{"id":125342,"date":"2021-09-14T21:08:39","date_gmt":"2021-09-14T15:38:39","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125342"},"modified":"2023-12-20T15:01:43","modified_gmt":"2023-12-20T09:31:43","slug":"rd-sharma-class-9-solutions-chapter-15-mcqs","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-15-mcqs\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 15 MCQS (Updated 2024-25)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-127698\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-15-MCQS.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 15 MCQS \" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-15-MCQS.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-15-MCQS-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 MCQS&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:268732,&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;14&quot;:[null,2,0],&quot;15&quot;:&quot;Arial&quot;,&quot;21&quot;:1}\"><strong>RD Sharma Class 9 Solutions Chapter 15 MCQS<\/strong>: These solutions are specifically created by our experts. Useful while preparing for the 9th grade exam. You can access <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> MCQS directly in the article below.<\/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-69d56826521fb\" 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-15-mcqs\/#faq-rd-sharma-class-9-solutions-chapter-15-mcqs\" title=\"FAQ: RD Sharma Class 9 Solutions Chapter 15 MCQS\">FAQ: RD Sharma Class 9 Solutions Chapter 15 MCQS<\/a><\/li><li class='ez-toc-page-1 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-15-mcqs\/#can-i-access-rd-sharma-class-9-solutions-chapter-15-mcqs-on-my-smartphone\" title=\"Can I access RD Sharma Class 9 Solutions Chapter 15 MCQs on my smartphone?\">Can I access RD Sharma Class 9 Solutions Chapter 15 MCQs on my smartphone?<\/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-15-mcqs\/#what-are-the-benefits-of-studying-rd-sharma-class-9-solutions-chapter-15-mcqs\" title=\"What are the benefits of studying RD Sharma Class 9 Solutions Chapter 15 MCQs?\">What are the benefits of studying RD Sharma Class 9 Solutions Chapter 15 MCQs?<\/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-15-mcqs\/#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=\"access-rd-sharma-class-9-solutions-chapter-15-mcqs\"><\/span>Access <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 15 MCQS&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:268732,&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;14&quot;:[null,2,0],&quot;15&quot;:&quot;Arial&quot;,&quot;21&quot;:1}\">RD Sharma Class 9 Solutions Chapter 15 MCQS<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Question 1.<br \/>If the length of a chord of a circle is 16 cm and is at a distance of 15 cm from the centre of the circle, then the radius of the circle is<br \/>(a) 15 cm<br \/>(b) 16 cm<br \/>(c) 17 cm<br \/>(d) 34 cm<br \/>Solution:<br \/>Length of chord AB of circle = 16 cm<br \/>Distance from the centre OL = 15 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1930\/44731149125_be6bdbcfca_o.png\" alt=\"Areas of Parallelograms and Triangles With Solutions PDF RD Sharma Class 9 Solutions\" width=\"181\" height=\"162\" \/><br \/>Let OA be the radius, then in right \u2206OAL,<br \/>OA<sup>2<\/sup>\u00a0= OL<sup>2<\/sup>\u00a0+ AL<sup>2<\/sup><br \/>16<br \/>= (15)<sup>2<\/sup>\u00a0+\u00a0<span id=\"MathJax-Element-54-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-338\" class=\"math\"><span id=\"MathJax-Span-339\" class=\"mrow\"><span id=\"MathJax-Span-340\" class=\"mfrac\"><span id=\"MathJax-Span-341\" class=\"mn\">16<\/span><span id=\"MathJax-Span-342\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 15<sup>2<\/sup>\u00a0+ 8<sup>2<\/sup><br \/>= 225 + 64 = 289 = (17)<sup>2<\/sup><br \/>\u2234 OA = 17 cm<br \/>Hence radius of the circle = 17 cm (c)<\/p>\n<p>Question 2.<br \/>The radius of a circle Js 6 cm. The perpendicular distance from the centre of the circle to the chord which is 8 cm in length, is<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1930\/45645337881_53f2daaa61_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 15 Areas of Parallelograms and Triangles\" width=\"284\" height=\"71\" \/><br \/>Solution:<br \/>Radius of the cirlce (r) = 6 cm<br \/>Perpendicular distance from centre = ?