332\" height=\"244\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u2220AOC = 120<sup>0<\/sup>\u00a0(given)<\/p>\n<p>By degree measure theorem: \u2220AOC = 2\u2220APC<\/p>\n<p>120<sup>0\u00a0<\/sup>= 2\u2220APC<\/p>\n<p>\u2220APC = 120<sup>0<\/sup>\/2 = 60<sup>0<\/sup><\/p>\n<p>Again, \u2220APC + \u2220ABC = 180<sup>0<\/sup>\u00a0[Sum of opposite angles of cyclic quadrilaterals = 180\u00a0<sup>o<\/sup>\u00a0]<\/p>\n<p>60<sup>0<\/sup>\u00a0+ \u2220ABC=180<sup>0<\/sup><\/p>\n<p>\u2220ABC=180<sup>0<\/sup>\u221260<sup>0<\/sup><\/p>\n<p>\u2220ABC = 120<sup>0<\/sup><\/p>\n<p>\u2220ABC + \u2220DBC = 180<sup>0<\/sup>\u00a0[Linear pair of angles]<\/p>\n<p>120<sup>0\u00a0<\/sup>+ x = 180<sup>0<\/sup><\/p>\n<p>x = 180<sup>0<\/sup>\u2212120<sup>0<\/sup>=60<sup>0<\/sup><\/p>\n<p><em>The value of x is 60<sup>0<\/sup><\/em><\/p>\n<p><strong>(iv)<\/strong><\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 4\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-4-questio-6.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 4\" width=\"321\" height=\"236\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u2220CBD = 65<sup>0<\/sup>\u00a0(given)<\/p>\n<p>From figure:<\/p>\n<p>\u2220ABC + \u2220CBD = 180<sup>0<\/sup>\u00a0[ Linear pair of angles]<\/p>\n<p>\u2220ABC + 65<sup>0<\/sup>\u00a0= 180<sup>0<\/sup><\/p>\n<p>\u2220ABC =180<sup>0<\/sup>\u221265<sup>0<\/sup>=115<sup>0<\/sup><\/p>\n<p>Again, reflex \u2220AOC = 2\u2220ABC [Degree measure theorem]<\/p>\n<p>x=2(115<sup>0<\/sup>) = 230<sup>0<\/sup><\/p>\n<p><em>The value of x is 230<sup>0<\/sup><\/em><\/p>\n<p><strong>(v)<\/strong><\/p>\n<p><strong><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 5\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-4-questio-7.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 5\" width=\"321\" height=\"236\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u2220OAB = 35<sup>0<\/sup>\u00a0(Given)<\/p>\n<p>From figure:<\/p>\n<p>\u2220OBA = \u2220OAB = 35<sup>0<\/sup>\u00a0[Angles opposite to equal radii]<\/p>\n<p>In\u0394AOB:<\/p>\n<p>\u2220AOB + \u2220OAB + \u2220OBA = 180<sup>0<\/sup>\u00a0[angle sum property]<\/p>\n<p>\u2220AOB + 35<sup>0<\/sup>\u00a0+ 35<sup>0<\/sup>\u00a0= 180<sup>0<\/sup><\/p>\n<p>\u2220AOB = 180<sup>0<\/sup>\u00a0\u2013 35<sup>0<\/sup>\u00a0\u2013 35<sup>0<\/sup>\u00a0= 110<sup>0<\/sup><\/p>\n<p>Now, \u2220AOB + reflex\u2220AOB = 360<sup>0<\/sup>\u00a0[Complex angle]<\/p>\n<p>110<sup>0<\/sup>\u00a0+ reflex\u2220AOB = 360<sup>0<\/sup><\/p>\n<p>reflex\u2220AOB = 360<sup>0<\/sup>\u00a0\u2013 110<sup>0<\/sup>\u00a0= 250<sup>0<\/sup><\/p>\n<p>By degree measure theorem: reflex \u2220AOB = 2\u2220ACB<\/p>\n<p>250<sup>0<\/sup>\u00a0= 2x<\/p>\n<p><em>x = 250<sup>0<\/sup>\/2=125<sup>0<\/sup><\/em><\/p>\n<p><strong>(vi)<\/strong><\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 6\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-4-questio-8.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 6\" width=\"300\" height=\"221\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u2220AOB = 60<sup>o<\/sup>\u00a0(given)<\/p>\n<p>By degree measure theorem: reflex\u2220AOB = 2\u2220OAC<\/p>\n<p>60<sup>\u00a0o<\/sup>\u00a0= 2\u2220 OAC<\/p>\n<p>\u2220OAC = 60<sup>\u00a0o<\/sup>\u00a0\/ 2 = 30<sup>\u00a0o<\/sup>\u00a0[Angles opposite to equal radii]<\/p>\n<p><em>Or x = 30<sup>0<\/sup><\/em><\/p>\n<p><strong>(vii)<\/strong><\/p>\n<p><strong><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 7\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-4-questio-9.