{"id":4258,"date":"2016-05-06T11:49:00","date_gmt":"2016-05-06T11:49:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-6-maths-practical-geometry-ex-14-6\/"},"modified":"2018-07-23T11:41:37","modified_gmt":"2018-07-23T06:11:37","slug":"ncert-solutions-class-6-maths-exercise-14-6","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-6-maths-exercise-14-6\/","title":{"rendered":"NCERT Solutions for Class 6 Maths Exercise 14.6"},"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-class-6-maths-exercise-14-6\/#NCERT_Solutions_for_Class_6_Maths_Practical_geometry\" >NCERT Solutions for Class 6 Maths\u00a0Practical 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-6-maths-exercise-14-6\/#Class_%E2%80%93VI_Mathematics\" >Class \u2013VI Mathematics<\/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-6-maths-exercise-14-6\/#Ex_146\" >(Ex. 14.6)<\/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-6-maths-exercise-14-6\/#Question_1Draw_texangle_texPOQ_of_measure_tex75circ_tex_and_find_its_line_of_symmetry\" >Question 1.Draw {tex}\\angle {\/tex}POQ of measure {tex}75^\\circ {\/tex} and find its line of symmetry.<\/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-6-maths-exercise-14-6\/#Question_2Draw_an_angle_of_measure_tex147circ_tex_and_construct_its_bisector\" >Question 2.Draw an angle of measure {tex}147^\\circ {\/tex} and construct its bisector.<\/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-6-maths-exercise-14-6\/#Question_3Draw_a_right_angle_and_construct_its_bisector\" >Question 3.Draw a right angle and construct its bisector.<\/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-class-6-maths-exercise-14-6\/#Question_4_Draw_an_angle_of_measure_tex153circ_tex_and_divide_it_into_four_equal_parts\" >Question 4. Draw an angle of measure {tex}153^\\circ {\/tex} and divide it into four equal parts.<\/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-class-6-maths-exercise-14-6\/#Question_5Construct_with_ruler_and_compasses_angles_of_following_measures\" >Question 5.Construct with ruler and compasses, angles of following measures:<\/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-class-6-maths-exercise-14-6\/#Question_6Draw_an_angle_of_measure_tex45_circ_tex_and_bisect_it\" >Question 6.Draw an angle of measure {tex}{45^ \\circ }{\/tex} and bisect it.<\/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-class-6-maths-exercise-14-6\/#Question_7Draw_an_angle_of_measure_tex135_circ_tex_and_bisect_it\" >Question 7.Draw an angle of measure {tex}{135^ \\circ }{\/tex} and bisect it.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-6-maths-exercise-14-6\/#Question_8Draw_an_angle_of_tex70circ_tex_Make_a_copy_of_it_using_only_a_straight_edge_and_compasses\" >Question 8.Draw an angle of {tex}70^\\circ .{\/tex} Make a copy of it using only a straight edge and compasses.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-6-maths-exercise-14-6\/#Question_9Draw_an_angle_of_tex40circ_tex_Copy_its_supplementary_angle\" >Question 9.Draw an angle of {tex}40^\\circ .{\/tex} Copy its supplementary angle.<\/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-13\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-6-maths-exercise-14-6\/#NCERT_Solutions_for_Class_6_Maths_Exercise_146\" >NCERT Solutions for Class 6 Maths Exercise 14.6<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-14\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-6-maths-exercise-14-6\/#NCERT_Solutions_for_Mathematics_Class_3rd_to_12th\" >NCERT Solutions for Mathematics Class 3rd to 