<br \/>Length of chord = 8 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1964\/30704328887_efee965913_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"173\" height=\"162\" \/><br \/>Let AB be chord, OL is the distance<br \/>In right \u2206OAL<br \/>OA<sup>2<\/sup>\u00a0= AL<sup>2<\/sup>\u00a0+ OL<sup>2<\/sup><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1979\/45645337671_a39943a457_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"349\" height=\"122\" \/><\/p>\n<p>Question 3.<br \/>If O is the centre of a circle of radius r and AB is a chord of the circle at a distance\u00a0<span id=\"MathJax-Element-55-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-343\" class=\"math\"><span id=\"MathJax-Span-344\" class=\"mrow\"><span id=\"MathJax-Span-345\" class=\"mfrac\"><span id=\"MathJax-Span-346\" class=\"mi\">r<\/span><span id=\"MathJax-Span-347\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0from O, then \u2220BAO =<br \/>(a) 60\u00b0<br \/>(b) 45\u00b0<br \/>(c) 30\u00b0<br \/>(d) 15\u00b0<br \/>Solution:<br \/>r is the radius of the circle with centre O<br \/>AB is the chord, at a distance of\u00a0<span id=\"MathJax-Element-56-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-348\" class=\"math\"><span id=\"MathJax-Span-349\" class=\"mrow\"><span id=\"MathJax-Span-350\" class=\"mfrac\"><span id=\"MathJax-Span-351\" class=\"mi\">r<\/span><span id=\"MathJax-Span-352\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0from the centre<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1950\/30704328817_c70c969d7c_o.png\" alt=\"RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"236\" height=\"278\" \/><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1911\/44731148765_f705629cf8_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"350\" height=\"127\" \/><\/p>\n<p>Question 4.<br \/>ABCD is a cyclic quadrilateral such that \u2220ADB = 30\u00b0 and \u2220DCA = 80\u00b0, then \u2220DAB=<br \/>(a) 70\u00b0<br \/>(b) 100\u00b0<br \/>(c) 125\u00b0<br \/>(d) 150\u00b0<br \/>Solution:<br \/>ABCD is a cyclic quadrilateral \u2220DCA = 80\u00b0 and \u2220ADB = 30\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1960\/45645337371_7b04fb0f93_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"217\" height=\"199\" \/><br \/>\u2235\u2220ADB = \u2220ACB (Angles in the same segment)<br \/>\u2234 \u2220ACB = 30\u00b0<br \/>\u2234 \u2220BCD = 80\u00b0 + 30\u00b0 = 110\u00b0<br \/>\u2235 ABCD is a cyclic quadrilateral<br \/>\u2234\u2220BAD + \u2220BCD = 180\u00b0<br \/>\u21d2 \u2220BAD + 110\u00b0= 180\u00b0<br \/>\u21d2 \u2220BAD = 180\u00b0- 110\u00b0 = 70\u00b0<br \/>or \u2220DAB = 70\u00b0 (a)<\/p>\n<p>Question 5.<br \/>A chord of length 14 cm is at a distance of 6 cm from the centre of a circle. The length of another chord at a distance of 2 cm from the centre of the circle is<br \/>(a) 12 cm<br \/>(b) 14 cm<br \/>(c) 16 cm<br \/>(d) 18 cm<br \/>Solution:<br \/>In a circle AB chord = 14 cm<br \/>and distance from centre OL = 6 cm<br \/>Let r be the radius of the circle, then OA<sup>2<\/sup>\u00a0= AL<sup>2<\/sup>\u00a0+ OL<sup>2<\/sup><br \/>\u21d2 r<sup>2<\/sup>\u00a0= (7)<sup>2<\/sup>\u00a0+ (6)<sup>2<\/sup>\u00a0= 49 + 36 = 85<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1962\/44731148635_9c25055e7b_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 15 Areas of Parallelograms and Triangles\" width=\"218\" height=\"207\" \/><br \/>In the same circle length of another chord CD = ?<br \/>Distance from centre = 2 cm<br \/>\u2234 r<sup>2<\/sup>\u00a0= OM<sup>2<\/sup>\u00a0+ MD<sup>2<\/sup><br \/>\u21d2 85 = (2)<sup>2<\/sup>\u00a0+ DM<sup>2<\/sup><br \/>\u21d2 85 = 4 + DM<sup>2<\/sup><br \/>\u21d2 DM<sup>2<\/sup>\u00a0= 85-4 = 81 = (9)<sup>2<\/sup><br \/>\u2234 DM = 9<br \/>\u2234 CD = 2 x DM = 2 x 9 = 18 cm<br \/>\u2234Length of another chord = 18 cm (d)<\/p>\n<p>Question 6.