{"id":4986,"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-13-1\/"},"modified":"2018-10-31T15:06:45","modified_gmt":"2018-10-31T09:36:45","slug":"ncert-solutions-for-class-9-maths-exercise-13-1","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-1\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Exercise 13.1"},"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-13-1\/#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-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-1\/#1_A_plastic_box_15_m_long_125_m_wide_and_65_cm_deep_is_to_be_made_It_is_to_be_open_at_the_top_Ignoring_the_thickness_of_the_plastic_sheet_determine\" >1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-1\/#2_The_length_breadth_and_height_of_a_room_are_5_m_4_m_and_3_m_respectively_Find_the_cost_of_white_washing_the_walls_of_the_room_and_the_ceiling_at_the_rate_of_Rs_750_per_m2\" >2. The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs. 7.50 per m2.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-1\/#3_The_floor_of_a_rectangular_hall_has_a_perimeter_250_m_If_the_cost_of_painting_the_four_walls_at_the_rate_of_Rs_10_per_m2_is_Rs_15000_find_the_height_of_the_hall\" >3. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs. 10 per m2 is Rs. 15000, find the height of the hall.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-1\/#4_The_paint_in_a_certain_container_is_sufficient_to_paint_an_area_equal_to_9375_m2_How_many_bricks_of_dimensions_225_cm_x_10_cm_x_75_cm_can_be_painted_out_of_this_container\" >4. The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm x 10 cm x 7.5 cm can be painted out of this container?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-1\/#5_A_cubical_box_has_each_edge_10_cm_and_a_cuboidal_box_is_10_cm_wide_125_cm_long_and_8_cm_high\" >5. A cubical box has each edge 10 cm and a cuboidal box is 10 cm wide, 12.5 cm long and 8 cm high.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-1\/#6_A_small_indoor_green_house_herbarium_is_made_entirely_of_glass_panes_including_base_held_together_with_tape_It_is_30_cm_long_25_cm_wide_and_25_cm_high\" >6. A small indoor green house (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-1\/#7_Shanti_Sweets_Stall_was_placing_an_order_for_making_cardboard_boxes_for_packing_their_sweets_Two_sizes_of_boxes_were_required_The_bigger_of_dimensions_25_cm_by_20_cm_by_5_cm_and_the_smaller_of_dimensions_15_cm_by_12_cm_by_5_cm_5_of_the_total_surface_area_is_required_extra_for_all_the_overlaps_If_the_cost_of_the_card_board_is_Rs_4_for_1000_cm2_find_the_cost_of_cardboard_required_for_supplying_250_boxes_of_each_kind\" >7. Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm by 20 cm by 5 cm and the smaller of dimensions 15 cm by 12 cm by 5 cm. 5% of the total surface area is required extra, for all the overlaps. If the cost of the card board is Rs. 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each kind.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-1\/#8_Parveen_wanted_to_make_a_temporary_shelter_for_her_car_by_making_a_box-like_structure_with_tarpaulin_that_covers_all_the_four_sides_and_the_top_of_the_car_with_the_front_face_as_a_flap_which_can_be_rolled_up_Assuming_that_the_stitching_margins_are_very_small_and_therefore_negligible_how_much_tarpaulin_would_be_required_to_make_the_shelter_of_height_25_m_with_base_simensions_4_m_x_3_m\" >8. Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m with base simensions 4 m x 3 m?<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-1\/#NCERT_Solutions_for_Class_9_Maths_Exercise_131\" >NCERT Solutions for Class 9 Maths Exercise 13.1<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-1\/#CBSE_app_for_Class_9\" >CBSE app for Class 9<\/a><\/li><\/ul><\/nav><\/div>\n<p>NCERT Solutions for Class 9 Maths Exercise 13.1 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE 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=\"alignright\" 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 13.1\" width=\"113\" height=\"145\" \/><\/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<h3><span class=\"ez-toc-section\" id=\"1_A_plastic_box_15_m_long_125_m_wide_and_65_cm_deep_is_to_be_made_It_is_to_be_open_at_the_top_Ignoring_the_thickness_of_the_plastic_sheet_determine\"><\/span><strong>1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine:<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>(i) <\/strong><strong>The area of the sheet required for making the box.