12th<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-15\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-6-maths-exercise-14-6\/#CBSE_app_for_Students\" >CBSE app for Students<\/a><\/li><\/ul><\/nav><\/div>\n<p>NCERT Solutions for Class 6 Maths Exercise 14.6 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 6 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\u00a0Practical geometry\u00a0<a class=\"button\" href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-06-mathematics-practical-geometry\/1626\/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\/06_Maths_Book.jpg\" alt=\"NCERT Solutions for Class 6 Maths Exercise 14.6\" width=\"167\" height=\"194\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_6_Maths_Practical_geometry\"><\/span>NCERT Solutions for Class 6 Maths\u00a0Practical geometry<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h6 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"Class_%E2%80%93VI_Mathematics\"><\/span><strong>Class \u2013VI Mathematics <\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<h6 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"Ex_146\"><\/span><strong>(Ex. 14.6)<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<h6><span class=\"ez-toc-section\" id=\"Question_1Draw_texangle_texPOQ_of_measure_tex75circ_tex_and_find_its_line_of_symmetry\"><\/span><strong>Question 1.<\/strong>Draw {tex}\\angle {\/tex}POQ of measure {tex}75^\\circ {\/tex} and find its line of symmetry.<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Answer: Steps of construction<\/strong>:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/06\/mathematics\/06_math_ncert_ch14_27.jpg\" \/><\/p>\n<p>(a) Draw a line {tex}l{\/tex} and mark a point O on it.<\/p>\n<p>(b) Place the pointer of the compasses at O and draw an arc of any radius which intersects the line {tex}l{\/tex} at A.<\/p>\n<p>(c) Taking same radius, with centre A, cut the previous arc at B.<\/p>\n<p>(d) Join OB, then {tex}\\angle {\/tex}BOA = {tex}{60^ \\circ }.{\/tex}<\/p>\n<p>(e) Taking same radius, with centre B, cut the previous arc at C.<\/p>\n<p>(f) Draw bisector of {tex}\\angle {\/tex}BOC. The angle is of {tex}{90^ \\circ }.{\/tex} Mark it at D. Thus, {tex}\\angle {\/tex}DOA = {tex}{90^ \\circ }{\/tex}<\/p>\n<p>(g) Draw {tex}\\overline {{\\text{OP}}} {\/tex} as bisector of {tex}\\angle {\/tex}DOB.<\/p>\n<p>Thus, {tex}\\angle {\/tex}POA = {tex}75^\\circ {\/tex}<\/p>\n<hr \/>\n<h6><span class=\"ez-toc-section\" id=\"Question_2Draw_an_angle_of_measure_tex147circ_tex_and_construct_its_bisector\"><\/span><strong>Question 2.<\/strong>Draw an angle of measure {tex}147^\\circ {\/tex} and construct its bisector.<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Answer: Steps of construction<\/strong>:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/06\/mathematics\/06_math_ncert_ch14_28.jpg\" \/><\/p>\n<p>(a) Draw a ray {tex}\\overleftrightarrow {{\\text{OA}}}.{\/tex}<\/p>\n<p>(b) With the help of protractor, construct {tex}\\angle {\/tex}AOB = {tex}147^\\circ .{\/tex}<\/p>\n<p>(c) Taking centre O and any convenient radius, draw an arc which intersects the arms {tex}\\overline {{\\text{OA}}} {\/tex} and {tex}\\overline {{\\text{OB}}} {\/tex} at P and Q respectively.<\/p>\n<p>(d) Taking P as centre and radius more than half of PQ, draw an arc.<\/p>\n<p>(e) Taking Q as centre and with the same radius, draw another arc which intersects the previous at R.<\/p>\n<p>(f) Join OR and produce it.<\/p>\n<p>Thus, {tex}\\overline {{\\text{OR}}} {\/tex} is the required bisector of {tex}\\angle {\/tex}AOB.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 6 Maths Exercise 14.6<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_3Draw_a_right_angle_and_construct_its_bisector\"><\/span><strong>Question 3.