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 7\" width=\"353\" height=\"260\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u2220BAC = 50<sup>0<\/sup>\u00a0and \u2220DBC = 70<sup>0<\/sup>\u00a0(given)<\/p>\n<p>From figure:<\/p>\n<p>\u2220BDC = \u2220BAC = 50<sup>0\u00a0<\/sup>[Angle on the same segment]<\/p>\n<p>Now,<\/p>\n<p>In \u0394BDC:<\/p>\n<p>Using the angle sum property, we have<\/p>\n<p>\u2220BDC+\u2220BCD+\u2220DBC=180<sup>0<\/sup><\/p>\n<p>Substituting given values, we get<\/p>\n<p>50<sup>0<\/sup>\u00a0+ x<sup>0<\/sup>\u00a0+ 70<sup>0<\/sup>\u00a0= 180<sup>0<\/sup><\/p>\n<p>x<sup>0\u00a0<\/sup>= 180<sup>0<\/sup>\u221250<sup>0<\/sup>\u221270<sup>0<\/sup>=60<sup>0<\/sup><\/p>\n<p><em>or x = 60<sup>o<\/sup><\/em><\/p>\n<p><strong>(viii)<\/strong><\/p>\n<p><strong><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 8\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-4-questio-10.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 8\" width=\"379\" height=\"279\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u2220DBO = 40<sup>0<\/sup>\u00a0(Given)<\/p>\n<p>Form figure:<\/p>\n<p>\u2220DBC = 90<sup>0<\/sup>\u00a0[Angle in a semicircle]<\/p>\n<p>\u2220DBO + \u2220OBC = 90<sup>0<\/sup><\/p>\n<p>40<sup>0<\/sup>+\u2220OBC=90<sup>0<\/sup><\/p>\n<p>or \u2220OBC=90<sup>0<\/sup>\u221240<sup>0<\/sup>=50<sup>0<\/sup><\/p>\n<p>Again, By degree measure theorem: \u2220AOC = 2\u2220OBC<\/p>\n<p><em>or x = 2\u00d750<sup>0<\/sup>=100<sup>0<\/sup><\/em><\/p>\n<p><strong>(ix)<\/strong><\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 9\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-4-questio-11.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 9\" width=\"307\" height=\"226\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u2220CAD = 28, \u2220ADB = 32 and \u2220ABC = 50 (Given)<\/p>\n<p>From figure:<\/p>\n<p>In \u0394DAB:<\/p>\n<p>Angle sum property: \u2220ADB + \u2220DAB + \u2220ABD = 180<sup>0<\/sup><\/p>\n<p>By substituting the given values, we get<\/p>\n<p>32<sup>0<\/sup>\u00a0+ \u2220DAB + 50<sup>0<\/sup>\u00a0= 180<sup>0<\/sup><\/p>\n<p>\u2220DAB=180<sup>0<\/sup>\u221232<sup>0<\/sup>\u221250<sup>0<\/sup><\/p>\n<p>\u2220DAB = 98<sup>0<\/sup><\/p>\n<p>Now,<\/p>\n<p>\u2220DAB+\u2220DCB=180<sup>0\u00a0<\/sup>[Opposite angles of cyclic quadrilateral, their sum = 180 degrees]<\/p>\n<p>98<sup>0<\/sup>+x=180<sup>0<\/sup><\/p>\n<p>or x = 180<sup>0<\/sup>\u221298<sup>0<\/sup>=82<sup>0<\/sup><\/p>\n<p><em>The value of x is 82 degrees.<\/em><\/p>\n<p><strong>(x)<\/strong><\/p>\n<p><strong><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 10\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-4-questio-12.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 10\" width=\"393\" height=\"289\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u2220BAC = 35<sup>0<\/sup>\u00a0and \u2220DBC = 65<sup>0<\/sup><\/p>\n<p>From figure:<\/p>\n<p>\u2220BDC = \u2220BAC = 35<sup>0<\/sup> [Angle in the same segment]<\/p>\n<p>In \u0394BCD:<\/p>\n<p>Angle sum property, we have<\/p>\n<p>\u2220BDC + \u2220BCD + \u2220DBC = 180<sup>0<\/sup><\/p>\n<p>35<sup>0<\/sup>\u00a0+ x + 65<sup>0<\/sup>\u00a0= 180<sup>0<\/sup><\/p>\n<p><em>or x = 180<sup>0<\/sup>\u00a0\u2013 35<sup>0<\/sup>\u00a0\u2013 65<sup>0<\/sup>\u00a0= 80<sup>0<\/sup><\/em><\/p>\n<p><strong>(xi)<\/strong><\/p>\n<p><strong><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 11\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-4-questio-13.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 11\" width=\"423\" height=\"311\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u2220ABD = 40<sup>0<\/sup>, \u2220CPD = 110<sup>0<\/sup>\u00a0(Given)<\/p>\n<p>Form figure:<\/p>\n<p>\u2220ACD = \u2220ABD = 40<sup>0<\/sup> [Angle in the same segment]<\/p>\n<p>In \u0394PCD,<\/p>\n<p>Angle sum property: \u2220PCD+\u2220CPO+\u2220PDC=180<sup>0<\/sup><\/p>\n<p>400 + 110<sup>0<\/sup>\u00a0+ x = 180<sup>0<\/sup><\/p>\n<p>x=180<sup>0<\/sup>\u2212150<sup>0\u00a0<\/sup>=30<sup>0<\/sup><\/p>\n<p><em>The value of x is 30 degrees.<\/em><\/p>\n<p><strong>(xii)<\/strong><\/p>\n<p><strong><img title=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 12\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-4-questio-14.