<br \/>One chord of a circle is known to be 10 cm. The radius of this circle must be<br \/>(a) 5 cm<br \/>(b) greater than 5 cm<br \/>(c) greater than or equal to 5 cm<br \/>(d) less than 5 cm<br \/>Solution:<br \/>Length of chord of a circle = 10 cm<br \/>Length of radius of the circle greater than half of the chord<br \/>More than\u00a0<span id=\"MathJax-Element-57-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-353\" class=\"math\"><span id=\"MathJax-Span-354\" class=\"mrow\"><span id=\"MathJax-Span-355\" class=\"mfrac\"><span id=\"MathJax-Span-356\" class=\"mn\">10<\/span><span id=\"MathJax-Span-357\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0= 5 cm (b)<\/p>\n<p>Question 7.<br \/>ABC is a triangle with B as right angle, AC = 5 cm and AB = 4 cm. A circle is drawn with O as centre and OC as radius. The length of the chord of this circle passing through C and B is<br \/>(a) 3 cm<br \/>(b) 4 cm<br \/>(c) 5 cm<br \/>(d) 6 cm<br \/>Solution:<br \/>In right \u2206ABC, \u2220B = 90\u00b0<br \/>AC = 5 cm, AB = 4 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1948\/44731148615_67b97fd8af_o.png\" alt=\"Areas of Parallelograms and Triangles Class 9 RD Sharma Solutions\" width=\"200\" height=\"188\" \/><br \/>\u2234 BC<sup>2<\/sup>\u00a0= AC<sup>2\u00a0<\/sup>-AB<sup>2<\/sup><br \/>= 5<sup>2<\/sup>\u00a0\u2013 4<sup>2<\/sup>\u00a0= 25 \u2013 16<br \/>= 9 = (3)<sup>2<\/sup><br \/>\u2234 BC = 3 cm<br \/>\u2234 Length of chord BC = 3 cm (a)<\/p>\n<p>Question 8.<br \/>If AB, BC and CD are equal chords of a circle with O as centre, and AD diameter then \u2220AOB =<br \/>(a) 60\u00b0<br \/>(b) 90\u00b0<br \/>(c) 120\u00b0<br \/>(d) none of these<br \/>Solution:<br \/>In a circle chords AB = BC = CD<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1929\/44920319434_87e60f18ba_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 15 Areas of Parallelograms and Triangles\" width=\"198\" height=\"198\" \/><br \/>O is the centre of the circle<br \/>\u2234 \u2220AOB = cannot be found (d)<\/p>\n<p>Question 9.<br \/>Let C be the mid-point of an arc AB of a circle such that m\u00a0<span id=\"MathJax-Element-58-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-358\" class=\"math\"><span id=\"MathJax-Span-359\" class=\"mrow\"><span id=\"MathJax-Span-360\" class=\"texatom\"><span id=\"MathJax-Span-361\" class=\"mrow\"><span id=\"MathJax-Span-362\" class=\"munderover\"><span id=\"MathJax-Span-363\" class=\"mrow\"><span id=\"MathJax-Span-364\" class=\"mi\">A<\/span><span id=\"MathJax-Span-365\" class=\"mi\">B<\/span><\/span><span id=\"MathJax-Span-366\" class=\"mo\">\u02d8<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u00a0= 183\u00b0. If the region bounded by the arc ACB and line segment AB is denoted by S, then the centre O of the circle lies<br \/>(a) in the interior of S<br \/>(b) in the exterior of S<br \/>(c) on the segment AB<br \/>(d) on AB and bisects AB<br \/>Solution:<br \/><span id=\"MathJax-Element-59-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-367\" class=\"math\"><span id=\"MathJax-Span-368\" class=\"mrow\"><span id=\"MathJax-Span-369\" class=\"texatom\"><span id=\"MathJax-Span-370\" class=\"mrow\"><span id=\"MathJax-Span-371\" class=\"munderover\"><span id=\"MathJax-Span-372\" class=\"mrow\"><span id=\"MathJax-Span-373\" class=\"mi\">A<\/span><span id=\"MathJax-Span-374\" class=\"mi\">B<\/span><\/span><span id=\"MathJax-Span-375\" class=\"mo\">\u02d8<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u00a0= 183\u00b0<br \/>\u2234 AB is the diameter of the circle with centre O and C is the mid point of arc AB<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1958\/44731148335_e3f024b9b2_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"226\" height=\"210\" \/><br \/>Line segment AB = S<br \/>\u2234 Centre will lie on AB (c<\/p>\n<p>Question 10.