<\/strong>Draw a right angle and construct its bisector.<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Answer: Steps of construction<\/strong>:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/06\/mathematics\/06_math_ncert_ch14_29.jpg\" \/><\/p>\n<p>(a) Draw a line PQ and take a point O on it.<\/p>\n<p>(b) Taking O as centre and convenient radius, draw an arc which intersects PQ at A and B.<\/p>\n<p>(c) Taking A and B as centers and radius more than half of AB, draw two arcs which intersect each other at C.<\/p>\n<p>(d) Join OC. Thus, {tex}\\angle {\/tex}COQ is the required right angle.<\/p>\n<p>(e) Taking B and E as centre and radius more than half of BE, draw two arcs which intersect each other at the point D.<\/p>\n<p>(f) Join OD. Thus, {tex}\\overline {{\\text{OD}}} {\/tex} is the required bisector of {tex}\\angle {\/tex}COQ.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 6 Maths Exercise 14.6<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_4_Draw_an_angle_of_measure_tex153circ_tex_and_divide_it_into_four_equal_parts\"><\/span><strong>Question 4.<\/strong> Draw an angle of measure {tex}153^\\circ {\/tex} and divide it into four equal parts.<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Answer: Steps of construction<\/strong>:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/06\/mathematics\/06_math_ncert_ch14_30.jpg\" \/><\/p>\n<p>(a) Draw a ray {tex}\\overleftrightarrow {{\\text{OA}}}.{\/tex}<\/p>\n<p>(b) At O, with the help of a protractor, construct {tex}\\angle {\/tex}AOB = {tex}153^\\circ .{\/tex}<\/p>\n<p>(c) Draw {tex}\\overline {{\\text{OC}}} {\/tex} as the bisector of {tex}\\angle {\/tex}AOB.<\/p>\n<p>(d) Again, draw {tex}\\overline {{\\text{OD}}} {\/tex} as bisector of {tex}\\angle {\/tex}AOC.<\/p>\n<p>(e) Again, draw {tex}\\overline {{\\text{OE}}} {\/tex} as bisector of {tex}\\angle {\/tex}BOC.<\/p>\n<p>(f) Thus, {tex}\\overline {{\\text{OC}}} {\/tex}, {tex}\\overline {{\\text{OD}}} {\/tex} and {tex}\\overline {{\\text{OE}}} {\/tex} divide {tex}\\angle {\/tex}AOB in four equal arts.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 6 Maths Exercise 14.6<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_5Construct_with_ruler_and_compasses_angles_of_following_measures\"><\/span><strong>Question 5.<\/strong>Construct with ruler and compasses, angles of following measures:<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p>(a) {tex}{60^ \\circ }{\/tex}<\/p>\n<p>(b) {tex}{30^ \\circ }{\/tex}<\/p>\n<p>(c) {tex}{90^ \\circ }{\/tex}<\/p>\n<p>(d) {tex}{120^ \\circ }{\/tex}<\/p>\n<p>(e) {tex}{45^ \\circ }{\/tex}<\/p>\n<p>(f) {tex}{135^ \\circ }{\/tex}<\/p>\n<p><strong>Answer:<\/strong><\/p>\n<p><strong>Steps of construction<\/strong>:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/06\/mathematics\/06_math_ncert_ch14_31.jpg\" \/><\/p>\n<p>(a) {tex}{60^ \\circ }{\/tex}<\/p>\n<p>(i) Draw a ray {tex}\\overleftrightarrow {{\\text{OA}}}.{\/tex}<\/p>\n<p>(ii) Taking O as centre and convenient radius, mark an arc, which intersects {tex}\\overleftrightarrow {{\\text{OA}}}{\/tex} at P.<\/p>\n<p>(iii) Taking P as centre and same radius, cut previous arc at Q.<\/p>\n<p>(iv) Join OQ.<\/p>\n<p>Thus, {tex}\\angle {\/tex}BOA is required angle of {tex}{60^ \\circ }.{\/tex}<\/p>\n<p>(b) {tex}{30^ \\circ }{\/tex}<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/06\/mathematics\/06_math_ncert_ch14_32.jpg\" \/><\/p>\n<p>(i) Draw a ray {tex}\\overleftrightarrow {{\\text{OA}}}.