<\/strong><\/p>\n<p><strong>(ii) <\/strong><strong>The cost of sheet for it, if a sheet measuring 1m<sup>2<\/sup> cost Rs.20.<\/strong><\/p>\n<p><strong>Ans.<\/strong> <strong>(i)<\/strong> Given: Length <img decoding=\"async\" style=\"height: 27px; width: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image001.png\" \/>= 1.5 m, Breadth <img decoding=\"async\" style=\"height: 27px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image002.png\" \/>= 1.25 m and Depth <img decoding=\"async\" style=\"height: 27px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image003.png\" \/><\/p>\n<p>= 65 cm = 0.65 m<\/p>\n<p>Area of the sheet required for making the box open at the top = <img decoding=\"async\" style=\"height: 27px; width: 97px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image004.png\" \/><\/p>\n<p>= <img decoding=\"async\" style=\"height: 27px; width: 236px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image005.png\" \/><\/p>\n<p>= <img decoding=\"async\" style=\"height: 27px; width: 169px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image006.png\" \/><\/p>\n<p>= <img decoding=\"async\" style=\"height: 19px; width: 116px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image007.png\" \/><\/p>\n<p>= 3.575 + 1.875 = 5.45 m<sup>2<\/sup><\/p>\n<p><strong>(ii)<\/strong> Since, Cost of 1 m<sup>2<\/sup> sheet = Rs. 20<\/p>\n<p><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image008.png\" \/>Cost of 5.45 m<sup>2<\/sup> sheet = 20 x 5.45 = Rs. 109<\/p>\n<hr \/>\n<h3><span class=\"ez-toc-section\" id=\"2_The_length_breadth_and_height_of_a_room_are_5_m_4_m_and_3_m_respectively_Find_the_cost_of_white_washing_the_walls_of_the_room_and_the_ceiling_at_the_rate_of_Rs_750_per_m2\"><\/span><strong>2. The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs. 7.50 per m<sup>2<\/sup>.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Ans.<\/strong> Given: Length <img decoding=\"async\" style=\"height: 27px; width: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image001.png\" \/>= 5 m, Breadth <img decoding=\"async\" style=\"height: 27px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image002.png\" \/>= 4 m and Height <img decoding=\"async\" style=\"height: 27px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image003.png\" \/>= 3 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_1\/image009.png\" \/>Area of the four walls = Lateral surface area = <img decoding=\"async\" style=\"height: 27px; width: 71px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image010.png\" \/>= <img decoding=\"async\" style=\"height: 27px; width: 63px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image011.png\" \/><\/p>\n<p>= 2 x 3 (4 + 5)<\/p>\n<p>= 2 x 9 x 3 = 54 m<sup>2<\/sup><\/p>\n<p>Area of ceiling = <img decoding=\"async\" style=\"height: 19px; width: 31px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image012.png\" \/>= 5 x 4 = 20 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_1\/image008.png\" \/>Total area of walls and ceiling of the room = 54 + 20 = 74 m<sup>2<\/sup><\/p>\n<p>Now Cost of white washing for 1 m<sup>2<\/sup> = Rs. 7.50<\/p>\n<p><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image008.png\" \/>Cost of white washing for 74 m<sup>2<\/sup> = 74 x 7.50 = Rs. 555<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 13.1<\/p>\n<h3><span class=\"ez-toc-section\" id=\"3_The_floor_of_a_rectangular_hall_has_a_perimeter_250_m_If_the_cost_of_painting_the_four_walls_at_the_rate_of_Rs_10_per_m2_is_Rs_15000_find_the_height_of_the_hall\"><\/span><strong>3. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs. 10 per m<sup>2<\/sup> is Rs. 15000, find the height of the hall.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Ans.<\/strong> Given: Perimeter of rectangular wall<\/p>\n<p>= <img decoding=\"async\" style=\"height: 27px; width: 55px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image013.png\" \/>= 250 m \u2026\u2026\u2026.(i)<\/p>\n<p>Now Area of the four walls of the room<\/p>\n<p>= <img decoding=\"async\" style=\"height: 44px; width: 239px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image014.png\" \/><\/p>\n<p>= <img decoding=\"async\" style=\"height: 41px; width: 45px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image015.png\" \/>= 1500 m<sup>2<\/sup> \u2026\u2026\u2026.