{"id":4971,"date":"2016-05-19T11:49:00","date_gmt":"2016-05-19T06:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-7-3\/"},"modified":"2018-06-04T13:58:06","modified_gmt":"2018-06-04T08:28:06","slug":"ncert-solutions-class-10-maths-exercise-7-3","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-7-3\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Exercise 7.3"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_76 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-toggle-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-7-3\/#NCERT_Solutions_for_Class_10_Maths_Coordinate_Geometry\" >NCERT Solutions for Class 10 Maths\u00a0Coordinate Geometry<\/a><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-7-3\/#1_Find_the_area_of_the_triangle_whose_vertices_are\" >1. Find the area of the triangle whose vertices are:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-7-3\/#2_In_each_of_the_following_find_the_value_of_%E2%80%98k_for_which_the_points_are_collinear\" >2. In each of the following find the value of &#8216;k&#8217;, for which the points are collinear.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-7-3\/#3_Find_the_area_of_the_triangle_formed_by_joining_the_mid%E2%80%93points_of_the_sides_of_the_triangle_whose_vertices_are_0_%E2%80%931_2_1_and_0_3_Find_the_ratio_of_this_area_to_the_area_of_the_given_triangle\" >3. Find the area of the triangle formed by joining the mid\u2013points of the sides of the triangle whose vertices are (0, \u20131), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-7-3\/#4_Find_the_area_of_the_quadrilateral_whose_vertices_taken_in_order_are_%E2%80%934_%E2%80%932_%E2%80%933_%E2%80%935_3_%E2%80%932_and_2_3\" >4. Find the area of the quadrilateral whose vertices taken in order are (\u20134, \u20132), (\u20133, \u20135), (3, \u20132) and (2, 3).<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-7-3\/#5_We_know_that_median_of_a_triangle_divides_it_into_two_triangles_of_equal_areas_Verify_this_result_for_%E2%96%B3ABC_whose_vertices_are_A_4_%E2%80%936_B_3_%E2%80%932_and_C_5_2\" >5. We know that median of a triangle divides it into two triangles of equal areas. Verify this result for \u25b3ABC whose vertices are A (4, \u20136), B (3, \u20132) and C (5, 2).<\/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-7\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-7-3\/#NCERT_Solutions_for_Class_10_Maths_Exercise_73\" >NCERT Solutions for Class 10 Maths Exercise 7.3<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-7-3\/#CBSE_app_for_Class_10\" >CBSE app for Class 10<\/a><\/li><\/ul><\/nav><\/div>\n<p>NCERT Solutions for Class 10 Maths Exercise 7.3 Class 10 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 10 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.<\/p>\n<p style=\"text-align: center;\"><strong>NCERT solutions for Maths\u00a0Coordinate Geometry\u00a0<\/strong><strong><a class=\"button\" href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-10-mathematics-coordinate-geometry\/1209\/ncert-solutions\/5\/\">Download as PDF<\/a><\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/10%20Maths%20Book.jpg\" alt=\"NCERT Solutions for Class 10 Maths Exercise 7.3\" width=\"177\" height=\"218\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_10_Maths_Coordinate_Geometry\"><\/span>NCERT Solutions for Class 10 Maths\u00a0Coordinate Geometry<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"1_Find_the_area_of_the_triangle_whose_vertices_are\"><\/span><strong>1. Find the area of the triangle whose vertices are:<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(i) (2, 3), (\u20131, 0), (2, \u20134)<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(ii) (\u20135, \u20131), (3, \u20135), (5, 2)<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. (i)<\/strong> (2, 3), (\u20131, 0), (2, \u20134)<\/p>\n<p style=\"text-align: justify;\">Area of Triangle = <img decoding=\"async\" style=\"height: 42px; width: 263px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image001.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/>[2 {0 \u2212 (\u22124)} \u2013 1 (\u22124 \u2212 3) + 2 (3 \u2212 0)]\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/>[2 (0 + 4) \u2013 1 (\u22127) + 2 (3)]\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/>(8 + 7 + 6) = <img decoding=\"async\" style=\"height: 42px; width: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image003.png\" \/>sq. units<\/p>\n<p style=\"text-align: justify;\"><strong>(ii)<\/strong> (\u20135, \u20131), (3, \u20135), (5, 2)<\/p>\n<p style=\"text-align: justify;\">Area of Triangle = <img decoding=\"async\" style=\"height: 42px; width: 263px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image001.png\" \/><\/p>\n<p style=\"text-align: justify;\">=<img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> [\u22125 (\u22125 \u2212 2) + 3 {2 \u2212 (\u22121)} + 5 {\u22121 \u2212 (\u22125)}]\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> [\u22125 (\u22127) + 3 (3) + 5 (4)]\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> (35 + 9 + 20)<\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> (64) = 32 sq. units<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 7.3<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"2_In_each_of_the_following_find_the_value_of_%E2%80%98k_for_which_the_points_are_collinear\"><\/span><strong>2. In each of the following find the value of &#8216;k&#8217;, for which the points are collinear.