{"id":4989,"date":"2016-05-19T11:49:00","date_gmt":"2016-05-19T06:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-9-maths-exercise-13-2\/"},"modified":"2019-08-20T18:12:54","modified_gmt":"2019-08-20T12:42:54","slug":"ncert-solutions-for-class-9-maths-exercise-13-2","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-2\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Exercise 13.2"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_76 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\" style=\"cursor:inherit\">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=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" 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: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-toggle-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-2\/#NCERT_Solutions_for_Class_9_Mathematics_Surface_Areas_and_Volumes\" >NCERT Solutions for Class 9 Mathematics\u00a0Surface Areas and Volumes<\/a><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-2\/#Assume_unless_stated_otherwise\" >Assume unless stated otherwise<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-2\/#1_The_curved_surface_area_of_a_right_circular_cylinder_of_height_14_cm_is_88_cm2_Find_the_diameter_of_the_base_of_the_cylinder\" >1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-2\/#2_It_is_required_to_make_a_closed_cylindrical_tank_of_height_1_m_and_base_diameter_140_cm_from_a_metal_sheet_How_many_square_meters_of_the_sheet_are_required_for_the_same\" >2. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square meters of the sheet are required for the same?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-2\/#3_A_metal_pipe_is_77_cm_long_The_inner_diameter_of_a_cross_section_is_4_cm_the_outer_diameter_being_44_cm_See_fig_Find_its\" >3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm. [See fig.]. Find its:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-2\/#4_The_diameter_of_a_roller_is_84_cm_and_its_length_is_120_cm_It_takes_500_complete_revolutions_to_move_once_over_to_level_a_playground_Find_the_area_of_the_playground_in_m2\" >4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-2\/#5_A_cylindrical_pillar_is_50_cm_in_diameter_and_35_m_in_height_Find_the_cost_of_white_washing_the_curved_surface_of_the_pillar_at_the_rate_of_Rs_1250_per_m2\" >5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of white washing the curved surface of the pillar at the rate of Rs. 12.50 per m2.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-2\/#6_Curved_surface_area_of_a_right_circular_cylinder_is_44_m2_If_the_radius_of_the_base_of_the_cylinder_is_07_m_find_its_height\" >6. Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-2\/#7_The_inner_diameter_of_a_circular_well_is_35_m_It_is_10_m_deep_Find\" >7. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-2\/#8_Find\" >8. Find:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-2\/#9_In_the_adjoining_figure_you_see_the_frame_of_a_lampshade_It_is_to_be_covered_with_a_decorative_cloth_The_frame_has_a_base_diameter_of_20_cm_and_height_of_30_cm_A_margin_of_25_cm_is_to_be_given_for_folding_it_over_the_top_and_bottom_of_the_frame_Find_how_much_cloth_is_required_for_covering_the_lampshade_See_fig\" >9. In the adjoining figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. [See fig.]<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-2\/#10_The_students_of_a_Vidyalayawere_asked_to_participate_in_a_competition_for_making_and_decorating_penholders_in_the_shape_of_a_cylinder_with_a_base_using_cardboard_Each_penholder_was_to_be_of_radius_3_cm_and_height_105_cm_The_Vidyalaya_was_to_supply_the_competitors_with_cardboard_If_there_were_35_competitors_how_much_cardboard_was_required_to_be_bought_for_the_competition\" >10. The students of a Vidyalayawere asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?<\/a><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-13\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-2\/#NCERT_Solutions_for_Class_9_Maths_Exercise_132\" >NCERT Solutions for Class 9 Maths Exercise 13.2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-14\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-2\/#CBSE_app_for_Class_9\" >CBSE app for Class 9<\/a><\/li><\/ul><\/nav><\/div>\n<p>NCERT Solutions for Class 9 Maths Exercise 13.2 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 9 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.