{"id":738,"date":"2022-08-23T04:23:06","date_gmt":"2022-08-23T04:23:06","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=738"},"modified":"2024-02-21T06:53:54","modified_gmt":"2024-02-21T06:53:54","slug":"ncert-solutions-class-6-maths-chapter-14-practical-geometry-ex-14-6","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-6-maths-chapter-14-practical-geometry-ex-14-6\/","title":{"rendered":"NCERT Solutions Class 6 Maths Chapter 14 Practical Geometry Ex 14.6"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"font-family: georgia, palatino, serif;\"><strong><span style=\"color: #000000;\">NCERT Solutions Class 6 Maths<\/span><\/strong><\/span><br \/>\n<span style=\"font-family: georgia, palatino, serif;\"><strong><span style=\"color: #000000;\">Chapter &#8211; 14 (Practical Geometry)<\/span><\/strong><\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 6 Mathematics <strong>Chapter &#8211; 14 Practical Geometry <\/strong>Exercise 14.6 has been provided here to help the students in solving the questions from this exercise.<\/span><\/p>\n<h4 id=\"chapter-14\"><span style=\"font-family: georgia, palatino, serif;\"><strong><span id=\"Chapter_14_Practical_Geometry\" style=\"color: #000000;\">Chapter 14: Practical Geometry<\/span><\/strong><\/span><\/h4>\n<ul>\n<li><span style=\"font-family: georgia, palatino, serif;\"><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-6-maths-chapter-14-practical-geometry-ex-14-1\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 6 Maths Exercise &#8211; 14.1<\/a><\/span><\/li>\n<li><span style=\"font-family: georgia, palatino, serif;\"><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-6-maths-chapter-14-practical-geometry-ex-14-2\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 6 Maths Exercise &#8211; 14.2<\/a><\/span><\/li>\n<li><span style=\"font-family: georgia, palatino, serif;\"><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-6-maths-chapter-14-practical-geometry-ex-14-3\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 6 Maths Exercise &#8211; 14.3<\/a><\/span><\/li>\n<li><span style=\"font-family: georgia, palatino, serif;\"><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-6-maths-chapter-14-practical-geometry-ex-14-4\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 6 Maths Exercise &#8211; 14.4<\/a><\/span><\/li>\n<li><span style=\"font-family: georgia, palatino, serif;\"><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-6-maths-chapter-14-practical-geometry-ex-14-5\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 6 Maths Exercise &#8211; 14.5<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"font-family: georgia, palatino, serif;\"><strong><span style=\"color: #000000;\">Exercise &#8211; 14.6<\/span><\/strong><\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>1. Draw \u2220POQ of measure 75\u00b0 and find its line of symmetry.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solutions:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-794 aligncenter\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A1.png\" alt=\"NCERT Class 6 Math Solution\" width=\"227\" height=\"175\" \/><\/strong><strong>Step 1:<\/strong> Draw a line <i>l<\/i>\u00a0and mark two points O and Q on it, as shown in the figure. Draw an arc of convenient radius, while taking point O as centre. Let it intersect line\u00a0<i>l<\/i>\u00a0at R.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Step 2: <\/strong>Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Step 3: <\/strong>Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Step 4: <\/strong>Taking S and T as centre, draw an arc of same radius to intersect each other at U.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Step 5: <\/strong>\u00a0Join OU. Let it intersect the arc at V. Now, taking S and V as centres, draw arcs with radius more than <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> SV. Let those intersect each other at P. Join OP, which is the ray making 75\u00b0 with the line <i>l<\/i>.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Step 6: <\/strong>Let this ray be intersecting our major arc at point W. Now, taking R and W as centres, draw arcs with radius more than <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> RW in the interior of angle of 75\u00ba. Let these be intersecting each other at X. Join OX.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">OX is the line of symmetry for \u2220POQ = 75\u00b0.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>2. Draw an angle of measure 147\u00b0 and construct its bisector.