{"id":125440,"date":"2021-09-15T12:37:23","date_gmt":"2021-09-15T07:07:23","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125440"},"modified":"2021-09-15T12:37:27","modified_gmt":"2021-09-15T07:07:27","slug":"rd-sharma-class-9-solutions-chapter-16-exercise-16-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-16-exercise-16-2\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2 (Updated 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-127786\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-16-Exercise-16.2.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-16-Exercise-16.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-16-Exercise-16.2-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\"><strong>RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2<\/strong>: Our experts have created these <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-16-circles\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 16<\/a> Exercise 16.2. You can easily download the free PDF. These solutions make it very useful for Grade 9 exams.<\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-for-maths\/\" target=\"_blank\" rel=\"noopener\"><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Maths Solutions&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Maths Solutions<\/span><\/a><\/li>\n<\/ul>\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-69d896501d7b4\" 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\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-16-exercise-16-2\/#access-rd-sharma-class-9-solutions-chapter-16-exercise-162\" title=\"Access RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2\">Access RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2<\/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-class-9-solutions-chapter-16-exercise-16-2\/#faq-rd-sharma-class-9-solutions-chapter-16-exercise-162\" title=\"FAQ: RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2\">FAQ: RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2<\/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-class-9-solutions-chapter-16-exercise-16-2\/#can-i-download-rd-sharma-class-9-solutions-chapter-16-exercise-162-pdf-free\" title=\"Can I download RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2 PDF free?\">Can I download RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2 PDF free?<\/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-class-9-solutions-chapter-16-exercise-16-2\/#what-are-the-benefits-of-studying-rd-sharma-class-9-solutions\" title=\"What are the benefits of studying RD Sharma Class 9 Solutions?\">What are the benefits of studying RD Sharma Class 9 Solutions?<\/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-class-9-solutions-chapter-16-exercise-16-2\/#is-rd-sharma-enough-for-class-12-maths\" title=\"Is RD Sharma enough for Class 12 Maths?\">Is RD Sharma enough for Class 12 Maths?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-9-solutions-chapter-16-exercise-162\"><\/span>Download <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2<\/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\/09\/RD-SHARMA-Solutions-Class-9-Maths-Chapter-16-Ex-16.2.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-SHARMA-Solutions-Class-9-Maths-Chapter-16-Ex-16.2.pdf\">RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-class-9-solutions-chapter-16-exercise-162\"><\/span>Access <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Question 1.<br \/>Draw an angle and label it as \u2220BAC. Construct another angle, equal to \u2220BAC.<br \/>Solution:<br \/>Steps of construction :<\/p>\n<p>(i) Draw an angle BAC.<br \/>(ii) Draw a line DF.<br \/>(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.<br \/>(iv) Cut off PQ = LM and join DQ and produce it to E, then<br \/>\u2220EDF = \u2220BAC.<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1904\/44732258075_b255f3febc_o.png\" alt=\"RD Sharma Class 9 Chapter 16 Circles\" width=\"343\" height=\"188\" \/><\/p>\n<p>\u00a0<\/p>\n<p>Question 2.<br \/>Draw an obtuse angle. Bisect it. Measure each of the angles so obtained.