{"id":125320,"date":"2021-09-14T20:57:54","date_gmt":"2021-09-14T15:27:54","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125320"},"modified":"2021-09-14T20:57:59","modified_gmt":"2021-09-14T15:27:59","slug":"rd-sharma-class-9-solutions-chapter-15-exercise-15-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-15-exercise-15-2\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 15 Exercise 15.2 (Updated 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-127674\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-15-Exercise-15.2.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Exercise 15.2 \" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-15-Exercise-15.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-15-Exercise-15.2-768x432.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 15 Exercise 15.2&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 15 Exercise 15.2<\/strong>: Our experts have created this PDF in the easy manner. You can easily understand these <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-15-areas-of-parallelograms-and-triangles\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 15<\/a>\u00a0Exercise 15.2 solutions while preparing for your Class 9 exams.\u00a0<\/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 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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-class-9-solutions-chapter-15-exercise-15-2\/#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-15-exercise-152\"><\/span>Download <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 15 Exercise 15.2&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 15 Exercise 15.2<\/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-15-Ex-15.2.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-15-Ex-15.2.pdf\">RD Sharma Class 9 Solutions Chapter 15 Exercise 15.2<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-class-9-solutions-chapter-15-exercise-152\"><\/span>Access <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 15 Exercise 15.2&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 15 Exercise 15.2<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Question 1.<\/p>\n<p>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.<\/p>\n<p>Solution:<\/p>\n<p>Radius of circle with centre O is OA = 8 cm<br \/>Length of chord AB = 12 cm<br \/>OC \u22a5 AB which bisects AB at C<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1934\/44920312954_433d96dc6a_o.png\" alt=\"RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"153\" height=\"142\" \/><\/p>\n<p>\u2234 AC = CB = 12 x\u00a0<span id=\"MathJax-Element-1-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-1\" class=\"math\"><span id=\"MathJax-Span-2\" class=\"mrow\"><span id=\"MathJax-Span-3\" class=\"mfrac\"><span id=\"MathJax-Span-4\" class=\"mn\">1<\/span><span id=\"MathJax-Span-5\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 6 cm<br \/>In \u2206OAC,<br \/>OA2 = OC2 + AC2 (Pythagoras Theorem)<br \/>\u21d2 (8)2 = OC2 + (6)2<br \/>\u21d2 64 = OC2 + 36<br \/>OC2 = 64 \u2013 36 = 28<br \/>\u2234 OC =\u00a0<span id=\"MathJax-Element-2-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-6\" class=\"math\"><span id=\"MathJax-Span-7\" class=\"mrow\"><span id=\"MathJax-Span-8\" class=\"msqrt\"><span id=\"MathJax-Span-9\" class=\"mrow\"><span id=\"MathJax-Span-10\" class=\"mn\">28<\/span><\/span>\u2212\u2212\u221a<\/span><\/span><\/span><\/span>\u00a0=\u00a0<span id=\"MathJax-Element-3-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-11\" class=\"math\"><span id=\"MathJax-Span-12\" class=\"mrow\"><span id=\"MathJax-Span-13\" class=\"msqrt\"><span id=\"MathJax-Span-14\" class=\"mrow\"><span id=\"MathJax-Span-15\" class=\"mn\">4<\/span><span id=\"MathJax-Span-16\" class=\"mo\">\u00d7<\/span><span id=\"MathJax-Span-17\" class=\"mn\">7<\/span><\/span>\u2212\u2212\u2212\u2212\u221a<\/span><\/span><\/span><\/span>\u00a0cm<br \/>= 2 x 2.6457 = 5.291 cm<\/p>\n<p>Question 2.