<\/p>\n<p style=\"text-align: center;\"><strong>NCERT solutions for Class 9 Maths\u00a0<\/strong><strong>Surface Areas and Volumes\u00a0<\/strong><strong><a class=\"button\" href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-09-mathematics-surface-areas-and-volumes\/1247\/ncert-solutions\/5\/\">Download as PDF<\/a><\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/09_Class_Maths_Book.jpg\" alt=\"NCERT Solutions for Class 9 Maths Exercise 13.2\" width=\"172\" height=\"221\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_9_Mathematics_Surface_Areas_and_Volumes\"><\/span>NCERT Solutions for Class 9 Mathematics\u00a0Surface Areas and Volumes<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h6><span class=\"ez-toc-section\" id=\"Assume_unless_stated_otherwise\"><\/span><strong>Assume <img decoding=\"async\" style=\"height: 41px; width: 49px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image001.png\" \/>unless stated otherwise<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<h6><span class=\"ez-toc-section\" id=\"1_The_curved_surface_area_of_a_right_circular_cylinder_of_height_14_cm_is_88_cm2_Find_the_diameter_of_the_base_of_the_cylinder\"><\/span><strong>1. <\/strong><strong>The curved surface area of a right circular cylinder of height 14 cm is 88 cm<sup>2<\/sup>. Find the diameter of the base of the cylinder.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Ans.<\/strong> Given: Height of cylinder <img decoding=\"async\" style=\"height: 27px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image002.png\" \/>= 14 cm, Curved Surface Area = 88 cm<sup>2<\/sup><\/p>\n<p>Let radius of base of right circular cylinder = <img decoding=\"async\" style=\"height: 13px; width: 12px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image003.png\" \/>cm<\/p>\n<p><img decoding=\"async\" id=\"Picture 12\" style=\"height: 164px; width: 129px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image004.jpg\" \/><\/p>\n<p><img decoding=\"async\" style=\"height: 19px; width: 37px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image005.png\" \/>= 88<\/p>\n<p><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image006.png\" \/><img decoding=\"async\" style=\"height: 41px; width: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image007.png\" \/><\/p>\n<p><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image006.png\" \/><img decoding=\"async\" style=\"height: 41px; width: 127px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image008.png\" \/><\/p>\n<p><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image006.png\" \/><img decoding=\"async\" style=\"height: 13px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image009.png\" \/>1 cm<\/p>\n<p>Diameter of the base of the cylinder = <img decoding=\"async\" style=\"height: 17px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image010.png\" \/>= 2 x 1 = 2 cm<\/p>\n<hr \/>\n<h6><span class=\"ez-toc-section\" id=\"2_It_is_required_to_make_a_closed_cylindrical_tank_of_height_1_m_and_base_diameter_140_cm_from_a_metal_sheet_How_many_square_meters_of_the_sheet_are_required_for_the_same\"><\/span><strong>2. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square meters of the sheet are required for the same?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><img decoding=\"async\" style=\"height: 162px; width: 312px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image011.png\" \/><\/p>\n<p><strong>Ans. <\/strong>Given: Diameter = 140 cm<\/p>\n<p><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image006.png\" \/>Radius <img decoding=\"async\" style=\"height: 27px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image012.png\" \/>= 70 cm = 0.7 m<\/p>\n<p>Height of the cylinder <img decoding=\"async\" style=\"height: 27px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image002.png\" \/>= 1 m<\/p>\n<p>Total Surface Area of the cylinder<\/p>\n<p>= <img decoding=\"async\" style=\"height: 27px; width: 75px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image013.png\" \/><\/p>\n<p>= <img decoding=\"async\" style=\"height: 41px; width: 129px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image014.png\" \/><\/p>\n<p>= 2 x 22 x 0.1 x 1.7 = 7.48 m<sup>2<\/sup><\/p>\n<p>Hence 7.48 m<sup>2<\/sup> metal sheet is required to make the close cylindrical tank.<\/p>\n<hr \/>\n<h6><span class=\"ez-toc-section\" id=\"3_A_metal_pipe_is_77_cm_long_The_inner_diameter_of_a_cross_section_is_4_cm_the_outer_diameter_being_44_cm_See_fig_Find_its\"><\/span><strong>3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm. [See fig.]. Find its:<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>(i) <\/strong><strong>Inner curved surface area<\/strong><\/p>\n<p><strong>(ii) <\/strong><strong>Outer curved surface area<\/strong><\/p>\n<p><strong>(iii) <\/strong><strong>Total surface area<\/strong><\/p>\n<p><img decoding=\"async\" style=\"height: 136px; width: 110px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image015.png\" \/><img decoding=\"async\" style=\"height: 110px; width: 218px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image016.png\" \/><\/p>\n<p><strong>Ans. (i)<\/strong> Length of the pipe = 77 cm, Inner diameter of cross-section = 4 cm<\/p>\n<p><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image006.png\" \/>Inner radius of cross-section = 2 cm<\/p>\n<p>Inner curved surface area of pipe = <img decoding=\"async\" style=\"height: 19px; width: 