NCERT Solutions for Class 6 Maths<\/a> Chapter 14 Practical Geometry InText Questions and Answers are prepared by our highly skilled subject experts.<\/p>\nNCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry InText Questions<\/h2>\n
NCERT In-text Question Page No. 286<\/span>
\nQuestion 1.
\nIn Step 2 of the construction using ruler and compasses, what would happen if we take the length of radius to be smaller than half the length of \\(\\overline{\\mathrm{AB}}\\) ?
\nAnswer:
\nIn case we take the radius smaller than half of the length of \\(\\overline{\\mathrm{AB}}\\), the arcs will not intersect each other at two points P and Q.<\/p>\nNCERT In-text Question Page No. 289<\/span>
\nQuestion 1.
\nIn Step 2 above, what would happen if we take radius to be smaller than half the length BC?
\nAnswer:
\nIf we take radius to be smaller than half of BC, the arcs drawn with centres B and C will not intersect each other.<\/p>\n<\/p>\n
NCERT In-text Question Page No. 290<\/span>
\nQuestion 1.
\nHow will your construct at 15\u00b0 angle?
\nAnswer:
\nSteps of construction:
\nStep I: Construct an angle \u2220ABC of 60\u00b0.
\nStep II: Bisect \u2220ABC to get an angle of 30\u00b0
\ni. e. \u2220ABD = 30\u00b0.
\n
\nStep III: Bisect \u2220ABD, such that \\(\\overline{\\mathrm{BE}}\\) is bisector of \u2220ABD.
\nThus, \u2220ABD = \\(\\frac { 1 }{ 2 }\\)(30\u00b0) = 15\u00b0.<\/p>\nNCERT In-text Question Page No. 291\u00a0<\/span>
\nQuestion 1.
\nHow will you construct a 150\u00b0 angle?
\nAnswer:
\nSteps of construction:
\nStep I: Draw a line I and mark a point O on it.
\nStep II: With centre O and a convenient radius, draw an arc intersecting I at A.
\nStep III: With the same radius and centre at A, draw an arc to cut the first arc at B.
\nStep IV: Again with the same radius and centre at B, draw another arc to intersect the
\nfirst arc at C.
\n
\nStep V : Once again with the same radius and centre at C, draw an arc to cut the first are at D.
\nStep VI: Now, bisect \u2220COD, such that \u2220COE = \u2220EOD = 30\u00b0.
\nStep VII: Since, 150\u00b0 = 120\u00b0 + 30\u00b0, therefore \u2220AOC + \u2220COE = \u2220AOE. Thus, \u2220AOE is the required angle whose measure is 150\u00b0.<\/p>\n<\/p>\n
Question 2.
\nHow will you construct a 45\u00b0 angle?
\nAnswer:
\nSteps of construction :
\nStep I: Construct an angle of 90\u00b0 as shown in
\nthe figure. \u2220POQ = 90\u00b0.
\n
\nStep II: Draw OR, the angle bisector of \u2220POQ such that \\(\\frac { 1 }{ 2 }\\)– [\u2220POQ] = \\(\\frac { 1 }{ 2 }\\) (90\u00b0) = 45\u00b0 or \u2220POR = 45\u00b0<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry InText Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry InText Questions NCERT In-text Question Page No. 286 Question 1. In Step 2 of the construction using ruler and compasses, what would …<\/p>\n
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry InText Questions<\/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 InText Questions - 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