{"id":4308,"date":"2016-05-13T11:49:00","date_gmt":"2016-05-13T06:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-7-maths-triangles-and-its-properties-ex-6-5\/"},"modified":"2018-07-07T11:13:20","modified_gmt":"2018-07-07T05:43:20","slug":"ncert-solutions-for-class-7-maths-exercise-6-5","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-7-maths-exercise-6-5\/","title":{"rendered":"NCERT Solutions for Class 7 Maths Exercise 6.5"},"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-7-maths-exercise-6-5\/#NCERT_Solutions_for_Class_7_Maths_Triangles_and_its_Properties\" >NCERT Solutions for Class 7 Maths\u00a0Triangles and its Properties<\/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-6-5\/#Class_%E2%80%93VII_Mathematics_Ex_65\" >Class \u2013VII Mathematics (Ex. 6.5)<\/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-6-5\/#Question_1PQR_is_a_triangle_right_angled_at_P_If_PQ_10_cm_and_PR_24_cm_find_QR\" >Question 1.PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.<\/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-6-5\/#Question_2ABC_is_a_triangle_right_angled_at_C_If_AB_25_cm_and_AC_7_cm_find_BC\" >Question 2.ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC.<\/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-6-5\/#Question_3A_15_m_long_ladder_reached_a_window_12_m_high_from_the_ground_on_placing_it_against_a_wall_at_a_distance_texatex_Find_the_distance_of_the_foot_of_the_ladder_from_the_wall\" >Question 3.A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance {tex}a.{\/tex} Find the distance of the foot of the ladder from the wall.<\/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-6-5\/#Question_4Which_of_the_following_can_be_the_sides_of_a_right_triangle\" >Question 4.Which of the following can be the sides of a right triangle?<\/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-6-5\/#Question_5A_tree_is_broken_at_a_height_of_5_m_from_the_ground_and_its_top_touches_the_ground_at_a_distance_of_12_m_from_the_base_of_the_tree_Find_the_original_height_of_the_tree\" >Question 5.A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.<\/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-6-5\/#Question_6Angles_Q_and_R_of_a_texDelta_texPQR_are_tex25circ_tex_and_tex65circ_tex\" >Question 6.Angles Q and R of a {tex}\\Delta {\/tex}PQR are {tex}25^\\circ {\/tex} and {tex}65^\\circ .{\/tex}<\/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-6-5\/#Question_7Find_the_perimeter_of_the_rectangle_whose_length_is_40_cm_and_a_diagonal_is_41_cm\" >Question 7.Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.<\/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-6-5\/#Question_8The_diagonals_of_a_rhombus_measure_16_cm_and_30_cm_Find_its_perimeter\" >Question 8.The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.<\/a><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-7-maths-exercise-6-5\/#NCERT_Solutions_for_Class_7_Maths_Exercise_65\" >NCERT Solutions for Class 7 Maths Exercise 6.5<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-7-maths-exercise-6-5\/#CBSE_app_for_Students\" >CBSE app for Students<\/a><\/li><\/ul><\/nav><\/div>\n<p>NCERT Solutions for Class 7 Maths Exercise 6.5 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\u00a0Triangles and its Properties\u00a0<\/strong><strong><a class=\"button\" href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-07-mathematics-the-triangle-and-its-properties\/1515\/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 6.5\" width=\"170\" height=\"216\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_7_Maths_Triangles_and_its_Properties\"><\/span>NCERT Solutions for Class 7 Maths\u00a0Triangles and its Properties<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_65\"><\/span><strong>Class \u2013VII Mathematics (Ex. 6.5)<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<h6><span class=\"ez-toc-section\" id=\"Question_1PQR_is_a_triangle_right_angled_at_P_If_PQ_10_cm_and_PR_24_cm_find_QR\"><\/span><strong>Question 1.