{"id":4323,"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-11-2\/"},"modified":"2018-07-07T16:12:03","modified_gmt":"2018-07-07T10:42:03","slug":"ncert-solutions-for-class-7-maths-exercise-11-2","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-7-maths-exercise-11-2\/","title":{"rendered":"NCERT Solutions for Class 7 Maths Exercise 11.2"},"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-7-maths-exercise-11-2\/#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-11-2\/#Class_%E2%80%93VII_Mathematics_Ex_112\" >Class \u2013VII Mathematics (Ex. 11.2)<\/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-11-2\/#Question_1Find_the_area_of_each_of_the_following_parallelograms\" >Question 1.Find the area of each of the following parallelograms:<\/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-11-2\/#Question_2Find_the_area_of_each_of_the_following_triangles\" >Question 2.Find the area of each of the following triangles:<\/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-11-2\/#Question_3Find_the_missing_values\" >Question 3.Find the missing values:<\/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-11-2\/#Question_4Find_the_missing_values\" >Question 4.Find the missing values:<\/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-11-2\/#Question_5PQRS_is_a_parallelogram_QM_is_the_height_from_Q_to_SR_and_QN_is_the_height_from_Q_to_PS_If_SR_12_cm_and_QM_76_cm_Find\" >Question 5.PQRS is a parallelogram. QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. 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-11-2\/#Question_7texDelta_texABC_is_right_angled_at_A_AD_is_perpendicular_to_BC_If_AB_5_cm_BC_13_cm_and_AC_12_cm_find_the_area_of_texDelta_texABC_Also_find_the_length_of_AD\" >Question 7.{tex}\\Delta {\/tex}ABC is right angled at A. AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, find the area of {tex}\\Delta {\/tex}ABC. Also, find the length of AD.<\/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-11-2\/#Question_8texDelta_texABC_is_isosceles_with_AB_AC_75_cm_and_BC_9_cm_The_height_AD_from_A_to_BC_is_6_cm_Find_the_area_of_texDelta_texABC_What_will_be_the_height_from_C_to_AB_ie_CE\" >Question 8.{tex}\\Delta {\/tex}ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm. The height AD from A to BC, is 6 cm. Find the area of {tex}\\Delta {\/tex}ABC. What will be the height from C to AB i.e., CE?<\/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-10\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-7-maths-exercise-11-2\/#NCERT_Solutions_for_Class_7_Maths_Exercise_112\" >NCERT Solutions for Class 7 Maths Exercise 11.2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-7-maths-exercise-11-2\/#CBSE_app_for_Students\" >CBSE app for Students<\/a><\/li><\/ul><\/nav><\/div>\n<p>NCERT Solutions for Class 7 Maths Exercise 11.2 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 11.2\" width=\"158\" height=\"200\" \/><\/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_112\"><\/span><strong>Class \u2013VII Mathematics (Ex. 11.2)<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<h6><span class=\"ez-toc-section\" id=\"Question_1Find_the_area_of_each_of_the_following_parallelograms\"><\/span><strong>Question 1.<\/strong>Find the area of each of the following parallelograms:<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>We know that the area of parallelogram = base x height<\/p>\n<p>(a) Here base = 7 cm and height = 4 cm<\/p>\n<p>{tex}\\therefore {\/tex} Area of parallelogram = 7 x 4 = 28{tex}c{m^2}{\/tex}<\/p>\n<p>(b) Here base = 5 cm and height = 3 cm<\/p>\n<p>{tex}\\therefore {\/tex} Area of parallelogram = 5 x 3 = 15 {tex}c{m^2}{\/tex}<\/p>\n<p>(c) Here base = 2.5 cm and height = 3.5 cm<\/p>\n<p>{tex}\\therefore {\/tex} Area of parallelogram = 2.5 x 3.5 = 8.75 {tex}c{m^2}{\/tex}<\/p>\n<p>(d) Here base = 5 cm and height = 4.8 cm<\/p>\n<p>{tex}\\therefore {\/tex} Area of parallelogram = 5 x 4.8 = 24 {tex}c{m^2}{\/tex}<\/p>\n<p>(e) Here base = 2 cm and height = 4.4 cm<\/p>\n<p>{tex}\\therefore {\/tex} Area of parallelogram = 2 x 4.4 = 8.8 {tex}c{m^2}{\/tex}<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 11.2<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_2Find_the_area_of_each_of_the_following_triangles\"><\/span><strong>Question 2.