{"id":5026,"date":"2016-05-16T11:49:00","date_gmt":"2016-05-16T06:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-9-maths-exercise-12-2\/"},"modified":"2018-06-18T18:23:16","modified_gmt":"2018-06-18T12:53:16","slug":"ncert-solutions-for-class-9-maths-exercise-12-2","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-12-2\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Exercise 12.2"},"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-12-2\/#NCERT_Solutions_for_Class_9_Mathematics_Herons_Formula\" >NCERT Solutions for Class 9 Mathematics\u00a0Herons Formula<\/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-12-2\/#1_A_park_in_the_shape_of_a_quadrilateral_ABCD_has_C_AB_9_m_BC_12_m_CD_5_m_and_AD_8_m_How_much_area_does_it_occupy\" >1. A park, in the shape of a quadrilateral ABCD has C =  AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?<\/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-12-2\/#2_Find_the_area_of_a_quadrilateral_ABCD_in_which_AB_3_cm_BC_4_cm_CD_4_cm_DA_5_cm_and_AC_5_cm\" >2. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.<\/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-12-2\/#3_Radha_made_a_picture_of_an_aeroplane_with_coloured_paper_as_shown_in_figure_Find_the_total_area_of_the_paper_used\" >3. Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used.<\/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-12-2\/#4_A_triangle_and_a_parallelogram_have_the_same_base_and_the_same_area_If_the_sides_of_the_triangle_are_26_cm_29_cm_and_30_cm_and_the_parallelogram_stands_on_the_base_28_cm_find_the_height_of_the_parallelogram\" >4. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 29 cm and 30 cm and the parallelogram stands on the base 28 cm, find the height of the parallelogram.<\/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-12-2\/#5_A_rhombus_shaped_field_has_green_grass_for_18_cows_to_graze_If_each_side_of_the_rhombus_is_30_m_and_its_longer_diagonal_is_48_m_grass_of_how_much_area_of_grass_field_will_each_cow_be_getting\" >5. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, grass of how much area of grass field will each cow be getting?<\/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-12-2\/#6_An_umbrella_is_made_by_stitching_10_triangular_pieces_of_cloth_of_two_different_colours_see_figure_each_piece_measuring_20_cm_50_cm_and_50_cm_How_much_cloth_of_each_colour_is_required_for_the_umbrella\" >6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?<\/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-12-2\/#7_A_kite_is_in_the_shape_of_a_square_with_a_diagonal_32_cm_and_an_isosceles_triangle_of_base_8_cm_and_sides_6_cm_each_is_to_be_made_of_three_different_shades_as_shown_in_figure\" >7. A kite is in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure.<\/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-12-2\/#8_A_floral_design_on_a_floor_is_made_up_of_16_tiles_which_are_triangular_the_sides_of_the_triangle_being_9_cm_28_cm_and_35_cm_see_figure_Find_the_cost_of_polishing_the_tiles_at_the_rate_of_50_paise_per_cm2\" >8. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). Find the cost of polishing the tiles at the rate of 50 paise per cm2.<\/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-12-2\/#9_A_field_is_in_the_shape_of_a_trapezium_whose_parallel_sides_are_25_m_and_10_m_The_non-parallel_sides_are_14_m_and_13_m_Find_the_area_of_the_field\" >9. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.<\/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-11\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-12-2\/#NCERT_Solutions_for_Class_9_Maths_Exercise_122\" >NCERT Solutions for Class 9 Maths Exercise 12.2<\/a><\/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-for-class-9-maths-exercise-12-2\/#CBSE_app_for_Class_9\" >CBSE app for Class 9<\/a><\/li><\/ul><\/nav><\/div>\n<p>NCERT Solutions for Class 9 Maths Exercise 12.2 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\u00a0Herons Formula\u00a0<\/strong><strong><a class=\"button\" href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-09-mathematics-herons-formula\/1246\/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 12.2\" width=\"164\" height=\"211\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_9_Mathematics_Herons_Formula\"><\/span>NCERT Solutions for Class 9 Mathematics\u00a0Herons Formula<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h6><span class=\"ez-toc-section\" id=\"1_A_park_in_the_shape_of_a_quadrilateral_ABCD_has_C_AB_9_m_BC_12_m_CD_5_m_and_AD_8_m_How_much_area_does_it_occupy\"><\/span><strong>1. A park, in the shape of a quadrilateral ABCD has <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image001.png\" \/>C = <img decoding=\"async\" style=\"height: 24px; width: 29px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image002.png\" \/> AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Since BD divides quadrilateral ABCD in two triangles:<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 164\" style=\"height: 144px; width: 179px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image003.jpg\" \/><\/p>\n<p style=\"text-align: justify;\">(i) Right triangle BCD and (ii) <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ABD.