<\/strong>PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Answer:<\/strong><\/p>\n<p>Given: PQ = 10 cm, PR = 24 cm<\/p>\n<p>Let QR be {tex}x{\/tex} cm.<\/p>\n<p>In right angled triangle QPR,<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch06_18.jpg\" \/><\/p>\n<p>{tex}{\\left( {Hypotenuse} \\right)^2} = {\\text{ }}{\\left( {Base} \\right)^2} + {\\text{ }}{\\left( {Perpendicular} \\right)^2}{\/tex}<\/p>\n[By Pythagoras theorem]\n<p>{tex} \\Rightarrow {\/tex} {tex}{\\left( {QR} \\right)^2} = {\\text{ }}{\\left( {PQ} \\right)^2} + {\\text{ }}{\\left( {PR} \\right)^2}{\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} {tex}{x^2} = {\\left( {10} \\right)^2} + {\\left( {24} \\right)^2}{\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} {tex}{x^2}{\/tex} = 100 + 576 = 676<\/p>\n<p>{tex} \\Rightarrow {\/tex} {tex}x = \\sqrt {676} {\/tex} = 26 cm<\/p>\n<p>Thus, the length of QR is 26 cm.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 6.5<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_2ABC_is_a_triangle_right_angled_at_C_If_AB_25_cm_and_AC_7_cm_find_BC\"><\/span><strong>Question 2.<\/strong>ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC.<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Answer:<\/strong><\/p>\n<p>Given: AB = 25 cm, AC = 7 cm<\/p>\n<p>Let BC be {tex}x{\/tex} cm.<\/p>\n<p>In right angled triangle ACB,<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch06_19.jpg\" \/><\/p>\n<p>{tex}{\\left( {Hypotenuse} \\right)^2} = {\\text{ }}{\\left( {Base} \\right)^2} + {\\text{ }}{\\left( {Perpendicular} \\right)^2}{\/tex}<\/p>\n[By Pythagoras theorem]\n<p>{tex} \\Rightarrow {\/tex} {tex}{\\left( {AB} \\right)^2} = {\\text{ }}{\\left( {AC} \\right)^2} + {\\text{ }}{\\left( {BC} \\right)^2}{\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} {tex}{\\left( {25} \\right)^2} = {\\left( 7 \\right)^2} + {x^2}{\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} 625 = 49 + {tex}{x^2}{\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} {tex}{x^2}{\/tex} = 625 \u2013 49 = 576<\/p>\n<p>{tex} \\Rightarrow {\/tex} {tex}x = \\sqrt {576} {\/tex} = 24 cm<\/p>\n<p>Thus, the length of BC is 24 cm.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 6.5<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_3A_15_m_long_ladder_reached_a_window_12_m_high_from_the_ground_on_placing_it_against_a_wall_at_a_distance_texatex_Find_the_distance_of_the_foot_of_the_ladder_from_the_wall\"><\/span><strong>Question 3.<\/strong>A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance {tex}a.{\/tex} Find the distance of the foot of the ladder from the wall.<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_ch06_11.jpg\" \/><\/p>\n<p><strong>Answer:<\/strong><\/p>\n<p>Let AC be the ladder and A be the window.<\/p>\n<p>Given: AC = 15 m, AB = 12 m, CB = {tex}a{\/tex} m<\/p>\n<p>In right angled triangle ACB,<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch06_20.jpg\" \/><\/p>\n<p>{tex}{\\left( {Hypotenuse} \\right)^2} = {\\text{ }}{\\left( {Base} \\right)^2} + {\\text{ }}{\\left( {Perpendicular} \\right)^2}{\/tex}<\/p>\n[By Pythagoras theorem]\n<p>{tex} \\Rightarrow {\/tex} {tex}{\\left( {AC} \\right)^2} = {\\text{ }}{\\left( {CB} \\right)^2} + {\\text{ }}{\\left( {AB} \\right)^2}{\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} {tex}{\\left( {15} \\right)^2} = {\\left( a \\right)^2} + {\\left( {12} \\right)^2}{\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} 225 = {tex}{a^2}{\/tex} + 144<\/p>\n<p>{tex} \\Rightarrow {\/tex} {tex}{a^2}{\/tex} = 225 \u2013 144 = 81<\/p>\n<p>{tex} \\Rightarrow {\/tex} {tex}a = \\sqrt {81} {\/tex} = 9 cm<\/p>\n<p>Thus, the distance of the foot of the ladder from the wall is 9 m.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 6.5<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_4Which_of_the_following_can_be_the_sides_of_a_right_triangle\"><\/span><strong>Question 4.<\/strong>Which of the following can be the sides of a right triangle?