<\/strong>Find the area of each of the following triangles:<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_03.png\" \/><\/p>\n<p><strong>Answer:<\/strong><\/p>\n<p>We know that the area of triangle = {tex}\\frac{1}{2}{\/tex} x base x height<\/p>\n<p>(a) Here, base = 4 cm and height = 3 cm<\/p>\n<p>{tex}\\therefore {\/tex} Area of triangle = {tex}\\frac{1}{2}{\/tex} x 4 x 3 = 6 {tex}c{m^2}{\/tex}<\/p>\n<p>(b) Here, base = 5 cm and height = 3.2 cm<\/p>\n<p>{tex}\\therefore {\/tex} Area of triangle = {tex}\\frac{1}{2}{\/tex} x 5 x 3.2 = 8 {tex}c{m^2}{\/tex}<\/p>\n<p>(c) Here, base = 3 cm and height = 4 cm<\/p>\n<p>{tex}\\therefore {\/tex} Area of triangle = {tex}\\frac{1}{2}{\/tex} x 3 x 4 = 6{tex}c{m^2}{\/tex}<\/p>\n<p>(d) Here, base = 3 cm and height = 2 cm<\/p>\n<p>{tex}\\therefore {\/tex} Area of triangle = {tex}\\frac{1}{2}{\/tex} x 3 x 2 = 3{tex}c{m^2}{\/tex}<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 11.2<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_3Find_the_missing_values\"><\/span><strong>Question 3.<\/strong>Find the missing values:<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<table class=\"responsive\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td><strong>S. No.<\/strong><\/td>\n<td><strong>Base<\/strong><\/td>\n<td><strong>Hieght<\/strong><\/td>\n<td><strong>Area of the parallelogram<\/strong><\/td>\n<\/tr>\n<tr>\n<td>a.<\/td>\n<td>20 cm<\/td>\n<td><\/td>\n<td>246 {tex}c{m^2}{\/tex}<\/td>\n<\/tr>\n<tr>\n<td>b.<\/td>\n<td><\/td>\n<td>15 cm<\/td>\n<td>154.5{tex}c{m^2}{\/tex}<\/td>\n<\/tr>\n<tr>\n<td>c.<\/td>\n<td><\/td>\n<td>84 cm<\/td>\n<td>48.72 {tex}c{m^2}{\/tex}<\/td>\n<\/tr>\n<tr>\n<td>d.<\/td>\n<td>15.6 cm<\/td>\n<td><\/td>\n<td>16.38{tex}{m^2}{\/tex}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Answer:<\/strong><\/p>\n<p>We know that the area of parallelogram = base x height<\/p>\n<p>(a) Here, base = 20 cm and area = 246 {tex}c{m^2}{\/tex}<\/p>\n<p>{tex}\\therefore {\/tex} Area of parallelogram = base x height<\/p>\n<p>{tex} \\Rightarrow {\/tex} 246 = 20 x height {tex} \\Rightarrow {\/tex} height = {tex}\\frac{{246}}{{20}}{\/tex} = 12.3 cm<\/p>\n<p>(b) Here, height = 15 cm and area = 154.5 cm2<\/p>\n<p>{tex}\\therefore {\/tex} Area of parallelogram = base x height<\/p>\n<p>{tex} \\Rightarrow {\/tex} 154.5 = base x 15 {tex} \\Rightarrow {\/tex} base = {tex}\\frac{{154.5}}{{15}}{\/tex} = 10.3 cm<\/p>\n<p>(c) Here, height = 8.4 cm and area = 48.72{tex}c{m^2}{\/tex}<\/p>\n<p>{tex}\\therefore {\/tex} Area of parallelogram = base x height<\/p>\n<p>{tex} \\Rightarrow {\/tex} 48.72 = base x 8.4 {tex} \\Rightarrow {\/tex} base = {tex}\\frac{{48.72}}{{8.4}}{\/tex} = 5.8 cm<\/p>\n<p>(d) Here, base = 15.6 cm and area = 16.38 {tex}c{m^2}{\/tex}<\/p>\n<p>{tex}\\therefore {\/tex} Area of parallelogram = base x height<\/p>\n<p>{tex} \\Rightarrow {\/tex} 16.38 = 15.6 x height {tex} \\Rightarrow {\/tex} height = {tex}\\frac{{16.38}}{{15.6}}{\/tex} = 1.05 cm<\/p>\n<p>Thus, the missing values are:<\/p>\n<table class=\"responsive\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td><strong>S. No.<\/strong><\/td>\n<td><strong>Base<\/strong><\/td>\n<td><strong>Hieght<\/strong><\/td>\n<td><strong>Area of the parallelogram<\/strong><\/td>\n<\/tr>\n<tr>\n<td>a.<\/td>\n<td>20 cm<\/td>\n<td><strong>12.3 cm<\/strong><\/td>\n<td>246 {tex}c{m^2}{\/tex}<\/td>\n<\/tr>\n<tr>\n<td>b.