{\/tex}<\/p>\n<p>(ii) Taking O as centre and convenient radius, mark an arc, which intersects {tex}\\overleftrightarrow {{\\text{OA}}}{\/tex} at P.<\/p>\n<p>(iii) Taking P as centre and same radius, cut previous arc at Q.<\/p>\n<p>(iv) Join OQ. Thus, {tex}\\angle {\/tex}BOA is required angle of {tex}{60^ \\circ }.{\/tex}<\/p>\n<p>(v) Put the pointer on P and mark an arc.<\/p>\n<p>(vi) Put the pointer on Q and with same radius, cut the previous arc at C.<\/p>\n<p>Thus, {tex}\\angle {\/tex}COA is required angle of {tex}{30^ \\circ }.{\/tex}<\/p>\n<p>(c){tex}{90^ \\circ }{\/tex}<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/06\/mathematics\/06_math_ncert_ch14_33.jpg\" \/><\/p>\n<p>(i) Draw a ray {tex}\\overleftrightarrow {{\\text{OA}}}.{\/tex}<\/p>\n<p>(ii) Taking O as centre and convenient radius, mark an arc, which intersects {tex}\\overleftrightarrow {{\\text{OA}}}{\/tex} at X.<\/p>\n<p>(iii) Taking X as centre and same radius, cut previous arc at Y.<\/p>\n<p>(iv) Taking Y as centre and same radius, draw another arc intersecting the same arc at Z.<\/p>\n<p>(v) Taking Y and Z as centers and same radius, draw two arcs intersecting each other at S.<\/p>\n<p>(vi) Join OS and produce it to form a ray OB.<\/p>\n<p>Thus, {tex}\\angle {\/tex}BOA is required angle of {tex}{90^ \\circ }.{\/tex}<\/p>\n<p>(d){tex}{120^ \\circ }{\/tex}<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/06\/mathematics\/06_math_ncert_ch14_34.jpg\" \/><\/p>\n<p>(i) Draw a ray {tex}\\overleftrightarrow {{\\text{OA}}}.{\/tex}<\/p>\n<p>(ii) Taking O as centre and convenient radius, mark an arc, which intersects {tex}\\overleftrightarrow {{\\text{OA}}}{\/tex} at P.<\/p>\n<p>(iii) Taking P as centre and same radius, cut previous arc at Q.<\/p>\n<p>(iv) Taking Q as centre and same radius cut the arc at S.<\/p>\n<p>(v) Join OS.<\/p>\n<p>Thus, {tex}\\angle {\/tex}AOD is required angle of {tex}{120^ \\circ }.{\/tex}<\/p>\n<p>(e){tex}{45^o}{\/tex}<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/06\/mathematics\/06_math_ncert_ch14_35.jpg\" \/><\/p>\n<p>(i) Draw a ray {tex}\\overrightarrow {OA} {\/tex}<\/p>\n<p>(ii) Taking O as centre and convenient radius, mark an arc, which intersects {tex}\\overrightarrow {OA} {\/tex} at X.<\/p>\n<p>(iii) Taking X as centre and same radius, cut previous arc at Y.<\/p>\n<p>(iv) Taking Y as centre and same radius, draw another arc intersecting the same arc at Z.<\/p>\n<p>(v) Taking Y and Z as centers and same radius, draw two arcs intersecting each other at S.<\/p>\n<p>(vi) Join OS and produce it to form a ray OB. Thus, {tex}\\angle {\/tex}BOA is required angle of {tex}{90^o}{\/tex}<\/p>\n<p>(vii) Draw the bisector of {tex}\\angle {\/tex}BOA.<\/p>\n<p>Thus, {tex}\\angle {\/tex}MOA is required angle of {tex}{45^o}{\/tex}<\/p>\n<p>(f){tex}{135^o}{\/tex}<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/06\/mathematics\/06_math_ncert_ch14_36.png\" \/><\/p>\n<p>(i) Draw a line PQ and take a point O on it.<\/p>\n<p>(ii) Taking O as centre and convenient radius, mark an arc, which intersects PQ at A and B.<\/p>\n<p>(iii) Taking A and B as centers and radius more than half of AB, draw two arcs intersecting each other at R.<\/p>\n<p>(iv) Join OR. Thus, {tex}\\angle {\/tex}QOR = {tex}\\angle {\/tex}POQ = {tex}{90^o}{\/tex}<\/p>\n<p>(v) Draw {tex}\\overrightarrow {OD} {\/tex} the bisector of {tex}\\angle {\/tex}POR.<\/p>\n<p>thus, {tex}\\angle {\/tex}QOD is required angle of {tex}{135^o}{\/tex}<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 6 Maths Exercise 14.6<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_6Draw_an_angle_of_measure_tex45_circ_tex_and_bisect_it\"><\/span><strong>Question 6.