(ii)<\/p>\n<p><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image009.png\" \/>Area of the four walls = Lateral surface area<\/p>\n<p>= <img decoding=\"async\" style=\"height: 27px; width: 71px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image010.png\" \/>= <img decoding=\"async\" style=\"height: 27px; width: 63px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image011.png\" \/>= 1500<\/p>\n<p><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image016.png\" \/><img decoding=\"async\" style=\"height: 19px; width: 96px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image017.png\" \/>[using eq. (i) and (ii)<\/p>\n<p><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image016.png\" \/><img decoding=\"async\" style=\"height: 41px; width: 61px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image018.png\" \/>= 6 m<\/p>\n<p>Hence required height of the hall is 6 m.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 13.1<\/p>\n<h3><span class=\"ez-toc-section\" id=\"4_The_paint_in_a_certain_container_is_sufficient_to_paint_an_area_equal_to_9375_m2_How_many_bricks_of_dimensions_225_cm_x_10_cm_x_75_cm_can_be_painted_out_of_this_container\"><\/span><strong>4. The paint in a certain container is sufficient to paint an area equal to 9.375 m<sup>2<\/sup>. How many bricks of dimensions 22.5 cm x 10 cm x 7.5 cm can be painted out of this container?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Ans.<\/strong> Given: Length of the brick <img decoding=\"async\" style=\"height: 27px; width: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image001.png\" \/><\/p>\n<p>= 22.5 cm, Breadth <img decoding=\"async\" style=\"height: 27px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image002.png\" \/><\/p>\n<p>= 10 cm and Height <img decoding=\"async\" style=\"height: 27px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image003.png\" \/>= 7.5 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_1\/image008.png\" \/>Surface area of the brick<\/p>\n<p>= <img decoding=\"async\" style=\"height: 27px; width: 97px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image019.png\" \/><\/p>\n<p>= 2 (22.5 x 10 + 10 x 7.5 + 7.5 x 22.5)<\/p>\n<p>= 2 (225 + 75 + 468.75)<\/p>\n<p>= 937.5 cm<sup>2<\/sup><\/p>\n<p>= 0.09375 m<sup>2<\/sup> [1 cm = 0.01 m]\n<p>Now No. of bricks to be painted<\/p>\n<p>= <img decoding=\"async\" style=\"height: 41px; width: 159px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image020.png\" \/>= <img decoding=\"async\" style=\"height: 41px; width: 59px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image021.png\" \/><\/p>\n<p>= 100<\/p>\n<p>Hence 100 bricks can be painted.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 13.1<\/p>\n<h3><span class=\"ez-toc-section\" id=\"5_A_cubical_box_has_each_edge_10_cm_and_a_cuboidal_box_is_10_cm_wide_125_cm_long_and_8_cm_high\"><\/span><strong>5. A cubical box has each edge 10 cm and a cuboidal box is 10 cm wide, 12.5 cm long and 8 cm high.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>(i) <\/strong><strong>Which box has the greater lateral surface area and by how much?<\/strong><\/p>\n<p><strong>(ii) <\/strong><strong>Which box has the smaller total surface area and how much?<\/strong><\/p>\n<p><strong>Ans.<\/strong> <strong>(i)<\/strong> Lateral surface area of a cube = 4 (side)<sup>2<\/sup> = 4 x (10)<sup>2<\/sup> = 400 cm<sup>2<\/sup><\/p>\n<p>Lateral surface area of a cuboid = <img decoding=\"async\" style=\"height: 27px; width: 63px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image022.png\" \/><\/p>\n<p>= 2 x 8 (12.5 + 10) = 16 x 22.5 = 360 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_1\/image008.png\" \/>Lateral surface area of cubical box is greater by (400 \u2013 360) = 40 cm<sup>2<\/sup><\/p>\n<p><strong>(ii)<\/strong> Total surface area of a cube = 6 (side)<sup>2<\/sup><\/p>\n<p>= 6 x (10)<sup>2<\/sup> = 600 cm<sup>2<\/sup><\/p>\n<p>Total surface area of cuboid = <img decoding=\"async\" style=\"height: 27px; width: 97px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image019.png\" \/><\/p>\n<p>= 2 (12.5 x 10 + 10 x 8 + 8 x 12.5)<\/p>\n<p>= 2 (125 + 80 + 100)<\/p>\n<p>= 2 x 305 = 610 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_1\/image008.png\" \/>Total surface area of cuboid box is greater by (610 \u2013 600) = 10 cm<sup>2<\/sup><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 13.1<\/p>\n<h3><span class=\"ez-toc-section\" id=\"6_A_small_indoor_green_house_herbarium_is_made_entirely_of_glass_panes_including_base_held_together_with_tape_It_is_30_cm_long_25_cm_wide_and_25_cm_high\"><\/span><strong>6. A small indoor green house (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>(i) <\/strong><strong>What is the surface area of the glass?