<br \/>Solution:<br \/>Steps of construction :<\/p>\n<p>(i) Draw an angle ABC which is an obtuse i.e. more than 90\u00b0.<br \/>(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.<br \/>(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.<br \/>(iv) Join BG and produce it to D.<br \/>ThenBD is die bisector of \u2220ABC.<br \/>On measuring each part, we find each angle = 53\u00b0.<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1927\/44732257755_8351697df1_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 16 Circles\" width=\"257\" height=\"207\" \/><\/p>\n<div class=\"google-auto-placed ap_container\">\u00a0<\/div>\n<p>Question 3.<br \/>Using your protractor, draw an angle of measure 108\u00b0. With this angle as given, draw an angle of 54\u00b0.<br \/>Solution:<br \/>Steps of construction :<\/p>\n<p>(i) With the help of protractor draw an angle ABC = 108\u00b0.<br \/>As 54\u00b0 =\u00a0<span id=\"MathJax-Element-5-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-21\" class=\"math\"><span id=\"MathJax-Span-22\" class=\"mrow\"><span id=\"MathJax-Span-23\" class=\"mfrac\"><span id=\"MathJax-Span-24\" class=\"mn\">1<\/span><span id=\"MathJax-Span-25\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 108\u00b0<br \/>\u2234 We shall bisect it.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1911\/44732257445_0280ecf39c_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 16 Circles\" width=\"267\" height=\"214\" \/><\/p>\n<p>(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.<br \/>(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G<br \/>(iv) Join BG and produce it to D.<br \/>Then \u2220DBC = 54\u00b0.<\/p>\n<p>Question 4.<br \/>Using protractor, draw a right angle. Bisect it to get an angle of measure 45\u00b0.<br \/>Solution:<br \/>Steps of construction :<\/p>\n<p>(i) Using protractor, draw a right angle \u2206ABC.<br \/>i. e. \u2220ABC = 90\u00b0.<br \/>(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.<br \/>(iii) With centre E and F, draw arcs intersecting each other at G.<br \/>(iv) Join BG and produce it to D.<br \/>Then BD is the bisector of \u2220ABC.<br \/>\u2234 \u2220DBC =<span id=\"MathJax-Element-6-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-26\" class=\"math\"><span id=\"MathJax-Span-27\" class=\"mrow\"><span id=\"MathJax-Span-28\" class=\"mfrac\"><span id=\"MathJax-Span-29\" class=\"mn\">1<\/span><span id=\"MathJax-Span-30\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 90\u00b0 = 45\u00b0.<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1934\/44732256975_8ac2797c67_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 16 Circles\" width=\"196\" height=\"185\" \/><\/p>\n<p>\u00a0<\/p>\n<p>Question 5.<br \/>Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.<br \/>Solution:<\/p>\n<p>Steps of construction :<br \/>(i) Draw a linear pair \u2220DCA and \u2220DCB.<br \/>(ii) Draw the bisectors of \u2220DCA and \u2220DCB. Forming \u2220ECF on measuring we get \u2220ECF = 90\u00b0.<br \/>Verification : \u2235\u2220DCA + \u2220DCB = 180\u00b0<br \/>\u21d2\u00a0<span id=\"MathJax-Element-7-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-31\" class=\"math\"><span id=\"MathJax-Span-32\" class=\"mrow\"><span id=\"MathJax-Span-33\" class=\"mfrac\"><span id=\"MathJax-Span-34\" class=\"mn\">1<\/span><span id=\"MathJax-Span-35\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220DCA +\u00a0<span id=\"MathJax-Element-8-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-36\" class=\"math\"><span id=\"MathJax-Span-37\" class=\"mrow\"><span id=\"MathJax-Span-38\" class=\"mfrac\"><span id=\"MathJax-Span-39\" class=\"mn\">1<\/span><span id=\"MathJax-Span-40\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220DCB = 180\u00b0 x\u00a0<span id=\"MathJax-Element-9-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-41\" class=\"math\"><span id=\"MathJax-Span-42\" class=\"mrow\"><span id=\"MathJax-Span-43\" class=\"mfrac\"><span id=\"MathJax-Span-44\" class=\"mn\">1<\/span><span id=\"MathJax-Span-45\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 90\u00b0<br \/>\u2234 \u2220ECF = 90\u00b0<br \/>i.e. EC and FC are perpendicular to each other.