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.4 question 4 part 12\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u2220BAC = 52<sup>0<\/sup>\u00a0(Given)<\/p>\n<p>From figure:<\/p>\n<p>\u2220BDC = \u2220BAC = 52<sup>0<\/sup> [Angle in the same segment]<\/p>\n<p>Since OD = OC (radii), then \u2220ODC = \u2220OCD [Opposite angle to equal radii]<\/p>\n<p><em>So, x = 52<sup>0<\/sup><\/em><\/p>\n<p><strong>Question 5: O is the circumcentre of the triangle ABC and OD is perpendicular on BC. Prove that \u2220BOD = \u2220A.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.4 Q5 solution\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-4-q5-solu.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.4 Q5 solution\" width=\"420\" height=\"309\" \/><\/p>\n<p>In \u0394OBD and \u0394OCD:<\/p>\n<p>OB = OC [Radius]<\/p>\n<p>\u2220ODB = \u2220ODC [Each 90<sup>0<\/sup>]<\/p>\n<p>OD = OD [Common]<\/p>\n<p>Therefore, By RHS Condition<\/p>\n<p>\u0394OBD \u2245 \u0394OCD<\/p>\n<p>So, \u2220BOD = \u2220COD\u2026..(i)[By CPCT]<\/p>\n<p>Again,<\/p>\n<p>By degree measure theorem: \u2220BOC = 2\u2220BAC<\/p>\n<p>2\u2220BOD = 2\u2220BAC [Using(i)]<\/p>\n<p><em>\u2220BOD = \u2220BAC<\/em><\/p>\n<p>Hence proved.<\/p>\n<p><strong>Question 6: In the figure, O is the centre of the circle, and BO is the bisector of \u2220ABC. Show that AB = AC.<\/strong><\/p>\n<p><strong><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.4 Q6 solution\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-4-q6-solu.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.4 Q6 solution\" width=\"395\" height=\"291\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Since, BO is the bisector of \u2220ABC, then,<\/p>\n<p>\u2220ABO = \u2220CBO \u2026..(i)<\/p>\n<p>From figure:<\/p>\n<p>Radius of circle = OB = OA = OB = OC<\/p>\n<p>\u2220OAB = \u2220OCB \u2026..(ii) [opposite angles to equal sides]<\/p>\n<p>\u2220ABO = \u2220DAB \u2026..(iii) [opposite angles to equal sides]<\/p>\n<p>From equations (i), (ii) and (iii), we get<\/p>\n<p>\u2220OAB = \u2220OCB \u2026..(iv)<\/p>\n<p>In \u0394OAB and \u0394OCB:<\/p>\n<p>\u2220OAB = \u2220OCB [From (iv)]<\/p>\n<p>OB = OB [Common]<\/p>\n<p>\u2220OBA = \u2220OBC [Given]<\/p>\n<p>Then, By AAS condition: \u0394OAB \u2245 \u0394OCB<\/p>\n<p><em>So, AB = BC [By CPCT]<\/em><\/p>\n<p><strong>Question 7: In the figure, O is the centre of the circle, then prove that \u2220x = \u2220y + \u2220z.<\/strong><\/p>\n<p><img title=\"RD sharma class 9 maths chapter 16 ex 16.4 Q7 solution\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-4-q7-solu.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.4 Q7 solution\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>From the figure:<\/p>\n<p>\u22203 = \u22204 \u2026.(i) [Angles in same segment]<\/p>\n<p>\u2220x = 2\u22203 [By degree measure theorem]<\/p>\n<p>\u2220x = \u22203 + \u22203<\/p>\n<p>\u2220x = \u22203 + \u22204 (Using (i) ) \u2026..(ii)<\/p>\n<p>Again, \u2220y = \u22203 + \u22201 [By exterior angle property]<\/p>\n<p>or \u22203 = \u2220y \u2212 \u22201 \u2026..(iii)<\/p>\n<p>\u22204 = \u2220z + \u22201 \u2026. (iv) [By exterior angle property]<\/p>\n<p>Now, from equations (ii) , (iii) and (iv), we get<\/p>\n<p>\u2220x = \u2220y \u2212 \u22201 + \u2220z + \u22201<\/p>\n<p>or \u2220x = \u2220y + \u2220z + \u22201 \u2212 \u22201<\/p>\n<p><em>or x = \u2220y + \u2220z<\/em><\/p>\n<p>Hence proved.<\/p>\n<hr \/>\n<h3><span class=\"ez-toc-section\" id=\"exercise-165-page-no-1683\"><\/span>Exercise 16.5 Page No: 16.83<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Question 1: In the figure, \u0394ABC is an equilateral triangle. Find m\u2220BEC.<\/strong><\/p>\n<p><strong><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.5 question 1\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-5-questio.