<br \/>In a circle, the major arc is 3 ti.nes the minor arc. The corresponding central angles and the degree measures of two arcs are<br \/>(a) 90\u00b0 and 270\u00b0<br \/>(b) 90\u00b0 and 90\u00b0<br \/>(c) 270\u00b0 adn 90\u00b0<br \/>(d) 60\u00b0 and 210\u00b0<br \/>Solution:<br \/>In a circle, major arc is 3 times the minor arc i.e. arc ACB = 3 arc ADB<br \/>\u2234 Reflex \u2220AOB = 3\u2220AOB<br \/>But angle at O = 360\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1910\/44731148235_4ee99f12d7_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"222\" height=\"213\" \/><br \/>and let \u2220AOB = x<br \/>Then reflex \u2220ADB = x<br \/>x + 3x \u2013 360\u00b0<br \/>\u21d2 4x = 360\u00b0<br \/>\u21d2 x =\u00a0<span id=\"MathJax-Element-60-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-376\" class=\"math\"><span id=\"MathJax-Span-377\" class=\"mrow\"><span id=\"MathJax-Span-378\" class=\"mfrac\"><span id=\"MathJax-Span-379\" class=\"msubsup\"><span id=\"MathJax-Span-380\" class=\"texatom\"><span id=\"MathJax-Span-381\" class=\"mrow\"><span id=\"MathJax-Span-382\" class=\"mn\">62<\/span><\/span><\/span><span id=\"MathJax-Span-383\" class=\"texatom\"><span id=\"MathJax-Span-384\" class=\"mrow\"><span id=\"MathJax-Span-385\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-386\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0 = 90\u00b0<br \/>\u2234 3x = 90\u00b0 x 3 = 270\u00b0<br \/>Here angles are 270\u00b0 and 90\u00b0 (c)<\/p>\n<p>Question 11.<br \/>If A and B are two points on a circle such that m(<span id=\"MathJax-Element-61-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-387\" class=\"math\"><span id=\"MathJax-Span-388\" class=\"mrow\"><span id=\"MathJax-Span-389\" class=\"texatom\"><span id=\"MathJax-Span-390\" class=\"mrow\"><span id=\"MathJax-Span-391\" class=\"munderover\"><span id=\"MathJax-Span-392\" class=\"mrow\"><span id=\"MathJax-Span-393\" class=\"mi\">A<\/span><span id=\"MathJax-Span-394\" class=\"mi\">B<\/span><\/span><span id=\"MathJax-Span-395\" class=\"mo\">\u02d8<\/span><\/span><\/span><\/span><\/span><\/span><\/span>) = 260\u00b0. A possible value for the angle subtended by arc BA at a point on the circle is<br \/>(a) 100\u00b0<br \/>(b) 75\u00b0<br \/>(c) 50\u00b0<br \/>(d) 25\u00b0<br \/>Solution:<br \/>A and B are two points on the circle such that reflex \u2220AOB = 260\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1908\/44731148115_d75e56e3eb_o.png\" alt=\"Class 9 Maths Chapter 15 Areas of Parallelograms and Triangles RD Sharma Solutions\" width=\"215\" height=\"204\" \/><br \/>\u2234 \u2220AOB = 360\u00b0 \u2013 260\u00b0 = 100\u00b0<br \/>C is a point on the circle<br \/>\u2234 By joining AC and BC,<br \/>\u2220ACB =\u00a0<span id=\"MathJax-Element-62-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-396\" class=\"math\"><span id=\"MathJax-Span-397\" class=\"mrow\"><span id=\"MathJax-Span-398\" class=\"mfrac\"><span id=\"MathJax-Span-399\" class=\"mn\">1<\/span><span id=\"MathJax-Span-400\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u2220AOB =\u00a0<span id=\"MathJax-Element-63-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-401\" class=\"math\"><span id=\"MathJax-Span-402\" class=\"mrow\"><span id=\"MathJax-Span-403\" class=\"mfrac\"><span id=\"MathJax-Span-404\" class=\"mn\">1<\/span><span id=\"MathJax-Span-405\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 100\u00b0 = 50\u00b0 (c)<\/p>\n<p>Question 12.<br \/>An equilateral triangle ABC is inscribed in a circle with centre O. The measures of \u2220BOC is<br \/>(a) 30\u00b0<br \/>(b) 60\u00b0<br \/>(c) 90\u00b0<br \/>(d) 120\u00b0<br \/>Solution:<br \/>\u2206ABC is an equilateral triangle inscribed in a circle with centre O<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1921\/44731148025_3646cccb2a_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 15 Areas of Parallelograms and Triangles\" width=\"222\" height=\"213\" \/><br \/>\u2234 Measure of \u2220BOC = 2\u2220BAC<br \/>= 2 x 60\u00b0 = 120\u00b0 (d)<\/p>\n<p>Question 13.<br \/>If two diameters of a circle intersect each other at right angles, then quadrilateral formed by joining their end points is a<br \/>(a) rhombus<br \/>(b) rectangle<br \/>(c) parallelogram<br \/>(d) square<br \/>Solution:<br \/>Two diameter of a circle AB and CD intersect each other at right angles<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1926\/44731147945_f878cd7255_o.png\" alt=\"RD Sharma Class 9 Book Chapter 15 Areas of Parallelograms and Triangles\" width=\"254\" height=\"228\" \/><br \/>AD, DB, BC and CA are joined forming a quad. ABCD.<br \/>\u2235 The diagonals are equal and bisect each other at right angles<br \/>\u2234 ACBD is a square (d)<\/p>\n<p>Question 14.