<\/p>\n<p style=\"text-align: justify;\">In right triangle BCD, right angled at C,<\/p>\n<p style=\"text-align: justify;\">therefore, Base = CD = 5 m and Altitude = BC = 12 m<\/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\/09\/maths\/ch12\/Ex12.2\/image005.png\" \/>Area of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>BCD = <img decoding=\"async\" style=\"height: 41px; width: 85px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image006.png\" \/> = <img decoding=\"async\" style=\"height: 41px; width: 125px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image007.png\" \/><\/p>\n<p style=\"text-align: justify;\">In <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ABD, AB = 9 m, AD = 8 m<\/p>\n<p style=\"text-align: justify;\">And BD = <img decoding=\"async\" style=\"height: 27px; width: 88px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image008.png\" \/> [Using Pythagoras theorem]\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image009.png\" \/>BD = <img decoding=\"async\" style=\"height: 35px; width: 91px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image010.png\" \/> = <img decoding=\"async\" style=\"height: 24px; width: 69px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image011.png\" \/> = <img decoding=\"async\" style=\"height: 24px; width: 39px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image012.png\" \/> = 13 m<\/p>\n<p style=\"text-align: justify;\">Now, Semi=perimeter of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ABD = <img decoding=\"async\" style=\"height: 41px; width: 64px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image013.png\" \/> = 15 m<\/p>\n<p style=\"text-align: justify;\">Using Heron\u2019s formula,<\/p>\n<p style=\"text-align: justify;\">Area of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ABD = <img decoding=\"async\" style=\"height: 31px; width: 153px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image014.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 31px; width: 187px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image015.png\" \/> = <img decoding=\"async\" style=\"height: 24px; width: 92px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image016.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 24px; width: 40px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image017.png\" \/> = <img decoding=\"async\" style=\"height: 22px; width: 73px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image018.png\" \/><img decoding=\"async\" style=\"height: 22px; width: 72px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image019.png\" \/>(approx.)<\/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\/09\/maths\/ch12\/Ex12.2\/image005.png\" \/>Area of quadrilateral ABCD = Area of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>BCD + Area of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ABD<\/p>\n<p style=\"text-align: justify;\">= 30 + 35.4<\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" style=\"height: 22px; width: 72px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image020.png\" \/><\/strong><\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"2_Find_the_area_of_a_quadrilateral_ABCD_in_which_AB_3_cm_BC_4_cm_CD_4_cm_DA_5_cm_and_AC_5_cm\"><\/span><strong>2. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>In quadrilateral ABCE, diagonal AC divides it in two triangles, <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ABC and <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ADC.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 167\" style=\"height: 130px; width: 132px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image021.jpg\" \/><\/p>\n<p style=\"text-align: justify;\">In <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ABC, Semi-perimeter of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ABC = <img decoding=\"async\" style=\"height: 41px; width: 59px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image022.png\" \/> = 6 cm<\/p>\n<p style=\"text-align: justify;\">Using Heron\u2019s formula,<\/p>\n<p style=\"text-align: justify;\">Area of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ABC = <img decoding=\"async\" style=\"height: 31px; width: 153px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image014.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 31px; width: 155px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image023.png\" \/> = <img decoding=\"async\" style=\"height: 24px; width: 83px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image024.png\" \/> <img decoding=\"async\" style=\"height: 22px; width: 59px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image025.png\" \/><\/p>\n<p style=\"text-align: justify;\">Again, In <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ADC, Semi-perimeter of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ADC = <img decoding=\"async\" style=\"height: 41px; width: 59px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image026.png\" \/> = 7 cm<\/p>\n<p style=\"text-align: justify;\">Using Heron\u2019s formula, Area of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ABC = <img decoding=\"async\" style=\"height: 31px; width: 153px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image014.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 31px; width: 156px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image027.png\" \/> = <img decoding=\"async\" style=\"height: 24px; width: 85px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image028.png\" \/> = 2<img decoding=\"async\" style=\"height: 23px; width: 32px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image029.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 22px; width: 132px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image030.png\" \/> (approx.)