{"id":4325,"date":"2016-05-13T11:49:00","date_gmt":"2016-05-13T06:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-7-maths-perimeter-and-area-ex-11-4\/"},"modified":"2018-07-07T16:51:29","modified_gmt":"2018-07-07T11:21:29","slug":"ncert-solutions-for-class-7-maths-exercise-11-4","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-7-maths-exercise-11-4\/","title":{"rendered":"NCERT Solutions for Class 7 Maths Exercise 11.4"},"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-7-maths-exercise-11-4\/#NCERT_Solutions_for_Class_7_Maths_Perimeter_and_Area\" >NCERT Solutions for Class 7 Maths\u00a0Perimeter and Area<\/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-7-maths-exercise-11-4\/#Class_%E2%80%93VII_Mathematics_Ex_114\" >Class \u2013VII Mathematics (Ex. 11.4)<\/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-7-maths-exercise-11-4\/#Question_1A_garden_is_90_m_long_and_75_m_broad_A_path_5_m_wide_is_to_be_built_outside_and_around_it_Find_the_area_of_the_path_Also_find_the_area_of_the_garden_in_hectares\" >Question 1.A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectares.<\/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-7-maths-exercise-11-4\/#Question_2A_3_m_wide_path_runs_outside_and_around_a_rectangular_park_of_length_125_m_and_breadth_65_m_Find_the_area_of_the_path\" >Question 2.A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path.<\/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-7-maths-exercise-11-4\/#Question_3A_picture_is_painted_on_a_cardboard_8_cm_long_and_5_cm_wide_such_that_there_is_a_margin_of_15_cm_along_each_of_its_sides_Find_the_total_area_of_the_margin\" >Question 3.A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.<\/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-7-maths-exercise-11-4\/#Question_4A_verandah_of_width_225_m_is_constructed_all_along_outside_a_room_which_is_55_m_long_and_4_m_wide_Find\" >Question 4.A verandah of width 2.25 m is constructed all along outside a room which is 5.5 m long and 4 m wide. Find:<\/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-7-maths-exercise-11-4\/#Question_5A_path_1_m_wide_is_built_along_the_border_and_inside_a_square_garden_of_side_30_m_Find\" >Question 5.A path 1 m wide is built along the border and inside a square garden of side 30 m. Find:<\/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-7-maths-exercise-11-4\/#Question_6Two_cross_roads_each_of_width_10_m_cut_at_right_angles_through_the_centre_of_a_rectangular_park_of_length_700_m_and_breadth_300_m_and_parallel_to_its_sides_Find_the_area_of_the_roads_Also_find_the_area_of_the_park_excluding_cross_roads_Give_the_answer_in_hectares\" >Question 6.Two cross roads, each of width 10 m, cut at right angles through the centre of a rectangular park of length 700 m and breadth 300 m and parallel to its sides. Find the area of the roads. Also find the area of the park excluding cross roads. Give the answer in hectares.<\/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-7-maths-exercise-11-4\/#Question_7Through_a_rectangular_field_of_length_90_m_and_breadth_60_m_two_roads_are_constructed_which_are_parallel_to_the_sides_and_cut_each_other_at_right_angles_through_the_centre_of_the_fields_If_the_width_of_each_road_is_3_m_find\" >Question 7.Through a rectangular field of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 m, find:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-7-maths-exercise-11-4\/#Question_8Pragya_wrapped_a_cord_around_a_circular_pipe_of_radius_4_cm_adjoining_figure_and_cut_off_the_length_required_of_the_cord_Then_she_wrapped_it_around_a_square_box_of_side_4_cm_also_shown_Did_she_have_any_cord_left_texleft_textTake_pi_text_3text14_righttex\" >Question 8.Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any cord left? {tex}\\left( {{\\text{Take }}\\pi {\\text{ = 3}}{\\text{.14}}} \\right){\/tex}<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-7-maths-exercise-11-4\/#Question_9The_adjoining_figure_represents_a_rectangular_lawn_with_a_circular_flower_bed_in_the_middle_Find\" >Question 9.The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-7-maths-exercise-11-4\/#Question_10In_the_following_figures_find_the_area_of_the_shaded_portions\" >Question 10.In the following figures, find the area of the shaded portions:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-13\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-7-maths-exercise-11-4\/#Question_11Find_the_area_of_the_equilateral_ABCD_Here_AC_22_cm_BM_3_cm_DN_3_cm_and_BM_tex_bot_tex_AC_DN_tex_bot_tex_AC\" >Question 11.Find the area of the equilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3 cm and BM {tex} \\bot {\/tex} AC, DN {tex} \\bot {\/tex} AC.<\/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-14\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-7-maths-exercise-11-4\/#NCERT_Solutions_for_Class_7_Maths_Exercise_114\" >NCERT Solutions for Class 7 Maths Exercise 11.4<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-15\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-7-maths-exercise-11-4\/#CBSE_app_for_Students\" >CBSE app for Students<\/a><\/li><\/ul><\/nav><\/div>\n<p>NCERT Solutions for Class 7 Maths Exercise 11.4 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 7 Maths chapter wise NCERT solution for Maths Book 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\u00a0Perimeter and Area\u00a0<\/strong><strong><a class=\"button\" href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-07-mathematics-perimeter-and-area\/1520\/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\/07_Maths_Book.jpg\" alt=\"NCERT Solutions for Class 7 Maths Exercise 11.4\" width=\"154\" height=\"195\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_7_Maths_Perimeter_and_Area\"><\/span>NCERT Solutions for Class 7 Maths\u00a0Perimeter and Area<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h6 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"Class_%E2%80%93VII_Mathematics_Ex_114\"><\/span><strong>Class \u2013VII Mathematics (Ex. 11.4)<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<h6><span class=\"ez-toc-section\" id=\"Question_1A_garden_is_90_m_long_and_75_m_broad_A_path_5_m_wide_is_to_be_built_outside_and_around_it_Find_the_area_of_the_path_Also_find_the_area_of_the_garden_in_hectares\"><\/span><strong>Question 1<\/strong>.A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectares.<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Answer:<\/strong><\/p>\n<p>Length of rectangular garden = 90 m and breadth of rectangular garden = 75 m<\/p>\n<p>Outer length of rectangular garden with path<\/p>\n<p>= 90 + 5 + 5 = 100 m<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch11_17.jpg\" \/><\/p>\n<p>Outer breadth of rectangular garden with path<\/p>\n<p>= 75 + 5 + 5 = 85 m<\/p>\n<p>Outer area of rectangular garden with path<\/p>\n<p>= length x breadth = 100 x 85 = 8,500 {tex}{m^2}{\/tex}<\/p>\n<p>Inner area of garden without path = length x breadth = 90 x 75 = 6,750 {tex}{m^2}{\/tex}<\/p>\n<p>Now Area of path = Area of garden with path \u2013 Area of garden without path<\/p>\n<p>= 8,500 \u2013 6,750<\/p>\n<p>= 1,750{tex}{m^2}{\/tex}<\/p>\n<p>Since, 1 m2 = {tex}\\frac{1}{{10000}}{\/tex} hectares<\/p>\n<p>Therefore, 6,750 {tex}{m^2}{\/tex}= {tex}\\frac{{6750}}{{10000}}{\/tex} = 0.675 hectares<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 11.4<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_2A_3_m_wide_path_runs_outside_and_around_a_rectangular_park_of_length_125_m_and_breadth_65_m_Find_the_area_of_the_path\"><\/span><strong>Question 2.<\/strong>A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path.<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Answer:<\/strong><\/p>\n<p>Length of rectangular park = 125 m, breadth of rectangular park = 65 m and width of the path = 3 m<\/p>\n<p>Length of rectangular park with path<\/p>\n<p>= 125 + 3 + 3 = 131 m<\/p>\n<p>Breadth of rectangular park with path<\/p>\n<p>= 65 + 3 + 3 = 71 m<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch11_18.jpg\" \/><\/p>\n<p>{tex}\\therefore {\/tex} Area of path<\/p>\n<p>= Area of park with path \u2013 Area of park without path<\/p>\n<p>= (AB x AD) \u2013 (EF x EH)<\/p>\n<p>= (131 x 71) \u2013 (125 x 65)<\/p>\n<p>= 9301 \u2013 8125<\/p>\n<p>= 1,176{tex}{m^2}{\/tex}<\/p>\n<p>Thus, area of path around the park is 1,176{tex}{m^2}{\/tex}.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 11.4<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_3A_picture_is_painted_on_a_cardboard_8_cm_long_and_5_cm_wide_such_that_there_is_a_margin_of_15_cm_along_each_of_its_sides_Find_the_total_area_of_the_margin\"><\/span><strong>Question 3.<\/strong>A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Answer:<\/strong><\/p>\n<p>Length of painted cardboard = 8 cm and breadth of painted card = 5 cm<\/p>\n<p>Since, there is a margin of 1.5 cm long from each of its side.<\/p>\n<p>Therefore reduced length<\/p>\n<p>= 8 \u2013 (1.5 + 1.5) = 8 \u2013 3 = 5 cm<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch11_19.jpg\" \/><\/p>\n<p>And reduced breadth<\/p>\n<p>= 5 \u2013 (1.5 + 1.5) = 5 \u2013 3 = 2 cm<\/p>\n<p>{tex}\\therefore {\/tex} Area of margin = Area of cardboard (ABCD) \u2013 Area of cardboard (EFGH)<\/p>\n<p>= (AB x AD) \u2013 (EF x EH)<\/p>\n<p>= (8 x 5) \u2013 (5 x 2)<\/p>\n<p>= 40 \u2013 10<\/p>\n<p>= 30 {tex}c{m^2}{\/tex}<\/p>\n<p>Thus, the total area of margin is 30 {tex}c{m^2}{\/tex}.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 11.4<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_4A_verandah_of_width_225_m_is_constructed_all_along_outside_a_room_which_is_55_m_long_and_4_m_wide_Find\"><\/span><strong>Question 4.<\/strong>A <em>verandah<\/em> of width 2.25 m is constructed all along outside a room which is 5.5 m long and 4 m wide. Find:<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p>(i) the area of the <em>verandah<\/em>.<\/p>\n<p>(ii) the cost of cementing the floor of the <em>verandah<\/em> at the rate of Rs. 200 per {tex}{m^2}{\/tex}.