<br \/>In ABC is an arc of a circle and \u2220ABC = 135\u00b0, then the ratio of arc\u00a0<span id=\"MathJax-Element-64-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-406\" class=\"math\"><span id=\"MathJax-Span-407\" class=\"mrow\"><span id=\"MathJax-Span-408\" class=\"texatom\"><span id=\"MathJax-Span-409\" class=\"mrow\"><span id=\"MathJax-Span-410\" class=\"munderover\"><span id=\"MathJax-Span-411\" class=\"mrow\"><span id=\"MathJax-Span-412\" class=\"mi\">A<\/span><span id=\"MathJax-Span-413\" class=\"mi\">B<\/span><\/span><span id=\"MathJax-Span-414\" class=\"mo\">\u02d8<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u00a0to the circumference is<br \/>(a) 1 : 4<br \/>(b) 3 : 4<br \/>(c) 3 : 8<br \/>(d) 1 : 2<br \/>Solution:<br \/>Arc ABC of a circle and \u2220ABC = 135\u00b0<br \/>Join OA and OC<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1910\/44920318664_10fdf94989_o.png\" alt=\"Areas of Parallelograms and Triangles With Solutions PDF RD Sharma Class 9 Solutions\" width=\"207\" height=\"206\" \/><br \/>\u2234 Angle subtended by arc ABC at the centre = 2 x \u2220ABC = 2 x 135\u00b0 = 270\u00b0<br \/>Angle at the centre of the circle = 360\u00b0<br \/>\u2234 Ratio with circumference = 270\u00b0 : 360\u00b0 = 3:4 (b)<\/p>\n<p>Question 15.<br \/>The chord of a circle is equal to its radius. The angle subtended by this chord at the minor arc of the circle is<br \/>(a) 60\u00b0<br \/>(b) 75\u00b0<br \/>(c) 120\u00b0<br \/>(d) 150\u00b0<br \/>Solution:<br \/>The chord of a circle = radius of the circle In the figure OA = OB = AB<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1964\/44731147735_bfaca94baa_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 15 Areas of Parallelograms and Triangles\" width=\"214\" height=\"189\" \/><br \/>\u2234 \u2220AOB = 60\u00b0<br \/>(Each angle of an equilateral = 60\u00b0) (a)<\/p>\n<p>Question 16.<br \/>PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If \u2220QPR = 67\u00b0 and \u2220SPR = 72\u00b0, then \u2220QRS =<br \/>(a) 41\u00b0<br \/>(b) 23\u00b0<br \/>(c) 67\u00b0<br \/>(d) 18\u00b0<br \/>Solution:<br \/>PQRS is a cyclic quadrilateral with centre O and \u2220QPR = 67\u00b0<br \/>\u2220SPR = 72\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1917\/44920318454_e3202dce06_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"211\" height=\"200\" \/><br \/>\u2234 \u2220QPS = 67\u00b0 + 72\u00b0 = 139\u00b0<br \/>\u2235 \u2220QPS + \u2220QRS = 180\u00b0 (Sum of opposite angles of a cyclic quad.)<br \/>\u21d2 139\u00b0 + \u2220QRS = 180\u00b0<br \/>\u21d2 \u2220QRS = 180\u00b0 \u2013 139\u00b0 = 41\u00b0 (a)<\/p>\n<p>Question 17.<br \/>If A, B, C are three points on a circle with centre O such that \u2220AOB = 90\u00b0 and \u2220BOC = 120\u00b0, then \u2220ABC =<br \/>(a) 60\u00b0<br \/>(b) 75\u00b0<br \/>(c) 90\u00b0<br \/>(d) 135\u00b0<br \/>Solution:<br \/>A, B and C are three points on a circle with centre O<br \/>\u2220AOB = 90\u00b0 and \u2220BOC = 120\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1920\/44731147665_2d11e20e1c_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"214\" height=\"212\" \/><br \/>\u2234 \u2220AOC = 360\u00b0 \u2013 (120\u00b0 + 90\u00b0)<br \/>= 360\u00b0 -210\u00b0= 150\u00b0<br \/>But \u2220AOC is at the centre made by arc AC and \u2220ABC at the remaining part of the circle<br \/>\u2234 \u2220ABC =\u00a0<span id=\"MathJax-Element-65-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-415\" class=\"math\"><span id=\"MathJax-Span-416\" class=\"mrow\"><span id=\"MathJax-Span-417\" class=\"mfrac\"><span id=\"MathJax-Span-418\" class=\"mn\">1<\/span><span id=\"MathJax-Span-419\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0\u2220AOC<br \/>=\u00a0<span id=\"MathJax-Element-66-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-420\" class=\"math\"><span id=\"MathJax-Span-421\" class=\"mrow\"><span id=\"MathJax-Span-422\" class=\"mfrac\"><span id=\"MathJax-Span-423\" class=\"mn\">1<\/span><span id=\"MathJax-Span-424\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 150\u00b0 = 75\u00b0 (b)<\/p>\n<p>Question 18.<br \/>The greatest chord of a circle is called its<br \/>(a) radius<br \/>(b) secant<br \/>(c) diameter<br \/>(d) none of these<br \/>Solution:<br \/>The greatest chord of a circle is called its diameter. (c)<\/p>\n<p>Question 19.