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solutions:<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Following steps are followed to construct an angle of measure 147<sup>0<\/sup>\u00a0and its bisector<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-796 aligncenter\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6Ans2.png\" alt=\"NCERT Class 6 Math Solution\" width=\"314\" height=\"125\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6Ans2.png 618w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6Ans2-300x119.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6Ans2-480x191.png 480w\" sizes=\"auto, (max-width: 314px) 100vw, 314px\" \/><strong>Step 1: <\/strong>Draw a line l and mark point O on it. Place the centre of protractor at point O and the zero edge along line l.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Step 2: <\/strong>Mark a point A at an angle of measure 147<sup>0<\/sup>. Join OA. Now OA is the required ray making 147<sup>0<\/sup> with line l.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Step 3: <\/strong>By taking point O as centre, draw an arc of convenient radius. Let this intersect both rays of angle 147<sup>0<\/sup> at points A and B.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Step 4: <\/strong>By taking A and B as centres draw arcs of radius more than 1 \/ 2 AB in the interior angle of 147<sup>0<\/sup>. Let these intersect each other at point C. Join OC.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">OC is the required bisector of 147<sup>0<\/sup>\u00a0angle<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>3. Draw a right angle and construct its bisector.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solutions:<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Following steps are followed to construct a right angle and its bisector.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-797 aligncenter\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A3.png\" alt=\"NCERT Class 6 Math Solution\" width=\"359\" height=\"175\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A3.png 646w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A3-300x146.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A3-480x234.png 480w\" sizes=\"auto, (max-width: 359px) 100vw, 359px\" \/><strong>Step 1:<\/strong> Draw a line l and mark a point P on it. Draw an arc of convenient radius by taking point P as centre. Let this intersect line l at R.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Step 2: <\/strong>Draw an arc by taking R as centre and with the same radius as before such that it is intersecting the previously drawn arc at S.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Step 3: <\/strong>Take S as centre and with the same radius as before, draw an arc intersecting the arc at T as shown in figure<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Step 4: <\/strong>By taking S and T as centres draw arcs of same radius such that they are intersecting each other at U.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Step 5: <\/strong>Join PU. PU is the required ray making a right angle with the line l. Let this intersect major arc at point V.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Step 6: <\/strong>Now take R and V as centres, draw arcs with radius more than 1 \/ 2 RV to intersect each other at point W. Join PW.<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">PW is the required bisector of this right angle.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>4. Draw an angle of measure 153\u00b0 and divide it into four equal parts.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solutions:<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Following steps are followed to construct an angle of measure 153<sup>0<\/sup>\u00a0and its bisector<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\" wp-image-808 aligncenter\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A4.png\" alt=\"NCERT Class 6 Math Solution\" width=\"401\" height=\"192\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A4.png 641w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A4-300x144.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A4-480x230.png 480w\" sizes=\"auto, (max-width: 401px) 100vw, 401px\" \/><strong>Step 1: <\/strong>Draw a line l and mark a point O on it. Place the centre of protractor at point O and the zero edge along line l.<br \/>\n<strong>Step 2: <\/strong>Mark a point A at the measure of angle 153<sup>0<\/sup>. Join OA. Now OA is the required ray making 153<sup>0<\/sup> with line l.<br \/>\n<strong>Step 3: <\/strong>Draw an arc of convenient radius by taking point O as centre. Let this intersect both rays of angle 153<sup>0<\/sup>\u00a0at points A and B.<br \/>\n<strong>Step 4: <\/strong>Take A and B as centres and draw arcs of radius more than <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/>AB in the interior of angle of 153<sup>0<\/sup>. Let these intersect each other at C. Join OC.