<\/strong><\/p>\n<p><strong>(ii) <\/strong><strong>How much of tape is needed for all the 12 edges?<\/strong><\/p>\n<p><strong>Ans.<\/strong> <strong>(i)<\/strong> Given: Length of glass herbarium <img decoding=\"async\" style=\"height: 27px; width: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image001.png\" \/>= 30 cm,<\/p>\n<p>Breadth <img decoding=\"async\" style=\"height: 27px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image002.png\" \/>= 25 cm and Height <img decoding=\"async\" style=\"height: 27px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image003.png\" \/>= 25 m<\/p>\n<p><img decoding=\"async\" style=\"float: left; height: 114px; width: 224px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image023.png\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>Total surface area of the glass<\/p>\n<p>= <img decoding=\"async\" style=\"height: 27px; width: 97px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image019.png\" \/><\/p>\n<p>= 2 (30 x 25 + 25 x 25 + 25 x 30)<\/p>\n<p>= 2 (750 + 625 + 750)<\/p>\n<p>= 2 x 2125 = 4250 cm<sup>2<\/sup><\/p>\n<p>Hence 4250 cm<sup>2<\/sup> of the glass is required to make a herbarium.<\/p>\n<p><strong>(ii)<\/strong> Tape is used at 12 edges.<\/p>\n<p><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image016.png\" \/>Tape is used at 4 lengths, 4 breadths and 4 heights.<\/p>\n<p><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image016.png\" \/>Total length of the tape = <img decoding=\"async\" style=\"height: 27px; width: 77px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image024.png\" \/><\/p>\n<p>= 2 (30 + 25 + 25) = 320 cm<\/p>\n<p>Hence 320 cm of the tape if needed to fix 12 edges of herbarium.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 13.1<\/p>\n<h3><span class=\"ez-toc-section\" id=\"7_Shanti_Sweets_Stall_was_placing_an_order_for_making_cardboard_boxes_for_packing_their_sweets_Two_sizes_of_boxes_were_required_The_bigger_of_dimensions_25_cm_by_20_cm_by_5_cm_and_the_smaller_of_dimensions_15_cm_by_12_cm_by_5_cm_5_of_the_total_surface_area_is_required_extra_for_all_the_overlaps_If_the_cost_of_the_card_board_is_Rs_4_for_1000_cm2_find_the_cost_of_cardboard_required_for_supplying_250_boxes_of_each_kind\"><\/span><strong>7. Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm by 20 cm by 5 cm and the smaller of dimensions 15 cm by 12 cm by 5 cm. 5% of the total surface area is required extra, for all the overlaps. If the cost of the card board is Rs. 4 for 1000 cm<sup>2<\/sup>, find the cost of cardboard required for supplying 250 boxes of each kind.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Ans.<\/strong> Given: Length of bigger cardboard box (L) = 25 cm<\/p>\n<p>Breadth (B) = 20 cm and Height (H) = 5cm<\/p>\n<p><img decoding=\"async\" style=\"float: left; height: 92px; width: 262px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image025.png\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>Total surface area of bigger cardboard box<\/p>\n<p>= 2 (LB + BH + HL)<\/p>\n<p>= 2 (25 x 20 + 20 x 5 + 5 x 25)<\/p>\n<p>= 2 (500 + 100 + 125)<\/p>\n<p>= 1450 cm<sup>2<\/sup><\/p>\n<p>5% extra surface of total surface area is required for all the overlaps.<\/p>\n<p><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image016.png\" \/>5% of 1450 = <img decoding=\"async\" style=\"height: 41px; width: 72px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image026.png\" \/>= 72.5 cm<sup>2<\/sup><\/p>\n<p>Now, total surface area of bigger cardboard box with extra overlaps<\/p>\n<p>= 1450 + 72.5 = 1522.5 cm<sup>2<\/sup><\/p>\n<p><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image016.png\" \/>Total surface area with extra overlaps of 250 such boxes<\/p>\n<p>= 250 x 1522.5 = 380625 cm<sup>2<\/sup><\/p>\n<p>Since, Cost of the cardboard for 1000 cm<sup>2<\/sup> = Rs. 4<\/p>\n<p><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image008.png\" \/>Cost of the cardboard for 1cm<sup>2<\/sup> = Rs. <img decoding=\"async\" style=\"height: 41px; width: 37px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image027.