{"id":4954,"date":"2016-05-19T09:49:00","date_gmt":"2016-05-19T04:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-13-1\/"},"modified":"2023-03-22T15:19:05","modified_gmt":"2023-03-22T09:49:05","slug":"ncert-solutions-class-10-maths-exercise-13-1","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-13-1\/","title":{"rendered":"NCERT Solutions for Class 10 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-class-10-maths-exercise-13-1\/#NCERT_Solutions_for_Class_10_Maths_Surface_Areas_and_Volumes\" >NCERT Solutions for Class 10 Maths\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-class-10-maths-exercise-13-1\/#Unless_stated_otherwise_take\" >Unless stated otherwise, take<\/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-class-10-maths-exercise-13-1\/#1_2_cubes_each_of_volume_64_cm3_are_joined_end_to_end_Find_the_surface_area_of_the_resulting_cuboid\" >1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.<\/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-class-10-maths-exercise-13-1\/#2_A_vessel_is_in_the_form_of_a_hollow_hemisphere_mounted_by_a_hollow_cylinder_The_diameter_of_the_hemisphere_is_14_cm_and_the_total_height_of_the_vessel_is_13_cm_Find_the_inner_surface_area_of_the_vessel\" >2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.<\/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-class-10-maths-exercise-13-1\/#3_A_toy_is_in_the_form_of_a_cone_of_radius_35_cm_mounted_on_a_hemisphere_of_same_radius_The_total_height_of_the_toy_is_155_cm_Find_the_total_surface_area_of_the_toy\" >3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.<\/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-class-10-maths-exercise-13-1\/#4_A_cubical_block_of_side_7_cm_is_surmounted_by_a_hemisphere_What_is_the_greatest_diameter_the_hemisphere_can_have_Find_the_surface_area_of_the_solid\" >4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.<\/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-class-10-maths-exercise-13-1\/#5_A_hemispherical_depression_is_cut_out_from_one_face_of_a_cubical_wooden_block_such_that_the_diameter_of_the_hemisphere_is_equal_to_the_edge_of_the_cube_Determine_the_surface_area_of_the_remaining_solid\" >5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter  of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.<\/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-class-10-maths-exercise-13-1\/#6_A_medicine_capsule_is_in_the_shape_of_a_cylinder_with_two_hemispheres_stuck_to_each_of_its_ends_see_figure_The_length_of_the_entire_capsule_is_14_mm_and_the_diameter_of_the_capsule_is_5_mm_Find_its_surface_area\" >6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.<\/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-class-10-maths-exercise-13-1\/#7_A_tent_is_in_the_shape_of_a_cylinder_surmounted_buy_a_conical_top_If_the_height_and_diameter_of_the_cylindrical_part_are_21_m_and_4_m_respectively_and_the_slant_height_of_the_top_is_28_m_find_the_area_of_the_canvas_used_for_making_the_tent_Also_find_the_cost_of_the_canvas_of_the_tent_at_the_rate_of_Rs_500_per_Note_that_the_base_of_the_tent_will_not_be_covered_with_canvas\" >7. A tent is in the shape of a cylinder surmounted buy a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs. 500 per. (Note that the base of the tent will not be covered with canvas.)<\/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-class-10-maths-exercise-13-1\/#8_From_a_solid_cylinder_whose_height_is_24_cm_and_diameter_14_cm_a_conical_cavity_of_the_same_height_and_same_diameter_is_hollowed_out_Find_the_total_surface_area_of_the_remaining_solid_to_the_nearest\" >8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest.<\/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-class-10-maths-exercise-13-1\/#9_A_wooden_article_was_made_by_scooping_out_a_hemisphere_from_each_end_of_a_solid_cylinder_as_shown_in_figure_If_the_height_of_the_cylinder_is_10_cm_and_its_base_is_of_radius_35_cm_find_the_total_surface_area_of_the_article\" >9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder as shown in figure. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the total surface area of the article.<\/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-12\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-13-1\/#NCERT_Solutions_for_Class_10_Maths_Exercise_131\" >NCERT Solutions for Class 10 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-13\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-13-1\/#CBSE_app_for_Class_10\" >CBSE app for Class 10<\/a><\/li><\/ul><\/nav><\/div>\n<p>NCERT Solutions for Class 10 Maths Exercise 13.1 Class 10 Maths 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 10 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 Maths\u00a0Surface Areas and Volumes\u00a0<\/strong><strong><a class=\"button\" href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-10-mathematics-surface-areas-and-volumes\/1213\/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\/10%20Maths%20Book.jpg\" alt=\"NCERT Solutions for Class 10 Maths Exercise 13.1\" width=\"176\" height=\"217\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_10_Maths_Surface_Areas_and_Volumes\"><\/span>NCERT Solutions for Class 10 Maths\u00a0Surface Areas and Volumes<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"Unless_stated_otherwise_take\"><\/span><strong>Unless stated otherwise, take <img decoding=\"async\" style=\"height: 41px; width: 53px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image001.png\" \/> <\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"1_2_cubes_each_of_volume_64_cm3_are_joined_end_to_end_Find_the_surface_area_of_the_resulting_cuboid\"><\/span><strong>1. 2 cubes each of volume 64 cm<sup>3<\/sup> are joined end to end. Find the surface area of the resulting cuboid.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Volume of cube = <img decoding=\"async\" style=\"height: 29px; width: 49px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image002.png\" \/><\/p>\n<p style=\"text-align: justify;\">According to question, <img decoding=\"async\" style=\"height: 29px; width: 49px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image002.png\" \/>= 64<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image003.png\" \/><img decoding=\"async\" style=\"height: 29px; width: 85px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image004.