{"id":5031,"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-7-5\/"},"modified":"2018-06-18T14:22:05","modified_gmt":"2018-06-18T08:52:05","slug":"ncert-solutions-for-class-9-maths-exercise-7-5","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-7-5\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Exercise 7.5"},"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-7-5\/#NCERT_Solutions_for_Class_9_Mathematics_Triangles\" >NCERT Solutions for Class 9 Mathematics\u00a0Triangles<\/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-7-5\/#1_ABC_is_a_triangle_Locate_a_point_in_the_interior_of_ABC_which_is_equidistant_from_all_the_vertices_of_ABC\" >1. ABC is a triangle. Locate a point in the interior of ABC which is equidistant from all the vertices of ABC.<\/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-7-5\/#2_In_a_triangle_locate_a_point_in_its_interior_which_is_equidistant_from_all_the_sides_of_the_triangle\" >2. In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.<\/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-7-5\/#In_BIK_and_BIJ\" >In BIK and BIJ,<\/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-7-5\/#3_In_a_huge_park_people_are_concentrated_at_three_points_See_figure\" >3. In a huge park, people are concentrated at three points (See figure).<\/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-7-5\/#Now_point_O_is_equidistant_from_points_A_B_and_C\" >Now point O is equidistant from points A, B and C.<\/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-7-5\/#4_Complete_the_hexagonal_rangoli_and_the_star_rangolies_See_figure_but_filling_them_with_as_many_equilateral_triangles_of_side_1_cm_as_you_can_Count_the_number_of_triangles_in_each_case_Which_has_more_triangles\" >4. Complete the hexagonal rangoli and the star rangolies (See figure) but filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?<\/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-7-5\/#Area_of_hexagonal_rangoli_6_x_Area_of_an_equilateral_triangle\" >Area of hexagonal rangoli = 6 x Area of an equilateral triangle<\/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-7-5\/#Therefore_total_area_of_star_rangoli_12_Area_of_an_equilateral_triangle_of_side_5_cm\" >Therefore, total area of star rangoli = 12  Area of an equilateral triangle of side 5 cm<\/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-9-maths-exercise-7-5\/#NCERT_Solutions_for_Class_9_Maths_Exercise_75\" >NCERT Solutions for Class 9 Maths Exercise 7.5<\/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-9-maths-exercise-7-5\/#CBSE_app_for_Class_9\" >CBSE app for Class 9<\/a><\/li><\/ul><\/nav><\/div>\n<p>NCERT Solutions for Class 9 Maths Exercise 7.5 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\u00a0<\/strong><strong>Triangles\u00a0<\/strong><strong><a class=\"button\" href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-09-mathematics-triangles\/1241\/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 7.5\" width=\"187\" height=\"241\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_9_Mathematics_Triangles\"><\/span>NCERT Solutions for Class 9 Mathematics\u00a0Triangles<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"1_ABC_is_a_triangle_Locate_a_point_in_the_interior_of_ABC_which_is_equidistant_from_all_the_vertices_of_ABC\"><\/span><strong>1. ABC is a triangle. Locate a point in the interior of <\/strong><img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image001.png\" \/><strong>ABC which is equidistant from all the vertices of <\/strong><img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image001.png\" \/><strong>ABC.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Let ABC be a triangle.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 88\" style=\"height: 180px; width: 161px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image002.jpg\" \/><\/p>\n<p style=\"text-align: justify;\">Draw perpendicular bisectors PQ and RS of sides AB and BC respectively of triangle ABC. Let PQ bisects AB at M and RS bisects BC at point N.<\/p>\n<p style=\"text-align: justify;\">Let PQ and RS intersect at point O.<\/p>\n<p style=\"text-align: justify;\">Join OA, OB and OC.<\/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\/ch07\/Ex7.5\/image001.png\" \/>AOM and <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image001.png\" \/> BOM,<\/p>\n<p style=\"text-align: justify;\">AM = MB [By construction]\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image003.png\" \/>AMO = <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image003.png\" \/>BMO = <img decoding=\"async\" style=\"height: 21px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image004.png\" \/>[By construction]\n<p style=\"text-align: justify;\">OM = OM [Common]\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\/ch07\/Ex7.5\/image005.png\" \/> <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image001.png\" \/>AOM <img decoding=\"async\" style=\"height: 17px; width: 28px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image006.png\" \/>BOM [By SAS congruency]\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\/ch07\/Ex7.5\/image007.png\" \/> OA = OB [By C.P.C.T.] \u2026..(i)<\/p>\n<p style=\"text-align: justify;\">Similarly, <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image001.png\" \/>BON <img decoding=\"async\" style=\"height: 17px; width: 28px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image006.png\" \/>CON<\/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\/ch07\/Ex7.5\/image007.png\" \/> OB = OC [By C.P.C.T.] \u2026..(ii)<\/p>\n<p style=\"text-align: justify;\">From eq. (i) and (ii),<\/p>\n<p style=\"text-align: justify;\">OA = OB = OC<\/p>\n<p style=\"text-align: justify;\">Hence O, the point of intersection of perpendicular bisectors of any two sides of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image001.png\" \/>ABC equidistant from its vertices.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 7.5<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"2_In_a_triangle_locate_a_point_in_its_interior_which_is_equidistant_from_all_the_sides_of_the_triangle\"><\/span><strong>2. In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Let ABC be a triangle.