<br \/>Angle formed in minor segment of a circle is<br \/>(a) acute<br \/>(b) obtuse<br \/>(c) right angle<br \/>(d) none of these<br \/>Solution:<br \/>The angle formed in minor segment of a circle is obtuse angle. (b)<\/p>\n<p>Question 20.<br \/>Number of circles that can be drawn through three non-collinear points is<br \/>(a) 1<br \/>(b) 0<br \/>(c) 2<br \/>(d) 3<br \/>Solution:<br \/>The number of circles that can pass through three non-collinear points is only one. (a)<\/p>\n<p>Question 21.<br \/>In the figure, if chords AB and CD of the circle intersect each other at right angles, then x + y =<br \/>(a) 45\u00b0<br \/>(b) 60\u00b0<br \/>(c) 75\u00b0<br \/>(d) 90\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1934\/44920318264_e06d621960_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 15 Areas of Parallelograms and Triangles\" width=\"231\" height=\"204\" \/><br \/>Solution:<br \/>In the circle, AB and CD are two chords which intersect each other at P at right angle i.e. \u2220CPB = 90\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1901\/44731147545_d6d55e4bbb_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"220\" height=\"208\" \/><br \/>\u2220CAB and \u2220CDB are in the same segment<br \/>\u2234 \u2220CDB = \u2220CAB = x<br \/>Now in \u2206PDB,<br \/>Ext. \u2220CPB = \u2220D + \u2220DBP<br \/>\u21d2 90\u00b0 = x + y (\u2235 CD \u22a5 AB)<br \/>Hence x + y = 90\u00b0 (d)<\/p>\n<p>Question 22.<br \/>In the figure, if \u2220ABC = 45\u00b0, then \u2220AOC=<br \/>(a) 45\u00b0<br \/>(b) 60\u00b0<br \/>(c) 75\u00b0<br \/>(d) 90\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1930\/44920317984_3af8533993_o.png\" alt=\"RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"184\" height=\"157\" \/><br \/>Solution:<br \/>\u2235 arc AC subtends<br \/>\u2220AOC at the centre of the circle and \u2220ABC<br \/>at the remaining part of the circle<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1949\/44731147405_0c352c52a4_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"196\" height=\"163\" \/><br \/>\u2234 \u2220AOC = 2\u2220ABC<br \/>= 2 x 45\u00b0 = 90\u00b0<br \/>Hence \u2220AOC = 90\u00b0 (d)<\/p>\n<p>Question 23.<br \/>In the figure, chords AD and BC intersect each other at right angles at a point P. If \u2220D AB = 35\u00b0, then \u2220ADC =<br \/>(a) 35\u00b0<br \/>(b) 45\u00b0<br \/>(c) 55\u00b0<br \/>(d) 65\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1932\/44920317854_428a2229d1_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"186\" height=\"171\" \/><br \/>Solution:<br \/>Two chords AD and BC intersect each other at right angles at P, \u2220DAB = 35\u00b0<br \/>AB and CD are joined<br \/>In \u2206ABP,<br \/>Ext. \u2220APC = \u2220B + \u2220A<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1960\/44731147225_36664d5533_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 15 Areas of Parallelograms and Triangles\" width=\"171\" height=\"163\" \/><br \/>\u21d2 90\u00b0 = \u2220B + 35\u00b0<br \/>\u2220B = 90\u00b0 \u2013 35\u00b0 = 55\u00b0<br \/>\u2235 \u2220ABC and \u2220ADC are in the same segment<br \/>\u2234 \u2220ADC = \u2220ABC = 55\u00b0 (c)<\/p>\n<p>Question 24.<br \/>In the figure, O is the centre of the circle and \u2220BDC = 42\u00b0. The measure of \u2220ACB is<br \/>(a) 42\u00b0<br \/>(b) 48\u00b0<br \/>(c) 58\u00b0<br \/>(d) 52\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1931\/44731147105_8ac72fc40f_o.png\" alt=\"Areas of Parallelograms and Triangles Class 9 RD Sharma Solutions\" width=\"203\" height=\"216\" \/><br \/>Solution:<br \/>In the figure, O is the centre of the circle<br \/>\u2220BDC = 42\u00b0<br \/>\u2220ABC = 90\u00b0 (Angle in a semicircle)<br \/>and \u2220BAC and \u2220BDC are in the same segment of the circle.<br \/>\u2234 \u2220BAC = \u2220BDC = 42\u00b0<br \/>Now in \u2206ABC,<br \/>\u2220A + \u2220ABC + \u2220ACB = 180\u00b0 (Sum of angles of a triangle)<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1962\/44920317574_5f9956dba3_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 15 Areas of Parallelograms and Triangles\" width=\"217\" height=\"216\" \/><br \/>\u21d2 42\u00b0 + 90\u00b0 + \u2220ACB = 180\u00b0<br \/>\u21d2 132\u00b0 + \u2220ACB \u2013 180\u00b0<br \/>\u21d2 \u2220ACB = 180\u00b0 \u2013 132\u00b0 = 48\u00b0 (b)<\/p>\n<p>Question 25.