<\/td>\n<td><strong>10.3 cm<\/strong><\/td>\n<td>15 cm<\/td>\n<td>154.5{tex}c{m^2}{\/tex}<\/td>\n<\/tr>\n<tr>\n<td>c.<\/td>\n<td><strong>5.8 cm<\/strong><\/td>\n<td>84 cm<\/td>\n<td>48.72{tex}c{m^2}{\/tex}<\/td>\n<\/tr>\n<tr>\n<td>d.<\/td>\n<td>15.6 cm<\/td>\n<td>1.05<\/td>\n<td>16.38 {tex}c{m^2}{\/tex}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 11.2<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_4Find_the_missing_values\"><\/span><strong>Question 4.<\/strong>Find the missing values:<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<table class=\"responsive\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td><strong>Base<\/strong><\/td>\n<td><strong>Hieght<\/strong><\/td>\n<td><strong>Area of triangle<\/strong><\/td>\n<\/tr>\n<tr>\n<td>15 cm<\/td>\n<td>&#8212;<\/td>\n<td>87 {tex}c{m^2}{\/tex}<\/td>\n<\/tr>\n<tr>\n<td>&#8212;<\/td>\n<td>31.4 mm<\/td>\n<td>1256 {tex}m{m^2}{\/tex}<\/td>\n<\/tr>\n<tr>\n<td>22 cm<\/td>\n<td>&#8212;<\/td>\n<td>170.5{tex}c{m^2}{\/tex}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Answer:<\/strong><\/p>\n<p>We know that the area of triangle = {tex}\\frac{1}{2}{\/tex} x base x height<\/p>\n<p>In first row, base = 15 cm and area = 87 {tex}c{m^2}{\/tex}<\/p>\n<p>{tex}\\therefore {\/tex} 87 = {tex}\\frac{1}{2}{\/tex} x 15 x height {tex} \\Rightarrow {\/tex} height = {tex}\\frac{{87 \\times 2}}{{15}}{\/tex} 11.6 cm<\/p>\n<p>In second row, height = 31.4 mm and area = 1256 {tex}m{m^2}{\/tex}<\/p>\n<p>{tex}\\therefore {\/tex} 1256 = {tex}\\frac{1}{2}{\/tex} x base x 31.4 {tex} \\Rightarrow {\/tex} base = {tex}\\frac{{1256 \\times 2}}{{31.4}}{\/tex} 80 mm<\/p>\n<p>In third row, base = 22 cm and area = 170.5 {tex}c{m^2}{\/tex}<\/p>\n<p>{tex}\\therefore {\/tex} 170.5 = {tex}\\frac{1}{2}{\/tex} x 22 x height {tex} \\Rightarrow {\/tex} height = {tex}\\frac{{170.5 \\times 2}}{{22}}{\/tex} 15.5 cm<\/p>\n<p>Thus, the missing values are:<\/p>\n<table class=\"responsive\" cellspacing=\"0\" cellpadding=\"0\">\n<tbody>\n<tr>\n<td><strong>Base<\/strong><\/td>\n<td><strong>Hieght<\/strong><\/td>\n<td><strong>Area of triangle<\/strong><\/td>\n<\/tr>\n<tr>\n<td>15 cm<\/td>\n<td><strong>11.6 cm<\/strong><\/td>\n<td>87 {tex}c{m^2}{\/tex}<\/td>\n<\/tr>\n<tr>\n<td><strong>80 mm<\/strong><\/td>\n<td>31.4 mm<\/td>\n<td>1256 {tex}m{m^2}{\/tex}<\/td>\n<\/tr>\n<tr>\n<td>22 cm<\/td>\n<td><strong>15.5 cm<\/strong><\/td>\n<td>170.5 {tex}c{m^2}{\/tex}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 11.2<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_5PQRS_is_a_parallelogram_QM_is_the_height_from_Q_to_SR_and_QN_is_the_height_from_Q_to_PS_If_SR_12_cm_and_QM_76_cm_Find\"><\/span><strong>Question 5.<\/strong>PQRS is a parallelogram. QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find:<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p>(a) the area of the parallelogram PRS<\/p>\n<p>(b) QN, if PS = 8 cm<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch11_04.png\" \/><\/p>\n<p><strong>Answer:<\/strong><\/p>\n<p>Given: SR = 12 cm, QM= 7.6 cm, PS = 8 cm.<\/p>\n<p>(a) Area of parallelogram = base x height<\/p>\n<p>= 12 x 7.6 = 91.2 {tex}c{m^2}{\/tex}<\/p>\n<p>(b) Area of parallelogram = base x height<\/p>\n<p>{tex} \\Rightarrow {\/tex} 91.2 = 8 x QN {tex} \\Rightarrow {\/tex} QN = {tex}\\frac{{91.2}}{8}{\/tex} = 11.4 cm<\/p>\n<p><strong>Question 6.<\/strong>DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD. If the area of the parallelogram is 1470{tex}c{m^2}{\/tex}, AB = 35 cm and AD = 49 cm, find the length of BM and DL.