{"id":24856,"date":"2021-06-21T14:31:05","date_gmt":"2021-06-21T09:01:05","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=24856"},"modified":"2022-03-02T10:38:09","modified_gmt":"2022-03-02T05:08:09","slug":"ncert-solutions-for-class-6-maths-chapter-14-ex-14-6","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-6-maths-chapter-14-ex-14-6\/","title":{"rendered":"NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6"},"content":{"rendered":"

These NCERT Solutions for Class 6 Maths<\/a> Chapter 14 Practical Geometry Ex 14.6 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Exercise 14.6<\/h2>\n

Question 1.
\nDraw \u2220POQ of measure 75\u00b0 and find its line of symmetry.
\nAnswer:
\n\"NCERT
\nSteps of construction:
\n(a) Draw a line l and mark a point O on it.
\n(b) Place the pointer of the compasses at O and draw an arc of any radius which intersects the line l at A.
\n(c) Taking same radius, with centre A, cut the previous arc at B.
\n(d) Join OB, then \u2220BOA = 60 .
\n(e) Taking same radius, with centre B, cut the previous arc at C.
\n(f) Draw bisector of \u2220BOC. The angle is of 90 . Mark it at D. Thus, \u2220DOA = 90\u00b0
\n(g) Draw \\(\\overline{\\mathrm{OP}}\\) as bisector of \u2220DOB. Thus, \u2220POA = 75\u00b0<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nDraw an angle of measure 147\u00b0 and construct its bisector.
\nAnswer:
\nSteps of construction:
\n(a) Draw a ray \\(\\overline{\\mathrm{OA}}\\) .
\n(b) With the help of protractor, construct \u2220AOB = 147\u00b0
\n(c) Taking centre O and any convenient radius, draw an arc which intersects the arms \\(\\overline{\\mathrm{OA}}\\) and \\(\\overline{\\mathrm{OB}}\\) at P and Q respectively.
\n(d) Taking P as centre and radius more than half of PQ, draw an arc.
\n(e) Taking Q as centre and with the same radius, draw another arc which intersects the previous at R.
\n(f) Join OR and produce it.
\n(g) Thus, \\(\\overline{\\mathrm{OR}}\\) is the required bisector of \u2220AOB.
\n\"NCERT<\/p>\n

Question 3.
\nDraw a right angle and construct its bisector.
\nAnswer:
\nSteps of construction:
\n(a) Draw a line PQ and take a point O on it.
\n(b) Taking O as centre and convenient radius, draw an arc which intersects PQ at A and B.
\n(c) Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C.
\n(d) Join OC. Thus, \u2220COQ is the required right angle.
\n(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.
\n(f) Join OD. Thus, \\(\\overline{\\mathrm{OD}}\\) is the required bisector of \u2220COQ.
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nDraw an angle of measure 153\u00b0 and divide it into four equal parts.
\nAnswer:
\nSteps of construction:
\n(a) Draw a ray \\(\\overline{\\mathrm{OA}}\\)
\n(b) At O, with the help of a protractor, construct \u2220AOB = 153\u00b0.
\n(c) Draw \\(\\overline{\\mathrm{OC}}\\) as the bisector of \u2220AOB.
\n(d) Again, draw \\(\\overline{\\mathrm{OD}}\\) as bisector of \u2220AOC.
\n(e) Again, draw \\(\\overline{\\mathrm{OE}}\\) as bisector of \u2220BOC.
\n(f) Thus, \\(\\overline{\\mathrm{OC}}\\), \\(\\overline{\\mathrm{OD}}\\) and \\(\\overline{\\mathrm{OE}}\\) divide \u2220AOB in four equal arts.
\n\"NCERT<\/p>\n

Question 5.
\nConstruct with ruler and compasses, angles of following measures:
\n(a) 60\u00b0
\n(b) 30\u00b0
\n(c) 90\u00b0
\n(d) 120\u00b0
\n(e) 45\u00b0
\n(f) 135\u00b0
\nAnswer:
\nSteps of construction:
\n(a) 60\u00b0
\n\"NCERT
\n(i) Draw a ray \\(\\overline{\\mathrm{OA}}\\).
\n(ii) Taking O as centre and convenient radius, mark an arc, which intersects \\(\\overline{\\mathrm{OA}}\\) at P.
\n(iii) Taking P as centre and same radius, cut previous arc at Q.
\n(iv) Join OQ.
\nThus, \u2220BOA is required angle of 60\u00b0.<\/p>\n

(b) 30\u00b0
\n\"NCERT
\n(i) Draw a ray \\(\\overline{\\mathrm{OA}}\\) .
\n(ii) Taking O as centre and convenient radius, mark an arc, which intersects \\(\\overline{\\mathrm{OA}}\\) at P.
\n(iii) Taking P as centre and same radius, cut previous arc at Q.
\n(iv) Join OQ. Thus, \u2220BOA is required angle of 60\u00b0.
\n(v) Put the pointer on P and mark an arc.
\n(vi) Put the pointer on Q and with same radius, cut the previous arc at C. Thus, \u2220COA is required angle of 30\u00b0.<\/p>\n

(c) 90\u00b0
\n\"NCERT<\/p>\n