<br \/>In a circle with centre O, AB and CD are two diameters perpendicular to each other. The length of chord AC is<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1916\/44731146945_0f87563cc1_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"292\" height=\"94\" \/><br \/>Solution:<br \/>AB and CD are two diameters of a circle with centre O<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1919\/44920317444_d97b1ca7aa_o.png\" alt=\"Class 9 Maths Chapter 15 Areas of Parallelograms and Triangles RD Sharma Solutions\" width=\"325\" height=\"391\" \/><\/p>\n<p>Question 26.<br \/>Two equal circles of radius r intersect such that each passes through the centre of the other. The length of the common chord of the circles is<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1967\/44731146725_9e52e4aa3d_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 15 Areas of Parallelograms and Triangles\" width=\"291\" height=\"92\" \/><br \/>Solution:<br \/>Two equal circles pass through the centre of the other and intersect each other at A and B<br \/>Let r be the radius of each circle<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1940\/44920317274_3a8c6aa131_o.png\" alt=\"RD Sharma Class 9 Book Chapter 15 Areas of Parallelograms and Triangles\" width=\"345\" height=\"534\" \/><\/p>\n<p>Question 27.<br \/>If AB is a chord of a circle, P and Q are the two points on the circle different from A and B,then<br \/>(a) \u2220APB = \u2220AQB<br \/>(b) \u2220APB + \u2220AQB = 180\u00b0 or \u2220APB = \u2220AQB<br \/>(c) \u2220APB + \u2220AQB = 90\u00b0<br \/>(d) \u2220APB + \u2220AQB = 180\u00b0<br \/>Solution:<br \/>AB is chord of a circle,<br \/>P and Q are two points other than from points A and B<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1959\/44920317064_988501c844_o.png\" alt=\"Areas of Parallelograms and Triangles With Solutions PDF RD Sharma Class 9 Solutions\" width=\"226\" height=\"198\" \/><br \/>\u2235 \u2220APB and \u2220AQB are in the same segment of the circle<br \/>\u2234 \u2220APB = \u2220AQB (a)<\/p>\n<p>Question 28.<br \/>AB and CD are two parallel chords of a circle with centre O such that AB = 6 cm and CD = 12 cm. The chords are on the same side of the centre and the distance between them is 3 cm. The radius of the circle is<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1939\/44731146465_d3b4af560a_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 15 Areas of Parallelograms and Triangles\" width=\"302\" height=\"74\" \/><br \/>Solution:<br \/>AB and CD are two parallel chords of a circle with centre O<br \/>Let r be the radius of the circle AB = 6 cm, CD = 12 cm<br \/>and distance between them = 3 cm<br \/>Join OC and OA, LM = 3 cm<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1905\/44920316944_3f8b6e5d15_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"227\" height=\"199\" \/><br \/>Let OM = x, then OL = x + 3<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1980\/44731146435_cde5a8713b_o.png\" alt=\"RD Sharma Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"354\" height=\"550\" \/><\/p>\n<p>Question 29.<br \/>In a circle of radius 17 cm, two parallel chords are drawn on opposite side of a diameter. This distance between the chords is 23 cm. If the length of one chord is 16 cm then the length of the other is<br \/>(a) 34 cm<br \/>(b) 15 cm<br \/>(c) 23 cm<br \/>(d) 30 cm<br \/>Solution:<br \/>Radius of a circle = 17 cm<br \/>The distance between two parallel chords = 23 cm<br \/>AB || CD and LM = 23 cm<br \/>Join OA and OC,<br \/>\u2234 OA = OC = 17 cm<br \/>Let OL = x, then OM = (23 \u2013 x) cm<br \/>AB = 16 cm<br \/>Now in right \u2206OAL,<br \/>OA<sup>2<\/sup>\u00a0= OL2<sup>2<\/sup>\u00a0+ AL<sup>2<\/sup><br \/>\u21d2 (17)<sup>2<\/sup>\u00a0= x<sup>2<\/sup>\u00a0+ AL<sup>2<\/sup><br \/>\u21d2 289 = x<sup>2<\/sup>\u00a0+ AL<sup>2<\/sup><br \/><img src=\"https:\/\/farm2.staticflickr.com\/1966\/44920316624_d8daf0c782_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles\" width=\"353\" height=\"494\" \/><\/p>\n<p>Question 30.