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.5 question 1\" width=\"402\" height=\"302\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>\u0394ABC is an equilateral triangle. (Given)<\/p>\n<p>Each angle of an equilateral triangle is 60 degrees.<\/p>\n<p>In quadrilateral ABEC:<\/p>\n<p>\u2220BAC + \u2220BEC = 180<sup>o<\/sup>\u00a0(Opposite angles of quadrilateral)<\/p>\n<p>60<sup>o<\/sup>\u00a0+ \u2220BEC = 180<sup>\u00a0o<\/sup><\/p>\n<p>\u2220BEC = 180<sup>\u00a0o<\/sup>\u00a0\u2013 60<sup>\u00a0o<\/sup><\/p>\n<p><em>\u2220BEC = 120<sup>\u00a0o<\/sup><\/em><\/p>\n<p><strong>Question 2: In the figure, \u0394 PQR is an isosceles triangle with PQ = PR and m\u2220PQR=35\u00b0. Find m\u2220QSR and m\u2220QTR.<\/strong><\/p>\n<p><strong><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.5 question 2\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-5-questio-1.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.5 question 2\" width=\"398\" height=\"299\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: \u0394PQR is an isosceles triangle with PQ = PR and m\u2220PQR = 35\u00b0<\/p>\n<p>In \u0394PQR:<\/p>\n<p>\u2220PQR = \u2220PRQ = 35<sup>o<\/sup>\u00a0(Angle opposite to equal sides)<\/p>\n<p>Again, by angle sum property<\/p>\n<p>\u2220P + \u2220Q + \u2220R = 180<sup>\u00a0o<\/sup><\/p>\n<p>\u2220P + 35<sup>\u00a0o<\/sup>\u00a0+ 35<sup>\u00a0o<\/sup>\u00a0= 180<sup>\u00a0o<\/sup><\/p>\n<p>\u2220P + 70<sup>\u00a0o<\/sup>\u00a0= 180<sup>\u00a0o<\/sup><\/p>\n<p>\u2220P = 180<sup>\u00a0o<\/sup>\u00a0\u2013 70<sup>\u00a0o<\/sup><\/p>\n<p>\u2220P = 110<sup>\u00a0o<\/sup><\/p>\n<p>Now, in quadrilateral SQTR,<\/p>\n<p>\u2220QSR + \u2220QTR = 180<sup>\u00a0o<\/sup>\u00a0(Opposite angles of quadrilateral)<\/p>\n<p>110<sup>\u00a0o<\/sup>\u00a0+ \u2220QTR = 180<sup>\u00a0o<\/sup><\/p>\n<p><em>\u2220QTR = 70<sup>\u00a0o<\/sup><\/em><\/p>\n<p><strong>Question 3: In the figure, O is the centre of the circle. If \u2220BOD = 160<sup>o<\/sup>, find the values of x and y.<\/strong><\/p>\n<p><strong><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.5 question 3\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-5-questio-2.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.5 question 3\" width=\"367\" height=\"276\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>From figure: \u2220BOD = 160<strong><sup>\u00a0o<\/sup><\/strong><\/p>\n<p>By degree measure theorem: \u2220BOD = 2 \u2220BCD<\/p>\n<p>160<strong><sup>\u00a0o<\/sup><\/strong>\u00a0= 2x<\/p>\n<p>or x = 80<strong><sup>\u00a0o<\/sup><\/strong><\/p>\n<p>Now, in quadrilateral ABCD,<\/p>\n<p>\u2220BAD + \u2220BCD = 180<strong><sup>\u00a0o<\/sup><\/strong>\u00a0(Opposite angles of Cyclic quadrilateral)<\/p>\n<p>y + x = 180<strong><sup>\u00a0o<\/sup><\/strong><\/p>\n<p>Putting the value of x,<\/p>\n<p>y + 80<strong><sup>\u00a0o<\/sup><\/strong>\u00a0= 180<strong><sup>\u00a0o<\/sup><\/strong><\/p>\n<p>y = 100<strong><sup>\u00a0o<\/sup><\/strong><\/p>\n<p><em>Answer: x = 80<strong><sup>\u00a0o<\/sup><\/strong>and y = 100<strong><sup>\u00a0o<\/sup><\/strong>.<\/em><\/p>\n<p><strong>Question 4: In the figure, ABCD is a cyclic quadrilateral. If \u2220BCD = 100<sup>o<\/sup>\u00a0and \u2220ABD = 70<sup>o<\/sup>, find \u2220ADB.<\/strong><\/p>\n<p><strong><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.5 question 4\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-5-questio-3.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.5 question 4\" width=\"394\" height=\"296\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>From figure:<\/p>\n<p>In quadrilateral ABCD,<\/p>\n<p>\u2220DCB + \u2220BAD = 180<strong><sup>o<\/sup><\/strong>\u00a0(Opposite angles of Cyclic quadrilateral)<\/p>\n<p>100<strong><sup>\u00a0o<\/sup><\/strong>\u00a0+ \u2220BAD = 180<strong><sup>o<\/sup><\/strong><\/p>\n<p>\u2220BAD = 80<sup>0<\/sup><\/p>\n<p>In \u0394 BAD:<\/p>\n<p>By angle sum property: \u2220ADB + \u2220DAB + \u2220ABD = 180<sup>\u00a0o<\/sup><\/p>\n<p>\u2220ADB + 80<sup>o<\/sup>\u00a0+ 70<sup>\u00a0o<\/sup>\u00a0= 180<sup>\u00a0o<\/sup><\/p>\n<p><em>\u2220ADB = 30<sup>o<\/sup><\/em><\/p>\n<p><strong>Question 5: If ABCD is a cyclic quadrilateral in which AD||BC (figure). Prove that \u2220B = \u2220C.