{"id":4321,"date":"2016-05-13T11:49:00","date_gmt":"2016-05-13T06:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-7-maths-practical-geometry-ex-10-6\/"},"modified":"2018-07-07T15:34:49","modified_gmt":"2018-07-07T10:04:49","slug":"ncert-solutions-for-class-7-maths-exercise-10-6","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-7-maths-exercise-10-6\/","title":{"rendered":"NCERT Solutions for Class 7 Maths Exercise 10.6"},"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-10-6\/#NCERT_Solutions_for_Class_7_Maths_Practical_Geometry\" >NCERT Solutions for Class 7 Maths Practical 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-for-class-7-maths-exercise-10-6\/#Class_%E2%80%93VII_Mathematics_Ex_106\" >Class \u2013VII Mathematics (Ex. 10.6)<\/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-10-6\/#Miscellaneous_Questions\" >Miscellaneous Questions<\/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-10-6\/#NCERT_Solutions_for_Class_7_Maths_Exercise_106\" >NCERT Solutions for Class 7 Maths Exercise 10.6<\/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-10-6\/#2To_construct_texDelta_texPQR_where_texmangle_texQ_tex30_circ_textexmangle_texR_tex60_circ_tex_and_QR_47_cm\" >2.To construct: {tex}\\Delta {\/tex}PQR where {tex}m\\angle {\/tex}Q = {tex}{30^ \\circ },{\/tex}{tex}m\\angle {\/tex}R = {tex}{60^ \\circ }{\/tex} and QR = 4.7 cm.<\/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-10-6\/#6_To_construct_texDelta_texPQR_where_PQ_35_cm_QR_4_cm_and_PR_35_cm\" >6. To construct: {tex}\\Delta {\/tex}PQR where PQ = 3.5 cm, QR = 4 cm and PR = 3.5 cm<\/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-10-6\/#7_To_construct_A_triangle_whose_sides_are_XY_3_cm_YZ_4_cm_and_XZ_5_cm\" >7. To construct: A triangle whose sides are XY = 3 cm, YZ = 4 cm and XZ = 5 cm.<\/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-10-6\/#8To_construct_A_triangle_DEF_whose_sides_are_DE_45_cm_EF_55_cm_and_DF_4_cm\" >8.To construct: A triangle DEF whose sides are DE = 4.5 cm, EF = 5.5 cm and DF = 4 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-9\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-7-maths-exercise-10-6\/#NCERT_Solutions_for_Class_7_Maths_Exercise_106-2\" >NCERT Solutions for Class 7 Maths Exercise 10.6<\/a><\/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-10-6\/#CBSE_app_for_Students\" >CBSE app for Students<\/a><\/li><\/ul><\/nav><\/div>\n<p>NCERT Solutions for Class 7 Maths Exercise 10.6 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 Practical Geometry\u00a0<\/strong><strong><a class=\"button\" href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-07-mathematics-practical-geometry\/1519\/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 10.6\" width=\"150\" height=\"191\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_7_Maths_Practical_Geometry\"><\/span>NCERT Solutions for Class 7 Maths Practical Geometry<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_106\"><\/span><strong>Class \u2013VII Mathematics (Ex. 10.6)<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p>Below are given the measures of certain sides and angles of triangles. Identify those which cannot be constructed and, say why you cannot construct them. Construct rest of the triangle.