37px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image005.png\" \/>= <img decoding=\"async\" style=\"height: 41px; width: 93px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image017.png\" \/><\/p>\n<p>= 2 x 22 x 2 x 11 = 968 cm<sup>2<\/sup><\/p>\n<p><strong>(ii)<\/strong> Length of pipe = 77 cm, Outer diameter of pipe = 4.4 cm<\/p>\n<p><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image006.png\" \/>Outer radius of the pipe = 2.2 cm<\/p>\n<p>Outer surface area of the pipe = <img decoding=\"async\" style=\"height: 19px; width: 37px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image005.png\" \/><\/p>\n<p>= <img decoding=\"async\" style=\"height: 41px; width: 105px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image018.png\" \/><\/p>\n<p>= 44 x 2.2 x 11 = 1064.8 cm<sup>2<\/sup><\/p>\n<p><strong>(iii)<\/strong> Now there are two circles of radii 2 cm and 2.2 cm at both the ends of the pipe.<\/p>\n<p><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image019.png\" \/>Area of two edges of the pipe = 2 (Area of outer circle \u2013 area of inner circle)<\/p>\n<p>= <img decoding=\"async\" style=\"height: 29px; width: 93px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image020.png\" \/>= <img decoding=\"async\" style=\"height: 29px; width: 84px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image021.png\" \/><\/p>\n<p>= <img decoding=\"async\" style=\"height: 41px; width: 140px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image022.png\" \/>= <img decoding=\"async\" style=\"height: 41px; width: 88px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image023.png\" \/><\/p>\n<p>= <img decoding=\"async\" style=\"height: 41px; width: 64px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image024.png\" \/>= 5.28 cm<sup>2<\/sup><\/p>\n<p><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image019.png\" \/>Total surface area of pipe<\/p>\n<p>= Inner curved surface area + Outer curved surface area + Area of two edges<\/p>\n<p>= 968 + 1064.8 + 5.28 = 2038.08 cm<sup>2<\/sup><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 13.2<\/p>\n<h6><span class=\"ez-toc-section\" id=\"4_The_diameter_of_a_roller_is_84_cm_and_its_length_is_120_cm_It_takes_500_complete_revolutions_to_move_once_over_to_level_a_playground_Find_the_area_of_the_playground_in_m2\"><\/span><strong>4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m<sup>2<\/sup>.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Ans.<\/strong> Diameter of roller = 84 cm<\/p>\n<p><img decoding=\"async\" id=\"Picture 21\" style=\"height: 125px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image025.jpg\" \/><br \/>\n<img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image006.png\" \/>Radius of the roller = 42 cm<\/p>\n<p>Length (Height) of the roller = 120 cm<\/p>\n<p>Curved surface area of the roller = <img decoding=\"async\" style=\"height: 19px; width: 37px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image005.png\" \/><\/p>\n<p>= <img decoding=\"async\" style=\"height: 41px; width: 107px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image026.png\" \/>= 31680 cm<sup>2<\/sup><\/p>\n<p>= 3.1680 m<sup>2<\/sup><\/p>\n<p><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image027.png\" \/>Now area leveled by roller in one revolution = 3.1680 m<sup>2<\/sup><\/p>\n<p><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image019.png\" \/>Area leveled by roller in 500 revolutions<\/p>\n<p>= 3.1680 x 500 = 1584.0000 = 1584 m<sup>2<\/sup><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 13.2<\/p>\n<h6><span class=\"ez-toc-section\" id=\"5_A_cylindrical_pillar_is_50_cm_in_diameter_and_35_m_in_height_Find_the_cost_of_white_washing_the_curved_surface_of_the_pillar_at_the_rate_of_Rs_1250_per_m2\"><\/span><strong>5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of white washing the curved surface of the pillar at the rate of Rs. 12.50 per m<sup>2<\/sup>.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Ans.<\/strong> Diameter of pillar = 50 cm<\/p>\n<p><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image006.png\" \/>Radius of pillar = 25 cm = <img decoding=\"async\" style=\"height: 41px; width: 56px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image028.png\" \/>m<\/p>\n<p>Height of the pillar = 3.5 m<\/p>\n<p>Now, Curved surface area of the pillar<\/p>\n<p>= <img decoding=\"async\" style=\"height: 19px; width: 37px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image005.png\" \/><\/p>\n<p>= <img decoding=\"async\" style=\"height: 41px; width: 88px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image029.png\" \/>= <img decoding=\"async\" style=\"height: 41px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image030.png\" \/>m<sup>2<\/sup><\/p>\n<p><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image027.png\" \/>Cost of white washing 1 m<sup>2<\/sup> = Rs. 12.50<\/p>\n<p><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image019.png\" \/>Cost of white washing <img decoding=\"async\" style=\"height: 41px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image030.png\" \/>m<sup>2<\/sup><\/p>\n<p>= 12.50 x <img decoding=\"async\" style=\"height: 41px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image030.png\" \/>= Rs. 68.75<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 13.2<\/p>\n<h6><span class=\"ez-toc-section\" id=\"6_Curved_surface_area_of_a_right_circular_cylinder_is_44_m2_If_the_radius_of_the_base_of_the_cylinder_is_07_m_find_its_height\"><\/span><strong>6. Curved surface area of a right circular cylinder is 4.4 m<sup>2<\/sup>. If the radius of the base of the cylinder is 0.7 m, find its height.