<\/p>\n<p>Find the length of a chord which is at a distance of 5 cm from the centre of a circle of radius 10 cm.<\/p>\n<p>Solution:<\/p>\n<p>Let AB be a chord of a circle with radius 10 cm. OC \u22a5 AB<br \/>\u2234 OA = 10 cm<br \/>OC = 5 cm<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1979\/43827187230_b6340c7749_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.2 - 2\" width=\"179\" height=\"165\" \/><\/p>\n<p>\u2235 OC divides AB into two equal parts<br \/>i.e. AC = CB<br \/>Now in right AOAC,<\/p>\n<p>OA2 = OC2 + AC2 (Pythagoras Theorem)<br \/>\u21d2 (10)2 = (5)2 + AC2<br \/>\u21d2 100 = 25 + AC2<br \/>\u21d2 AC2 = 100 \u2013 25 = 75<br \/>\u2234 AC =\u00a0<span id=\"MathJax-Element-4-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-18\" class=\"math\"><span id=\"MathJax-Span-19\" class=\"mrow\"><span id=\"MathJax-Span-20\" class=\"msqrt\"><span id=\"MathJax-Span-21\" class=\"mrow\"><span id=\"MathJax-Span-22\" class=\"mn\">75<\/span><\/span>\u2212\u2212\u221a<\/span><\/span><\/span><\/span>=\u00a0<span id=\"MathJax-Element-5-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-23\" class=\"math\"><span id=\"MathJax-Span-24\" class=\"mrow\"><span id=\"MathJax-Span-25\" class=\"msqrt\"><span id=\"MathJax-Span-26\" class=\"mrow\"><span id=\"MathJax-Span-27\" class=\"mn\">25<\/span><span id=\"MathJax-Span-28\" class=\"mo\">\u00d7<\/span><span id=\"MathJax-Span-29\" class=\"mn\">3<\/span><\/span>\u2212\u2212\u2212\u2212\u2212\u221a<\/span><\/span><\/span><\/span>\u00a0= 5 x 1.732<br \/>\u2234 AB = 2 x AC = 2 x 5 x 1.732 = 10 x 1.732 = 17.32 cm<\/p>\n<p>Question 3.<br \/>Find the length of a chord which is at a distance of 4 cm from the centre of the circle of radius 6 cm.<br \/>Solution:<\/p>\n<p>In a circle with centre O and radius 6 cm and a chord AB at a distance of 4 cm from the centre of the circle<br \/>i.e. OA = 6 cm and OL \u22a5 AB, OL = 4 cm<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1902\/44920312744_60f5001b41_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"186\" height=\"173\" \/><br \/>\u2235 Perpendicular OL bisects the chord AB at L 1<br \/>\u2234 AL = LB=<span id=\"MathJax-Element-6-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-30\" class=\"math\"><span id=\"MathJax-Span-31\" class=\"mrow\"><span id=\"MathJax-Span-32\" class=\"mfrac\"><span id=\"MathJax-Span-33\" class=\"mn\">1<\/span><span id=\"MathJax-Span-34\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0AB<br \/>Now in right \u2206OAL,<\/p>\n<p>OA2 = OL2 + AL2 (Pythagoras Theorem)<br \/>(6)2 = (4)2 + AL2<br \/>\u21d2 36=16+AL2<br \/>\u21d2 AL2 = 36 \u2013 16 = 20<br \/>\u2234 AL =\u00a0<span id=\"MathJax-Element-7-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-35\" class=\"math\"><span id=\"MathJax-Span-36\" class=\"mrow\"><span id=\"MathJax-Span-37\" class=\"msqrt\"><span id=\"MathJax-Span-38\" class=\"mrow\"><span id=\"MathJax-Span-39\" class=\"mn\">20<\/span><\/span>\u2212\u2212\u221a<\/span><\/span><\/span><\/span>\u00a0=\u00a0<span id=\"MathJax-Element-8-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-40\" class=\"math\"><span id=\"MathJax-Span-41\" class=\"mrow\"><span id=\"MathJax-Span-42\" class=\"msqrt\"><span id=\"MathJax-Span-43\" class=\"mrow\"><span id=\"MathJax-Span-44\" class=\"mn\">4<\/span><span id=\"MathJax-Span-45\" class=\"mo\">\u00d7<\/span><span id=\"MathJax-Span-46\" class=\"mn\">5<\/span><\/span>\u2212\u2212\u2212\u2212\u221a<\/span><\/span><\/span><\/span>\u00a0= 2 x 2.236 = 4.472 cm<br \/>\u2234 Chord AB = 4.472 x 2 = 8.944 = 8.94 cm<\/p>\n<p>Question 4.<\/p>\n<p>Give a method to find the centre of a given circle.