{"id":126602,"date":"2023-09-12T18:18:00","date_gmt":"2023-09-12T12:48:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126602"},"modified":"2023-11-30T10:27:47","modified_gmt":"2023-11-30T04:57:47","slug":"rd-sharma-class-9-solutions-chapter-14-mcqs","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-14-mcqs\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 14 MCQS (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-126608\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-14-MCQS.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 14 MCQS\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-14-MCQS.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-14-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 14 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 14 MCQs<\/strong> are available here. <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-14-quadrilaterals\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 14<\/a> MCQs are created by our team of experts in a detailed manner.\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 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-69d059d4c0768\" 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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MCQs?\">What are the benefits of studying RD Sharma Class 9 Solutions Chapter 14 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-14-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-14-mcqs\"><\/span>Access <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 14 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 14 MCQS<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Mark the correct alternative in each of the following:<br \/>Question 1.<br \/>Two parallelograms are on the same base and between the same parallels. The ratio of their areas is<br \/>(a) 1 : 2<br \/>(b) 2 : 1<br \/>(c) 1 : 1<br \/>(d) 3 : 1<br \/>Solution:<br \/>Two parallelograms which are on the same base and between the same parallels are equal in area<br \/>\u2234 Ratio in their areas =1 : 1 (c)<\/p>\n<div class=\"code-block code-block-7\">\n<div id=\"div-gpt-ad-1699162392598-0\" data-google-query-id=\"COz66ov36oIDFexGnQkd3o4IsA\">\n<div id=\"google_ads_iframe_\/22987339172\/insta_DSK_INTERSTITIAL_1x1_051123_0__container__\"><span style=\"font-size: inherit; background-color: initial;\">Question 2.<\/span><\/div>\n<\/div>\n<\/div>\n<p>A triangle and a parallelogram are on the same base and between the same parallels. The ratio of the areas of the triangle and parallelogram is<br \/>(a) 1 : 1<br \/>(b) 1 : 2<br \/>(c) 2 : 1<br \/>(d) 1 : 3<br \/>Solution:<br \/>A triangle and a parallelogram are on the same base and between the same parallels, then area of a triangle is half the area of the parallelogram<br \/>\u2234 Their ratio =1:2 (c)<\/p>\n<p>Question 3.<br \/>Let ABC be a triangle of area 24 sq. units and PQR be the triangle formed by the mid-points of sides of \u2206ABC. Then the area of \u2206PQR is<br \/>(a) 12 sq. units<br \/>(b) 6 sq. units<br \/>(c) 4 sq. units<br \/>(d) 3 sq. units<br \/>Solution:<br \/>Area of \u2206ABC = 24 sq. units<br \/><img class=\"alignnone size-full wp-image-67337\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q4.1.png\" sizes=\"(max-width: 317px) 100vw, 317px\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q4.1.png 317w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q4.1-233x300.png 233w\" alt=\"RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q4.1\" width=\"317\" height=\"408\" \/><br \/><img class=\"alignnone size-full wp-image-67338\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q4.2.png\" sizes=\"(max-width: 349px) 100vw, 349px\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q4.2.png 349w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q4.2-300x98.png 300w\" alt=\"RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q4.2\" width=\"349\" height=\"114\" \/><\/p>\n<p>Question 4.<br \/>The median of a triangle divides it into two<br \/>(a) congruent triangle<br \/>(b) isosceles triangles<br \/>(c) right triangles<br \/>(d) triangles of equal areas<br \/>Solution:<br \/>The median of a triangle divides it into two triangles equal in area (d)<\/p>\n<p>Question 5.