<\/p>\n<p><strong>Answer:<\/strong><\/p>\n<p>(i) The length of room = 5.5 m and width of the<\/p>\n<p>room = 4 m<\/p>\n<p>The length of room with verandah<\/p>\n<p>= 5.5 + 2.25 + 2.25 = 10 m<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch11_20.png\" \/><\/p>\n<p>The width of room with verandah<\/p>\n<p>= 4 + 2.25 + 2.25 = 8.5 m<\/p>\n<p>Area of verandah<\/p>\n<p>= Area of room with verandah \u2013 Area of room without verandah<\/p>\n<p>= Area of ABCD \u2013 Area of EFGH<\/p>\n<p>= (AB x AD) \u2013 (EF x EH)<\/p>\n<p>= (10 x 8.5) \u2013 (5.5 x 4)<\/p>\n<p>= 85 \u2013 22<\/p>\n<p>= 63 {tex}{m^2}{\/tex}<\/p>\n<p>(ii) The cost of cementing 1 {tex}{m^2}{\/tex} the floor of verandah = Rs. 200<\/p>\n<p>The cost of cementing 63 {tex}{m^2}{\/tex} the floor of verandah = 200 x 63 = Rs. 12,600<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 11.4<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_5A_path_1_m_wide_is_built_along_the_border_and_inside_a_square_garden_of_side_30_m_Find\"><\/span><strong>Question 5.<\/strong>A path 1 m wide is built along the border and inside a square garden of side 30 m. Find:<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p>(i) the area of the path.<\/p>\n<p>(ii) the cost of planting grass in the remaining portion of the garden at the rate of Rs. 40 per {tex}{m^2}{\/tex}.<\/p>\n<p><strong>Answer:<\/strong><\/p>\n<p>(i) Side of the square garden = 30 m and width of the path<\/p>\n<p>along the border = 1 m<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch11_21.png\" \/><\/p>\n<p>Side of square garden without path<\/p>\n<p>= 30 \u2013 (1 + 1) = 30 \u2013 2 = 28 m<\/p>\n<p>Now Area of path = Area of ABCD \u2013 Area of EFGH<\/p>\n<p>= (AB x AD) \u2013 (EF x EH)<\/p>\n<p>= (30 x 30) \u2013 (28 x 28)<\/p>\n<p>= 900 \u2013 784<\/p>\n<p>= 116 {tex}{m^2}{\/tex}<\/p>\n<p>(ii) Area of remaining portion = 28 x 28 = 784 {tex}{m^2}{\/tex}<\/p>\n<p>The cost of planting grass in 1 {tex}{m^2}{\/tex}of the garden = Rs. 40<\/p>\n<p>The cost of planting grass in 784 {tex}{m^2}{\/tex}of the garden = Rs. 40 x 784 = Rs. 31,360<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 11.4<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_6Two_cross_roads_each_of_width_10_m_cut_at_right_angles_through_the_centre_of_a_rectangular_park_of_length_700_m_and_breadth_300_m_and_parallel_to_its_sides_Find_the_area_of_the_roads_Also_find_the_area_of_the_park_excluding_cross_roads_Give_the_answer_in_hectares\"><\/span><strong>Question 6.<\/strong>Two cross roads, each of width 10 m, cut at right angles through the centre of a rectangular park of length 700 m and breadth 300 m and parallel to its sides. Find the area of the roads. Also find the area of the park excluding cross roads. Give the answer in hectares.<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Answer:<\/strong><\/p>\n<p>Here, PQ = 10 m and PS = 300 m, EH = 10 m and EF = 700 m<\/p>\n<p>And KL = 10 m and KN = 10 m<\/p>\n<p>Area of roads<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch11_22.jpg\" \/><\/p>\n<p>= Area of PQRS + Area of EFGH \u2013 Area of KLMN<\/p>\n[{tex}\\because {\/tex} KLMN is taken twice, which is to be subtracted]\n<p>= PS x PQ + EF x EH \u2013 KL x KN<\/p>\n<p>= (300 x 10) + (700 x 10) \u2013 (10 x 10)<\/p>\n<p>= 3000 + 7000 \u2013 100<\/p>\n<p>= 9,900 {tex}{m^2}{\/tex}<\/p>\n<p>Area of road in hectares, 1 {tex}{m^2}{\/tex}= {tex}\\frac{1}{{10000}}{\/tex} hectares<\/p>\n<p>{tex}\\therefore {\/tex} 9,900{tex}{m^2}{\/tex}= {tex}\\frac{{9900}}{{10000}}{\/tex} = 0.99 hectares<\/p>\n<p>Now, Area of park excluding cross roads = Area of park \u2013 Area of road<\/p>\n<p>= (AB x AD) \u2013 9,900<\/p>\n<p>= (700 x 300) \u2013 9,900<\/p>\n<p>= 2,10,000 \u2013 9,900<\/p>\n<p>= 2,00,100 {tex}{m^2}{\/tex}<\/p>\n<p>= {tex}\\frac{{200100}}{{10000}}{\/tex} hectares = 20.01 hectares<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 11.4<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_7Through_a_rectangular_field_of_length_90_m_and_breadth_60_m_two_roads_are_constructed_which_are_parallel_to_the_sides_and_cut_each_other_at_right_angles_through_the_centre_of_the_fields_If_the_width_of_each_road_is_3_m_find\"><\/span><strong>Question 7.