<\/strong><\/p>\n<p><strong><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.5 question 5\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-5-questio-4.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.5 question 5\" width=\"352\" height=\"265\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: ABCD is a cyclic quadrilateral with AD \u2016 BC<\/p>\n<p>=&gt; \u2220A + \u2220C = 180<sup>o<\/sup>\u00a0\u2026\u2026\u2026(1)<\/p>\n<p>[Opposite angles of cyclic quadrilateral]<\/p>\n<p>and \u2220A + \u2220B = 180<sup>o<\/sup>\u00a0\u2026\u2026\u2026(2)<\/p>\n<p>[Co-interior angles]<\/p>\n<p>Forms (1) and (2), we have<\/p>\n<p><em>\u2220B = \u2220C<\/em><\/p>\n<p>Hence proved.<\/p>\n<p><strong>Question 6: In the figure, O is the centre of the circle. Find \u2220CBD.<\/strong><\/p>\n<p><strong><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.5 question 6\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-5-questio-5.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.5 question 6\" width=\"422\" height=\"317\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: \u2220BOC = 100<sup>o<\/sup><\/p>\n<p>By degree measure theorem: \u2220AOC = 2 \u2220APC<\/p>\n<p>100<sup>\u00a0o<\/sup>\u00a0= 2 \u2220APC<\/p>\n<p>or \u2220APC = 50<sup>\u00a0o<\/sup><\/p>\n<p>Again,<\/p>\n<p>\u2220APC + \u2220ABC = 180<sup>\u00a0o<\/sup>\u00a0(Opposite angles of a cyclic quadrilateral)<\/p>\n<p>50<sup>o<\/sup>\u00a0+ \u2220ABC = 180<sup>\u00a0o<\/sup><\/p>\n<p>or \u2220ABC = 130<sup>\u00a0o<\/sup><\/p>\n<p>Now, \u2220ABC + \u2220CBD = 180<sup>\u00a0o<\/sup>\u00a0(Linear pair)<\/p>\n<p>130<sup>o<\/sup>\u00a0+ \u2220CBD = 180<sup>\u00a0o<\/sup><\/p>\n<p>or\u00a0<em>\u2220CBD = 50<sup>\u00a0o<\/sup><\/em><\/p>\n<p><strong>Question 7: In the figure, AB and CD are diameters of a circle with centre O. If \u2220OBD = 50<sup>0<\/sup>, find \u2220AOC.<\/strong><\/p>\n<p><strong><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.5 question 7\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-5-questio-6.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.5 question 7\" width=\"414\" height=\"311\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: \u2220OBD = 50<strong><sup>0<\/sup><\/strong><\/p>\n<p>Here, AB and CD are the diameters of the circles with centre O.<\/p>\n<p>\u2220DBC = 90<strong><sup>0<\/sup><\/strong>\u00a0\u2026.(i)<\/p>\n<p>[Angle in the semi-circle]<\/p>\n<p>Also, \u2220DBC = 50<strong><sup>0<\/sup><\/strong>\u00a0+ \u2220OBC<\/p>\n<p>90<strong><sup>0<\/sup><\/strong>\u00a0= 50<strong><sup>0<\/sup><\/strong>\u00a0+ \u2220OBC<\/p>\n<p>or \u2220OBC = 40<strong><sup>0<\/sup><\/strong><\/p>\n<p>Again, By degree measure theorem: \u2220AOC = 2 \u2220ABC<\/p>\n<p><em>\u2220AOC = 2\u2220OBC = 2 x 40<strong><sup>0<\/sup><\/strong>\u00a0= 80<strong><sup>0<\/sup><\/strong><\/em><\/p>\n<p><strong>Question 8: On a semi-circle with AB as diameter, a point C is taken, so that m(\u2220CAB) = 30<sup>0<\/sup>. Find m(\u2220ACB) and m(\u2220ABC).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: m(\u2220CAB)= 30<strong><sup>0<\/sup><\/strong><\/p>\n<p>To Find: m(\u2220ACB) and m(\u2220ABC).<\/p>\n<p>Now,<\/p>\n<p>\u2220ACB = 90<strong><sup>0<\/sup><\/strong> (Angle in a semi-circle)<\/p>\n<p>Now,<\/p>\n<p>In \u25b3ABC, by angle sum property: \u2220CAB + \u2220ACB + \u2220ABC = 180<strong><sup>0<\/sup><\/strong><\/p>\n<p>30<strong><sup>0<\/sup><\/strong>\u00a0+ 90<strong><sup>0<\/sup><\/strong>\u00a0+ \u2220ABC = 180<strong><sup>0<\/sup><\/strong><\/p>\n<p>\u2220ABC = 60<strong><sup>0<\/sup><\/strong><\/p>\n<p><em>Answer: \u2220ACB = 90<strong><sup>0<\/sup><\/strong>\u00a0and \u2220ABC = 60<strong><sup>0<\/sup><\/strong><\/em><\/p>\n<p><strong>Question 9: In a cyclic quadrilateral ABCD if AB||CD and \u2220B = 70<sup>o<\/sup>\u00a0, find the remaining angles.