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Ans.<\/strong> Curved surface area of the cylinder<\/p>\n<p>= 4.4 m<sup>2<\/sup>, Radius of cylinder = 0.7 m<\/p>\n<p>Let height of the cylinder = <img decoding=\"async\" style=\"height: 19px; width: 13px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image031.png\" \/><\/p>\n<p><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image019.png\" \/><img decoding=\"async\" style=\"height: 19px; width: 37px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image005.png\" \/>= 4.4<\/p>\n<p><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image006.png\" \/><img decoding=\"async\" style=\"height: 41px; width: 133px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image032.png\" \/><\/p>\n<p><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image006.png\" \/><img decoding=\"async\" style=\"height: 41px; width: 124px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image033.png\" \/><\/p>\n<p><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image006.png\" \/><img decoding=\"async\" style=\"height: 19px; width: 13px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image031.png\" \/>= 1 m<\/p>\n<hr \/>\n<p style=\"text-align: center;\"><strong>NCERT Solutions for Class 9 Maths Exercise 13.2<\/strong><\/p>\n<h6><span class=\"ez-toc-section\" id=\"7_The_inner_diameter_of_a_circular_well_is_35_m_It_is_10_m_deep_Find\"><\/span><strong>7. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find:<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>(i) <\/strong><strong>its inner curved surface area.<\/strong><\/p>\n<p><strong>(ii) <\/strong><strong>thecosr of plastering this curved surface at the rate of Rs. 40 per m<sup>2<\/sup>.<\/strong><\/p>\n<p><strong><img decoding=\"async\" id=\"Picture 24\" style=\"height: 148px; width: 197px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image034.jpg\" \/><\/strong><\/p>\n<p><strong>Ans.<\/strong> Inner diameter of circular well = 3.5 m<\/p>\n<p><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image019.png\" \/>Inner radius of circular well<\/p>\n<p>= <img decoding=\"async\" style=\"height: 41px; width: 27px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image035.png\" \/>= 1.75 m<\/p>\n<p>And Depth of the well = 10 m<\/p>\n<p><strong>(i)<\/strong> Inner surface area of the well = <img decoding=\"async\" style=\"height: 19px; width: 37px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image005.png\" \/><\/p>\n<p>= <img decoding=\"async\" style=\"height: 41px; width: 109px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image036.png\" \/>= 110 m<sup>2<\/sup><\/p>\n<p><strong>(ii)<\/strong> Cost of plastering 1 m<sup>2<\/sup> = Rs. 40<\/p>\n<p>Cost of plastering 100 m<sup>2<\/sup> = 40 x 110 = Rs. 4400<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 13.2<\/p>\n<h6><span class=\"ez-toc-section\" id=\"8_Find\"><\/span><strong>8. Find:<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>(i) <\/strong><strong>the lateral or curved surface area of a petrol storage tank that is 4.2 m in diameter and 4.5 m high.<\/strong><\/p>\n<p><strong>(ii) <\/strong><strong>how much steel was actually used if <\/strong><strong><img decoding=\"async\" style=\"height: 41px; width: 21px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image037.png\" \/><\/strong><strong>of the steel actually used was wasted in making the tank?<\/strong><\/p>\n<p>Ans. <strong>(i)<\/strong> Diameter of cylindrical petrol tank = 4.2 m<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image019.png\" data-cke-saved-src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image019.png\" \/>Radius of the cylindrical petrol tank = <img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image041.png\" data-cke-saved-src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image041.png\" \/>= 2.1 m<\/p>\n<p>And Height of the tank = 4.5 m<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image019.png\" data-cke-saved-src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image019.png\" \/>Total\u00a0surface area of the cylindrical tank =\u00a0<span class=\"cke_widget_wrapper cke_widget_inline