<br \/>Solution:<br \/>Steps of construction :<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1906\/44920312714_8ee3bab8bc_o.png\" alt=\"Areas of Parallelograms and Triangles Class 9 RD Sharma Solutions\" width=\"270\" height=\"223\" \/><br \/>(i) Take three distinct points on the circle say A, B and C.<br \/>(ii) Join AB and AC.<br \/>(iii) Draw the perpendicular bisectors of AB and AC which intersect each other at O.<br \/>O is the required centre of the given circle<\/p>\n<div class=\"google-auto-placed ap_container\"><ins class=\"adsbygoogle adsbygoogle-noablate\" data-ad-format=\"auto\" data-ad-client=\"ca-pub-7398766921532682\" data-adsbygoogle-status=\"done\" data-ad-status=\"unfilled\"><ins id=\"aswift_4_expand\" tabindex=\"0\" title=\"Advertisement\" aria-label=\"Advertisement\"><ins id=\"aswift_4_anchor\"><iframe id=\"aswift_4\" src=\"https:\/\/googleads.g.doubleclick.net\/pagead\/ads?client=ca-pub-7398766921532682&amp;output=html&amp;h=280&amp;adk=1731550593&amp;adf=152193507&amp;pi=t.aa~a.1381849204~i.16~rp.4&amp;w=750&amp;fwrn=4&amp;fwrnh=100&amp;lmt=1631070007&amp;num_ads=1&amp;rafmt=1&amp;armr=3&amp;sem=mc&amp;pwprc=3683862639&amp;psa=1&amp;ad_type=text_image&amp;format=750x280&amp;url=https%3A%2F%2Fwww.learncbse.in%2Frd-sharma-solutions-class-9-chapter-15-areas-parallelograms-triangles%2F&amp;flash=0&amp;fwr=0&amp;pra=3&amp;rh=188&amp;rw=750&amp;rpe=1&amp;resp_fmts=3&amp;wgl=1&amp;fa=27&amp;adsid=ChAI8L7hiQYQ5Ym05uvakKJPEiYAZAgb5yDnKkZM1GkcQGIjNeMejma9wC9vAQh81yi8pJXmTQJrLg&amp;uach=WyJXaW5kb3dzIiwiNi4xIiwieDg2IiwiIiwiOTIuMC40NTE1LjE1OSIsW10sbnVsbCxudWxsLG51bGxd&amp;tt_state=W3siaXNzdWVyT3JpZ2luIjoiaHR0cHM6Ly9hdHRlc3RhdGlvbi5hbmRyb2lkLmNvbSIsInN0YXRlIjo3fV0.&amp;dt=1631085666034&amp;bpp=4&amp;bdt=3142&amp;idt=4&amp;shv=r20210831&amp;mjsv=m202109010101&amp;ptt=9&amp;saldr=aa&amp;abxe=1&amp;cookie=ID%3D012e0a2ac7d78731-227695d5fcc800a3%3AT%3D1622201370%3ART%3D1622201370%3AS%3DALNI_MYEk6qCV_FwHZ96mF38v9WuNwhDKQ&amp;prev_fmts=0x0%2C728x280%2C750x280%2C750x280&amp;nras=4&amp;correlator=1536611512632&amp;frm=20&amp;pv=1&amp;ga_vid=1704677381.1622201373&amp;ga_sid=1631085665&amp;ga_hid=1877462570&amp;ga_fc=0&amp;u_tz=330&amp;u_his=7&amp;u_java=0&amp;u_h=768&amp;u_w=1366&amp;u_ah=728&amp;u_aw=1366&amp;u_cd=24&amp;u_nplug=3&amp;u_nmime=4&amp;adx=105&amp;ady=4557&amp;biw=1349&amp;bih=657&amp;scr_x=0&amp;scr_y=2042&amp;eid=44747621%2C31060566%2C21067496%2C31062297%2C31062094&amp;oid=3&amp;pvsid=1261128252710953&amp;pem=814&amp;ref=https%3A%2F%2Fwww.google.com%2F&amp;eae=0&amp;fc=1408&amp;brdim=0%2C0%2C0%2C0%2C1366%2C0%2C1366%2C728%2C1366%2C657&amp;vis=1&amp;rsz=%7C%7Cs%7C&amp;abl=NS&amp;cms=2&amp;alvm=r20210901&amp;fu=128&amp;bc=31&amp;jar=2021-09-08-07&amp;ifi=5&amp;uci=a!5&amp;btvi=3&amp;fsb=1&amp;xpc=t1jd3IG76h&amp;p=https%3A\/\/www.learncbse.in&amp;dtd=1071\" name=\"aswift_4\" width=\"750\" height=\"0\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\" sandbox=\"allow-forms allow-popups allow-popups-to-escape-sandbox allow-same-origin allow-scripts allow-top-navigation-by-user-activation\" allowfullscreen=\"allowfullscreen\" data-google-container-id=\"a!5\" data-google-query-id=\"CJjAo87r7vICFf3GFgUd31sM0w\" data-load-complete=\"true\" data-mce-fragment=\"1\"><\/iframe><\/ins><\/ins><\/ins><\/div>\n<p>Question 5.<br \/>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.<br \/>Solution:<\/p>\n<p>Given : In circle with centre O<br \/>CD is the diameter and AB is the chord<br \/>which is bisected by diameter at E<br \/>OA and OB are joined<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1942\/44920312504_ceb1804c9f_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 15 Areas of Parallelograms and Triangles\" width=\"226\" height=\"181\" \/><br \/>To prove : \u2220AOB = \u2220BOA<br \/>Proof : In \u2206OAE and \u2206OBE<br \/>OA = OB (Radii of the circle)<br \/>OE = OE (Common)<br \/>AE = EB (Given)<br \/>\u2234 \u2206OAE = \u2206OBE (SSS criterian)<br \/>\u2234 \u2220AOE = \u2220BOE (c.p.c.t.)