<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<ol>\n<li>2.5 cm, 6.5 cm, 6 cm<\/li>\n<li>2 cm, 2 cm, 5 cm<\/li>\n<li>1.5 cm, 2 cm, 2.5 cm<\/li>\n<\/ol>\n<p>In the case of right angled triangles, identify the right angles.<\/p>\n<p><strong>Answer:<\/strong><\/p>\n<p>Let us consider, the larger side be the hypotenuse and also using Pythagoras theorem,<\/p>\n<p>{tex}{\\left( {Hypotenuse} \\right)^2} = {\\text{ }}{\\left( {Base} \\right)^2} + {\\text{ }}{\\left( {Perpendicular} \\right)^2}{\/tex}<\/p>\n<p>(i) 2.5 cm, 6.5 cm, 6 cm<\/p>\n<p>In {tex}\\Delta {\\text{ABC}},{\/tex} {tex}{\\left( {{\\text{AC}}} \\right)^2} = {\\left( {{\\text{AB}}} \\right)^2} + {\\left( {{\\text{BC}}} \\right)^2}{\/tex}<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch06_21.jpg\" \/><\/p>\n<p>L.H.S. = {tex}{\\left( {6.5} \\right)^2}{\/tex} = 42.25 cm<\/p>\n<p>R.H.S. = {tex}{\\left( 6 \\right)^2} + {\\left( {2.5} \\right)^2}{\/tex} = 36 + 6.25 = 42.25 cm<\/p>\n<p>Since, L.H.S. = R.H.S.<\/p>\n<p>Therefore, the given sides are of the right angled triangle.<\/p>\n<p>Right angle lies on the opposite to the greater side 6.5 cm, i.e., at B.<\/p>\n<p>(ii) 2 cm, 2 cm, 5 cm<\/p>\n<p>{tex}{\\left( 5 \\right)^2} = {\\left( 2 \\right)^2} + {\\left( 2 \\right)^2}{\/tex}<\/p>\n<p>L.H.S. = {tex}{\\left( 5 \\right)^2}{\/tex} = 25<\/p>\n<p>R.H.S. = {tex}{\\left( 2 \\right)^2} + {\\left( 2 \\right)^2}{\/tex} = 4 + 4 = 8<\/p>\n<p>Since, L.H.S. {tex} \\ne {\/tex} R.H.S.<\/p>\n<p>Therefore, the given sides are not of the right angled triangle.<\/p>\n<p>(iii) 1.5 cm, 2 cm, 2.5 cm<\/p>\n<p>In {tex}\\Delta {\/tex}PQR, {tex}{\\left( {{\\text{PR}}} \\right)^2} = {\\left( {{\\text{PQ}}} \\right)^2} + {\\left( {{\\text{RQ}}} \\right)^2}{\/tex}<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch06_22.jpg\" \/><\/p>\n<p>L.H.S. = {tex}{\\left( {2.5} \\right)^2}{\/tex} = 6.25 cm<\/p>\n<p>R.H.S. = {tex}{\\left( {1.5} \\right)^2} + {\\left( 2 \\right)^2}{\/tex} = 2.25 + 4 = 6.25 cm<\/p>\n<p>Since, L.H.S. = R.H.S.<\/p>\n<p>Therefore, the given sides are of the right angled triangle.<\/p>\n<p>Right angle lies on the opposite to the greater side 2.5 cm, i.e., at Q.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 6.5<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_5A_tree_is_broken_at_a_height_of_5_m_from_the_ground_and_its_top_touches_the_ground_at_a_distance_of_12_m_from_the_base_of_the_tree_Find_the_original_height_of_the_tree\"><\/span><strong>Question 5.<\/strong>A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Answer:<\/strong><\/p>\n<p>Let A\u2019CB represents the tree before it broken at the point C and let the top A\u2019 touches the ground at A after it broke. Then {tex}\\Delta {\\text{ABC}}{\/tex} is a right angled triangle, right angled at B.<\/p>\n<p>AB = 12 m and BC = 5 m<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch06_23.jpg\" \/><\/p>\n<p>Using Pythagoras theorem, In {tex}\\Delta {\\text{ABC}}{\/tex}<\/p>\n<p>{tex}{\\left( {{\\text{AC}}} \\right)^2} = {\\left( {{\\text{AB}}} \\right)^2} + {\\left( {{\\text{BC}}} \\right)^2}{\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} {tex}{\\left( {{\\text{AC}}} \\right)^2} = {\\left( {12} \\right)^2} + {\\left( 5 \\right)^2}{\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} {tex}{\\left( {{\\text{AC}}} \\right)^2} = 144 + 25{\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} {tex}{\\left( {{\\text{AC}}} \\right)^2} = 169{\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} AC = 13 m<\/p>\n<p>Hence, the total height of the tree = AC + CB = 13 + 5 = 18 m.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 6.5<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_6Angles_Q_and_R_of_a_texDelta_texPQR_are_tex25circ_tex_and_tex65circ_tex\"><\/span><strong>Question 6.<\/strong>Angles Q and R of a {tex}\\Delta {\/tex}PQR are {tex}25^\\circ {\/tex} and {tex}65^\\circ .