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image003.png\" \/> Side = 4 cm<\/p>\n<p style=\"text-align: justify;\">For the resulting cubiod, length <img decoding=\"async\" style=\"height: 27px; width: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image005.png\" \/> = 4 + 4 = 8 cm, breadth <img decoding=\"async\" style=\"height: 27px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image006.png\" \/> = 4 cm and height <img decoding=\"async\" style=\"height: 27px; width: 25px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image007.png\" \/> = 4 cm<\/p>\n<p style=\"text-align: justify;\">Surface area of resulting cuboid = <img decoding=\"async\" style=\"height: 27px; width: 97px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image008.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 26px; width: 139px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image009.png\" \/><\/p>\n<p style=\"text-align: justify;\">= 2 (32 + 16 + 32)<\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 22px; width: 108px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image010.png\" \/><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 13.1<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"2_A_vessel_is_in_the_form_of_a_hollow_hemisphere_mounted_by_a_hollow_cylinder_The_diameter_of_the_hemisphere_is_14_cm_and_the_total_height_of_the_vessel_is_13_cm_Find_the_inner_surface_area_of_the_vessel\"><\/span><strong>2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image011.png\" \/>Diameter of the hollow hemisphere = 14 cm<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image012.png\" \/>Radius of the hollow hemisphere = <img decoding=\"async\" style=\"height: 41px; width: 21px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image013.png\" \/> = 7 cm<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 166px; width: 168px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image014.png\" \/><\/p>\n<p style=\"text-align: justify;\">Total height of the vessel = 13 cm<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image012.png\" \/> Height of the hollow cylinder = 13 \u2013 7 = 6 cm<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image012.png\" \/> Inner surface area of the vessel<\/p>\n<p style=\"text-align: justify;\">= Inner surface area of the hollow hemisphere + Inner surface area of the hollow cylinder<\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 29px; width: 129px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image015.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 19px; width: 120px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image016.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 41px; width: 59px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image017.png\" \/> = <img decoding=\"async\" style=\"height: 22px; width: 118px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image018.png\" \/><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 13.1<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"3_A_toy_is_in_the_form_of_a_cone_of_radius_35_cm_mounted_on_a_hemisphere_of_same_radius_The_total_height_of_the_toy_is_155_cm_Find_the_total_surface_area_of_the_toy\"><\/span><strong>3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Radius of the cone = 3.5 cm<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image012.png\" \/> Radius of the hemisphere = 3.5 cm<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 182px; width: 196px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image019.png\" \/><\/p>\n<p style=\"text-align: justify;\">Total height of the toy = 15.5 cm<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image012.png\" \/>Height of the cone = 15.5 \u2013 3.5 = 12 cm<\/p>\n<p style=\"text-align: justify;\">Slant height of the cone = <img decoding=\"async\" style=\"height: 35px; width: 103px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image020.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 24px; width: 88px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image021.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 24px; width: 59px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image022.png\" \/> = 12.5 cm<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image012.png\" \/> TSA of the toy = CSA of hemisphere + CSA of cone<\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 21px; width: 73px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image023.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 29px; width: 163px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image024.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 19px; width: 104px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image025.png\" \/> = <img decoding=\"async\" style=\"height: 19px; width: 51px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image026.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 41px; width: 72px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image027.png\" \/> = <img decoding=\"async\" style=\"height: 22px; width: 70px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image028.png\" \/><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 13.1<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"4_A_cubical_block_of_side_7_cm_is_surmounted_by_a_hemisphere_What_is_the_greatest_diameter_the_hemisphere_can_have_Find_the_surface_area_of_the_solid\"><\/span><strong>4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Greatest diameter of the hemisphere = Side of the cubical block = 7 cm<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image012.png\" \/> TSA of the solid = External surface area of the cubical block + CSA of hemisphere<\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 53px; width: 184px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image029.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 45px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image030.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 41px; width: 73px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image031.png\" \/> = <img decoding=\"async\" style=\"height: 41px; width: 93px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image032.