<\/p>\n<p>Triangle Given measurements<\/p>\n<p>1. {tex}\\Delta {\/tex}ABC {tex}m\\angle {\/tex}A = {tex}85^\\circ {\\text{ ;}}{\/tex} {tex}m\\angle {\/tex}B = {tex}115^\\circ {\\text{ ;}}{\/tex} AB = 5 cm<\/p>\n<p>2. {tex}\\Delta {\/tex}PQR {tex}m\\angle {\/tex}Q = {tex}30^\\circ {\\text{ ;}}{\/tex} {tex}m\\angle {\/tex}R = {tex}60^\\circ {\\text{ ;}}{\/tex} QR = 4.7 cm<\/p>\n<p>3. {tex}\\Delta {\/tex}ABC {tex}m\\angle {\/tex}A = {tex}70^\\circ {\\text{ ;}}{\/tex} {tex}m\\angle {\/tex}B = {tex}50^\\circ {\\text{ ;}}{\/tex} AC = 3 cm<\/p>\n<p>4. {tex}\\Delta {\/tex}LMN {tex}m\\angle {\/tex}L = {tex}60^\\circ {\\text{ ;}}{\/tex} {tex}m\\angle {\/tex}N = {tex}120^\\circ {\\text{ ;}}{\/tex} LM = 5 cm<\/p>\n<p>5. {tex}\\Delta {\/tex}ABC BC = 2 cm; AB = 4 cm; AC = 2 cm<\/p>\n<p>6. {tex}\\Delta {\/tex}PQR PQ = 3.5 cm; QR = 4 cm; PR = 3.5 cm<\/p>\n<p>7. {tex}\\Delta {\/tex}XYZ XY = 3 cm; YZ = 4 cm; XZ = 5 cm<\/p>\n<p>8. {tex}\\Delta {\/tex}DEF DE = 4.5 cm; EF = 5.5 cm; DF = 4 cm<\/p>\n<p><strong>Answers<\/strong><\/p>\n<h6><span class=\"ez-toc-section\" id=\"Miscellaneous_Questions\"><\/span><strong>Miscellaneous Questions<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p>1.In {tex}\\Delta {\/tex}ABC, {tex}m\\angle {\/tex}A = {tex}85^\\circ ,m\\angle {\/tex}B = {tex}115^\\circ ,{\/tex} AB = 5 cm Construction of {tex}\\Delta {\/tex}ABC is not possible because {tex}m\\angle {\/tex}A = {tex}85^\\circ + m\\angle {\/tex}B = {tex}200^\\circ ,{\/tex} and we know that the sum of angles of a triangle should be {tex}180^\\circ .{\/tex}<\/p>\n<hr \/>\n<h6 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_7_Maths_Exercise_106\"><\/span>NCERT Solutions for Class 7 Maths Exercise 10.6<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<h6><span class=\"ez-toc-section\" id=\"2To_construct_texDelta_texPQR_where_texmangle_texQ_tex30_circ_textexmangle_texR_tex60_circ_tex_and_QR_47_cm\"><\/span><strong>2.To construct<\/strong>: {tex}\\Delta {\/tex}PQR where {tex}m\\angle {\/tex}Q = {tex}{30^ \\circ },{\/tex}{tex}m\\angle {\/tex}R = {tex}{60^ \\circ }{\/tex} and QR = 4.7 cm.<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_ch10_15.jpg\" \/><\/p>\n<p><strong>Steps of construction<\/strong>:<\/p>\n<ol>\n<li>Draw a line segment QR = 4.7 cm.<\/li>\n<\/ol>\n<p>(a) At point Q, draw {tex}\\angle {\/tex}XQR = {tex}{30^ \\circ }{\/tex} with the help of compass.<\/p>\n<p>(b) At point R, draw {tex}\\angle {\/tex}YRQ = {tex}{60^ \\circ }{\/tex} with the help of compass.<\/p>\n<p>(c) QX and RY intersect at point P.<\/p>\n<p>It is the required triangle PQR.<\/p>\n<p>3. We know that the sum of angles of a triangle is {tex}180^\\circ .{\/tex}<\/p>\n<p>{tex}\\therefore {\/tex} {tex}m\\angle {\/tex}A + {tex}m\\angle {\/tex}B + {tex}m\\angle {\/tex}C = {tex}180^\\circ {\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} {tex}70^\\circ + 50^\\circ + m\\angle {\/tex}C = {tex}180^\\circ {\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} {tex}120^\\circ {\/tex} + {tex}m\\angle {\/tex}C = {tex}180^\\circ {\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} {tex}m\\angle {\/tex}C = {tex}180^\\circ {\/tex} \u2013 {tex}120^\\circ {\/tex}<\/p>\n<p>{tex} \\Rightarrow {\/tex} {tex}m\\angle {\/tex}C = {tex}60^\\circ {\/tex}<\/p>\n<p><strong>To construct<\/strong>: {tex}\\Delta {\/tex}ABC where {tex}m\\angle {\/tex}A = {tex}70^\\circ {\/tex}, {tex}m\\angle {\/tex}C = {tex}60^\\circ {\/tex} and AC = 3 cm.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch10_16.jpg\" \/><\/p>\n<p><strong>Steps of construction<\/strong>:<\/p>\n<p>(a) Draw a line segment AC = 3 cm.