<br \/>Hence diameter bisect the angle subtended by the chord AB.<\/p>\n<div class=\"google-auto-placed ap_container\"><ins class=\"adsbygoogle adsbygoogle-noablate\" data-ad-format=\"auto\" data-ad-client=\"ca-pub-7398766921532682\" data-adsbygoogle-status=\"done\" data-ad-status=\"unfilled\"><ins id=\"aswift_5_expand\" tabindex=\"0\" title=\"Advertisement\" aria-label=\"Advertisement\"><ins id=\"aswift_5_anchor\"><iframe id=\"aswift_5\" src=\"https:\/\/googleads.g.doubleclick.net\/pagead\/ads?client=ca-pub-7398766921532682&amp;output=html&amp;h=280&amp;adk=1731550593&amp;adf=1295003436&amp;pi=t.aa~a.1381849204~i.18~rp.4&amp;w=750&amp;fwrn=4&amp;fwrnh=100&amp;lmt=1631070007&amp;num_ads=1&amp;rafmt=1&amp;armr=3&amp;sem=mc&amp;pwprc=3683862639&amp;psa=1&amp;ad_type=text_image&amp;format=750x280&amp;url=https%3A%2F%2Fwww.learncbse.in%2Frd-sharma-solutions-class-9-chapter-15-areas-parallelograms-triangles%2F&amp;flash=0&amp;fwr=0&amp;pra=3&amp;rh=188&amp;rw=750&amp;rpe=1&amp;resp_fmts=3&amp;wgl=1&amp;fa=27&amp;adsid=ChAI8L7hiQYQ5Ym05uvakKJPEiYAZAgb5yDnKkZM1GkcQGIjNeMejma9wC9vAQh81yi8pJXmTQJrLg&amp;uach=WyJXaW5kb3dzIiwiNi4xIiwieDg2IiwiIiwiOTIuMC40NTE1LjE1OSIsW10sbnVsbCxudWxsLG51bGxd&amp;tt_state=W3siaXNzdWVyT3JpZ2luIjoiaHR0cHM6Ly9hdHRlc3RhdGlvbi5hbmRyb2lkLmNvbSIsInN0YXRlIjo3fV0.&amp;dt=1631085666049&amp;bpp=4&amp;bdt=3157&amp;idt=4&amp;shv=r20210831&amp;mjsv=m202109010101&amp;ptt=9&amp;saldr=aa&amp;abxe=1&amp;cookie=ID%3D012e0a2ac7d78731-227695d5fcc800a3%3AT%3D1622201370%3ART%3D1622201370%3AS%3DALNI_MYEk6qCV_FwHZ96mF38v9WuNwhDKQ&amp;prev_fmts=0x0%2C728x280%2C750x280%2C750x280%2C750x280&amp;nras=5&amp;correlator=1536611512632&amp;frm=20&amp;pv=1&amp;ga_vid=1704677381.1622201373&amp;ga_sid=1631085665&amp;ga_hid=1877462570&amp;ga_fc=0&amp;u_tz=330&amp;u_his=7&amp;u_java=0&amp;u_h=768&amp;u_w=1366&amp;u_ah=728&amp;u_aw=1366&amp;u_cd=24&amp;u_nplug=3&amp;u_nmime=4&amp;adx=105&amp;ady=5188&amp;biw=1349&amp;bih=657&amp;scr_x=0&amp;scr_y=2894&amp;eid=44747621%2C31060566%2C21067496%2C31062297%2C31062094&amp;oid=3&amp;pvsid=1261128252710953&amp;pem=814&amp;ref=https%3A%2F%2Fwww.google.com%2F&amp;eae=0&amp;fc=1408&amp;brdim=0%2C0%2C0%2C0%2C1366%2C0%2C1366%2C728%2C1366%2C657&amp;vis=1&amp;rsz=%7C%7Cs%7C&amp;abl=NS&amp;alvm=r20210901&amp;fu=128&amp;bc=31&amp;jar=2021-09-08-07&amp;ifi=6&amp;uci=a!6&amp;btvi=4&amp;fsb=1&amp;xpc=CcnSG3O71K&amp;p=https%3A\/\/www.learncbse.in&amp;dtd=1953\" name=\"aswift_5\" width=\"750\" height=\"0\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\" sandbox=\"allow-forms allow-popups allow-popups-to-escape-sandbox allow-same-origin allow-scripts allow-top-navigation-by-user-activation\" allowfullscreen=\"allowfullscreen\" data-google-container-id=\"a!6\" data-google-query-id=\"CLfVw87r7vICFQ20lgodanQBSA\" data-load-complete=\"true\" data-mce-fragment=\"1\"><\/iframe><\/ins><\/ins><\/ins><\/div>\n<p>Question 6.<br \/>A line segment AB is of length 5 cm. Draw a circle of radius 4 cm passing through A and B. Can you draw a circle of radius 2 cm passing through A and B? Give reason in support of your answer.<\/p>\n<p>Solution:<\/p>\n<p>Steps of construction :<br \/>(i) Draw a line segment AB = 5 cm.<br \/>(ii) Draw a perpendicular bisector of AB.<br \/>(iii) With centre A and radius 4 cm, draw an arc which intersects the perpendicular bisector at O.<\/p>\n<p>(iv) With centre O and radius 4 cm, draw a circle which passes through A and B.<br \/>With radius 2 cm, we cannot draw the circle passing through A and B as diameter<br \/>i. e. 2 + 2 = 4 cm is shorder than 5 cm.