cke_widget_selected\" tabindex=\"-1\" contenteditable=\"false\" data-cke-widget-wrapper=\"1\" data-cke-filter=\"off\" data-cke-display-name=\"math\" data-cke-widget-id=\"29\"><span class=\"math-tex cke_widget_element\" data-cke-survive=\"1\" data-cke-widget-data=\"%7B%22math%22%3A%22%7Btex%7D2%5C%5Cpi%20rh%20(r%2Bh)%7B%2Ftex%7D%22%2C%22classes%22%3A%7B%22math-tex%22%3A1%7D%7D\" data-cke-widget-upcasted=\"1\" data-cke-widget-keep-attr=\"0\" data-widget=\"mathjaxlatex\"><img decoding=\"async\" src=\"https:\/\/elpiscart.com\/cgi-bin\/mathtex.cgi?2\\pi rh (r+h)\" \/><\/span><img decoding=\"async\" class=\"cke_reset cke_widget_mask\" src=\"data:image\/gif;base64,R0lGODlhAQABAPABAP\/\/\/wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==\" \/><span class=\"cke_reset cke_widget_drag_handler_container\"><img loading=\"lazy\" decoding=\"async\" class=\"cke_reset cke_widget_drag_handler\" title=\"Click and drag to move\" draggable=\"true\" src=\"data:image\/gif;base64,R0lGODlhAQABAPABAP\/\/\/wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==\" width=\"15\" height=\"15\" data-cke-widget-drag-handler=\"1\" \/><\/span><\/span><\/p>\n<p>= <span class=\"cke_widget_wrapper cke_widget_inline cke_widget_selected\" tabindex=\"-1\" contenteditable=\"false\" data-cke-widget-wrapper=\"1\" data-cke-filter=\"off\" data-cke-display-name=\"math\" data-cke-widget-id=\"28\"><span class=\"math-tex cke_widget_element\" data-cke-survive=\"1\" data-cke-widget-data=\"%7B%22math%22%3A%22%7Btex%7D2%20%5C%5Ctimes%20%7B%7B22%7D%20%5C%5Cover%207%7D%20%5C%5Ctimes%202.1%20(%202.1%20%2B%204.5)%7B%2Ftex%7D%22%2C%22classes%22%3A%7B%22math-tex%22%3A1%7D%7D\" data-cke-widget-upcasted=\"1\" data-cke-widget-keep-attr=\"0\" data-widget=\"mathjaxlatex\"><img decoding=\"async\" src=\"https:\/\/elpiscart.com\/cgi-bin\/mathtex.cgi?2 \\times {{22} \\over 7} \\times 2.1 ( 2.1 + 4.5)\" \/><\/span><img decoding=\"async\" class=\"cke_reset cke_widget_mask\" src=\"data:image\/gif;base64,R0lGODlhAQABAPABAP\/\/\/wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==\" \/><span class=\"cke_reset cke_widget_drag_handler_container\"><img loading=\"lazy\" decoding=\"async\" class=\"cke_reset cke_widget_drag_handler\" title=\"Click and drag to move\" draggable=\"true\" src=\"data:image\/gif;base64,R0lGODlhAQABAPABAP\/\/\/wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==\" width=\"15\" height=\"15\" data-cke-widget-drag-handler=\"1\" \/><\/span><\/span>\u00a0= 87.12 m<sup>2<\/sup><\/p>\n<p><strong>(ii)<\/strong> Let the actual area of steel used be <img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image043.png\" data-cke-saved-src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image043.png\" \/>meters<\/p>\n<p>Since <img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image037.png\" data-cke-saved-src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image037.png\" \/>of the actual steel used was wasted, the area of steel which has gone into the tank.<\/p>\n<p>Then, steel actually used = <img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image044.png\" data-cke-saved-src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image044.png\" \/>= <img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image045.png\" data-cke-saved-src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image045.png\" \/>\u00a0of its Total Surface area<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image019.png\" data-cke-saved-src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image019.png\" \/><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image045.png\" data-cke-saved-src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image045.png\" \/>= <span class=\"cke_widget_wrapper cke_widget_inline cke_widget_selected\" tabindex=\"-1\" contenteditable=\"false\" data-cke-widget-wrapper=\"1\" data-cke-filter=\"off\" data-cke-display-name=\"math\" data-cke-widget-id=\"27\"><span class=\"math-tex cke_widget_element\" data-cke-survive=\"1\" data-cke-widget-data=\"%7B%22math%22%3A%22%7Btex%7D2%5C%5Cpi%20r%5C%5Cleft(%20%7Br%20%2B%20h%7D%20%5C%5Cright)%7B%2Ftex%7D%22%2C%22classes%22%3A%7B%22math-tex%22%3A1%7D%7D\" data-cke-widget-upcasted=\"1\" data-cke-widget-keep-attr=\"0\" data-widget=\"mathjaxlatex\"><img decoding=\"async\" src=\"https:\/\/elpiscart.com\/cgi-bin\/mathtex.cgi?2\\pi r\\left( {r + h} \\right)\" \/><\/span><img decoding=\"async\" class=\"cke_reset cke_widget_mask\" src=\"data:image\/gif;base64,R0lGODlhAQABAPABAP\/\/\/wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==\" \/><span class=\"cke_reset cke_widget_drag_handler_container\"><img loading=\"lazy\" decoding=\"async\" class=\"cke_reset cke_widget_drag_handler\" title=\"Click and drag to move\" draggable=\"true\" src=\"data:image\/gif;base64,R0lGODlhAQABAPABAP\/\/\/wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==\" width=\"15\" height=\"15\" data-cke-widget-drag-handler=\"1\" \/><\/span><\/span><\/p>\n<p><span class=\"cke_widget_wrapper