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 41px; width: 61px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image033.png\" \/> = 294 + 38.5 = <img decoding=\"async\" style=\"height: 22px; width: 70px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image034.png\" \/><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 13.1<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"5_A_hemispherical_depression_is_cut_out_from_one_face_of_a_cubical_wooden_block_such_that_the_diameter_of_the_hemisphere_is_equal_to_the_edge_of_the_cube_Determine_the_surface_area_of_the_remaining_solid\"><\/span><strong>5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter <img decoding=\"async\" style=\"height: 19px; width: 9px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image035.png\" \/> of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image011.png\" \/>Diameter of the hemisphere = <img decoding=\"async\" style=\"height: 19px; width: 9px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image035.png\" \/>, therefore radius of the hemisphere = <img decoding=\"async\" style=\"height: 41px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image036.png\" \/><\/p>\n<p style=\"text-align: justify;\">Also, length of the edge of the cube = <img decoding=\"async\" style=\"height: 19px; width: 9px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image035.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image012.png\" \/> Surface area of the remaining solid<\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 49px; width: 151px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image037.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 49px; width: 83px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image038.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 44px; width: 61px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image039.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 41px; width: 84px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image040.png\" \/><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 13.1<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"6_A_medicine_capsule_is_in_the_shape_of_a_cylinder_with_two_hemispheres_stuck_to_each_of_its_ends_see_figure_The_length_of_the_entire_capsule_is_14_mm_and_the_diameter_of_the_capsule_is_5_mm_Find_its_surface_area\"><\/span><strong>6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area. <\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" id=\"Picture 2\" style=\"height: 60px; width: 192px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image041.jpg\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Radius of the hemisphere = <img decoding=\"async\" style=\"height: 41px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image042.png\" \/> mm<\/p>\n<p style=\"text-align: justify;\">Let radius = <img decoding=\"async\" style=\"height: 13px; width: 12px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image043.png\" \/> = 2.5 mm<\/p>\n<p style=\"text-align: justify;\">Cylindrical height = Total height \u2013 Diameter of sphere = <img decoding=\"async\" style=\"height: 19px; width: 13px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image044.png\" \/> = 14 \u2013 (2.5 + 2.5) = 9 mm<\/p>\n<p style=\"text-align: justify;\">Surface area of the capsule = CSA of cylinder + Surface area of the hemisphere<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 6\" style=\"height: 115px; width: 220px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image045.jpg\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 29px; width: 107px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image046.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 53px; width: 169px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image047.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 19px; width: 72px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image048.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 41px; width: 93px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image049.png\" \/> = <img decoding=\"async\" style=\"height: 22px; width: 62px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image050.png\" \/><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 13.1<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"7_A_tent_is_in_the_shape_of_a_cylinder_surmounted_buy_a_conical_top_If_the_height_and_diameter_of_the_cylindrical_part_are_21_m_and_4_m_respectively_and_the_slant_height_of_the_top_is_28_m_find_the_area_of_the_canvas_used_for_making_the_tent_Also_find_the_cost_of_the_canvas_of_the_tent_at_the_rate_of_Rs_500_per_Note_that_the_base_of_the_tent_will_not_be_covered_with_canvas\"><\/span><strong>7. A tent is in the shape of a cylinder surmounted buy a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs. 500 per<img decoding=\"async\" style=\"height: 20px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image051.png\" \/>. (Note that the base of the tent will not be covered with canvas.)<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Diameter of the cylindrical part = 4 cm<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image012.png\" \/>Radius of the cylindrical part = 2 cm<\/p>\n<p style=\"text-align: justify;\">TSA of the tent = CSA of the cylindrical part + CSA of conical cap<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 7\" style=\"height: 242px; width: 142px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image052.jpg\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 27px; width: 160px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image053.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 19px; width: 79px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image054.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 19px; width: 29px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image055.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 41px; width: 51px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image056.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 22px; width: 43px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image057.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image012.png\" \/> Cost of the canvas of the tent at the rate of Rs. 500 per <img decoding=\"async\" style=\"height: 22px; width: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image058.