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>\nNCERT 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 \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
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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 <\/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 <\/p>\n
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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 <\/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 \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 \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 <\/p>\n
\nDraw a ray \\(\\overline{\\mathrm{OA}}\\).<\/li>\n Taking O as centre and convenient radius, mark an arc, which intersects \\(\\overline{\\mathrm{OA}}\\) at X.<\/li>\n Taking X as centre and same radius, cut previous arc at Y.<\/li>\n Taking Y as centre and same radius, draw another arc intersecting the same arc at Z.<\/li>\n Taking Y and Z as centres and same radius, draw two arcs intersecting each other at S.<\/li>\n Join OS and produce it to form a ray OB. \nThus, \u2220BOA is required angle of 90.<\/li>\n<\/ul>\n <\/p>\n
(d) 120\u00b0<\/p>\n
\nDraw a ray \\(\\overline{\\mathrm{OA}}\\).<\/li>\n Taking O as centre and convenient radius, mark an arc, which intersects \\(\\overline{\\mathrm{OA}}\\) at P.<\/li>\n Taking P as centre and same radius, cut previous arc at Q.<\/li>\n Taking Q as centre and same radius cut the arc at S.<\/li>\n Join OS. \nThus, \u2220AOD is required angle of 120\u00b0.<\/li>\n<\/ul>\n <\/p>\n
(e) 45\u00b0<\/p>\n
\nDraw a ray \\(\\overline{\\mathrm{OA}}\\) .<\/li>\n Taking O as centre and convenient radius, mark an arc, which intersects \\(\\overline{\\mathrm{OA}}\\) at X.<\/li>\n Taking X as centre and same radius, cut previous arc at Y.<\/li>\n Taking Y as centre and same radius, draw another arc intersecting the same arc at Z.<\/li>\n Taking Y and Z as centres and same radius, draw two arcs intersecting each other at S.<\/li>\n Join OS and produce it to form a ray OB. Thus, \u2220BOA is required angle of 90\u00b0.<\/li>\n Draw the bisector of \u2220BOA. \nThus, \u2220MOA is required angle of 45\u00b0.<\/li>\n<\/ul>\n <\/p>\n
(f) 135\u00b0<\/p>\n
\nDraw a line PQ and take a point O on it.<\/li>\n Taking O as centre and convenient radius, mark an arc, which intersects PQ at A and B.<\/li>\n Taking A and B as centres and radius more than half of AB, draw two arcs intersecting each other at R.<\/li>\n Join OR. Thus, \u2220QOR = \u2220POQ = 90\u00b0 .<\/li>\n Draw OD the bisector of \u2220POR. \nThus, \u2220QOD is required angle of 135\u00b0.<\/li>\n<\/ul>\n <\/p>\n
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Question 6. \nDraw an angle of measure 45\u00b0 and bisect it. \nAnswer: \nSteps of construction: \n(a) Draw a line PQ and take a point O on it. \n(b) Taking O as centre and a convenient radius, draw an arc which intersects PQ at two points 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. Then \u2220COQ is an angle of 90. \n(e) Draw \\(\\overline{\\mathrm{OE}}\\) as the bisector of \u2220COE. Thus, \u2220QOE = 45\u00b0 \n(f) Again draw \\(\\overline{\\mathrm{OG}}\\) as the bisector of \u2220QOE. \nThus, \u2220QOG = \u2220EOG = 22\\(\\frac { 1 }{ 2 }\\)\u00b0 \n <\/p>\n
Question 7. \nDraw an angle of measure 135\u00b0 and bisect it. \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, mark 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 intersecting each other at R. \n(d) Join OR. Thus, \u2220QOR = \u2220POQ = 90\u00b0. \n(e) Draw \\(\\overline{\\mathrm{OD}}\\) the bisector of \u2220POR. \nThus, \u2220QOD is required angle of 135\u00b0. \n(f) Now, draw \\(\\overline{\\mathrm{OE}}\\) as the bisector of \u2220QOD. \nThus, \u2220QOE = \u2220DOE =67\\(\\frac { 1 }{ 2 }\\)\u00b0 \n <\/p>\n
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Question 8. \nDraw an angle of 70\u00b0. Make a copy of it using only a straight edge and compasses. \nAnswer: \nSteps of construction: \nStep I: Draw a line 1 and mark a point O on it. \nStep II: Using a protractor construct \u2220AOB = 70\u00b0. \nStep III: With centre O and a suitable radius, draw an arc which intersects \\(\\overline{\\mathrm{OA}}\\) and \\(\\overline{\\mathrm{OB}}\\) at E and F respectively. \n \nStep IV: Draw a ray \\(\\overline{\\mathrm{PQ}}\\). \n \nStep V: Keeping the same radius and centre P, draw an arc intersecting \\(\\overline{\\mathrm{PQ}}\\) at R. \nStep VI: With centre R and radius equal to EF, draw an arc intersecting the previous arc at S. \nStep VII: Join PS and produce it.<\/p>\n
Question 9. \nDraw an angle of 40\u00b0 Copy its supplementary angle. \nAnswer: \nSteps of construction: \nStep I: (a) By using protractor draw \u2220AOB = 40\u00b0 \n\u2220COF is the supplementary angle. \nStep II: With centre O and a convenient radius, draw an arc which intersects \\(\\overline{\\mathrm{OC}}\\) and \\(\\overline{\\mathrm{OB}}\\) and E and F respectively. \n \nStep III: Draw a ray \\(\\overline{\\mathrm{OC}}\\). \n \nStep IV: With centre Q and same radius, draw an arc intersecting \\(\\overline{\\mathrm{OC}}\\) at L. \nStep V: With centre L and radius equal to EF draw an arc which intersects the previous arc at S. \nStep VI: Join QS and produce it. \nThus, ZPQS is the copy of the supplementary angle ZCOB.<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Exercise 14.6 Question 1. Draw \u2220POQ of measure 75\u00b0 and find its line of symmetry. Answer: Steps of construction: (a) Draw …<\/p>\n
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[10],"tags":[],"yoast_head":"\nNCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n