<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1958\/43828262680_5c5321378a_o.png\" alt=\"Circles Class 9 RD Sharma Solutions\" width=\"272\" height=\"197\" \/><\/p>\n<p>Question 6.<br \/>Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line<br \/>Solution:<\/p>\n<p>Steps of construction :<br \/>(i) Draw two lines AB and CD intersecting each other at O.<br \/>(ii) Draw the bisector of \u2220AOD and also the bisector of \u2220BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1913\/44732256625_1b706ac295_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 16 Circles\" width=\"284\" height=\"168\" \/><\/p>\n<p>Question 7.<br \/>Using ruler and compasses only, draw a right angle.<br \/>Solution:<br \/>Steps of construction :<\/p>\n<p>(i) Draw a line segment BC.<br \/>(ii) With centre B and a suitable radius, draw an arc meeting BC at E.<br \/>(iii) With centre E and same radius, cut off arcs EF and FG.<br \/>(iv) Bisect arc FG at H.<br \/>(v) Join BH and produce it to A.<br \/>Then \u2220ABC = 90\u00b0.<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1949\/43828262410_4930c96df2_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 16 Circles\" width=\"185\" height=\"205\" \/><\/p>\n<p>Question 8.<br \/>Using ruler and compasses only, draw an angle of measure 135\u00b0.<br \/>Solution:<br \/>Steps of construction :<\/p>\n<p>(i) Draw a line DC and take a point B on it.<br \/>(ii) With centre B and a suitable radius draw an arc meeting BC at P.<br \/>(iii) With centre P, cut off arcs PQ, QR and RS.<br \/>(iv) Bisect as QR at T and join BT and produce it to E.<br \/>(v) Now bisect the arc KS at RL.<br \/>(vi) Join BL and produce it to A.<br \/>Now \u2220ABC = 135\u00b0.<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1957\/30705520537_fca206f615_o.png\" alt=\"Class 9 Maths Chapter 16 Circles RD Sharma Solutions\" width=\"281\" height=\"206\" \/><\/p>\n<p>Question 9.<br \/>Using a protractor, draw an angle of measure 72\u00b0. With this angle as given, draw angles of measure 36\u00b0 and 54\u00b0.<br \/>Solution:<\/p>\n<p>Steps of construction :<\/p>\n<p>(i) Draw an angle ABC = 12\u00b0 with the help of protractor.<br \/>(ii) With centre B and a suitable radius, draw an arc EF.<br \/>(iii) With centre E and F, draw arcs intersecting<br \/>each other at G and produce it to D.<br \/>Then BD is the bisector of \u2220ABC.<br \/>\u2234 \u2220DBC = 72\u00b0 x\u00a0<span id=\"MathJax-Element-10-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-46\" class=\"math\"><span id=\"MathJax-Span-47\" class=\"mrow\"><span id=\"MathJax-Span-48\" class=\"mfrac\"><span id=\"MathJax-Span-49\" class=\"mn\">1<\/span><span id=\"MathJax-Span-50\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 36\u00b0.<\/p>\n<p>(iv) Again bisect \u2220ABD in the same way then PB is the bisector of \u2220ABD.<br \/>\u2234 \u2220PBC = 36\u00b0 +\u00a0<span id=\"MathJax-Element-11-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-51\" class=\"math\"><span id=\"MathJax-Span-52\" class=\"mrow\"><span id=\"MathJax-Span-53\" class=\"mfrac\"><span id=\"MathJax-Span-54\" class=\"mn\">1<\/span><span id=\"MathJax-Span-55\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 36\u00b0<br \/>= 36\u00b0 + 18\u00b0 = 54\u00b0<br \/>Hence \u2220PBC = 54\u00b0<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1952\/43828262240_1b0d0f4297_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 16 Circles\" width=\"254\" height=\"272\" \/><\/p>\n<p>Question 10.<br \/>Construct the following angles at the initial point of a given ray and justify the construction:<br \/>(i) 45\u00b0<br \/>(ii) 90\u00b0<br \/>Solution:<br \/>(i) 45\u00b0<\/p>\n<p>Steps of construction :<br \/>(a) Draw a line segment BC.<br \/>(b) With centre B and a suitable radius draw an arc meeting BC at E.<br \/>(c) With centre E, cut off equal arcs EF and FG.<br \/>(d) Bisect FG at H.<br \/>(e) Join BH and produce to X so that \u2220XBC = 90\u00b0.<br \/>(f) Bisect \u2220XBC so that \u2220ABC = 45\u00b0.<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1916\/43828262080_9813322e16_o.png\" alt=\"RD Sharma Class 9 Book Chapter 16 Circles\" width=\"207\" height=\"203\" \/><br \/>(ii) 90\u00b0<\/p>\n<p>Steps of construction :<\/p>\n<p>(a) Draw a line segment BC.