cke_widget_inline cke_widget_selected\" tabindex=\"-1\" contenteditable=\"false\" data-cke-widget-wrapper=\"1\" data-cke-filter=\"off\" data-cke-display-name=\"math\" data-cke-widget-id=\"26\"><span class=\"math-tex cke_widget_element\" data-cke-survive=\"1\" data-cke-widget-data=\"%7B%22math%22%3A%22%7Btex%7D%20%5C%5CRightarrow%20%7B%2Ftex%7D%22%2C%22classes%22%3A%7B%22math-tex%22%3A1%7D%7D\" data-cke-widget-upcasted=\"1\" data-cke-widget-keep-attr=\"0\" data-widget=\"mathjaxlatex\"><img decoding=\"async\" src=\"https:\/\/elpiscart.com\/cgi-bin\/mathtex.cgi? \\Rightarrow \" \/><\/span><img decoding=\"async\" class=\"cke_reset cke_widget_mask\" src=\"data:image\/gif;base64,R0lGODlhAQABAPABAP\/\/\/wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==\" \/><span class=\"cke_reset cke_widget_drag_handler_container\"><img loading=\"lazy\" decoding=\"async\" class=\"cke_reset cke_widget_drag_handler\" title=\"Click and drag to move\" draggable=\"true\" src=\"data:image\/gif;base64,R0lGODlhAQABAPABAP\/\/\/wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==\" width=\"15\" height=\"15\" data-cke-widget-drag-handler=\"1\" \/><\/span><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <span class=\"cke_widget_wrapper cke_widget_inline cke_widget_selected\" tabindex=\"-1\" contenteditable=\"false\" data-cke-widget-wrapper=\"1\" data-cke-filter=\"off\" data-cke-display-name=\"math\" data-cke-widget-id=\"25\"><span class=\"math-tex cke_widget_element\" data-cke-survive=\"1\" data-cke-widget-data=\"%7B%22math%22%3A%22%7Btex%7D%7B%7B11%7D%20%5C%5Cover%20%7B12%7D%7Dx%20%3D%202%20%5C%5Ctimes%20%7B%7B22%7D%20%5C%5Cover%207%7D%20%5C%5Ctimes%202.1%5C%5Cleft(%20%7B2.1%20%2B%204.5%7D%20%5C%5Cright)%7B%2Ftex%7D%22%2C%22classes%22%3A%7B%22math-tex%22%3A1%7D%7D\" data-cke-widget-upcasted=\"1\" data-cke-widget-keep-attr=\"0\" data-widget=\"mathjaxlatex\"><img decoding=\"async\" src=\"https:\/\/elpiscart.com\/cgi-bin\/mathtex.cgi?{{11} \\over {12}}x = 2 \\times {{22} \\over 7} \\times 2.1\\left( {2.1 + 4.5} \\right)\" \/><\/span><img decoding=\"async\" class=\"cke_reset cke_widget_mask\" src=\"data:image\/gif;base64,R0lGODlhAQABAPABAP\/\/\/wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==\" \/><span class=\"cke_reset cke_widget_drag_handler_container\"><img loading=\"lazy\" decoding=\"async\" class=\"cke_reset cke_widget_drag_handler\" title=\"Click and drag to move\" draggable=\"true\" src=\"data:image\/gif;base64,R0lGODlhAQABAPABAP\/\/\/wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==\" width=\"15\" height=\"15\" data-cke-widget-drag-handler=\"1\" \/><\/span><\/span><\/p>\n<p><span class=\"cke_widget_wrapper cke_widget_inline cke_widget_selected\" tabindex=\"-1\" contenteditable=\"false\" data-cke-widget-wrapper=\"1\" data-cke-filter=\"off\" data-cke-display-name=\"math\" data-cke-widget-id=\"24\"><span class=\"math-tex cke_widget_element\" data-cke-survive=\"1\" data-cke-widget-data=\"%7B%22math%22%3A%22%7Btex%7D%20%5C%5CRightarrow%20%7B%2Ftex%7D%22%2C%22classes%22%3A%7B%22math-tex%22%3A1%7D%7D\" data-cke-widget-upcasted=\"1\" data-cke-widget-keep-attr=\"0\" data-widget=\"mathjaxlatex\"><img decoding=\"async\" src=\"https:\/\/elpiscart.com\/cgi-bin\/mathtex.cgi? \\Rightarrow \" \/><\/span><img decoding=\"async\" class=\"cke_reset cke_widget_mask\" src=\"data:image\/gif;base64,R0lGODlhAQABAPABAP\/\/\/wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==\" \/><span class=\"cke_reset cke_widget_drag_handler_container\"><img loading=\"lazy\" decoding=\"async\" class=\"cke_reset cke_widget_drag_handler\" title=\"Click and drag to move\" draggable=\"true\" src=\"data:image\/gif;base64,R0lGODlhAQABAPABAP\/\/\/wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==\" width=\"15\" height=\"15\" data-cke-widget-drag-handler=\"1\" \/><\/span><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <span class=\"cke_widget_wrapper cke_widget_inline cke_widget_selected\" tabindex=\"-1\" contenteditable=\"false\" data-cke-widget-wrapper=\"1\" data-cke-filter=\"off\" data-cke-display-name=\"math\" data-cke-widget-id=\"23\"><span class=\"math-tex cke_widget_element\" data-cke-survive=\"1\" data-cke-widget-data=\"%7B%22math%22%3A%22%7Btex%7D%7B%7B11%7D%20%5C%5Cover%20%7B12%7D%7Dx%20%3D%2044%20%5C%5Ctimes%200.3%20%5C%5Ctimes%206.6%7B%2Ftex%7D%22%2C%22classes%22%3A%7B%22math-tex%22%3A1%7D%7D\" data-cke-widget-upcasted=\"1\" data-cke-widget-keep-attr=\"0\" data-widget=\"mathjaxlatex\"><img decoding=\"async\" src=\"https:\/\/elpiscart.com\/cgi-bin\/mathtex.cgi?