<\/strong>Through a rectangular field of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 m, find:<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p>(i) the area covered by the roads.<\/p>\n<p>(ii) the cost of constructing the roads at the rate of Rs. 110 per{tex}{m^2}{\/tex}.<\/p>\n<p><strong>Answer:<\/strong><\/p>\n<p>(i) Here, PQ = 3 m and PS = 60 m, EH = 3 m and<\/p>\n<p>EF = 90 m and KL = 3 m and KN = 3 m<\/p>\n<p>Area of roads<\/p>\n<p>= Area of PQRS + Area of EFGH \u2013 Area of KLMN<\/p>\n[{tex}\\because {\/tex} KLMN is taken twice, which is to be subtracted]\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch11_23.png\" \/><\/p>\n<p>= PS x PQ + EF x EH \u2013 KL x KN<\/p>\n<p>= (60 x 3) + (90 x 3) \u2013 (3 x 3)<\/p>\n<p>= 180 + 270 \u2013 9<\/p>\n<p>= 441 {tex}{m^2}{\/tex}<\/p>\n<p>(ii) The cost of 1 {tex}{m^2}{\/tex} constructing the roads = Rs. 110<\/p>\n<p>The cost of 441 {tex}{m^2}{\/tex} constructing the roads = Rs. 110 x 441 = Rs. 48,510<\/p>\n<p>Therefore, the cost of constructing the roads = Rs. 48,510<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 11.4<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_8Pragya_wrapped_a_cord_around_a_circular_pipe_of_radius_4_cm_adjoining_figure_and_cut_off_the_length_required_of_the_cord_Then_she_wrapped_it_around_a_square_box_of_side_4_cm_also_shown_Did_she_have_any_cord_left_texleft_textTake_pi_text_3text14_righttex\"><\/span><strong>Question 8.<\/strong>Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any cord left? {tex}\\left( {{\\text{Take }}\\pi {\\text{ = 3}}{\\text{.14}}} \\right){\/tex}<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch11_12.jpg\" \/><\/p>\n<p><strong>Answer:<\/strong><\/p>\n<p>Radius of pipe = 4 cm<\/p>\n<p>Wrapping cord around circular pipe = {tex}2\\pi r{\/tex}<\/p>\n<p>= 2 x 3.14 x 4 = 25.12 cm<\/p>\n<p>Again, wrapping cord around a square = 4 x side<\/p>\n<p>= 4 x 4 = 16 cm<\/p>\n<p>Remaining cord = Cord wrapped on pipe \u2013 Cord wrapped on square<\/p>\n<p>= 25.12 \u2013 16 = 9.12 cm<\/p>\n<p>Thus, she has left 9.12 cm cord.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 11.4<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_9The_adjoining_figure_represents_a_rectangular_lawn_with_a_circular_flower_bed_in_the_middle_Find\"><\/span><strong>Question 9.<\/strong>The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find:<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch11_13.png\" \/><\/p>\n<p>(i) the area of the whole land.<\/p>\n<p>(ii) the area of the flower bed.<\/p>\n<p>(iii) the area of the lawn excluding the area of the flower bed.<\/p>\n<p>(iv) the circumference of the flower bed.<\/p>\n<p><strong>Answer:<\/strong><\/p>\n<p>Length of rectangular lawn = 10 m, breadth of the rectangular lawn = 5 m<\/p>\n<p>And radius of the circular flower bed = 2 m<\/p>\n<p>(i) Area of the whole land = length x breadth<\/p>\n<p>= 10 x 5 = 50{tex}{m^2}{\/tex}<\/p>\n<p>(ii) Area of flower bed = {tex}\\pi {r^2}{\/tex}<\/p>\n<p>= 3.14 x 2 x 2 = 12.56{tex}{m^2}{\/tex}<\/p>\n<p>(iii) Area of lawn excluding the area of the flower bed = area of lawn \u2013 area of flower bed<\/p>\n<p>= 50 \u2013 12.56<\/p>\n<p>= 37.44 {tex}{m^2}{\/tex}<\/p>\n<p>(iv) The circumference of the flower bed = {tex}2\\pi r{\/tex}<\/p>\n<p>= 2 x 3.14 x 2 = 12.56 m<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 11.4<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_10In_the_following_figures_find_the_area_of_the_shaded_portions\"><\/span><strong>Question 10.