<\/p>\n<p style=\"text-align: justify;\">Now area of quadrilateral ABCD = Area of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ABC + Area of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ADC<\/p>\n<p style=\"text-align: justify;\">= 6 + 9.2<\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" style=\"height: 22px; width: 77px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image031.png\" \/><\/strong><\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"3_Radha_made_a_picture_of_an_aeroplane_with_coloured_paper_as_shown_in_figure_Find_the_total_area_of_the_paper_used\"><\/span><strong>3. Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used. <\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" id=\"Picture 170\" style=\"height: 247px; width: 243px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image032.jpg\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. Area of triangular part I<\/strong>: Here, Semi-perimeter <img decoding=\"async\" style=\"height: 41px; width: 91px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image033.png\" \/> = 5.5 cm<\/p>\n<p style=\"text-align: justify;\">Therefore, Area = <img decoding=\"async\" style=\"height: 31px; width: 153px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image014.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 31px; width: 199px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image034.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 24px; width: 132px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image035.png\" \/> = <img decoding=\"async\" style=\"height: 23px; width: 57px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image036.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" style=\"height: 22px; width: 188px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image037.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Area of triangular part II<\/strong> = Length x Breadth <img decoding=\"async\" style=\"height: 22px; width: 135px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image038.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>Area of triangular part III<\/strong> (trapezium): <img decoding=\"async\" style=\"height: 41px; width: 132px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image039.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 41px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image040.png\" \/> (AB + DC) <img decoding=\"async\" style=\"height: 25px; width: 97px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image041.png\" \/> = <img decoding=\"async\" style=\"height: 41px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image040.png\" \/> (1 + 2) <img decoding=\"async\" style=\"height: 24px; width: 71px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image042.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 41px; width: 45px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image043.png\" \/><img decoding=\"async\" style=\"height: 45px; width: 27px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image044.png\" \/> = <img decoding=\"async\" style=\"height: 41px; width: 61px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image045.png\" \/> <img decoding=\"async\" style=\"height: 22px; width: 85px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image046.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>Area of triangular parts IV &amp; V<\/strong>: <img decoding=\"async\" style=\"height: 45px; width: 92px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image047.png\" \/> <img decoding=\"async\" style=\"height: 22px; width: 59px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image048.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\/09\/maths\/ch12\/Ex12.2\/image005.png\" \/> <strong>Total area<\/strong> = 2.4825 + 6.2 + 1.299 + 9 <img decoding=\"async\" style=\"height: 22px; width: 85px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image049.png\" \/><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 12.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"4_A_triangle_and_a_parallelogram_have_the_same_base_and_the_same_area_If_the_sides_of_the_triangle_are_26_cm_29_cm_and_30_cm_and_the_parallelogram_stands_on_the_base_28_cm_find_the_height_of_the_parallelogram\"><\/span><strong>4. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 29 cm and 30 cm and the parallelogram stands on the base 28 cm, find the height of the parallelogram.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Semi-perimeter of triangle <img decoding=\"async\" style=\"height: 27px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image050.png\" \/> = <img decoding=\"async\" style=\"height: 41px; width: 83px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image051.png\" \/> = 42 cm<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 173\" style=\"height: 111px; width: 203px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image052.jpg\" \/><\/p>\n<p style=\"text-align: justify;\">Using Heron\u2019s formula,<\/p>\n<p style=\"text-align: justify;\">Area of triangle = <img decoding=\"async\" style=\"height: 31px; width: 153px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image014.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 31px; width: 213px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image053.