{{11} \\over {12}}x = 44 \\times 0.3 \\times 6.6\" \/><\/span><img decoding=\"async\" class=\"cke_reset cke_widget_mask\" src=\"data:image\/gif;base64,R0lGODlhAQABAPABAP\/\/\/wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==\" \/><span class=\"cke_reset cke_widget_drag_handler_container\"><img loading=\"lazy\" decoding=\"async\" class=\"cke_reset cke_widget_drag_handler\" title=\"Click and drag to move\" draggable=\"true\" src=\"data:image\/gif;base64,R0lGODlhAQABAPABAP\/\/\/wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==\" width=\"15\" height=\"15\" data-cke-widget-drag-handler=\"1\" \/><\/span><\/span><\/p>\n<p><span class=\"cke_widget_wrapper cke_widget_inline cke_widget_selected\" tabindex=\"-1\" contenteditable=\"false\" data-cke-widget-wrapper=\"1\" data-cke-filter=\"off\" data-cke-display-name=\"math\" data-cke-widget-id=\"22\"><span class=\"math-tex cke_widget_element\" data-cke-survive=\"1\" data-cke-widget-data=\"%7B%22math%22%3A%22%7Btex%7D%5C%5CRightarrow%7B%2Ftex%7D%22%2C%22classes%22%3A%7B%22math-tex%22%3A1%7D%7D\" data-cke-widget-upcasted=\"1\" data-cke-widget-keep-attr=\"0\" data-widget=\"mathjaxlatex\"><img decoding=\"async\" src=\"https:\/\/elpiscart.com\/cgi-bin\/mathtex.cgi?\\Rightarrow\" \/><\/span><img decoding=\"async\" class=\"cke_reset cke_widget_mask\" src=\"data:image\/gif;base64,R0lGODlhAQABAPABAP\/\/\/wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==\" \/><span class=\"cke_reset cke_widget_drag_handler_container\"><img loading=\"lazy\" decoding=\"async\" class=\"cke_reset cke_widget_drag_handler\" title=\"Click and drag to move\" draggable=\"true\" src=\"data:image\/gif;base64,R0lGODlhAQABAPABAP\/\/\/wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==\" width=\"15\" height=\"15\" data-cke-widget-drag-handler=\"1\" \/><\/span><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <span class=\"cke_widget_wrapper cke_widget_inline cke_widget_selected\" tabindex=\"-1\" contenteditable=\"false\" data-cke-widget-wrapper=\"1\" data-cke-filter=\"off\" data-cke-display-name=\"math\" data-cke-widget-id=\"21\"><span class=\"math-tex cke_widget_element\" data-cke-survive=\"1\" data-cke-widget-data=\"%7B%22math%22%3A%22%7Btex%7D%7B%7B11%7D%20%5C%5Cover%20%7B12%7D%7Dx%20%3D%2087.12%7B%2Ftex%7D%22%2C%22classes%22%3A%7B%22math-tex%22%3A1%7D%7D\" data-cke-widget-upcasted=\"1\" data-cke-widget-keep-attr=\"0\" data-widget=\"mathjaxlatex\"><img decoding=\"async\" src=\"https:\/\/elpiscart.com\/cgi-bin\/mathtex.cgi?{{11} \\over {12}}x = 87.12\" \/><\/span><img decoding=\"async\" class=\"cke_reset cke_widget_mask\" src=\"data:image\/gif;base64,R0lGODlhAQABAPABAP\/\/\/wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==\" \/><span class=\"cke_reset cke_widget_drag_handler_container\"><img loading=\"lazy\" decoding=\"async\" class=\"cke_reset cke_widget_drag_handler\" title=\"Click and drag to move\" draggable=\"true\" src=\"data:image\/gif;base64,R0lGODlhAQABAPABAP\/\/\/wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==\" width=\"15\" height=\"15\" data-cke-widget-drag-handler=\"1\" \/><\/span><\/span><\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image006.png\" data-cke-saved-src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image006.png\" \/>\u00a0<span class=\"cke_widget_wrapper cke_widget_inline cke_widget_selected\" tabindex=\"-1\" contenteditable=\"false\" data-cke-widget-wrapper=\"1\" data-cke-filter=\"off\" data-cke-display-name=\"math\" data-cke-widget-id=\"20\"><span class=\"math-tex cke_widget_element\" data-cke-survive=\"1\" data-cke-widget-data=\"%7B%22math%22%3A%22%7Btex%7Dx%20%3D%2087.12%20%5C%5Ctimes%20%7B%7B12%7D%20%5C%5Cover%20%7B11%7D%7D%7B%2Ftex%7D%22%2C%22classes%22%3A%7B%22math-tex%22%3A1%7D%7D\" data-cke-widget-upcasted=\"1\" data-cke-widget-keep-attr=\"0\" data-widget=\"mathjaxlatex\"><img decoding=\"async\" src=\"https:\/\/elpiscart.com\/cgi-bin\/mathtex.cgi?x = 87.12 \\times {{12} \\over {11}}\" \/><\/span><img decoding=\"async\" class=\"cke_reset cke_widget_mask\" src=\"data:image\/gif;base64,R0lGODlhAQABAPABAP\/\/\/wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==\" \/><span class=\"cke_reset cke_widget_drag_handler_container\"><img loading=\"lazy\" decoding=\"async\" class=\"cke_reset cke_widget_drag_handler\" title=\"Click and drag to move\" draggable=\"true\" src=\"data:image\/gif;base64,R0lGODlhAQABAPABAP\/\/\/wAAACH5BAEKAAAALAAAAAABAAEAAAICRAEAOw==\" width=\"15\" height=\"15\" data-cke-widget-drag-handler=\"1\" \/><\/span><\/span>\u00a0= 95.04 m<sup>2<\/sup><\/p>\n<p>Hence steel actually used is 95.04 m<sup>2<\/sup>.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 13.2<\/p>\n<h6><span class=\"ez-toc-section\" id=\"9_In_the_adjoining_figure_you_see_the_frame_of_a_lampshade_It_is_to_be_covered_with_a_decorative_cloth_The_frame_has_a_base_diameter_of_20_cm_and_height_of_30_cm_A_margin_of_25_cm_is_to_be_given_for_folding_it_over_the_top_and_bottom_of_the_frame_Find_how_much_cloth_is_required_for_covering_the_lampshade_See_fig\"><\/span><strong>9. In the adjoining figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. [See fig.]<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong><img decoding=\"async\" id=\"Picture 27\" style=\"height: 210px; width: 263px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image047.jpg\" \/><\/strong><\/p>\n<p><strong>Ans.