<\/p>\n<p>(b) At point C, draw {tex}\\angle {\/tex}YCA = {tex}{60^ \\circ }.{\/tex}<\/p>\n<p>(c) At point A, draw {tex}\\angle {\/tex}XAC = {tex}70^\\circ .{\/tex}<\/p>\n<p>(d) Rays XA and YC intersect at point B<\/p>\n<p>It is the required triangle ABC.<\/p>\n<p>4. In {tex}\\Delta {\/tex}LMN , {tex}m\\angle {\/tex}L = {tex}{60^ \\circ },{\/tex}{tex}m\\angle {\/tex}N = {tex}120^\\circ ,{\/tex} LM = 5 cm<\/p>\n<p>This {tex}\\Delta {\/tex}LMN is not possible to construct because {tex}m\\angle {\/tex}L + {tex}m\\angle {\/tex}N = {tex}60^\\circ + 120^\\circ = 180^\\circ {\/tex} which forms a linear pair.<\/p>\n<p>5. {tex}\\Delta {\/tex}ABC, BC = 2 cm, AB = 4 cm and AC = 2 cm<\/p>\n<p>This {tex}\\Delta {\/tex}ABC is not possible to construct because the condition is<\/p>\n<p>Sum of lengths of two sides of a triangle should be greater than the third side.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/07\/mathematics\/07_math_ncert_ch10_17.jpg\" \/><\/p>\n<p>AB &lt; BC + AC {tex} \\Rightarrow {\/tex} 4 &lt; 2 + 2 {tex} \\Rightarrow {\/tex} 4 = 4,<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 10.6<\/p>\n<h6><span class=\"ez-toc-section\" id=\"6_To_construct_texDelta_texPQR_where_PQ_35_cm_QR_4_cm_and_PR_35_cm\"><\/span><strong>6. To construct<\/strong>: {tex}\\Delta {\/tex}PQR where PQ = 3.5 cm, QR = 4 cm and PR = 3.5 cm<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Steps of construction<\/strong>:<\/p>\n<p>(a) Draw a line segment QR = 4 cm.<\/p>\n<p>(b) Taking Q as centre and radius 3.5 cm, draw an arc.<\/p>\n<p>(c) Similarly, taking R as centre and radius 3.5 cm, draw an another arc which intersects the first arc at point P.<\/p>\n<p>It is the required triangle PQR.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 10.6<\/p>\n<h6><span class=\"ez-toc-section\" id=\"7_To_construct_A_triangle_whose_sides_are_XY_3_cm_YZ_4_cm_and_XZ_5_cm\"><\/span><strong>7. To construct<\/strong>: A triangle whose sides are XY = 3 cm, YZ = 4 cm and XZ = 5 cm.<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_ch10_18.jpg\" \/><\/p>\n<p><strong>Steps of construction<\/strong>:<\/p>\n<p>(a) Draw a line segment ZY = 4 cm.<\/p>\n<p>(b) Taking Z as centre and radius 5 cm, draw an arc.<\/p>\n<p>(c) Taking Y as centre and radius 3 cm, draw another arc.<\/p>\n<p>(d) Both arcs intersect at point X.<\/p>\n<p>It is the required triangle XYZ.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 7 Maths Exercise 10.6<\/p>\n<h6><span class=\"ez-toc-section\" id=\"8To_construct_A_triangle_DEF_whose_sides_are_DE_45_cm_EF_55_cm_and_DF_4_cm\"><\/span><strong>8.To construct<\/strong>: A triangle DEF whose sides are DE = 4.5 cm, EF = 5.5 cm and DF = 4 cm.<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_ch10_19.jpg\" \/><\/p>\n<p><strong>Steps of construction<\/strong>:<\/p>\n<p>(a) Draw a line segment EF = 5.5 cm.<\/p>\n<p>(b) Taking E as centre and radius 4.5 cm, draw an arc.<\/p>\n<p>(a) Taking F as centre and radius 4 cm, draw an another arc which intersects the first arc at point D.<\/p>\n<p>It is the required triangle DEF.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_7_Maths_Exercise_106-2\"><\/span>NCERT Solutions for Class 7 Maths Exercise 10.6<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 10.6 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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