{"id":25886,"date":"2021-06-24T14:57:38","date_gmt":"2021-06-24T09:27:38","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=25886"},"modified":"2021-06-24T14:57:38","modified_gmt":"2021-06-24T09:27:38","slug":"ncert-solutions-for-class-9-maths-chapter-11-ex-11-1","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-11-ex-11-1\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 11 Constructions Ex 11.1 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 11 Constructions Exercise 11.1<\/h2>\n

Question 1.
\nConstruct an angle of 90\u00b0 at the initial point of a given ray and justify the construction.
\nSolution:
\nSteps of construction:
\n1. Draw a straight line AB.
\n2. Taking A as centre, draw an arc, which intersects AB at a point D.
\n\"NCERT
\n3. Now, taking D as a centre, the same radius AD, intersect at E and F on the succession arc. (here AD = \\(\\widehat{\\text { DE }}\\) = \\(\\widehat{\\text { EF }}\\))
\n4. From point E take a radius (which should be greater than \\(\\frac {1}{2}\\) \\(\\widehat{\\text { EF }}\\)) then with the same radius from F intersect the arc at point C.
\n5. Now, join C to A.
\n6. The required angle \u2220ABC = 90\u00b0.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nConstruct an angle of 45\u00b0 at the initial point of a given ray and justify the construction.
\nSolution:
\n\"NCERT
\nSteps of Constructions:
\n1. Follow the instruction of the previous question upto \u2220ABC = 90\u00b0.
\n2. Take an arc from the points H and G each which Intersect at I.
\n3. Here \u2220FBC is half of \u2220ABC, FB is the angle bisector.
\n4. So, the required \u2220FBC = 45\u00b0<\/p>\n

Question 3.
\nConstruct the angles of the following measurements:
\n(i) 30\u00b0
\n(ii) 22\u00bd\u00b0
\n(iii) 15\u00b0
\nSolution:
\n(i) Steps of Construction:
\n\"NCERT
\n1. Take a straight line AB.
\n2. Draw an arc, taking A as centre which intersects AB at C.
\n3. From C take another arc \\(\\widehat{\\mathrm{CD}}\\) such that AB = \\(\\widehat{\\mathrm{CD}}\\)
\n4. From C and D take similar arc which intersects at E.
\n5. The required \u2220FAB = 30\u00b0.<\/p>\n

(ii) Steps of construction:
\n\"NCERT
\n1. Follow the instruction of question (2) i.e. \u2220FAB = 45\u00b0.
\n2. Take the same arc from points C and K, intersect at G.
\n3. The required \u2220HAB = 22\u00bd\u00b0<\/p>\n

(iii) Step of construction:
\n\"NCERT
\n1. Follow the instruction of 3:
\n(i) \u2220EAB = 30\u00b0.
\n2. Take a small arc from the points G and C that intersects at F.
\n3. The required angle \u2220FAB = 15\u00b0.<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nConstruct the following angles and verify by measuring them by a protractor:
\n(i) 75\u00b0
\n(ii) 105\u00b0
\n(iii) 135\u00b0.
\nSolution:
\n(i) Steps of construction:
\n1. Follow as in question 1. i.e. \u2220EAC = 90\u00b0.
\n2. Take the same arc from points D and E which intersect at G.
\n3. The required \u2220KAB = 75\u00b0.
\n\"NCERT<\/p>\n

(ii) Steps of construction:
\n1. Follow the instruction as in the previous question upto \u2220EAB = 90\u00b0.
\n2. Take the same arc from points D and E intersect at F.
\n3. Join F to A.
\n4. The required \u2220GAB = 105\u00b0.
\n\"NCERT<\/p>\n

(iii) Steps of construction:
\n1. Follow the instruction as \u2220EAB = 90\u00b0.
\n2. Take the same arc from the points E and F, Intersect at G.
\n3. Join G to A.
\n4. The required \u2220HAB = 135\u00b0.
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nConstruct an equilateral triangle, given its side, and justify the construction.
\nSolution:
\nSteps of construction:
\n\"NCERT
\n1. Take a straight line XY.
\n2. By measuring 6 cm on the scale, cut one point C on XY line the reverse the same to get B point. Here, BC = 6 cm
\n3. From B and C points draw an arc of the same length intersect at A.
\n4. We find here AB = BC = CA = 6 cm. Thus ABC is an equilateral triangle.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 11 Constructions Exercise 11.1 Question 1. Construct an angle of 90\u00b0 at the initial point of a given ray and justify the construction. Solution: Steps …<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1<\/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":{"spay_email":""},"categories":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-11-ex-11-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1 Questions and Answers are prepared by our highly skilled subject experts. 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