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_1\/image008.png\" \/>Cost of the cardboard for 380625 cm<sup>2<\/sup><\/p>\n<p>= Rs. <img decoding=\"async\" style=\"height: 41px; width: 97px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image028.png\" \/>= Rs. 1522.50<\/p>\n<p>Now length of the smaller box <img decoding=\"async\" style=\"height: 27px; width: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image001.png\" \/>= 15 cm,<\/p>\n<p><img decoding=\"async\" id=\"Picture 9\" style=\"height: 83px; width: 246px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image029.jpg\" \/><\/p>\n<p>Breadth <img decoding=\"async\" style=\"height: 27px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image002.png\" \/>= 12 cm and Height <img decoding=\"async\" style=\"height: 27px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image003.png\" \/>= 5 cm<\/p>\n<p>Total surface area of the smaller cardboard box<\/p>\n<p>= <img decoding=\"async\" style=\"height: 27px; width: 97px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image019.png\" \/><\/p>\n<p>= 2 (15 x 12 + 12 x 5 + 5 x 15)<\/p>\n<p>= 2 (180 + 60 +75) = 2 x 315 = 630 cm<sup>2<\/sup><\/p>\n<p>5% of extra surface of total surface area is required for all the overlaps.<\/p>\n<p><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image008.png\" \/>5% of 630 = <img decoding=\"async\" style=\"height: 41px; width: 65px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image030.png\" \/>= 31.5 cm<sup>2<\/sup><\/p>\n<p>Total surface area with extra overlaps = 630 + 31.5 = 661.5 cm<sup>2<\/sup><\/p>\n<p>Now Total surface area with extra overlaps of 250 such smaller boxes<\/p>\n<p>= 661.5 x 250 = 165375 cm<sup>2<\/sup><\/p>\n<p>Cost of the cardboard for 1000 cm<sup>2<\/sup> = Rs. 4<\/p>\n<p>Cost of the cardboard for 1cm<sup>2<\/sup> = Rs. <img decoding=\"async\" style=\"height: 41px; width: 37px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image027.png\" \/><\/p>\n<p>Cost of the cardboard for 165375 cm<sup>2<\/sup> = Rs. <img decoding=\"async\" style=\"height: 41px; width: 96px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image031.png\" \/>= Rs. 661.50<\/p>\n<p><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image008.png\" \/>Total cost of the cardboard required for supplying 250 boxes of each kind<\/p>\n<p>= Total cost of bigger boxes + Total cost of smaller boxes<\/p>\n<p>= Rs. 1522.50 + Rs. 661.50<\/p>\n<p>= Rs. 2184<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 13.1<\/p>\n<h3><span class=\"ez-toc-section\" id=\"8_Parveen_wanted_to_make_a_temporary_shelter_for_her_car_by_making_a_box-like_structure_with_tarpaulin_that_covers_all_the_four_sides_and_the_top_of_the_car_with_the_front_face_as_a_flap_which_can_be_rolled_up_Assuming_that_the_stitching_margins_are_very_small_and_therefore_negligible_how_much_tarpaulin_would_be_required_to_make_the_shelter_of_height_25_m_with_base_simensions_4_m_x_3_m\"><\/span><strong>8. Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m with base simensions 4 m x 3 m?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Ans.<\/strong> Given: Length of base <img decoding=\"async\" style=\"height: 27px; width: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image001.png\" \/>= 4 m, Breadth <img decoding=\"async\" style=\"height: 27px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image002.png\" \/><\/p>\n<p>= 3 m and Height <img decoding=\"async\" style=\"height: 27px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image003.png\" \/>= 2.5 m<\/p>\n<p>Ans. Tarpaulin required to make shelter<\/p>\n<p>= Surface area of 4 walls + Area of roof<\/p>\n<p>= <img decoding=\"async\" style=\"height: 27px; width: 89px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/13_1\/image032.png\" \/><\/p>\n<p>= 2 (4 + 3) 2.5 + 4 x 3<\/p>\n<p>= 35 + 12 = 47 m<sup>2<\/sup><\/p>\n<p>Hence 47 m<sup>2<\/sup> of the tarpaulin is required to make the shelter for the car.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_9_Maths_Exercise_131\"><\/span>NCERT Solutions for Class 9 Maths Exercise 13.1<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 13.1 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 9 Maths chapter wise &#8230; <a title=\"NCERT Solutions for Class 9 Maths Exercise 13.1\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-13-1\/\" aria-label=\"More on NCERT Solutions for Class 9 Maths Exercise 13.1\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":-1,"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-4986","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 13.1 |<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for Class 9 Maths Exercise 13.1 in PDF format for free download. 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