<br \/>(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG<br \/>(c) Now bisect the arc EG at H.<br \/>(d) Join BH and produce it to A.<br \/>\u2234 \u2220ABC = 90\u00b0.<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1938\/44732255705_12b4859420_o.png\" alt=\"Circles With Solutions PDF RD Sharma Class 9 Solutions\" width=\"182\" height=\"204\" \/><\/p>\n<p>Question 11.<br \/>Construct the angles of the following measurements:<br \/>(i) 30\u00b0<br \/>(ii) 75\u00b0<br \/>(iii) 105\u00b0<br \/>(iv) 135\u00b0<br \/>(v) 15\u00b0<br \/>(vi) 22\u00a0<span id=\"MathJax-Element-12-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-56\" class=\"math\"><span id=\"MathJax-Span-57\" class=\"mrow\"><span id=\"MathJax-Span-58\" class=\"mfrac\"><span id=\"MathJax-Span-59\" class=\"mn\">1<\/span><span id=\"MathJax-Span-60\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00b0<\/p>\n<p>Solution:<\/p>\n<p>(i) 30\u00b0<br \/>Steps of construction :<br \/>(a) Draw a line segment BC.<br \/>(b) With centre B and a suitable radius draw an arc meeting BC at E.<br \/>(c) Cut off arcs EF and bisect it at G.<br \/>So that \u2220ABC = 30\u00b0.<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1926\/44732255575_9672fb304d_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 16 Circles\" width=\"211\" height=\"161\" \/><br \/>(ii) 75\u00b0<br \/>Steps of construction :<\/p>\n<p>(a) Draw a line segment BC.<br \/>(b) With centre B and a suitable radius, draw an arc meeting BC at E.<br \/>(c) Cut off arc EF = FG from E.<br \/>(d) Bisect the arc FG at K and join BK so that \u2220KBC = 90\u00b0.<br \/>(e) Now bisect arc HF at L and join BL and produce it to A so that \u2220ABC = 75\u00b0.<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1973\/44732255425_38b7cf12b0_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 16 Circles\" width=\"183\" height=\"193\" \/><\/p>\n<p>(iii) 105\u00b0<br \/>Steps of construction :<\/p>\n<p>(a) Draw a line segment BC.<br \/>(b) With centre B and a suitable radius draw an arc meeting BC at E.<br \/>(c) From E, cut off arc EF = FG and divide FG at H.<br \/>(d) Join BH meeting the arc at K.<br \/>(e) Now bisect the arc KG at L.<br \/>Join BL and produce it to A.<br \/>Then \u2220ABC = 105\u00b0.<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1915\/44732255285_06e5e4f6e0_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 16 Circles\" width=\"201\" height=\"199\" \/><br \/>(iv) 135\u00b0<\/p>\n<p>Steps of construction :<br \/>(a) Draw a line segment BC.<br \/>(b) With centre B and a suitable radius, draw an arc meeting BC at E.<br \/>(c) From E, cut off EF = FG = GH.<br \/>(d) Bisect arc FG at K, and join them.<br \/>(e) Bisect arc KH at L.<br \/>(f) Join BL and produce it to A, then \u2220ABC = 135\u00b0.<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1975\/43828261650_3d1e745969_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 16 Circles\" width=\"256\" height=\"197\" \/><br \/>(v) 15\u00b0<br \/>Steps of construction :<\/p>\n<p>(a) Draw a line segment BC.<br \/>(b) With centre B and a suitable radius, draw an arc meeting BC at E.<br \/>(c) Cut off arc EF from E and bisect it at G. Then \u2220GBC = 30\u00b0.<br \/>(d) Again bisect the arc EJ at H.<br \/>(e) Join BH and produce it to A.<br \/>Then \u2220ABC = 15\u00b0.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1919\/44732254965_f19f4fd30b_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 16 Circles\" width=\"197\" height=\"149\" \/><br \/>(vi) 22\u00a0<span id=\"MathJax-Element-13-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-61\" class=\"math\"><span id=\"MathJax-Span-62\" class=\"mrow\"><span id=\"MathJax-Span-63\" class=\"mfrac\"><span id=\"MathJax-Span-64\" class=\"mn\">1<\/span><span id=\"MathJax-Span-65\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00b0<br \/>Steps of construction :<br \/>(a) Draw a line segment BC.<br \/>(b) With centre B and a suitable radius, draw an arc and from E, cut off EF = FG<br \/>(c) Bisect FG at H so that \u2220HBC = 90\u00b0.