<br \/>\n<strong>Step 5: <\/strong>Let OC intersect major arc at point D. Draw arcs of radius more than <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> AD with A and D as centres and also D and B as centres. Let these are intersecting each other at points E and F, respectively. Now join OE and OF<br \/>\nOF, OC, OE are the rays dividing 153<sup>0<\/sup>\u00a0angle into four equal parts.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>5. Construct with ruler and compasses, angles of following measures:<br \/>\n<\/strong>(a) 60\u00b0<br \/>\n(b) 30\u00b0<br \/>\n(c) 90\u00b0<br \/>\n(d) 120\u00b0<br \/>\n(e) 45\u00b0<br \/>\n(f) 135\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solutions:<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(a) 60<sup>0<br \/>\n<\/sup><\/strong>Following steps are followed to construct an angle of 60<sup>0<\/sup><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-806 aligncenter\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5A.png\" alt=\"NCERT Class 6 Math Solution\" width=\"369\" height=\"168\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5A.png 698w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5A-300x137.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5A-480x219.png 480w\" sizes=\"auto, (max-width: 369px) 100vw, 369px\" \/><strong>Step 1: <\/strong>Draw a line l and mark a point P on it. Take P as centre and with convenient radius, draw an arc of a circle such that it intersects the line l at Q.<br \/>\n<strong>Step 2: <\/strong>Take Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point R.<br \/>\n<strong>Step 3: <\/strong>Join PR. PR is the required ray making 60<sup>0<\/sup>\u00a0with the line l.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(b) 30<sup>0<br \/>\n<\/sup><\/strong>Following steps are followed to construct an angle of 30<sup>0<br \/>\n<\/sup><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-807 aligncenter\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5B.png\" alt=\"NCERT Class 6 Math Solution\" width=\"330\" height=\"133\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5B.png 664w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5B-300x121.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5B-480x194.png 480w\" sizes=\"auto, (max-width: 330px) 100vw, 330px\" \/><strong><br \/>\nStep 1: <\/strong>Draw a line l and mark a point P on it. By taking P as centre and with convenient radius, draw an arc of a circle such that it is intersecting the line l at Q.<br \/>\n<strong>Step 2: <\/strong>Take Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point R.<br \/>\n<strong>Step 3: <\/strong>By taking Q and R as centres and with radius more than <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> RQ draw arcs such that they are intersecting each other at S. Join PS which is the required ray making 30<sup>0<\/sup>\u00a0with the line l.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(c) 90<sup>0<br \/>\n<\/sup><\/strong>Following steps are followed to construct an angle of measure 90<sup>0<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-805 aligncenter\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5C.png\" alt=\"NCERT Class 6 Math Solution\" width=\"315\" height=\"143\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5C.png 669w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5C-300x136.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5C-480x218.png 480w\" sizes=\"auto, (max-width: 315px) 100vw, 315px\" \/><\/sup><strong>Step 1: <\/strong>Draw a line l and mark a point P on it. Take P as centre and with convenient radius, draw an arc of a circle such that it is intersecting the line l at Q.<br \/>\n<strong>Step 2: <\/strong>Take Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at R.<br \/>\n<strong>Step 3: <\/strong>By taking R as centre and with the same radius as before, draw an arc intersecting the arc at S as shown in figure.<br \/>\n<strong>Step 4: <\/strong>Now take R and S as centre, draw arc of same radius to intersect each other at T.<br \/>\n<strong>Step 5: <\/strong>Join PT, which is the required ray making 90<sup>0<\/sup>\u00a0with the line l.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(d) 120<sup>0<br \/>\n<\/sup><\/strong>Following steps are followed to construct an angle of measure 120<sup>0<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-804 aligncenter\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5D.png\" alt=\"NCERT Class 6 Math Solution\" width=\"345\" height=\"161\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5D.png 653w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5D-300x140.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5D-480x224.png 480w\" sizes=\"auto, (max-width: 345px) 100vw, 345px\" \/><\/sup><strong>Step 1: <\/strong>Draw a line l and mark a point P on it. Taking P as centre and with convenient radius, draw an arc of circle such that it is intersecting the line l at Q.