<br \/>In\u00a0 \u2206ABC, D, E, and F are the mid-points of sides BC, CA, and AB respectively. If<br \/>ar(\u2206ABC) = 16 cm2, then ar(trapezium FBCE) =<br \/>(a) 4 cm\u00b2<br \/>(b) 8 cm\u00b2<br \/>(c) 12 cm\u00b2<br \/>(d) 10 cm\u00b2<br \/>Solution:<br \/>In \u2206ABC, D, E, and F are the midpoints of sides BC, CA, and AB respectively<br \/>ar(\u2206ABC) = 16 cm\u00b2<br \/><img class=\"alignnone size-full wp-image-67339\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q5.1.png\" sizes=\"(max-width: 358px) 100vw, 358px\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q5.1.png 358w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q5.1-278x300.png 278w\" alt=\"RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q5.1\" width=\"358\" height=\"387\" \/><\/p>\n<p>Question 6.<br \/>ABCD is a parallelogram. P is any point on CD. If ar(\u2206DPA) = 15 cm\u00b2 and ar(\u2206APC) = 20 cm\u00b2, then ar(\u2206APB) =<br \/>(a) 15 cm\u00b2<br \/>(b) 20 cm\u00b2<br \/>(c) 35 cm\u00b2<br \/>(d) 30 cm\u00b2<br \/>Solution:<br \/>In ||gm ABCD, P is any point on CD<br \/>AP, AC and PB are joined<br \/>ar(\u2206DPA) =15 cm\u00b2<br \/>ar(\u2206APC) = 20 cm\u00b2<br \/>Adding, ar(\u2206ADC) = 15 + 20 = 35 cm\u00b2<br \/><img class=\"alignnone size-full wp-image-67340\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q6.1.png\" alt=\"RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q6.1\" width=\"212\" height=\"172\" \/><br \/>\u2235 AC divides it into two triangles equal in area<br \/>\u2234 ar(\u2206ACB) = ar(\u2206ADC) = 35 cm\u00b2<br \/>\u2235 \u2206APB and \u2206ACB are on the same base<br \/>AB and between the same parallels<br \/>\u2234 ar(\u2206APB) = ar(\u2206ACB) = 35 cm\u00b2(c)<\/p>\n<p>Question 7.<br \/>The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 16 cm and 12 cm is<br \/>(a) 28 cm\u00b2<br \/>(b) 48 cm\u00b2<br \/>(c) 96 cm\u00b2<br \/>(d) 24 cm\u00b2<br \/>Solution:<br \/>In rhombus ABCD,<br \/>P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively and are joined in order to get a quad. PQRS<br \/><img class=\"alignnone size-full wp-image-67341\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q7.1.png\" sizes=\"(max-width: 360px) 100vw, 360px\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q7.1.png 360w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q7.1-283x300.png 283w\" alt=\"RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q7.1\" width=\"360\" height=\"382\" \/><\/p>\n<p>Question 8.<br \/>A, B, C, D are mid points of sides of parallelogram PQRS. If ar(PQRS) = 36 cm\u00b2,then ar(ABCD) =<br \/>(a) 24 cm\u00b2<br \/>(b) 18 cm\u00b2<br \/>(c) 30 cm\u00b2<br \/>(d) 36 cm\u00b2<br \/>Solution:<br \/>A, B, C and D are the mid points of a ||gm PQRS<br \/>Area of PQRS = 36 cm\u00b2<br \/><img class=\"alignnone size-full wp-image-67342\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q8.1.png\" alt=\"RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q8.1\" width=\"244\" height=\"189\" \/><br \/>The area of ||gm formed by joining AB, BC, CD and DA<br \/><img class=\"alignnone size-full wp-image-67343\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q8.2.png\" sizes=\"(max-width: 332px) 100vw, 332px\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q8.2.png 332w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q8.2-300x101.png 300w\" alt=\"RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q8.2\" width=\"332\" height=\"112\" \/><\/p>\n<p>Question 9.<br \/>The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is<br \/>(a) a rhombus of area 24 cm\u00b2<br \/>(b) a rectangle of area 24 cm\u00b2<br \/>(c) a square of area 26 cm\u00b2<br \/>(d) a trapezium of area 14 cm\u00b2<br \/>Solution:<br \/>Let P, Q, R, S be the mid points of sides of a rectangle ABCD. Whose sides 8 cm and 6 cm<br \/><img class=\"alignnone size-full wp-image-67344\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q9.1.png\" alt=\"RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q9.1\" width=\"266\" height=\"226\" \/><br \/>Their PQRS is a rhombus<br \/><img class=\"alignnone size-full wp-image-67345\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q9.2.png\" sizes=\"(max-width: 330px) 100vw, 330px\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q9.2.png 330w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q9.2-300x97.png 300w\" alt=\"RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q9.2\" width=\"330\" height=\"107\" \/><\/p>\n<p>Question 10.