<br \/>(d) Now bisect \u2220HBC at K. So that \u2220YBC = 45\u00b0.<br \/>(e) Again bisect \u2220YBC at J. So thttt \u2220ABC = 22\u00a0<span id=\"MathJax-Element-14-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-66\" class=\"math\"><span id=\"MathJax-Span-67\" class=\"mrow\"><span id=\"MathJax-Span-68\" class=\"mfrac\"><span id=\"MathJax-Span-69\" class=\"mn\">1<\/span><span id=\"MathJax-Span-70\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1968\/30705519967_c4f9dd788e_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 16 Circles\" width=\"197\" height=\"217\" \/><\/p>\n<p>We have included all the information regarding <a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2<\/span><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 20 Exercise 20.1&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">. If you have any query feel free to ask in the comment section.\u00a0<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"faq-rd-sharma-class-9-solutions-chapter-16-exercise-162\"><\/span>FAQ: <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4540,&quot;5&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;6&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;7&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;8&quot;:{&quot;1&quot;:[{&quot;1&quot;:2,&quot;2&quot;:0,&quot;5&quot;:[null,2,0]},{&quot;1&quot;:0,&quot;2&quot;:0,&quot;3&quot;:3},{&quot;1&quot;:1,&quot;2&quot;:0,&quot;4&quot;:1}]},&quot;10&quot;:2,&quot;11&quot;:0,&quot;15&quot;:&quot;Arial&quot;}\">RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2<\/span><span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631092823509\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-download-rd-sharma-class-9-solutions-chapter-16-exercise-162-pdf-free\"><\/span>Can I download RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2 PDF free?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, you can download RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2 PDF free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631092874239\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-studying-rd-sharma-class-9-solutions\"><\/span>What are the benefits of studying RD Sharma Class 9 Solutions?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>By practicing RD Sharma Class 9 Solutions, students can earn higher academic grades. Our experts solve these solutions with utmost accuracy to help students in their studies.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631092875384\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-rd-sharma-enough-for-class-12-maths\"><\/span>Is RD Sharma enough for Class 12 Maths?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>RD Sharma is a good book that gives you thousands of questions to practice.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2: Our experts have created these RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2. You can easily download the free PDF. These solutions make it very useful for Grade 9 exams. RD Sharma Class 9 Maths Solutions Download RD Sharma Class 9 Solutions Chapter 16 Exercise &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2 (Updated 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-16-exercise-16-2\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 16 Exercise 16.2 (Updated 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":127786,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73719],"tags":[76786,76827],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125440"}],"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\/244"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=125440"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125440\/revisions"}],"predecessor-version":[{"id":127826,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125440\/revisions\/127826"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/127786"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125440"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125440"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125440"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}