<\/strong>Draw an angle of measure {tex}{45^ \\circ }{\/tex} and bisect it.<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Answer: Steps of construction<\/strong>:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/06\/mathematics\/06_math_ncert_ch14_37.png\" \/><\/p>\n<p>(a)Draw a line PQ and take a point O on it.<\/p>\n<p>(b)Taking O as centre and a convenient radius, draw an arc which intersects PQ at two points A and B.<\/p>\n<p>(c)Taking A and B as centers and radius more than half of AB, draw two arcs which intersect each other at C.<\/p>\n<p>(d)Join OC. Then {tex}\\angle {\/tex}COQ is an angle of {tex}{90^o}{\/tex}<\/p>\n<p>(e)Draw {tex}\\overrightarrow {OE} {\/tex} as the bisector of {tex}\\angle {\/tex}COE. Thus, {tex}\\angle {\/tex}QOE = {tex}{45^o}{\/tex}<\/p>\n<p>(f)Again draw {tex}\\overrightarrow {OG} {\/tex} as the bisector of {tex}\\angle {\/tex}QOE.<\/p>\n<p>Thus, {tex}\\angle {\/tex}QOG = {tex}\\angle {\/tex}EOG = {tex}22\\frac{1}{2}{\\,^o}{\/tex}<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 6 Maths Exercise 14.6<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_7Draw_an_angle_of_measure_tex135_circ_tex_and_bisect_it\"><\/span><strong>Question 7.<\/strong>Draw an angle of measure {tex}{135^ \\circ }{\/tex} and bisect it.<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Answer: Steps of construction<\/strong>:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/06\/mathematics\/06_math_ncert_ch14_38.png\" \/><\/p>\n<p>(a) Draw a line PQ and take a point O on it.<\/p>\n<p>(b) Taking O as centre and convenient radius, mark an arc, which intersects PQ at A and B.<\/p>\n<p>(c) Taking A and B as centers and radius more than half of AB, draw two arcs intersecting each other at R.<\/p>\n<p>(d) Join OR. Thus, {tex}\\angle {\/tex}QOR = {tex}\\angle {\/tex}POQ = {tex}{90^o}{\/tex}<\/p>\n<p>(e) Draw {tex}\\overrightarrow {OD} {\/tex} the bisector of {tex}\\angle {\/tex}POR. Thus, {tex}\\angle {\/tex}QOD is required angle of {tex}{135^o}{\/tex}<\/p>\n<p>(f) Now, draw {tex}\\overrightarrow {OE} {\/tex} as the bisector of {tex}\\angle {\/tex}QOD.<\/p>\n<p>Thus, {tex}\\angle {\/tex}QOE = {tex}\\angle {\/tex}DOE = {tex}67\\frac{1}{2}{\\,^o}{\/tex}<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 6 Maths Exercise 14.6<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_8Draw_an_angle_of_tex70circ_tex_Make_a_copy_of_it_using_only_a_straight_edge_and_compasses\"><\/span><strong>Question 8.<\/strong>Draw an angle of {tex}70^\\circ .{\/tex} Make a copy of it using only a straight edge and compasses.<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Answer: Steps of construction<\/strong>:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/06\/mathematics\/06_math_ncert_ch14_39.png\" \/><\/p>\n<p>(a) Draw an angle {tex}{70^o}{\/tex} with protractor, i.e., {tex}\\angle POQ = {70^o}{\/tex}<\/p>\n<p>(b) Draw a ray {tex}\\overrightarrow {AB} {\/tex}<\/p>\n<p>(c) Place the compasses at O and draw an arc to cut the rays of {tex}\\angle {\/tex}POQ at L and M.<\/p>\n<p>(d) Use the same compasses, setting to draw an arc with A as centre, cutting AB at X.<\/p>\n<p>(e) Set your compasses setting to the length LM with the same radius.<\/p>\n<p>(f) Place the compasses pointer at X and draw the arc to cut the arc drawn earlier at Y.<\/p>\n<p>(g) Join AY.<\/p>\n<p>Thus, {tex}\\angle YAX = {70^o}{\/tex}<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 6 Maths Exercise 14.6<\/p>\n<h6><span class=\"ez-toc-section\" id=\"Question_9Draw_an_angle_of_tex40circ_tex_Copy_its_supplementary_angle\"><\/span><strong>Question 9.<\/strong>Draw an angle of {tex}40^\\circ .