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch11_05.jpg\" \/><\/p>\n<p><strong>Answer:<\/strong><\/p>\n<p>Given: Area of parallelogram = 1470{tex}c{m^2}{\/tex}<\/p>\n<p>Base (AB) = 35 cm and base (AD) = 49 cm<\/p>\n<p>Since Area of parallelogram = base x height<\/p>\n<p>{tex} \\Rightarrow {\/tex} 1470 = 35 x DL {tex} \\Rightarrow {\/tex} DL = {tex}\\frac{{1470}}{{35}}{\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} DL = 42 cm<\/p>\n<p>Again, Area of parallelogram = base x height<\/p>\n<p>{tex} \\Rightarrow {\/tex} 1470 = 49 x BM {tex} \\Rightarrow {\/tex} BM = {tex}\\frac{{1470}}{{49}}{\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} BM = 30 cm<\/p>\n<p>Thus, the lengths of DL and BM are 42 cm and 30 cm respectively.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 11.2<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_7texDelta_texABC_is_right_angled_at_A_AD_is_perpendicular_to_BC_If_AB_5_cm_BC_13_cm_and_AC_12_cm_find_the_area_of_texDelta_texABC_Also_find_the_length_of_AD\"><\/span><strong>Question 7.<\/strong>{tex}\\Delta {\/tex}ABC is right angled at A. AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, find the area of {tex}\\Delta {\/tex}ABC. Also, find the length of AD.<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_06.jpg\" \/><\/p>\n<p><strong>Answer:<\/strong><\/p>\n<p>In right angles triangle BAC, AB = 5 cm and AC = 12 cm<\/p>\n<p>Area of triangle = {tex}\\frac{1}{2}{\/tex} x base x height = {tex}\\frac{1}{2}{\/tex} x AB x AC<\/p>\n<p>= {tex}\\frac{1}{2}{\/tex} x 5 x 12 = 30{tex}c{m^2}{\/tex}<\/p>\n<p>Now, in {tex}\\Delta {\/tex}ABC,<\/p>\n<p>Area of triangle ABC = {tex}\\frac{1}{2}{\/tex} x BC x AD<\/p>\n<p>{tex} \\Rightarrow {\/tex} 30 = {tex}\\frac{1}{2}{\/tex} x 13 x AD {tex} \\Rightarrow {\/tex} AD = {tex}\\frac{{30 \\times 2}}{{13}}{\/tex} = {tex}\\frac{{60}}{{13}}{\/tex} cm<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 11.2<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_8texDelta_texABC_is_isosceles_with_AB_AC_75_cm_and_BC_9_cm_The_height_AD_from_A_to_BC_is_6_cm_Find_the_area_of_texDelta_texABC_What_will_be_the_height_from_C_to_AB_ie_CE\"><\/span><strong>Question 8.<\/strong>{tex}\\Delta {\/tex}ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm. The height AD from A to BC, is 6 cm. Find the area of {tex}\\Delta {\/tex}ABC. What will be the height from C to AB i.e., CE?<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_07.jpg\" \/><\/p>\n<p><strong>Answer:<\/strong><\/p>\n<p>In {tex}\\Delta {\/tex}ABC, AD = 6 cm and BC = 9 cm<\/p>\n<p>Area of triangle = {tex}\\frac{1}{2}{\/tex} x base x height = {tex}\\frac{1}{2}{\/tex} x BC x AD<\/p>\n<p>= {tex}\\frac{1}{2}{\/tex} x 9 x 6 = 27{tex}c{m^2}{\/tex}<\/p>\n<p>Again, Area of triangle = {tex}\\frac{1}{2}{\/tex} x base x height = {tex}\\frac{1}{2}{\/tex} x AB x CE<\/p>\n<p>{tex} \\Rightarrow {\/tex} 27 = {tex}\\frac{1}{2}{\/tex} x 7.5 x CE {tex} \\Rightarrow {\/tex} CE = {tex}\\frac{{27 \\times 2}}{{7.5}}{\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} CE = 7.2 cm<\/p>\n<p>Thus, height from C to AB i.e., CE is 7.2 cm.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_7_Maths_Exercise_112\"><\/span>NCERT Solutions for Class 7 Maths Exercise 11.2<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 11.2 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 &#8230; <a title=\"NCERT Solutions for Class 7 Maths Exercise 11.2\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-7-maths-exercise-11-2\/\" aria-label=\"More on NCERT Solutions for Class 7 Maths Exercise 11.2\">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":[1388,281],"tags":[283,1535,321,1485,216],"class_list":["post-4323","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematics-cbse-class-07","category-ncert-solutions","tag-cbse-study-material","tag-class-7","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 7 Maths Exercise 11.2 | myCBSEguide<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for Class 7 Maths Exercise 11.2 in PDF format for free download. 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