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 91\" style=\"height: 167px; width: 153px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image008.jpg\" \/><\/p>\n<p style=\"text-align: justify;\">Draw bisectors of <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image003.png\" \/>B and <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image003.png\" \/>C.<\/p>\n<p style=\"text-align: justify;\">Let these angle bisectors intersect each other at point I.<\/p>\n<p style=\"text-align: justify;\">Draw IK <img decoding=\"async\" style=\"height: 17px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image009.png\" \/> BC<\/p>\n<p style=\"text-align: justify;\">Also draw IJ <img decoding=\"async\" style=\"height: 17px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image009.png\" \/> AB and IL <img decoding=\"async\" style=\"height: 17px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image009.png\" \/> AC.<\/p>\n<p style=\"text-align: justify;\">Join AI.<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"In_BIK_and_BIJ\"><\/span>In <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image001.png\" \/>BIK and <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image001.png\" \/>BIJ,<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image003.png\" \/>IKB = <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image003.png\" \/>IJB = <img decoding=\"async\" style=\"height: 21px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image004.png\" \/> [By construction]\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image003.png\" \/>IBK = <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image003.png\" \/>IBJ<\/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\/ch07\/Ex7.5\/image010.png\" \/> BI is the bisector of <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image003.png\" \/>B (By construction)]\n<p style=\"text-align: justify;\">BI = BI [Common]\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\/ch07\/Ex7.5\/image005.png\" \/> <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image001.png\" \/>BIK <img decoding=\"async\" style=\"height: 17px; width: 28px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image006.png\" \/>BIJ [ASA criteria of congruency]\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\/ch07\/Ex7.5\/image005.png\" \/> IK = IJ [By C.P.C.T.] \u2026\u2026\u2026.(i)<\/p>\n<p style=\"text-align: justify;\">Similarly,<img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image001.png\" \/>CIK <img decoding=\"async\" style=\"height: 17px; width: 28px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image006.png\" \/>CIL<\/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\/ch07\/Ex7.5\/image005.png\" \/> IK = IL [By C.P.C.T.] \u2026\u2026\u2026.(ii)<\/p>\n<p style=\"text-align: justify;\">From eq (i) and (ii),<\/p>\n<p style=\"text-align: justify;\">IK = IJ = IL<\/p>\n<p style=\"text-align: justify;\">Hence, I is the point of intersection of angle bisectors of any two angles of <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image001.png\" \/>ABC equidistant from its sides.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 7.5<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"3_In_a_huge_park_people_are_concentrated_at_three_points_See_figure\"><\/span><strong>3. In a huge park, people are concentrated at three points (See figure).<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>A: where there are different slides and swings for children.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>B: near which a man-made lake is situated.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>C: which is near to a large parking and exit.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Where should an ice cream parlour be set up so that maximum number of persons can approach it?<\/strong><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 16\" style=\"height: 106px; width: 162px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image011.jpg\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>The parlour should be equidistant from A, B and C.<\/p>\n<p style=\"text-align: justify;\">For this let we draw perpendicular bisector say <img decoding=\"async\" style=\"height: 19px; width: 9px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image012.png\" \/> of line joining points B and C also draw perpendicular bisector say <img decoding=\"async\" style=\"height: 15px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image013.png\" \/> of line joining points A and C.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 94\" style=\"height: 182px; width: 164px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image014.jpg\" \/><\/p>\n<p style=\"text-align: justify;\">Let <img decoding=\"async\" style=\"height: 19px; width: 9px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image012.png\" \/> and <img decoding=\"async\" style=\"height: 15px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image013.png\" \/> intersect each other at point O.<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"Now_point_O_is_equidistant_from_points_A_B_and_C\"><\/span>Now point O is equidistant from points A, B and C.<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\">Join OA, OB and OC.<\/p>\n<p style=\"text-align: justify;\">Proof: In <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image001.png\" \/>BOP and <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image001.png\" \/>COP,<\/p>\n<p style=\"text-align: justify;\">OP = OP [Common]\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image003.png\" \/>OPB = <img decoding=\"async\" style=\"height: 16px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image003.png\" \/>OPC = <img decoding=\"async\" style=\"height: 21px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image004.png\" \/><\/p>\n<p style=\"text-align: justify;\">BP = PC [P is the mid-point of BC]\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\/ch07\/Ex7.5\/image005.png\" \/> <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image001.png\" \/>BOP <img decoding=\"async\" style=\"height: 17px; width: 28px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image006.png\" \/>COP [By SAS congruency]\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\/ch07\/Ex7.5\/image007.png\" \/> OB = OC [By C.P.C.T.] \u2026..