<\/strong>In the following figures, find the area of the shaded portions:<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch11_14.png\" \/><\/p>\n<p><strong>Answer:<\/strong><\/p>\n<p>(i) Here, AB = 18 cm, BC = 10 cm, AF = 6 cm, AE = 10 cm and BE = 8 cm<\/p>\n<p>Area of shaded portion = Area of rectangle ABCD \u2013 (Area of {tex}\\Delta {\/tex}FAE + area of {tex}\\Delta {\/tex}EBC)<\/p>\n<p>= (AB x BC) \u2013 ({tex}\\frac{1}{2}{\/tex} x AE x AF + {tex}\\frac{1}{2}{\/tex} x BE x BC)<\/p>\n<p>= (18 x 10) \u2013 ({tex}\\frac{1}{2}{\/tex} x 10 x 6 + {tex}\\frac{1}{2}{\/tex} x 8 x 10)<\/p>\n<p>= 180 \u2013 (30 + 40)<\/p>\n<p>= 180 \u2013 70 = 110 cm2<\/p>\n<p>(ii) Here, SR = SU + UR = 10 + 10 = 20 cm, QR = 20 cm<\/p>\n<p>PQ = SR = 20 cm, PT = PS \u2013 TS = 20 \u2013 10 cm<\/p>\n<p>TS = 10 cm, SU = 10 cm, QR = 20 cm and UR = 10 cm<\/p>\n<p>Area of shaded region<\/p>\n<p>= Area of square PQRS \u2013 Area of {tex}\\Delta {\/tex}QPT \u2013 Area of {tex}\\Delta {\/tex}TSU \u2013 Area of {tex}\\Delta {\/tex}UQR<\/p>\n<p>= (SR x QR) &#8211; {tex}\\frac{1}{2}{\/tex} x PQ x PT \u2013 {tex}\\frac{1}{2}{\/tex} x ST x SU \u2013 {tex}\\frac{1}{2}{\/tex}<\/p>\n<p>= 20 x 20 \u2013 {tex}\\frac{1}{2}{\/tex} x 20 x 10 \u2013 {tex}\\frac{1}{2}{\/tex} x 10 x 10 \u2013 {tex}\\frac{1}{2}{\/tex} x 20 x 10<\/p>\n<p>= 400 \u2013 100 \u2013 50 \u2013 100 = 150 cm2<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 11.4<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_11Find_the_area_of_the_equilateral_ABCD_Here_AC_22_cm_BM_3_cm_DN_3_cm_and_BM_tex_bot_tex_AC_DN_tex_bot_tex_AC\"><\/span><strong>Question 11.<\/strong>Find the area of the equilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3 cm and BM {tex} \\bot {\/tex} AC, DN {tex} \\bot {\/tex} AC.<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch11_15.png\" \/><\/p>\n<p><strong>Answer:<\/strong><\/p>\n<p>Here, AC = 22 cm, BM = 3 cm, DN = 3 cm<\/p>\n<p>Area of quadrilateral ABCDF = Area of {tex}\\Delta {\/tex}ABC + Area of {tex}\\Delta {\/tex}ADC<\/p>\n<p>= {tex}\\frac{1}{2}{\/tex} x AC x BM + {tex}\\frac{1}{2}{\/tex} x AC x DN<\/p>\n<p>= {tex}\\frac{1}{2}{\/tex} x 22 x 3 + {tex}\\frac{1}{2}{\/tex} x 22 x 3<\/p>\n<p>= 3 x 11 + 3 x 11<\/p>\n<p>= 33 + 33<\/p>\n<p>= 66 {tex}c{m^2}{\/tex}<\/p>\n<p>Thus, the area of quadrilateral ABCD is{tex}c{m^2}{\/tex}.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_7_Maths_Exercise_114\"><\/span>NCERT Solutions for Class 7 Maths Exercise 11.4<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>NCERT Solutions Class 7 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 7 Maths includes text book solutions from Class 7 Maths Book . NCERT Solutions for CBSE Class 7 Maths have total 15 chapters. 7 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 7 solutions PDF and Maths ncert class 7 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_Students\"><\/span>CBSE app for Students<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>To download NCERT Solutions for Class 7 Maths, Social Science Computer Science, Home Science, Hindi English, Maths 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 7 Maths Exercise 11.4 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. 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