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 24px; width: 113px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image054.png\" \/> <img decoding=\"async\" style=\"height: 22px; width: 74px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image055.png\" \/><\/p>\n<p style=\"text-align: justify;\">According to question, Area of parallelogram = Area of triangle<\/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\/09\/maths\/ch12\/Ex12.2\/image009.png\" \/>Base x Corresponding height = 336<\/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\/09\/maths\/ch12\/Ex12.2\/image009.png\" \/> <img decoding=\"async\" style=\"height: 22px; width: 79px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image056.png\" \/>= 336<\/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\/09\/maths\/ch12\/Ex12.2\/image009.png\" \/> Height = 12 cm<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 12.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"5_A_rhombus_shaped_field_has_green_grass_for_18_cows_to_graze_If_each_side_of_the_rhombus_is_30_m_and_its_longer_diagonal_is_48_m_grass_of_how_much_area_of_grass_field_will_each_cow_be_getting\"><\/span><strong>5. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, grass of how much area of grass field will each cow be getting?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Here, AB = BC = CD = DA = 30 m and Diagonal AC = 48 m which divides the rhombus ABCD in two congruent triangle.<\/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\/09\/maths\/ch12\/Ex12.2\/image005.png\" \/>Area of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ABC = Area of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ACD<\/p>\n<p style=\"text-align: justify;\">Now, Semi-perimeter of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ABC <img decoding=\"async\" style=\"height: 27px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image050.png\" \/> = <img decoding=\"async\" style=\"height: 41px; width: 83px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image057.png\" \/> = 54 m<\/p>\n<p style=\"text-align: justify;\">Now Area of rhombus ABCD = Area of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ABC + Area of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ACD<\/p>\n<p style=\"text-align: justify;\">= 2 <img decoding=\"async\" style=\"height: 14px; width: 12px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image058.png\" \/> Area of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ABC [<img decoding=\"async\" style=\"height: 13px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image059.png\" \/> Area of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ABC = Area of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>ACD]\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 31px; width: 161px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image060.png\" \/>[ Using Heron\u2019s formula]\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 31px; width: 231px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image061.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 24px; width: 131px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image062.png\" \/>= <img decoding=\"async\" style=\"height: 19px; width: 61px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image063.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 22px; width: 68px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image064.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\/09\/maths\/ch12\/Ex12.2\/image059.png\" \/> Field available for 18 cows to graze the grass <img decoding=\"async\" style=\"height: 22px; width: 68px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image064.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\/09\/maths\/ch12\/Ex12.2\/image005.png\" \/>Field available for 1 cow to graze the grass = <img decoding=\"async\" style=\"height: 41px; width: 31px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image065.png\" \/> <img decoding=\"async\" style=\"height: 22px; width: 60px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image066.png\" \/><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 12.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"6_An_umbrella_is_made_by_stitching_10_triangular_pieces_of_cloth_of_two_different_colours_see_figure_each_piece_measuring_20_cm_50_cm_and_50_cm_How_much_cloth_of_each_colour_is_required_for_the_umbrella\"><\/span><strong>6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella? <\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" id=\"Picture 176\" style=\"height: 152px; width: 168px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image067.jpg\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Here, sides of each of 10 triangular pieces of two different colours are 20 cm, 50 cm and 50 cm.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 179\" style=\"height: 138px; width: 315px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image068.jpg\" \/><\/p>\n<p style=\"text-align: justify;\">Semi-perimeter of each triangle <img decoding=\"async\" style=\"height: 27px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image050.png\" \/> = <img decoding=\"async\" style=\"height: 41px; width: 83px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image069.png\" \/> = 60 cm<\/p>\n<p style=\"text-align: justify;\">Now, Area of each triangle = <img decoding=\"async\" style=\"height: 31px; width: 153px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image014.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 31px; width: 212px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image070.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 24px; width: 115px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image071.png\" \/> = <img decoding=\"async\" style=\"height: 24px; width: 72px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image072.png\" \/><\/p>\n<p style=\"text-align: justify;\">According to question, there are 5 pieces of red colour and 5 pieces of green colour.