<br \/>\n<strong>Step 2: <\/strong>By taking Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at R.<br \/>\n<strong>Step 3: <\/strong>Take R as centre and with the same radius as before, draw an arc such that it is intersecting the arc at S as shown in figure.<br \/>\n<strong>Step 4: <\/strong>Join PS, which is the required ray making 120<sup>0<\/sup>\u00a0with the line l<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(e) 45<sup>0<br \/>\n<\/sup><\/strong>Following steps are followed to construct an angle of measure 45<sup>0<br \/>\n<\/sup><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-803 aligncenter\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5E.png\" alt=\"NCERT Class 6 Math Solution\" width=\"384\" height=\"187\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5E.png 664w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5E-300x146.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5E-480x233.png 480w\" sizes=\"auto, (max-width: 384px) 100vw, 384px\" \/><strong>Step 1: <\/strong>Draw a line l and mark a point P on it. Take P as centre and with convenient radius, draw an arc of a circle such that it is intersecting the line l at Q.<br \/>\n<strong>Step 2: <\/strong>Take Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at R.<br \/>\n<strong>Step 3: <\/strong>By taking R as centre and with the same radius as before, draw an arc such that it is intersecting the arc at S as shown in figure.<br \/>\n<strong>Step 4: <\/strong>Take R and S as centres and draw arcs of same radius such that they are intersecting each other at T.<br \/>\n<strong>Step 5: <\/strong>Join PT. Let this intersect the major arc at point U.<br \/>\n<strong>Step 6: <\/strong>Now take Q and U as centres and draw arcs with radius more than <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> QU to intersect each other at point V. Join PV.<br \/>\nPV is the required ray making 45<sup>0<\/sup>\u00a0with the line l<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>(f) 135<sup>0<br \/>\n<\/sup><\/strong>Following steps are followed to construct an angle of measure 135<sup>0<br \/>\n<\/sup><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-802 aligncenter\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5F.png\" alt=\"NCERT Class 6 Math Solution\" width=\"373\" height=\"163\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5F.png 672w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5F-300x131.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A5F-480x210.png 480w\" sizes=\"auto, (max-width: 373px) 100vw, 373px\" \/><strong>Step 1: <\/strong>Draw a line l and mark a point P on it. Taking P as centre and with convenient radius, draw a semicircle which intersects the line l at Q and R, respectively.<br \/>\n<strong>Step 2: <\/strong>By taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.<br \/>\n<strong>Step 3: <\/strong>Taking S as centre and with the same radius as before, draw an arc such that it is intersecting the arc at T as shown in figure.<br \/>\n<strong>Step 4: <\/strong>Take S and T as centres, draw arcs of same radius to intersect each other at U.<br \/>\n<strong>Step 5: <\/strong>Join PU. Let this intersect the arc at V. Now take Q and V as centres and with radius more than <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> QV, draw arcs to intersect each other at W.<br \/>\n<strong>Step 6: <\/strong>Join PW which is the required ray making 135<sup>0<\/sup>\u00a0with the line l<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>6. Draw an angle of measure 45\u00b0 and bisect it.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solutions:<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Following steps are followed to construct an angle of measure 45<sup>0<\/sup>\u00a0and its bisector.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-800 aligncenter\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A6.png\" alt=\"NCERT Class 6 Math Solution\" width=\"288\" height=\"131\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A6.png 645w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A6-300x136.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A6-480x218.png 480w\" sizes=\"auto, (max-width: 288px) 100vw, 288px\" \/><strong>Step 1: <\/strong>Using the protractor \u2220POQ of 45<sup>0<\/sup> measure may be formed on a line l.<br \/>\n<strong>Step 2: <\/strong>Draw an arc of convenient radius with centre as O. Let this intersect both rays of angle 45<sup>0<\/sup> at points A and B.<br \/>\n<strong>Step 3: <\/strong>Take A and B as centres, draw arcs of radius more than 1 \/ 2 AB in the interior of angle of 45<sup>0<\/sup>. Let these intersect each other at C. Join OC<br \/>\nOC is the required bisector of 45<sup>0<\/sup>\u00a0angle<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>7. Draw an angle of measure 135\u00b0 and bisect it.