<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1938\/44920312434_44eb9ac1d2_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"216\" height=\"289\" \/><\/p>\n<div class=\"google-auto-placed ap_container\"><ins class=\"adsbygoogle adsbygoogle-noablate\" data-ad-format=\"auto\" data-ad-client=\"ca-pub-7398766921532682\" data-adsbygoogle-status=\"done\" data-ad-status=\"unfilled\"><ins id=\"aswift_6_expand\" tabindex=\"0\" title=\"Advertisement\" aria-label=\"Advertisement\"><ins id=\"aswift_6_anchor\"><iframe id=\"aswift_6\" src=\"https:\/\/googleads.g.doubleclick.net\/pagead\/ads?client=ca-pub-7398766921532682&amp;output=html&amp;h=280&amp;adk=1731550593&amp;adf=2534421946&amp;pi=t.aa~a.1381849204~i.20~rp.4&amp;w=750&amp;fwrn=4&amp;fwrnh=100&amp;lmt=1631070007&amp;num_ads=1&amp;rafmt=1&amp;armr=3&amp;sem=mc&amp;pwprc=3683862639&amp;psa=1&amp;ad_type=text_image&amp;format=750x280&amp;url=https%3A%2F%2Fwww.learncbse.in%2Frd-sharma-solutions-class-9-chapter-15-areas-parallelograms-triangles%2F&amp;flash=0&amp;fwr=0&amp;pra=3&amp;rh=188&amp;rw=750&amp;rpe=1&amp;resp_fmts=3&amp;wgl=1&amp;fa=27&amp;adsid=ChAI8L7hiQYQ5Ym05uvakKJPEiYAZAgb5yDnKkZM1GkcQGIjNeMejma9wC9vAQh81yi8pJXmTQJrLg&amp;uach=WyJXaW5kb3dzIiwiNi4xIiwieDg2IiwiIiwiOTIuMC40NTE1LjE1OSIsW10sbnVsbCxudWxsLG51bGxd&amp;tt_state=W3siaXNzdWVyT3JpZ2luIjoiaHR0cHM6Ly9hdHRlc3RhdGlvbi5hbmRyb2lkLmNvbSIsInN0YXRlIjo3fV0.&amp;dt=1631085666063&amp;bpp=4&amp;bdt=3170&amp;idt=4&amp;shv=r20210831&amp;mjsv=m202109010101&amp;ptt=9&amp;saldr=aa&amp;abxe=1&amp;cookie=ID%3D012e0a2ac7d78731-227695d5fcc800a3%3AT%3D1622201370%3ART%3D1622201370%3AS%3DALNI_MYEk6qCV_FwHZ96mF38v9WuNwhDKQ&amp;prev_fmts=0x0%2C728x280%2C750x280%2C750x280%2C750x280%2C750x280%2C1349x657%2C1005x90&amp;nras=8&amp;correlator=1536611512632&amp;frm=20&amp;pv=1&amp;ga_vid=1704677381.1622201373&amp;ga_sid=1631085665&amp;ga_hid=1877462570&amp;ga_fc=0&amp;u_tz=330&amp;u_his=7&amp;u_java=0&amp;u_h=768&amp;u_w=1366&amp;u_ah=728&amp;u_aw=1366&amp;u_cd=24&amp;u_nplug=3&amp;u_nmime=4&amp;adx=105&amp;ady=5798&amp;biw=1349&amp;bih=657&amp;scr_x=0&amp;scr_y=3380&amp;eid=44747621%2C31060566%2C21067496%2C31062297%2C31062094&amp;oid=3&amp;pvsid=1261128252710953&amp;pem=814&amp;ref=https%3A%2F%2Fwww.google.com%2F&amp;eae=0&amp;fc=1408&amp;brdim=0%2C0%2C0%2C0%2C1366%2C0%2C1366%2C728%2C1366%2C657&amp;vis=1&amp;rsz=%7C%7Cs%7C&amp;abl=NS&amp;alvm=r20210901&amp;fu=128&amp;bc=31&amp;jar=2021-09-08-07&amp;ifi=7&amp;uci=a!7&amp;btvi=6&amp;fsb=1&amp;xpc=xyGiO9suYX&amp;p=https%3A\/\/www.learncbse.in&amp;dtd=2383\" name=\"aswift_6\" width=\"750\" height=\"0\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\" sandbox=\"allow-forms allow-popups allow-popups-to-escape-sandbox allow-same-origin allow-scripts allow-top-navigation-by-user-activation\" allowfullscreen=\"allowfullscreen\" data-google-container-id=\"a!7\" data-google-query-id=\"CJmw7s7r7vICFZCylgodKyIOgw\" data-load-complete=\"true\" data-mce-fragment=\"1\"><\/iframe><\/ins><\/ins><\/ins><\/div>\n<p>Question 7.<br \/>An equilateral triangle of side 9 cm is inscribed in a circle. Find the radius of the circle.<br \/>Solution:<\/p>\n<p>Steps of construction :<br \/>(i) Draw a line segment BC = 9 cm.<br \/>(ii) With centres B and C, draw arcs of 9 cm radius which intersect each other at A.<br \/>(iii) Join AB and AC.<br \/>\u2206ABC is the required triangle.<\/p>\n<p>(iv) Draw perpendicular bisectors of sides AB and BC which intersect each other at O.<br \/>(v) With centre O and radius OB, draw a circle which passes through A, B and C.