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 19px; width: 57px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image059.png\" \/>= Rs. 22000<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 13.1<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"8_From_a_solid_cylinder_whose_height_is_24_cm_and_diameter_14_cm_a_conical_cavity_of_the_same_height_and_same_diameter_is_hollowed_out_Find_the_total_surface_area_of_the_remaining_solid_to_the_nearest\"><\/span><strong>8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest<img decoding=\"async\" style=\"height: 22px; width: 31px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image060.png\" \/>.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Diameter of the solid cylinder = 1.4 cm<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image012.png\" \/> Radius of the solid cylinder = 0.7 cm<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image012.png\" \/> Radius of the base of the conical cavity = 0.7 cm<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 228px; width: 176px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image061.png\" \/><\/p>\n<p style=\"text-align: justify;\">Height of the solid cylinder = 2.4 cm<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image012.png\" \/> Height of the conical cavity = 2.4 cm<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image012.png\" \/> Slant height of the conical cavity = <img decoding=\"async\" style=\"height: 35px; width: 109px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image062.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 24px; width: 87px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image063.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 24px; width: 44px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image064.png\" \/> = 2.5 cm<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image012.png\" \/> TSA of remaining solid<\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 29px; width: 251px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image065.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 19px; width: 147px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image066.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 19px; width: 35px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image067.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 41px; width: 56px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image068.png\" \/> = 17.6 <img decoding=\"async\" style=\"height: 22px; width: 31px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image060.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 22px; width: 48px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image069.png\" \/>(to the nearest <img decoding=\"async\" style=\"height: 22px; width: 31px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image060.png\" \/>)<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 13.1<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"9_A_wooden_article_was_made_by_scooping_out_a_hemisphere_from_each_end_of_a_solid_cylinder_as_shown_in_figure_If_the_height_of_the_cylinder_is_10_cm_and_its_base_is_of_radius_35_cm_find_the_total_surface_area_of_the_article\"><\/span><strong>9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder as shown in figure. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the total surface area of the article. <\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" id=\"Picture 3\" style=\"height: 111px; width: 80px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image070.jpg\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>TSA of the article = <img decoding=\"async\" style=\"height: 29px; width: 112px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image071.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 35px; width: 183px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image072.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 19px; width: 72px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image073.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 19px; width: 37px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image074.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 41px; width: 59px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image075.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 22px; width: 57px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image076.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 9\" style=\"height: 127px; width: 141px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch13\/Ex13.1\/image077.jpg\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_10_Maths_Exercise_131\"><\/span>NCERT Solutions for Class 10 Maths Exercise 13.1<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>NCERT Solutions Class 10 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 10 Maths includes text book solutions from Mathematics Book. NCERT Solutions for CBSE Class 10 Maths have total 15 chapters. 10 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 10 solutions PDF and Maths ncert class 10 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_10\"><\/span>CBSE app for Class 10<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>To download NCERT Solutions for Class 10 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<strong><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\">the best app for CBSE\u00a0<\/a><\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions for Class 10 Maths Exercise 13.1 Class 10 Maths 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 &#8230; <a title=\"NCERT Solutions for Class 10 Maths Exercise 13.1\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-13-1\/\" aria-label=\"More on NCERT Solutions for Class 10 Maths Exercise 13.1\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":30011,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1347,281],"tags":[1042,283,438,321,1485,216],"class_list":["post-4954","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematics","category-ncert-solutions","tag-cbse-class-10-mathematics","tag-cbse-study-material","tag-class-10","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 10 Maths Exercise 13.1 | myCBSEguide<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for Class 10 Maths Exercise 13.1 in PDF format for free download. 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