<br \/>If AD is median of \u2206ABC and P is a point on AC such that ar(\u2206ADP) : ar(\u2206ABD) = 2:3, then ar(\u2206PDC) : ar(\u2206ABC) is<br \/>(a) 1 : 5<br \/>(b) 1 : 5<br \/>(c) 1 : 6<br \/>(d) 3 : 5<br \/>Solution:<br \/>AD is the median of \u2206ABC,<br \/>P is a point on AC such that<br \/>ar(\u2206ADP) : ar(\u2206ABD) = 2:3<br \/>Let area of \u2206ADP = 2\u00d72<br \/>Then area of \u2206ABD = 3\u00d72<br \/><img class=\"alignnone size-full wp-image-67346\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q10.1.png\" alt=\"RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q10.1\" width=\"204\" height=\"193\" \/><br \/>But area of AABD =\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>\u00a0area AABC<br \/>\u2234 Area \u2206ABC = 2 x area of \u2206ABD<br \/>= 2 x 3x\u00b2 = 6x\u00b2<br \/>and area of \u2206PDC = area \u2206ADC \u2013 (ar\u2206ADP) = area \u2206ABD \u2013 ar \u2206ADP<br \/>= 3x\u00b2 \u2013 2x\u00b2 = x\u00b2<br \/>\u2234 Ratio = x\u00b2 : 6x\u00b2<br \/>= 1 : 6 (c)<\/p>\n<p>Question 11.<br \/>Medians of AABC, intersect at G. If ar(\u2206ABC) = 27 cm2, then ar(\u2206BGC) =<br \/>(a) 6 cm2<br \/>(b) 9 cm2<br \/>(c) 12 cm2<br \/>(d) 18 cm2<br \/>Solution:<br \/>In \u2206ABC, AD, BE and CF are the medians which intersect each other at G<br \/><img class=\"alignnone size-full wp-image-67347\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q11.1.png\" sizes=\"(max-width: 345px) 100vw, 345px\" srcset=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q11.1.png 345w, https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q11.1-279x300.png 279w\" alt=\"RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q11.1\" width=\"345\" height=\"371\" \/><\/p>\n<p>Question 12.<br \/>In a \u2206ABC if D and E are mid-points of BC and AD respectively such that ar(\u2206AEC) = 4 cm\u00b2, then ar(\u2206BEC) =<br \/>(a) 4 cm\u00b2<br \/>(b) 6 cm\u00b2<br \/>(c) 8 cm\u00b2<br \/>(d) 12 cm\u00b2<br \/>Solution:<br \/>In \u2206ABC, D and E are the mid points of BC and AD<br \/>Join BE and CE ar(\u2206AEC) = 4 cm\u00b2<br \/><img class=\"alignnone size-full wp-image-67348\" src=\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/07\/RD-Sharma-Class-9-Solutions-Chapter-14-Quadrilaterals-MCQS-Q12.1.png\" alt=\"RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS Q12.1\" width=\"245\" height=\"188\" \/><br \/>In \u2206ABC,<br \/>\u2235 AD is the median of BC<br \/>\u2234 ar(\u2206ABD) = ar(\u2206ACD)<br \/>Similarly in \u2206EBC,<br \/>ED is the median<br \/>\u2234 ar(\u2206EBD) = ar(\u2206ECD)<br \/>and in \u2206ADC, CE is the median<br \/>\u2234 ar(\u2206FDC) = ar(\u2206AEC)<br \/>= 4 cm<br \/>\u2234ar\u2206EBC = 2 x ar(\u2206EDC)<br \/>= 2 x 4 = 8 cm (c)<\/p>\n<p>We have included all the information regarding\u00a0<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 14 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 14 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 queries 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-14-mcqs\"><\/span>FAQ: <span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;RD Sharma Class 9 Solutions Chapter 14 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 14 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-1631276603474\" 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-14-mcqs-for-free\"><\/span>Can I access RD Sharma Class 9 Solutions Chapter 14 MCQS for free?<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 14 MCQS for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631276877641\" 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-14-mcqs\"><\/span>What are the benefits of studying RD Sharma Class 9 Solutions Chapter 14 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, 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-1631276878628\" 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 14 MCQs are available here. RD Sharma Class 9 Solutions Chapter 14 MCQs are created by our team of experts in a detailed manner.\u00a0 RD Sharma Class 9 Maths Solutions Access RD Sharma Class 9 Solutions Chapter 14 MCQS Mark the correct alternative in each of the following:Question 1.Two &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 14 MCQS (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-14-mcqs\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 14 MCQS (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":126608,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73719],"tags":[76786,76852],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126602"}],"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=126602"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126602\/revisions"}],"predecessor-version":[{"id":514502,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126602\/revisions\/514502"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126608"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126602"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126602"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126602"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}