{\/tex}<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p>Write which of the following is true:<\/p>\n<ol>\n<li>{tex}P{Q^2} + {\\text{ }}Q{R^2} = {\\text{ }}R{P^2}{\/tex}<\/li>\n<li>{tex}P{Q^2} + {\\text{ }}R{P^2} = {\\text{ }}Q{R^2}{\/tex}<\/li>\n<li>{tex}R{P^2} + {\\text{ }}Q{R^2} = {\\text{ }}P{Q^2}{\/tex}<\/li>\n<\/ol>\n<p>{tex}25^\\circ {\/tex} {tex}65^\\circ {\/tex}<\/p>\n<p><strong>Answer:<\/strong><\/p>\n<p>In {tex}\\Delta {\/tex}PQR,<\/p>\n<p>{tex}\\angle {\/tex}PQR + {tex}\\angle {\/tex}QRP + {tex}\\angle {\/tex}RPQ = {tex}180^\\circ {\/tex}<\/p>\n[By Angle sum property of a {tex}\\Delta {\/tex} ]\n<p>{tex} \\Rightarrow {\/tex} {tex}25^\\circ + 65^\\circ + \\angle {\\text{RPQ = 180}}^\\circ {\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} {tex}90^\\circ + \\angle {\\text{RPQ = 180}}^\\circ {\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} {tex}\\angle {\/tex}RPQ = {tex}180^\\circ &#8211; 90^\\circ = 90^\\circ {\/tex}<\/p>\n<p>Thus, {tex}\\Delta {\/tex}PQR is a right angled triangle, right angled at P.<\/p>\n<p>{tex}\\therefore {\/tex} (Hypotenuse)2 = (Base)2 + (Perpendicular)2 [By Pythagoras theorem]\n<p>{tex} \\Rightarrow {\/tex} {tex}{\\left( {{\\text{QR}}} \\right)^2} = {\\left( {{\\text{PR}}} \\right)^2} + {\\left( {{\\text{QP}}} \\right)^2}{\/tex}<\/p>\n<p>Hence, Option (ii) is correct.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 6.5<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_7Find_the_perimeter_of_the_rectangle_whose_length_is_40_cm_and_a_diagonal_is_41_cm\"><\/span><strong>Question 7.<\/strong>Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Answer:<\/strong><\/p>\n<p>Given diagonal (PR) = 41 cm, length (PQ) = 40 cm<\/p>\n<p>Let breadth (QR) be {tex}x{\/tex} cm.<\/p>\n<p>Now, in right angled triangle PQR,<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch06_24.png\" \/><\/p>\n<p>{tex}{\\left( {{\\text{PR}}} \\right)^2} = {\\left( {{\\text{RQ}}} \\right)^2} + {\\left( {{\\text{PQ}}} \\right)^2}{\/tex}<\/p>\n[By Pythagoras theorem]\n<p>{tex} \\Rightarrow {\/tex} {tex}{\\left( {41} \\right)^2} = {x^2} + {\\left( {40} \\right)^2}{\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} 1681 = {tex}{x^2}{\/tex} + 1600 {tex} \\Rightarrow {\/tex} {tex}{x^2}{\/tex} = 1681 \u2013 1600<\/p>\n<p>{tex} \\Rightarrow {\/tex} {tex}{x^2}{\/tex} = 81 {tex} \\Rightarrow {\/tex} {tex}x = \\sqrt {81} = 9{\/tex} cm<\/p>\n<p>Therefore the breadth of the rectangle is 9 cm.<\/p>\n<p>Perimeter of rectangle = 2(length + breadth)<\/p>\n<p>= 2 (9 + 49)<\/p>\n<p>= 2 x 49 = 98 cm<\/p>\n<p>Hence the perimeter of the rectangle is 98 cm.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 6.5<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_8The_diagonals_of_a_rhombus_measure_16_cm_and_30_cm_Find_its_perimeter\"><\/span><strong>Question 8.<\/strong>The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Answer:<\/strong><\/p>\n<p>Given: Diagonals AC = 30 cm and DB = 16 cm.<\/p>\n<p>Since the diagonals of the rhombus bisect at right angle to each other.<\/p>\n<p>Therefore, OD = {tex}\\frac{{{\\text{DB}}}}{2} = \\frac{{16}}{2}{\/tex} = 8 cm<\/p>\n<p>And OC = {tex}\\frac{{{\\text{AC}}}}{2} = \\frac{{30}}{2}{\/tex} = 15 cm<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch06_25.jpg\" \/><\/p>\n<p>Now, In right angle triangle DOC,<\/p>\n<p>{tex}{\\left( {{\\text{DC}}} \\right)^2} = {\\left( {{\\text{OD}}} \\right)^2} + {\\left( {{\\text{OC}}} \\right)^2}{\/tex} [By Pythagoras theorem]\n<p>{tex} \\Rightarrow {\/tex} {tex}{\\left( {{\\text{DC}}} \\right)^2} = {\\left( 8 \\right)^2} + {\\left( {15} \\right)^2}{\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} {tex}{\\left( {{\\text{DC}}} \\right)^2}{\/tex} = 64 + 225 = 289<\/p>\n<p>{tex} \\Rightarrow {\/tex} DC = {tex}\\sqrt {289} {\/tex} = 17 cm<\/p>\n<p>Perimeter of rhombus = 4 x side<\/p>\n<p>= 4 x 17 = 68 cm<\/p>\n<p>Thus, the perimeter of rhombus is 68 cm.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_7_Maths_Exercise_65\"><\/span>NCERT Solutions for Class 7 Maths Exercise 6.5<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 6.5 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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