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>A cyclic quadrilateral ABCD with AB||CD and \u2220B = 70<sup>o<\/sup>.<\/p>\n<p>\u2220B + \u2220C = 180<sup>o<\/sup>\u00a0(Co-interior angle)<\/p>\n<p>70<sup>0<\/sup>\u00a0+ \u2220C = 180<sup>0<\/sup><\/p>\n<p>\u2220C = 110<sup>0<\/sup><\/p>\n<p>And,<\/p>\n<p>=&gt; \u2220B + \u2220D = 180<sup>0<\/sup>\u00a0(Opposite angles of Cyclic quadrilateral)<\/p>\n<p>70<sup>0<\/sup>\u00a0+ \u2220D = 180<sup>0<\/sup><\/p>\n<p>\u2220D = 110<sup>0<\/sup><\/p>\n<p>Again, \u2220A + \u2220C = 180<sup>0<\/sup> (Opposite angles of a cyclic quadrilateral)<\/p>\n<p>\u2220A + 110<sup>0<\/sup>\u00a0= 180<sup>0<\/sup><\/p>\n<p>\u2220A = 70<sup>0<\/sup><\/p>\n<p><em>Answer: \u2220A = 70<sup>0<\/sup>\u00a0, \u2220C = 110<sup>0\u00a0<\/sup>and \u2220D = 110<sup>0<\/sup><\/em><\/p>\n<p><strong>Question 10: In a cyclic quadrilateral ABCD, if m \u2220A = 3(m\u2220C). Find m \u2220A.<br \/>Solution:<\/strong><\/p>\n<p>\u2220A + \u2220C = 180<sup>o<\/sup>\u00a0\u2026..(1)<\/p>\n<p>[Opposite angles of cyclic quadrilateral]<\/p>\n<p>Since m \u2220A = 3(m\u2220C) (given)<\/p>\n<p>=&gt; \u2220A = 3\u2220C \u2026(2)<\/p>\n<p>Equation (1) =&gt; 3\u2220C + \u2220C = 180<sup>\u00a0o<\/sup><\/p>\n<p>or 4\u2220C = 180<sup>o<\/sup><\/p>\n<p>or \u2220C = 45<sup>o<\/sup><\/p>\n<p>From equation (2)<\/p>\n<p><em>\u2220A = 3 x 45<sup>o<\/sup>\u00a0= 135<sup>o<\/sup><\/em><\/p>\n<p><strong>Question 11: In the figure, O is the centre of the circle \u2220DAB = 50\u00b0. Calculate the values of x and y.<\/strong><\/p>\n<p><strong><img class=\"\" title=\"RD sharma class 9 maths chapter 16 ex 16.5 question 11\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-class-9-maths-chapter-16-ex-16-5-questio-7.png\" alt=\"RD sharma class 9 maths chapter 16 ex 16.5 question 11\" width=\"370\" height=\"278\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given : \u2220DAB = 50<sup>o<\/sup><\/p>\n<p>By degree measure theorem: \u2220BOD = 2 \u2220BAD<\/p>\n<p>so,\u00a0<em>x = 2( 50<sup>0<\/sup>) = 100<sup>0<\/sup><\/em><\/p>\n<p>Since ABCD is a cyclic quadrilateral, we have<\/p>\n<p>\u2220A + \u2220C = 180<sup>0<\/sup><\/p>\n<p>50<sup>0<\/sup>\u00a0+ y = 180<sup>0<\/sup><\/p>\n<p><em>y = 130<sup>0<\/sup><\/em><\/p>\n<hr \/>\n<h3><span class=\"ez-toc-section\" id=\"exercise-vsaqs-page-no-1689\"><\/span>Exercise VSAQs Page No: 16.89<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Question 1: In the figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If \u2220APB = 70\u00b0, find \u2220ACB.<\/strong><\/p>\n<p><img class=\"\" title=\"rd sharma solution class 9 chapter 16 vsaq 1\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-solution-class-9-chapter-16-vsaq-1.png\" alt=\"rd sharma solution class 9 chapter 16 vsaq 1\" width=\"441\" height=\"275\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>By degree measure theorem: \u2220AOB = 2 \u2220APB<\/p>\n<p>so, \u2220AOB = 2 \u00d7 70\u00b0 = 140\u00b0<\/p>\n<p>Since AOBC is a cyclic quadrilateral, we have<\/p>\n<p>\u2220ACB + \u2220AOB = 180\u00b0<\/p>\n<p>\u2220ACB + 140\u00b0 = 180\u00b0<\/p>\n<p><em>\u2220ACB = 40\u00b0<\/em><\/p>\n<p><strong>Question 2: In the figure, two congruent circles with centres O and O\u2019 intersect at A and B. If \u2220AO\u2019B = 50\u00b0, then find \u2220APB.<\/strong><\/p>\n<p><img class=\"\" title=\"rd sharma solution class 9 chapter 16 vsaq 2\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-solution-class-9-chapter-16-vsaq-2.png\" alt=\"rd sharma solution class 9 chapter 16 vsaq 2\" width=\"449\" height=\"280\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>As we are given, both triangles are congruent which means their corresponding angles are equal.