<br \/>In the figure, O is the centre of the circle such that \u2220AOC = 130\u00b0, then \u2220ABC =<br \/>(a) 130\u00b0<br \/>(b) 115\u00b0<br \/>(c) 65\u00b0<br \/>(d) 165\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1919\/44920316304_eddfca0079_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 15 Areas of Parallelograms and Triangles\" width=\"215\" height=\"191\" \/><br \/>Solution:<br \/>O is the centre of the circle and \u2220AOC = 130\u00b0<br \/>Reflex \u2220AOC = 360\u00b0 \u2013 130\u00b0 = 230\u00b0<br \/><img src=\"https:\/\/farm2.staticflickr.com\/1901\/44731146025_fa114dc527_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 15 Areas of Parallelograms and Triangles\" width=\"226\" height=\"210\" \/><br \/>Now arc ADB subtends \u2220AOC at the centre and \u2220ABC at the remaining part of the circle<br \/>\u2234 \u2220ABC =\u00a0<span id=\"MathJax-Element-67-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-425\" class=\"math\"><span id=\"MathJax-Span-426\" class=\"mrow\"><span id=\"MathJax-Span-427\" class=\"mfrac\"><span id=\"MathJax-Span-428\" class=\"mn\">1<\/span><span id=\"MathJax-Span-429\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>reflex \u2220AOC<br \/>=\u00a0<span id=\"MathJax-Element-68-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-430\" class=\"math\"><span id=\"MathJax-Span-431\" class=\"mrow\"><span id=\"MathJax-Span-432\" class=\"mfrac\"><span id=\"MathJax-Span-433\" class=\"mn\">1<\/span><span id=\"MathJax-Span-434\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>\u00a0x 230\u00b0= 115\u00b0 (b)<\/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 MCQS&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:268732,&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;14&quot;:[null,2,0],&quot;15&quot;:&quot;Arial&quot;,&quot;21&quot;:1}\">RD Sharma Class 9 Solutions Chapter 15 MCQS<\/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-mcqs\"><\/span>FAQ: <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 15 MCQS&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:268732,&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;14&quot;:[null,2,0],&quot;15&quot;:&quot;Arial&quot;,&quot;21&quot;:1}\">RD Sharma Class 9 Solutions Chapter 15 MCQS<\/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-1631088261123\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-rd-sharma-class-9-solutions-chapter-15-mcqs-on-my-smartphone\"><\/span>Can I access RD Sharma Class 9 Solutions Chapter 15 MCQs on my smartphone?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, you can access RD Sharma Class 9 Solutions Chapter 15 MCQs on any device.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631088372702\" 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-chapter-15-mcqs\"><\/span>What are the benefits of studying RD Sharma Class 9 Solutions Chapter 15 MCQs?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>By practicing RD Sharma Class 9 Solutions Chapter 15 MCQS, 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-1631088401084\" 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 MCQS: These solutions are specifically created by our experts. Useful while preparing for the 9th grade exam. You can access RD Sharma Class 9 Solutions Chapter 15 MCQS directly in the article below. RD Sharma Class 9 Maths Solutions Access RD Sharma Class 9 Solutions Chapter 15 MCQS &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 15 MCQS (Updated 2024-25)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-15-mcqs\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 15 MCQS (Updated 2024-25)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":127698,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73719],"tags":[76786,76826],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125342"}],"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=125342"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125342\/revisions"}],"predecessor-version":[{"id":524878,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125342\/revisions\/524878"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/127698"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125342"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125342"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125342"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}