<br \/>This is the require circle in which \u2206ABC is inscribed<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1923\/44920312334_ae53a67336_o.png\" alt=\"Class 9 Maths Chapter 15 Areas of Parallelograms and Triangles RD Sharma Solutions\" width=\"225\" height=\"220\" \/><br \/>On measuring its radius, it is 5.2 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1907\/44920312254_b75c2417b4_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 15 Areas of Parallelograms and Triangles\" width=\"288\" height=\"150\" \/><\/p>\n<div class=\"google-auto-placed ap_container\"><ins class=\"adsbygoogle adsbygoogle-noablate\" data-ad-format=\"auto\" data-ad-client=\"ca-pub-7398766921532682\" data-adsbygoogle-status=\"done\" data-ad-status=\"unfilled\"><ins id=\"aswift_7_expand\" tabindex=\"0\" title=\"Advertisement\" aria-label=\"Advertisement\"><ins id=\"aswift_7_anchor\"><iframe id=\"aswift_7\" src=\"https:\/\/googleads.g.doubleclick.net\/pagead\/ads?client=ca-pub-7398766921532682&amp;output=html&amp;h=280&amp;adk=1731550593&amp;adf=1201277335&amp;pi=t.aa~a.1381849204~i.22~rp.4&amp;w=750&amp;fwrn=4&amp;fwrnh=100&amp;lmt=1631070007&amp;num_ads=1&amp;rafmt=1&amp;armr=3&amp;sem=mc&amp;pwprc=3683862639&amp;psa=1&amp;ad_type=text_image&amp;format=750x280&amp;url=https%3A%2F%2Fwww.learncbse.in%2Frd-sharma-solutions-class-9-chapter-15-areas-parallelograms-triangles%2F&amp;flash=0&amp;fwr=0&amp;pra=3&amp;rh=188&amp;rw=750&amp;rpe=1&amp;resp_fmts=3&amp;wgl=1&amp;fa=27&amp;uach=WyJXaW5kb3dzIiwiNi4xIiwieDg2IiwiIiwiOTIuMC40NTE1LjE1OSIsW10sbnVsbCxudWxsLG51bGxd&amp;tt_state=W3siaXNzdWVyT3JpZ2luIjoiaHR0cHM6Ly9hdHRlc3RhdGlvbi5hbmRyb2lkLmNvbSIsInN0YXRlIjo3fV0.&amp;dt=1631085666077&amp;bpp=3&amp;bdt=3154&amp;idt=4&amp;shv=r20210831&amp;mjsv=m202109010101&amp;ptt=9&amp;saldr=aa&amp;abxe=1&amp;cookie=ID%3D012e0a2ac7d78731-227695d5fcc800a3%3AT%3D1622201370%3ART%3D1622201370%3AS%3DALNI_MYEk6qCV_FwHZ96mF38v9WuNwhDKQ&amp;prev_fmts=0x0%2C728x280%2C750x280%2C750x280%2C750x280%2C750x280%2C1349x657%2C1005x90%2C750x280&amp;nras=9&amp;correlator=1536611512632&amp;frm=20&amp;pv=1&amp;ga_vid=1704677381.1622201373&amp;ga_sid=1631085665&amp;ga_hid=1877462570&amp;ga_fc=0&amp;u_tz=330&amp;u_his=7&amp;u_java=0&amp;u_h=768&amp;u_w=1366&amp;u_ah=728&amp;u_aw=1366&amp;u_cd=24&amp;u_nplug=3&amp;u_nmime=4&amp;adx=105&amp;ady=6242&amp;biw=1349&amp;bih=657&amp;scr_x=0&amp;scr_y=3620&amp;eid=44747621%2C31060566%2C21067496%2C31062297%2C31062094&amp;oid=3&amp;pvsid=1261128252710953&amp;pem=814&amp;ref=https%3A%2F%2Fwww.google.com%2F&amp;eae=0&amp;fc=1408&amp;brdim=0%2C0%2C0%2C0%2C1366%2C0%2C1366%2C728%2C1366%2C657&amp;vis=1&amp;rsz=%7C%7Cs%7C&amp;abl=NS&amp;alvm=r20210901&amp;fu=128&amp;bc=31&amp;ifi=8&amp;uci=a!8&amp;btvi=7&amp;fsb=1&amp;xpc=n63yTCYKRC&amp;p=https%3A\/\/www.learncbse.in&amp;dtd=M\" name=\"aswift_7\" width=\"750\" height=\"0\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\" sandbox=\"allow-forms allow-popups allow-popups-to-escape-sandbox allow-same-origin allow-scripts allow-top-navigation-by-user-activation\" allowfullscreen=\"allowfullscreen\" data-google-container-id=\"a!8\" data-google-query-id=\"CNit3abw7vICFUYHlgodle8IJQ\" data-load-complete=\"true\" data-mce-fragment=\"1\"><\/iframe><\/ins><\/ins><\/ins><\/div>\n<p>Question 8.<br \/>Given an arc of a circle, complete the circle.<br \/>Solution:<br \/>Steps of construction :<br \/>(i) Take three points A, B and C on the arc and join AB and BC.<br \/>(ii) Draw the perpendicular bisector of AB and BC which intersect each other at O.<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1976\/44920312174_c69f5e47b7_o.png\" alt=\"RD Sharma Class 9 Book Chapter 15 Areas of Parallelograms and Triangles\" width=\"260\" height=\"246\" \/><br \/>(iii) With centre O and radius OA or OB, complete the circle.<br \/>This is the required circle.<\/p>\n<p>Question 9.<br \/>Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?<br \/>Solution:<br \/>Below, three different pairs of circles are drawn:<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1915\/44920312034_829981073d_o.png\" alt=\"Areas of Parallelograms and Triangles With Solutions PDF RD Sharma Class 9 Solutions\" width=\"286\" height=\"176\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1919\/44920311994_81b081cf6c_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 15 Areas of Parallelograms and Triangles\" width=\"273\" height=\"383\" \/><\/p>\n<p>(i) In the first pair, two circles do not intersect each other. Therefore they have no point in common.