<\/p>\n<p>Therefore, \u2220AOB = AO\u2019B = 50\u00b0<\/p>\n<p>Now, by the degree measure theorem, we have<\/p>\n<p><em>\u2220APB = \u2220AOB\/2 = 25<sup>0<\/sup><\/em><\/p>\n<p><strong>Question 3: In the figure, ABCD is a cyclic quadrilateral in which \u2220BAD=75\u00b0, \u2220ABD=58\u00b0 and \u2220ADC=77\u00b0, AC and BD intersect at P. Then, find \u2220DPC.<\/strong><\/p>\n<p><img class=\"\" title=\"rd sharma solution class 9 chapter 16 vsaq 3\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-solution-class-9-chapter-16-vsaq-3.png\" alt=\"rd sharma solution class 9 chapter 16 vsaq 3\" width=\"401\" height=\"295\" \/><\/p>\n<p><strong>Solution<\/strong>:<\/p>\n<p>\u2220DBA = \u2220DCA = 58<sup>0<\/sup>\u00a0\u2026(1)<\/p>\n<p>[Angles in the same segment]<\/p>\n<p>ABCD is a cyclic quadrilateral :<\/p>\n<p>The sum of opposite angles = 180 degrees<\/p>\n<p>\u2220A +\u2220C = 180<sup>0<\/sup><\/p>\n<p>75<sup>0<\/sup>\u00a0+ \u2220C = 180<sup>0<\/sup><\/p>\n<p>\u2220C = 105<sup>0<\/sup><\/p>\n<p>Again, \u2220ACB + \u2220ACD = 105<sup>0<\/sup><\/p>\n<p>\u2220ACB + 58<sup>0<\/sup>\u00a0= 105<sup>0<\/sup><\/p>\n<p>or \u2220ACB = 47<sup>0<\/sup>\u00a0\u2026(2)<\/p>\n<p>Now, \u2220ACB = \u2220ADB = 47<sup>0<\/sup><\/p>\n<p>[Angles in the same segment]<\/p>\n<p>Also, \u2220D = 77<sup>0<\/sup>\u00a0(Given)<\/p>\n<p>Again From figure, \u2220BDC + \u2220ADB = 77<sup>0<\/sup><\/p>\n<p>\u2220BDC + 47<sup>0<\/sup>\u00a0= 77<sup>0<\/sup><\/p>\n<p>\u2220BDC = 30<sup>0<\/sup><\/p>\n<p>In triangle DPC<\/p>\n<p>\u2220PDC + \u2220DCP + \u2220DPC = 180<sup>0<\/sup><\/p>\n<p>30<sup>0<\/sup>\u00a0+ 58<sup>0<\/sup>\u00a0+ \u2220DPC = 180<sup>0<\/sup><\/p>\n<p><em>or \u2220DPC = 92<sup>0\u00a0<\/sup><\/em><\/p>\n<p><strong>Question 4: In the figure, if \u2220AOB = 80\u00b0 and \u2220ABC=30\u00b0, then find \u2220CAO.<\/strong><\/p>\n<p><img class=\"\" title=\"rd sharma solution class 9 chapter 16 vsaq 4\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/rd-sharma-solution-class-9-chapter-16-vsaq-4.png\" alt=\"rd sharma solution class 9 chapter 16 vsaq 4\" width=\"329\" height=\"242\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: \u2220AOB = 80<sup>0<\/sup>\u00a0and \u2220ABC = 30<sup>0<\/sup><\/p>\n<p>To find: \u2220CAO<\/p>\n<p>Join OC.<\/p>\n<p>Central angle subtended by arc AC = \u2220COA<\/p>\n<p>then \u2220COA = 2 x \u2220ABC = 2 x 30<sup>0<\/sup>\u00a0= 60<sup>0<\/sup>\u00a0\u2026(1)<\/p>\n<p>In triangle OCA,<\/p>\n<p>OC = OA<\/p>\n<p>[same radii]<\/p>\n<p>\u2220OCA = \u2220CAO \u2026(2)<\/p>\n<p>[Angle opposite to equal sides]<\/p>\n<p>In triangle COA,<\/p>\n<p>\u2220OCA + \u2220CAO + \u2220COA = 180<sup>0<\/sup><\/p>\n<p>From (1) and (2), we get<\/p>\n<p>2\u2220CAO + 60<sup>0\u00a0<\/sup>= 180<sup>0<\/sup><\/p>\n<p><em>\u2220CAO = 60<sup>0<\/sup><\/em><\/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<ul>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\"><strong>RD Sharma class 9 chapter 16 exercise 16a<\/strong>: This exercise is based on the topics related to the position of a point with respect to a circle, Concentric Circles, Length of an arc, Circular Disc, Minor &amp; Major Arc, &amp; many more.\u00a0<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\"><strong>RD Sharma class 9 chapter 16 exercise 16b:<\/strong> This exercise includes topics related to the congruence of circles &amp; arcs. Practising RD Sharma&#8217;s class 9 chapter 16 exercise 16b questions enable the students to build an in-depth understanding of the basics of a circle.\u00a0<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\"><strong>RD Sharma class 9 chapter 16 exercise 16c:<\/strong> This exercise includes topics based on a few important results on equal chords. The solutions of RD Sharma class 9 chapter 16 exercise 16c are helpful to solve several circle problems. The students will study more circles concepts in detail in order to obtain better marks in the exam.