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(i) (7, \u20132), (5, 1), (3, k)<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(ii) (8, 1), (k, \u20134), (2, \u20135)<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. (i)<\/strong> (7, \u20132), (5, 1), (3, k)<\/p>\n<p style=\"text-align: justify;\">Since, the given points are collinear, it means the area of triangle formed by them is equal to zero.<\/p>\n<p style=\"text-align: justify;\">Area of Triangle = <img decoding=\"async\" style=\"height: 42px; width: 301px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image004.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> [7 (1 \u2212 k) + 5 {k \u2212 (\u22122)} + 3 (\u22122 \u2212 1)]\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> (7 \u2212 7k + 5k + 10 \u2212 9) = 0<\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> (7 \u2212 7k + 5k + 1) = 0<\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> (8 \u2212 2k) = 0<\/p>\n<p style=\"text-align: justify;\">\u21d2 8 \u2212 2k = 0<\/p>\n<p style=\"text-align: justify;\">\u21d2 2k = 8<\/p>\n<p style=\"text-align: justify;\">\u21d2 k = 4<\/p>\n<p style=\"text-align: justify;\"><strong>(ii)<\/strong> (8, 1), (k, \u20134), (2, \u20135)<\/p>\n<p style=\"text-align: justify;\">Since, the given points are collinear, it means the area of triangle formed by them is equal to zero.<\/p>\n<p style=\"text-align: justify;\">Area of Triangle = <img decoding=\"async\" style=\"height: 42px; width: 292px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image005.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> [8 {\u22124 \u2212 (\u22125)} + k (\u22125 \u2212 1) + 2 {1 \u2212 (\u22124)}]\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> (8 \u2212 6k + 10) = 0<\/p>\n<p style=\"text-align: justify;\">\u21d2 <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/>(18 \u2212 6k) = 0<\/p>\n<p style=\"text-align: justify;\">\u21d218 \u2212 6k = 0<\/p>\n<p style=\"text-align: justify;\">\u21d2 18 = 6k<\/p>\n<p style=\"text-align: justify;\">\u21d2 k = 3<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 7.3<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"3_Find_the_area_of_the_triangle_formed_by_joining_the_mid%E2%80%93points_of_the_sides_of_the_triangle_whose_vertices_are_0_%E2%80%931_2_1_and_0_3_Find_the_ratio_of_this_area_to_the_area_of_the_given_triangle\"><\/span><strong>3. Find the area of the triangle formed by joining the mid\u2013points of the sides of the triangle whose vertices are (0, \u20131), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Let A = (0, \u20131) =<img decoding=\"async\" style=\"height: 26px; width: 52px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image006.png\" \/>, B = (2, 1) = <img decoding=\"async\" style=\"height: 26px; width: 54px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image007.png\" \/>and<\/p>\n<p style=\"text-align: justify;\">C = (0, 3) = <img decoding=\"async\" style=\"height: 26px; width: 53px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image008.png\" \/><\/p>\n<p style=\"text-align: justify;\">Area of \u25b3ABC = <img decoding=\"async\" style=\"height: 42px; width: 263px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image001.png\" \/><\/p>\n<p style=\"text-align: justify;\">\u21d2 Area of \u25b3ABC<\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> [0 (1 \u2212 3) + 2 {3 \u2212 (\u22121)} + 0 (\u22121 \u2212 1)] = <img decoding=\"async\" style=\"height: 42px; width: 34px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image009.png\" \/><\/p>\n<p style=\"text-align: justify;\">= 4 sq. units<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 13\" style=\"height: 145px; width: 192px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image010.jpg\" alt=\"\" \/><\/p>\n<p style=\"text-align: justify;\">P, Q and R are the mid\u2013points of sides AB, AC and BC respectively.<\/p>\n<p style=\"text-align: justify;\">Applying Section Formula to find the vertices of P, Q and R, we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 141px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image011.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 155px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image012.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 147px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image013.png\" \/><\/p>\n<p style=\"text-align: justify;\">Applying same formula, Area of \u25b3PQR = <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> [1 (1 \u2212 2) + 0 (2 \u2212 0) + 1 (0 \u2212 1)] = <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> <img decoding=\"async\" style=\"height: 27px; width: 27px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image014.png\" \/><\/p>\n<p style=\"text-align: justify;\">= 1 sq. units (numerically)<\/p>\n<p style=\"text-align: justify;\">Now, <img decoding=\"async\" style=\"height: 45px; width: 169px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image015.png\" \/><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 7.3<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"4_Find_the_area_of_the_quadrilateral_whose_vertices_taken_in_order_are_%E2%80%934_%E2%80%932_%E2%80%933_%E2%80%935_3_%E2%80%932_and_2_3\"><\/span><strong>4. Find the area of the quadrilateral whose vertices taken in order are (\u20134, \u20132), (\u20133, \u20135), (3, \u20132) and (2, 3).