<\/strong> Height of each of the folding at the top and bottom <img decoding=\"async\" style=\"height: 27px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image002.png\" \/>= 2.5 cm<\/p>\n<p>Height of the frame (H) = 30 cm<\/p>\n<p>Diameter = 20 cm<\/p>\n<p><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image006.png\" \/>Radius = 10 cm<\/p>\n<p>Now cloth required for covering the lampshade<\/p>\n<p>= CSA of top part + CSA of middle part + CSA of bottom part<\/p>\n<p>= <img decoding=\"async\" style=\"height: 19px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image048.png\" \/><\/p>\n<p>= <img decoding=\"async\" style=\"height: 27px; width: 101px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image049.png\" \/><\/p>\n<p>= <img decoding=\"async\" style=\"height: 27px; width: 88px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image050.png\" \/><\/p>\n<p>= <img decoding=\"async\" style=\"height: 41px; width: 144px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image051.png\" \/><\/p>\n<p>= 2200 cm<sup>2<\/sup><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 13.2<\/p>\n<h6><span class=\"ez-toc-section\" id=\"10_The_students_of_a_Vidyalayawere_asked_to_participate_in_a_competition_for_making_and_decorating_penholders_in_the_shape_of_a_cylinder_with_a_base_using_cardboard_Each_penholder_was_to_be_of_radius_3_cm_and_height_105_cm_The_Vidyalaya_was_to_supply_the_competitors_with_cardboard_If_there_were_35_competitors_how_much_cardboard_was_required_to_be_bought_for_the_competition\"><\/span><strong>10. The students of a Vidyalayawere asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Ans.<\/strong> Radius of a cylindrical pen holder <img decoding=\"async\" style=\"height: 27px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image012.png\" \/>= 3 cm<\/p>\n<p>Height of the cylindrical pen holder <img decoding=\"async\" style=\"height: 27px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image002.png\" \/><\/p>\n<p>= 10.5 cm<\/p>\n<p>Cardboard required for pen holder = CSA of pen holder + Area of circular base<\/p>\n<p>= <img decoding=\"async\" style=\"height: 21px; width: 75px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image052.png\" \/>= <img decoding=\"async\" style=\"height: 27px; width: 75px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image053.png\" \/><\/p>\n<p>= <img decoding=\"async\" style=\"height: 41px; width: 125px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image054.png\" \/>= 226.28 cm<sup>2<\/sup><\/p>\n<p>Since Cardboard required for making 1 pen holder = 226.28 cm<sup>2<\/sup><\/p>\n<p><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_2\/image019.png\" \/>Cardboard required for making 35 pen holders = 226.28 x 35 = 7919.8 cm<sup>2<\/sup><\/p>\n<p>= 7920 cm<sup>2<\/sup> (approx.)<\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_9_Maths_Exercise_132\"><\/span>NCERT Solutions for Class 9 Maths Exercise 13.2<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>NCERT Solutions for Class 9 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 9 Maths includes text book solutions from Mathematics Book. NCERT Solutions for CBSE Class 9 Maths have total 15 chapters. 9 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 9 solutions PDF and Maths ncert class 9 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_app_for_Class_9\"><\/span>CBSE app for Class 9<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>To download NCERT Solutions for Class 9 Maths, Computer Science, Home Science,Hindi ,English, Social Science do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through\u00a0<a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide&amp;referrer=utm_source%3Dmycbse_bottom%26utm_medium%3Dtext%26utm_campaign%3Dmycbseads\"><strong>the best app for CBSE students<\/strong><\/a>\u00a0and myCBSEguide website.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions for Class 9 Maths Exercise 13.2 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 9 Maths chapter &#8230; <a title=\"NCERT Solutions for Class 9 Maths Exercise 13.2\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-2\/\" aria-label=\"More on NCERT Solutions for Class 9 Maths Exercise 13.2\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":26525,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1376,281],"tags":[283,1344,321,1485,216],"class_list":["post-4989","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematics-cbse-class-09","category-ncert-solutions","tag-cbse-study-material","tag-class-9","tag-mathematics","tag-maths","tag-ncert-solutions"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.0 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>NCERT Solutions for Class 9 Maths Exercise 13.2 | myCBSEguide<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for Class 9 Maths Exercise 13.2 in PDF format for free download. 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