<\/p>\n<p>(ii) In the second pair, two circles intersect (touch) each other at one point P. Therefore they have one point in common.<\/p>\n<p>(iii) In the third pair, two circles intersect each other at two points. Therefore they have two points in common.<br \/>There is no other possibility of two circles intersecting each other.<br \/>Therefore, two circles have at the most two points in common.<\/p>\n<p>Question 10.<br \/>Suppose you are given a circle. Give a construction to find its centre.<br \/>Solution:<br \/>See Q. No. 4 of this exercise.<\/p>\n<p>Question 11.<br \/>The length of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance of 4 cm from the centre, what is the distance of the other chord from the centre? [NCERT]<\/p>\n<p>Solution:<\/p>\n<p>A circle with centre O and two parallel chords<br \/>AB and CD are AB = 6 cm, CD = 8 cm<br \/>Let OL \u22a5 AB and OM \u22a5 CD<br \/>\u2234 OL = 4 cm<br \/>Let OM = x cm<br \/>Let r be the radius of the circle<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1906\/44920311954_dc00545b3c_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"359\" height=\"601\" \/><\/p>\n<p>Question 12.<br \/>Two chords AB, 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.<br \/>Solution:<\/p>\n<p>Let two chords AB and CD of length 5 cm and 11 cm are parallel to each other AB = 5 cm, CD = 11 cm<br \/>Distance between AB and LM = 3 cm<br \/>Join OB and OD<\/p>\n<p>OL and OM are the perpendicular on CD and AB respectively. Which bisects AB and CD.<br \/>Let OL = x, then OM = (x + 3)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1949\/44920311834_3e6a6e5810_o.png\" alt=\"Solution Of Rd Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"228\" height=\"209\" \/><br \/>Now in right \u2206OLD,<br \/>OD2 = OL2 + LD2<br \/>= x2 + (5.5)2<\/p>\n<p>Similarly in right \u2206OMB,<br \/>OB2 = OM2 + MB2 = (x + 3)2 + (2.5)2<br \/>But OD = OB (Radii of the circle)<br \/>\u2234 (x + 3)2 + (2.5)2 = x2 + (5.5)2<br \/>x2 + 6x + 9 + 6.25 = x2 + 30.25<br \/>6x = 30.25 \u2013 6.25 \u2013 9 = 15<\/p>\n<p><img src=\"https:\/\/farm2.staticflickr.com\/1919\/44920311774_7feb40e6a7_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"354\" height=\"248\" \/><\/p>\n<p>Question 13.<br \/>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.<br \/>Solution:<\/p>\n<p>Given : A circle with centre O and a chord AB<br \/>Let M be the mid point of AB and OM is joined and produced to meet the minor arc AB at N<br \/>To prove : M is the mid point of arc AB<br \/>Construction : Join OA, OB<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1940\/44920311664_c6d905ac62_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 15 Areas of Parallelograms and Triangles\" width=\"220\" height=\"207\" \/><br \/>Proof: \u2235 M is mid point of AB<br \/>\u2234 OM \u22a5 AB<br \/>In AOAM and OBM,<br \/>OA = OB (Radii of the circle)<br \/>OM = OM (common)<br \/>AM = BM (M is mid point of AB)<\/p>\n<p>\u2234 \u2206OAM = \u2206OBM (SSS criterian)<br \/>\u2234 \u2220AOM = \u2220BOM (c.p.c.t.)<br \/>\u21d2 \u2220AOM = \u2220BOM<br \/>But these are centre angles at the centre made by arcs AN and BN<br \/>\u2234 Arc AN = Arc BN<br \/>Hence N divides the arc in two equal parts<\/p>\n<p>Question 14.<br \/>Prove that two different circles cannot intersect each other at more than two points.