{\/tex} Copy its supplementary angle.<span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p><strong>Answer:<\/strong> Steps of construction:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/06\/mathematics\/06_math_ncert_ch14_40.png\" \/><\/p>\n<p>(a) Draw an angle of {tex}{40^o}{\/tex} with the help of protractor, naming {tex}\\angle {\/tex}AOB.<\/p>\n<p>(b) Draw a line PQ.<\/p>\n<p>(c) Take any point M on PQ.<\/p>\n<p>(d) Place the compasses at O and draw an arc to cut the rays of {tex}\\angle {\/tex}AOB at L and N.<\/p>\n<p>(e) Use the same compasses setting to draw an arc O as centre, cutting MQ at X.<\/p>\n<p>(f) Set your compasses to length LN with the same radius.<\/p>\n<p>(g) Place the compasses at X and draw the arc to cut the arc drawn earlier Y.<\/p>\n<p>(h) Join MY.<\/p>\n<p>Thus, {tex}\\angle {\/tex}QMY = {tex}{40^o}{\/tex} and {tex}\\angle {\/tex}PMY is supplementary of it.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_6_Maths_Exercise_146\"><\/span>NCERT Solutions for Class 6 Maths Exercise 14.6<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>NCERT Solutions Class 6 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 6 Maths includes text book solutions from Class 6 Maths Book . NCERT Solutions for CBSE Class 6 Maths have total 14 chapters. 6 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 6 solutions PDF and Maths ncert class 6 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=\"NCERT_Solutions_for_Mathematics_Class_3rd_to_12th\"><\/span>NCERT Solutions for Mathematics Class 3rd to 12th<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-12-mathematics\/\">NCERT Solutions for Class 12 Mathematics<\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-11-mathematics\/\">NCERT Solutions for Class 11 Mathematics<\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-10-maths\/\">NCERT Solutions for Class 10 Mathematics<\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-maths\/\">NCERT Solutions for Class 9 Mathematics<\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-8-mathematics\/\">NCERT Solutions for Class 8 Mathematics<\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-7-mathematics\/\">NCERT Solutions for Class 7 Mathematics<\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-6-maths\/\">NCERT Solutions for Class 6 Mathematics<\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-5-mathematics\/\">NCERT Solutions for Class 5 Mathematics<\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-4-mathematics\/\">NCERT Solutions for Class 4 Mathematics<\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-3-mathematics\/\">NCERT Solutions for Class 3 Mathematics<\/a><\/li>\n<\/ul>\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 6 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 6 Maths Exercise 14.6 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 6 Maths chapter wise &#8230; <a title=\"NCERT Solutions for Class 6 Maths Exercise 14.6\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-6-maths-exercise-14-6\/\" aria-label=\"More on NCERT Solutions for Class 6 Maths Exercise 14.6\">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":[1395,281],"tags":[283,196,321,1485,216],"class_list":["post-4258","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematics-cbse-class-06","category-ncert-solutions","tag-cbse-study-material","tag-class-6","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 6 Maths Exercise 14.6 | myCBSEguide<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for Class 6 Maths Exercise 14.6 in PDF format for free download. 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