<\/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\/09\/maths\/ch12\/Ex12.2\/image005.png\" \/>Cloth required for 5 red pieces = <img decoding=\"async\" style=\"height: 24px; width: 69px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image073.png\" \/> = <img decoding=\"async\" style=\"height: 24px; width: 79px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image074.png\" \/><\/p>\n<p style=\"text-align: justify;\">And Cloth required to 5 green pieces = <img decoding=\"async\" style=\"height: 24px; width: 69px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image073.png\" \/> = <img decoding=\"async\" style=\"height: 24px; width: 79px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image074.png\" \/><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 12.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"7_A_kite_is_in_the_shape_of_a_square_with_a_diagonal_32_cm_and_an_isosceles_triangle_of_base_8_cm_and_sides_6_cm_each_is_to_be_made_of_three_different_shades_as_shown_in_figure\"><\/span><strong>7. A kite is in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. <\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" id=\"Picture 169\" style=\"height: 135px; width: 121px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image075.jpg\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>How much paper of each side has been used in it?<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Let ABCD is a square of side <img decoding=\"async\" style=\"height: 15px; width: 13px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image076.png\" \/> cm and diagonals AC = BD = 32 cm<\/p>\n<p style=\"text-align: justify;\">In right triangle ABC, <img decoding=\"async\" style=\"height: 22px; width: 130px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image077.png\" \/>[Using Pythagoras theorem]\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image009.png\" \/><img decoding=\"async\" style=\"height: 29px; width: 99px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image078.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\/09\/maths\/ch12\/Ex12.2\/image009.png\" \/> <img decoding=\"async\" style=\"height: 21px; width: 88px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image079.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\/09\/maths\/ch12\/Ex12.2\/image009.png\" \/> <img decoding=\"async\" style=\"height: 41px; width: 83px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image080.png\" \/> = 512<\/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\/09\/maths\/ch12\/Ex12.2\/image009.png\" \/> Area of square <img decoding=\"async\" style=\"height: 22px; width: 77px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image081.png\" \/>[Area of square =<img decoding=\"async\" style=\"height: 19px; width: 70px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image082.png\" \/>]\n<p style=\"text-align: justify;\">Diagonal BD divides the square in two equal triangular parts I and II.<\/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\/09\/maths\/ch12\/Ex12.2\/image005.png\" \/>Area of shaded part I = Area of shaded part II = <img decoding=\"async\" style=\"height: 41px; width: 130px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image083.png\" \/><\/p>\n<p style=\"text-align: justify;\">Now, semi-perimeter of shaded part III <img decoding=\"async\" style=\"height: 41px; width: 93px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image084.png\" \/> = 10 cm<\/p>\n<p style=\"text-align: justify;\">Area of shaded part III = <img decoding=\"async\" style=\"height: 31px; width: 153px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image014.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 31px; width: 181px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image085.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 24px; width: 93px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image086.png\" \/> = <img decoding=\"async\" style=\"height: 24px; width: 32px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image087.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" style=\"height: 22px; width: 163px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image088.png\" \/><\/strong><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 12.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"8_A_floral_design_on_a_floor_is_made_up_of_16_tiles_which_are_triangular_the_sides_of_the_triangle_being_9_cm_28_cm_and_35_cm_see_figure_Find_the_cost_of_polishing_the_tiles_at_the_rate_of_50_paise_per_cm2\"><\/span><strong>8. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). Find the cost of polishing the tiles at the rate of 50 paise per cm<sup>2<\/sup>. <\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" id=\"Picture 185\" style=\"height: 207px; width: 219px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image089.jpg\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Here, Sides of a triangular shaped tile area 9 cm, 28 cm and 35 cm.<\/p>\n<p style=\"text-align: justify;\">Semi-perimeter of tile <img decoding=\"async\" style=\"height: 27px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image050.png\" \/> = <img decoding=\"async\" style=\"height: 41px; width: 75px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image090.png\" \/> = 36 cm<\/p>\n<p style=\"text-align: justify;\">Area of triangular shaped tile = <img decoding=\"async\" style=\"height: 31px; width: 153px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image014.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 31px; width: 203px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image091.