<br \/>Solution:<br \/>Given : Two circles<br \/>To prove : They cannot intersect each other more than two points<br \/>Construction : Let two circles intersect each other at three points A, B and C<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1980\/44920311614_37175c683a_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"236\" height=\"162\" \/><br \/>Proof : Since two circles with centres O and O\u2019 intersect at A, B and C<br \/>\u2234 A, B and C are non-collinear points<br \/>\u2234 Circle with centre O passes through three points A, B and C<br \/>and circle with centre O\u2019 also passes through three points A, B and C<br \/>But one and only one circle can be drawn through three points<br \/>\u2234Our supposition is wrong<br \/>\u2234 Two circle cannot intersect each other not more than two points.<\/p>\n<p>Question 15.<br \/>Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are opposite side of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle. [NCERT]<br \/>Solution:<br \/>Let r be the radius of the circle with centre O.<br \/>Two parallel chords AB = 5 cm, CD = 11 cm<br \/>Let OL \u22a5 AB and OM \u22a5CD<br \/>\u2234 LM = 6 cm<br \/>Let OM = x, then<br \/>OL = 6 \u2013 x<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1937\/44920311564_0bb5877b7c_o.png\" alt=\"RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"329\" height=\"463\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1977\/44920311374_fceb4821bc_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"305\" height=\"587\" \/><\/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 15 Exercise 15.2&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 15 Exercise 15.2<\/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-15-exercise-152\"><\/span>FAQ: <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 15 Exercise 15.2&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 15 Exercise 15.2<\/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-1631088161237\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-open-rd-sharma-class-9-solutions-chapter-15-exercise-152-pdf-on-my-smartphone\"><\/span>Can I open RD Sharma Class 9 Solutions Chapter 15 Exercise 15.2 PDF on my smartphone?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, you can open RD Sharma Class 9 Solutions Chapter 15 Exercise 15.2 PDF on any device.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631088326136\" 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-15-exercise-152-pdf-free\"><\/span>Can I download RD Sharma Class 9 Solutions Chapter 15 Exercise 15.2 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 15 Exercise 15.2 PDF free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631088327296\" 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-1631088330408\" 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 15 Exercise 15.2: Our experts have created this PDF in the easy manner. You can easily understand these RD Sharma Class 9 Solutions Chapter 15\u00a0Exercise 15.2 solutions while preparing for your Class 9 exams.\u00a0 RD Sharma Class 9 Maths Solutions Download RD Sharma Class 9 Solutions Chapter 15 Exercise &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 15 Exercise 15.2 (Updated 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-15-exercise-15-2\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 15 Exercise 15.2 (Updated 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":127674,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73719],"tags":[76786,76820],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125320"}],"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=125320"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125320\/revisions"}],"predecessor-version":[{"id":127692,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125320\/revisions\/127692"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/127674"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125320"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125320"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125320"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}