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solutions:<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Following steps are followed to construct an angle of measure 135<sup>0<\/sup>\u00a0and its bisector.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-801 aligncenter\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A7.png\" alt=\"NCERT Class 6 Math Solution\" width=\"298\" height=\"140\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A7.png 653w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A7-300x141.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A7-480x226.png 480w\" sizes=\"auto, (max-width: 298px) 100vw, 298px\" \/><strong>Step 1: <\/strong>By using a protractor \u2220POQ of 135<sup>0<\/sup> measure may be formed on a line l.<br \/>\n<strong>Step 2: <\/strong>Draw an arc of convenient radius by taking O as centre. Let this intersect both rays of angle 135<sup>0<\/sup>\u00a0at points A and B, respectively.<br \/>\n<strong>Step 3: <\/strong>Take A and B as centres, draw arcs of radius more than 1 \/ 2 AB in the interior of angle of 135<sup>0<\/sup>. Let these intersect each other at C. Join OC.<br \/>\nOC is the required bisector of 135<sup>0<\/sup>\u00a0angle<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>8. Draw an angle of 70<sup>0<\/sup>. Make a copy of it using only a straight edge and compasses.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solutions:<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Following steps are followed to construct an angle of measure 70<sup>0<\/sup>\u00a0and its copy.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-798 aligncenter\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A8.png\" alt=\"NCERT Class 6 Math Solution\" width=\"417\" height=\"125\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A8.png 661w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A8-300x90.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A8-480x144.png 480w\" sizes=\"auto, (max-width: 417px) 100vw, 417px\" \/><strong>Step 1: <\/strong>Draw a line l and mark a point O on it. Now place the centre of protractor at point O and the zero edge along line l.<br \/>\n<strong>Step 2: <\/strong>Mark a point A at an angle of measure 70<sup>0<\/sup>. Join OA. Now OA is the ray making 70<sup>0<\/sup>\u00a0with line l. With point O as centre, draw an arc of convenient radius in the interior of 70<sup>0<\/sup>\u00a0angle. Let this intersect both rays of angle 70<sup>0<\/sup>\u00a0at points B and C, respectively<br \/>\n<strong>Step 3: <\/strong>Draw a line m and mark a point P on it. Again draw an arc with same radius as before and P as centre. Let it cut the line m at point D.<br \/>\n<strong>Step 4: <\/strong>Adjust the compasses up to the length of BC. With this radius draw an arc taking D as centre which intersects the previously drawn arc at point E.<br \/>\n<strong>Step 5: <\/strong>Join PE. Here PE is the required ray which makes same angle of measure 70<sup>0<\/sup>\u00a0with the line m<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>9. Draw an angle of 40<sup>0<\/sup>. Copy its supplementary angle.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\"><strong>Solutions:<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Following steps are followed to construct an angle of measure 45<sup>0<\/sup>\u00a0and a copy of its supplementary angle<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-799 aligncenter\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A9.png\" alt=\"NCERT Class 6 Math Solution\" width=\"316\" height=\"298\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A9.png 685w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A9-300x283.png 300w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/07\/NCERT-Class-6-Math-Ex-13.6A9-480x453.png 480w\" sizes=\"auto, (max-width: 316px) 100vw, 316px\" \/><br \/>\n<strong>Step 1: <\/strong>Draw a line segment <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\overline{PQ}\" alt=\"\\overline{PQ}\" align=\"absmiddle\" \/>\u00a0and mark a point O on it. Place the centre of protractor at point O and the zero edge along line segment <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\overline{PQ}\" alt=\"\\overline{PQ}\" align=\"absmiddle\" \/>.<br \/>\n<strong>Step 2: <\/strong>Mark a point A at an angle of measure 40<sup>0<\/sup>. Join OA. Here OA is the required ray making 40<sup>0<\/sup> with <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\overline{PQ}\" alt=\"\\overline{PQ}\" align=\"absmiddle\" \/>. \u2220POA is the supplementary angle of 40<sup>0<\/sup>.<br \/>\n<strong>Step 3: <\/strong>With point O as centre, draw an arc of convenient radius in the interior of \u2220POA. Let this intersect both rays of \u2220POA at points B and C, respectively<strong>.