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\"><strong>RD Sharma class 9 chapter 16 exercise 16d:<\/strong> This exercise includes topics based on arcs &amp; angles subtended by them. Each answer is created by subject experts that assist the students to understand this concept in a better and more fun way. The students can practice RD Sharma class 9 chapter 16 exercise 16d &amp; clear their doubts to secure excellent marks in the final mathematics exam.<\/span><\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"important-topics-in-the-exercise\"><\/span><span style=\"font-weight: 400;\">Important Topics in the Exercise<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>RD Sharma Solutions Class 9 Maths Chapter 16 Circles <span style=\"font-weight: 400;\">include some important topics that are listed below:<\/span><\/p>\n<ul>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Introduction of Circle<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Position of a point with respect to a circle<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Circular Disc<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Concentric Circles<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">The degree measure of an arc<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Chord and segment of a circle<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Congruence of circles and arcs<\/span><\/li>\n<\/ul>\n<p>This is the complete blog on RD Sharma Solutions Class 9 Maths Chapter 16. If <span style=\"font-weight: 400;\">you have any doubts regarding the <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> Class 9 Maths exam, ask in the comments.<\/span><\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-9-maths-chapter-16-circles\"><\/span><strong>FAQs on RD Sharma Solutions Class 9 Maths Chapter 16 Circles<\/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-1630755111220\" 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-16\"><\/span>From where can I download the PDF of RD Sharma Solutions Class 9 Maths Chapter 16?<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-1630755173830\" 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-16\"><\/span>How much does it cost to download the PDF of RD Sharma Solutions for Class 9 Maths Chapter 16?<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-1630755205219\" 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-16-pdf-offline\"><\/span>Can I access the RD Sharma Solutions for Class 9 Maths Chapter 16\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 16 &#8211; Circles: RD Sharma Solutions Class 9 Maths Chapter 16 enables them also to study all topics more effectively &amp; clear their doubts quickly. The students who want to obtain higher marks in the exams must regularly practice mathematics subject exercises with the help of RD Sharma &#8230; <a title=\"RD Sharma Solutions Class 9 Maths Chapter 16 &#8211; Circles (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-16-circles\/\" aria-label=\"More on RD Sharma Solutions Class 9 Maths Chapter 16 &#8211; Circles (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":124557,"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\/63793"}],"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=63793"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63793\/revisions"}],"predecessor-version":[{"id":511189,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/63793\/revisions\/511189"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/124557"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=63793"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=63793"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=63793"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}