<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Area of Quadrilateral ABCD<\/p>\n<p style=\"text-align: justify;\">= Area of Triangle ABD +<\/p>\n<p style=\"text-align: justify;\">Area of Triangle BCD \u2026 (1)<\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" id=\"Picture 14\" style=\"height: 154px; width: 260px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image016.jpg\" alt=\"\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\">Using formula to find area of triangle:<\/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\/10\/mathematics\/ch07\/Ex7.3\/image017.png\" \/>ABD<\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 42px; width: 263px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image001.png\" \/><\/p>\n<p style=\"text-align: justify;\">=<img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> [\u22124 (\u22125 \u2212 3) \u2013 3 {3 \u2212 (\u22122)} + 2 {\u22122 \u2212 (\u22125)}]\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> (32 \u2013 15 + 6)<\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> (23) = 11.5 sq units \u2026 (2)<\/p>\n<p style=\"text-align: justify;\">Again using formula to find area of triangle:<\/p>\n<p style=\"text-align: justify;\">Area of \u25b3BCD = <img decoding=\"async\" style=\"height: 42px; width: 263px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image001.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> [\u22123 (\u22122 \u2212 3) + 3 {3 \u2212 (\u22125)} + 2 {\u22125 \u2212 (\u22122)}]\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> (15 + 24 \u2212 6)<\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> (33) = 16.5 sq units \u2026 (3)<\/p>\n<p style=\"text-align: justify;\">Putting (2) and (3) in (1), we get<\/p>\n<p style=\"text-align: justify;\">Area of Quadrilateral ABCD = 11.5 + 16.5 = 28 sq units<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 7.3<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"5_We_know_that_median_of_a_triangle_divides_it_into_two_triangles_of_equal_areas_Verify_this_result_for_%E2%96%B3ABC_whose_vertices_are_A_4_%E2%80%936_B_3_%E2%80%932_and_C_5_2\"><\/span><strong>5. We know that median of a triangle divides it into two triangles of equal areas. Verify this result for \u25b3ABC whose vertices are A (4, \u20136), B (3, \u20132) and C (5, 2).<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>We have \u25b3ABC whose vertices are given.<\/p>\n<p style=\"text-align: justify;\">We need to show that ar(\u25b3ABD) = ar(\u25b3ACD).<\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" id=\"Picture 15\" style=\"height: 137px; width: 231px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image018.jpg\" alt=\"\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\">Let coordinates of point D are (x, y)<\/p>\n<p style=\"text-align: justify;\">Using section formula to find coordinates of D, we get<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 111px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image019.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 131px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image020.png\" \/><\/p>\n<p style=\"text-align: justify;\">Therefore, coordinates of point D are (4, 0)<\/p>\n<p style=\"text-align: justify;\">Using formula to find area of triangle:<\/p>\n<p style=\"text-align: justify;\">Area of \u25b3ABD = <img decoding=\"async\" style=\"height: 42px; width: 263px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image001.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> [4 (\u22122 \u2212 0) + 3 {0 \u2212 (\u22126)} + 4 {\u22126 \u2212 (\u22122)}]\n<p style=\"text-align: justify;\">=<img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> (\u22128 + 18 \u221216)<\/p>\n<p style=\"text-align: justify;\">=<img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> (\u22126) = \u22123 sq units<\/p>\n<p style=\"text-align: justify;\">Area cannot be in negative.<\/p>\n<p style=\"text-align: justify;\">Therefore, we just consider its numerical value.<\/p>\n<p style=\"text-align: justify;\">Therefore, area of \u25b3ABD = 3 sq units \u2026 (1)<\/p>\n<p style=\"text-align: justify;\">Again using formula to find area of triangle:<\/p>\n<p style=\"text-align: justify;\">Area of \u25b3ACD = <img decoding=\"async\" style=\"height: 42px; width: 263px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image001.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> [4 (2 \u2212 0) + 5 {0 \u2212 (\u22126)} + 4 {\u22126 \u22122 )}]\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 42px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch07\/Ex7.3\/image002.png\" \/> (8 + 30 \u2212 32) = \u00bd (6) = 3 sq units \u2026 (2)<\/p>\n<p style=\"text-align: justify;\">From (1) and (2), we get ar(\u25b3ABD) = ar(\u25b3ACD)<\/p>\n<p style=\"text-align: justify;\">Hence Proved.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_10_Maths_Exercise_73\"><\/span>NCERT Solutions for Class 10 Maths Exercise 7.3<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>NCERT Solutions Class 10 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 10 Maths includes text book solutions from Mathematics Book. NCERT Solutions for CBSE Class 10 Maths have total 15 chapters. 10 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 10 solutions PDF and Maths ncert class 10 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_app_for_Class_10\"><\/span>CBSE app for Class 10<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>To download NCERT Solutions for Class 10 Maths, Computer Science, Home Science,Hindi ,English, Social Science do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through\u00a0<strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide&amp;referrer=utm_source%3Dmycbse_bottom%26utm_medium%3Dtext%26utm_campaign%3Dmycbseads\">the best app for CBSE\u00a0<\/a><\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions for Class 10 Maths Exercise 7.3 Class 10 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. 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