<br \/>\n<\/strong><strong>Step 4: <\/strong>Draw a line m and mark a point S on it. Again draw an arc by taking S as centre with the same radius as used before. Let it cut the line m at point T.<br \/>\n<strong>Step 5: <\/strong>Now adjust the compasses up to the length of BC. Taking T as centre draw an arc with this radius which will intersect the previously drawn arc at point R<strong>.<br \/>\n<\/strong><strong>Step 6: <\/strong>Join RS. Here RS is the required ray which makes same angle with the line m, as the supplementary of 40<sup>0<\/sup>\u00a0[i.e 140<sup>0<\/sup>]<\/span><\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 100%; background-color: #f2e079; text-align: center;\"><span style=\"color: #000000; font-size: 14pt; font-family: georgia, palatino, serif;\"><strong>NCERT Class 6<sup>th\u00a0<\/sup>Solution\u00a0<\/strong><\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%; text-align: left;\"><span style=\"color: #ff0000; font-size: 14pt; font-family: georgia, palatino, serif;\"><strong><span style=\"color: #0000ff; font-size: 12pt;\"><a class=\"row-title\" style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-6-english\/\" target=\"_blank\" rel=\"noopener\" aria-label=\"\u201cNCERT Solutions Class 6 English\u201d (Edit)\">NCERT Solutions Class 6 English<\/a><\/span><\/strong><\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%; text-align: left;\"><span style=\"color: #ff0000; font-size: 14pt; font-family: georgia, palatino, serif;\"><strong><span style=\"color: #0000ff; font-size: 12pt;\"><a class=\"row-title\" style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-6-hindi\/\" aria-label=\"\u201cNCERT Solutions Class 6 Maths\u201d (Edit)\">NCERT Solutions Class 6 Hindi<\/a><\/span><\/strong><\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%; text-align: left;\"><span style=\"color: #ff0000; font-size: 14pt; font-family: georgia, palatino, serif;\"><strong><span style=\"color: #0000ff; font-size: 12pt;\"><a class=\"row-title\" style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-6-maths\/\" target=\"_blank\" rel=\"noopener\" aria-label=\"\u201cNCERT Solutions Class 6 Maths\u201d (Edit)\">NCERT Solutions Class 6 Maths<\/a><\/span><\/strong><\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%; text-align: left;\"><span style=\"font-family: georgia, palatino, serif;\"><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-6-sanskrit\/\" target=\"_blank\" rel=\"noopener\"><span style=\"color: #ff0000; font-size: 14pt;\"><strong><span style=\"color: #0000ff; font-size: 12pt;\">NCERT Solutions Class 6 Sanskrit<\/span><\/strong><\/span><\/a><\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%; text-align: left;\"><span style=\"color: #ff0000; font-size: 14pt; font-family: georgia, palatino, serif;\"><strong><span style=\"color: #0000ff; font-size: 12pt;\"><a class=\"row-title\" style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-6-science\/\" target=\"_blank\" rel=\"noopener\" aria-label=\"\u201cNCERT Solutions Class 6 Social Science\u201d (Edit)\">NCERT Solutions Class 6 Science<\/a><\/span><\/strong><\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%; text-align: left;\"><span style=\"color: #ff0000; font-size: 14pt; font-family: georgia, palatino, serif;\"><strong><span style=\"color: #0000ff; font-size: 12pt;\"><a class=\"row-title\" style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-6-social-science\/\" target=\"_blank\" rel=\"noopener\" aria-label=\"\u201cNCERT Solutions Class 6 Social Science\u201d (Edit)\">NCERT Solutions Class 6 Social Science<\/a><\/span><\/strong><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 6 Maths Chapter &#8211; 14 (Practical Geometry) The NCERT Solutions in English Language for Class 6 Mathematics Chapter &#8211; 14 Practical Geometry Exercise 14.6 has been provided here to help the students in solving the questions from this exercise. Chapter 14: Practical Geometry NCERT Solution Class 6 Maths Exercise &#8211; 14.1 NCERT [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[31,32,4,5],"class_list":["post-738","post","type-post","status-publish","format-standard","hentry","category-class-6-maths","tag-chapter-14-practical-geometry-in-english","tag-ncert-solutions-class-6-maths-chapter-14-in-english","tag-ncert-solutions-class-6-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT Solutions Class 6 Maths Chapter 14 Practical Geometry Ex 14.6 | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 6 Maths Chapter - 14 (Practical Geometry) The NCERT Solutions in English Language for Class 6 Mathematics Chapter - 14 Practical Geometry Exercise 14.6 has been provided here to help the students in solving the questions from this exercise. 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