(i)<\/p>\n<p style=\"text-align: justify;\">Similarly, <img decoding=\"async\" style=\"height: 17px; width: 15px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image001.png\" \/>AOQ <img decoding=\"async\" style=\"height: 17px; width: 28px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image006.png\" \/>COQ<\/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\/ch07\/Ex7.5\/image007.png\" \/> OA = OC [By C.P.C.T.] \u2026..(ii)<\/p>\n<p style=\"text-align: justify;\">From eq. (i) and (ii),<\/p>\n<p style=\"text-align: justify;\">OA = OB = OC<\/p>\n<p style=\"text-align: justify;\">Therefore, ice cream parlour should be set up at point O, the point of intersection of perpendicular bisectors of any two sides out of three formed by joining these points.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Maths Exercise 7.5<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"4_Complete_the_hexagonal_rangoli_and_the_star_rangolies_See_figure_but_filling_them_with_as_many_equilateral_triangles_of_side_1_cm_as_you_can_Count_the_number_of_triangles_in_each_case_Which_has_more_triangles\"><\/span><strong>4. Complete the hexagonal rangoli and the star rangolies (See figure) but filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles? <\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" id=\"Picture 100\" style=\"height: 177px; width: 313px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image015.jpg\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>In hexagonal rangoli, Number of equilateral triangles each of side 5 cm are 6.<\/p>\n<p style=\"text-align: justify;\">Area of equilateral triangle = <img decoding=\"async\" style=\"height: 45px; width: 72px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image016.png\" \/> = <img decoding=\"async\" style=\"height: 45px; width: 53px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image017.png\" \/> = <img decoding=\"async\" style=\"height: 45px; width: 55px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image018.png\" \/> sq. cm<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"Area_of_hexagonal_rangoli_6_x_Area_of_an_equilateral_triangle\"><\/span>Area of hexagonal rangoli = 6 x Area of an equilateral triangle<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 45px; width: 75px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image019.png\" \/> = <img decoding=\"async\" style=\"height: 45px; width: 61px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image020.png\" \/> sq. cm \u2026\u2026\u2026.(i)<\/p>\n<p style=\"text-align: justify;\">Now area of equilateral triangle of side 1 cm = = <img decoding=\"async\" style=\"height: 45px; width: 72px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image016.png\" \/> = <img decoding=\"async\" style=\"height: 45px; width: 51px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image021.png\" \/> = <img decoding=\"async\" style=\"height: 45px; width: 27px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image022.png\" \/> sq. cm \u2026..(ii)<\/p>\n<p style=\"text-align: justify;\">Number of equilateral triangles each of side 1 cm in hexagonal rangoli<\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 45px; width: 61px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image020.png\" \/> <img decoding=\"async\" style=\"height: 13px; width: 13px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image023.png\" \/> <img decoding=\"async\" style=\"height: 45px; width: 27px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image022.png\" \/> = <img decoding=\"async\" style=\"height: 45px; width: 61px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image020.png\" \/> <img decoding=\"async\" style=\"height: 44px; width: 36px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image024.png\" \/> = 150 \u2026..(iii)<\/p>\n<p style=\"text-align: justify;\">Now in Star rangoli,<\/p>\n<p style=\"text-align: justify;\">Number of equilateral triangles each of side 5 cm = 12<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 97\" style=\"height: 179px; width: 172px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image025.jpg\" \/><\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"Therefore_total_area_of_star_rangoli_12_Area_of_an_equilateral_triangle_of_side_5_cm\"><\/span>Therefore, total area of star rangoli = 12 <img decoding=\"async\" style=\"height: 14px; width: 12px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image026.png\" \/> Area of an equilateral triangle of side 5 cm<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 53px; width: 98px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image027.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 45px; width: 81px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image028.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 45px; width: 52px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image029.png\" \/> sq. cm \u2026\u2026\u2026.(iv)<\/p>\n<p style=\"text-align: justify;\">Number of equilateral triangles each of side 1 cm in star rangoli<\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 45px; width: 52px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image029.png\" \/> <img decoding=\"async\" style=\"height: 45px; width: 37px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image030.png\" \/><\/p>\n<p style=\"text-align: justify;\">= <img decoding=\"async\" style=\"height: 45px; width: 52px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image029.png\" \/> <img decoding=\"async\" style=\"height: 44px; width: 36px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/maths\/ch07\/Ex7.5\/image024.png\" \/><\/p>\n<p style=\"text-align: justify;\">= 300 \u2026\u2026\u2026.(v)<\/p>\n<p style=\"text-align: justify;\">From eq. (iii) and (v), we observe that star rangoli has more equilateral triangles each of side 1 cm.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_9_Maths_Exercise_75\"><\/span>NCERT Solutions for Class 9 Maths Exercise 7.5<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 7.5 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 7.5\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths-exercise-7-5\/\" aria-label=\"More on NCERT Solutions for Class 9 Maths Exercise 7.5\">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-5031","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 7.5 | myCBSEguide<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for Class 9 Maths Exercise 7.5 in PDF format for free download. 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