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 24px; width: 99px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image092.png\" \/> = <img decoding=\"async\" style=\"height: 24px; width: 40px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image093.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 22px; width: 161px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image094.png\" \/>(approx.)<\/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\/09\/maths\/ch12\/Ex12.2\/image005.png\" \/>Area of 16 such tiles <img decoding=\"async\" style=\"height: 22px; width: 169px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image095.png\" \/>(Approx.)<\/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\/09\/maths\/ch12\/Ex12.2\/image059.png\" \/>Cost of polishing <img decoding=\"async\" style=\"height: 22px; width: 39px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image096.png\" \/>of tile = Rs. 0.50<\/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\/09\/maths\/ch12\/Ex12.2\/image005.png\" \/>Cost of polishing <img decoding=\"async\" style=\"height: 22px; width: 76px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image097.png\" \/>of tile = <img decoding=\"async\" style=\"height: 19px; width: 112px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image098.png\" \/>= Rs. 705.60 (Approx.)<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 12.2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"9_A_field_is_in_the_shape_of_a_trapezium_whose_parallel_sides_are_25_m_and_10_m_The_non-parallel_sides_are_14_m_and_13_m_Find_the_area_of_the_field\"><\/span><strong>9. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Let ABCD be a field in the shape of trapezium and parallel side AB = 10 m &amp; CD = 25 m<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 188\" style=\"height: 166px; width: 246px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image099.jpg\" \/><\/p>\n<p style=\"text-align: justify;\">And Non-parallel sides AD = 13 m and BC = 14 m<\/p>\n<p style=\"text-align: justify;\">Draw BM <img decoding=\"async\" style=\"height: 17px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image100.png\" \/> DC and BE<img decoding=\"async\" style=\"height: 21px; width: 11px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image101.png\" \/> AD so that ABED is a parallelogram.<\/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\/09\/maths\/ch12\/Ex12.2\/image005.png\" \/>BE = AD = 13 m and DE = AB = 10 m<\/p>\n<p style=\"text-align: justify;\">Now in <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>BEC, Semi-perimeter <img decoding=\"async\" style=\"height: 41px; width: 113px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image102.png\" \/><\/p>\n<p style=\"text-align: justify;\">= 21 m<\/p>\n<p style=\"text-align: justify;\">Area of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>BEC = <img decoding=\"async\" style=\"height: 31px; width: 153px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image014.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 31px; width: 203px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image103.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 24px; width: 92px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image104.png\" \/> = <img decoding=\"async\" style=\"height: 22px; width: 42px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image105.png\" \/><\/p>\n<p style=\"text-align: justify;\">And Area of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image004.png\" \/>BEC = <img decoding=\"async\" style=\"height: 22px; width: 42px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image105.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\/09\/maths\/ch12\/Ex12.2\/image009.png\" \/><img decoding=\"async\" style=\"height: 41px; width: 90px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image106.png\" \/> = 84<\/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\/09\/maths\/ch12\/Ex12.2\/image009.png\" \/> <img decoding=\"async\" style=\"height: 41px; width: 82px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image107.png\" \/> = 84<\/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\/09\/maths\/ch12\/Ex12.2\/image009.png\" \/> BM = <img decoding=\"async\" style=\"height: 41px; width: 43px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image108.png\" \/> = 11.2 m<\/p>\n<p style=\"text-align: justify;\">Now area of trapezium ABCD = <img decoding=\"async\" style=\"height: 41px; width: 136px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image109.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 41px; width: 122px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image110.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 22px; width: 67px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch12\/Ex12.2\/image111.png\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_9_Maths_Exercise_122\"><\/span>NCERT Solutions for Class 9 Maths Exercise 12.2<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 12.2 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 12.2\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-12-2\/\" aria-label=\"More on NCERT Solutions for Class 9 Maths Exercise 12.2\">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-5026","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 12.2 | myCBSEguide<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for Class 9 Maths Exercise 12.2 in PDF format for free download. 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