NCERT Solutions for Class 9 Maths<\/a> Chapter 11 Constructions Ex 11.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\nNCERT Solutions for Class 9 Maths Chapter 11 Constructions Exercise 11.2<\/h2>\n Question 1. \nConstruct a triangle ABC in which BC = 7 cm, \u2220B = 75\u00b0 and AB + AC = 13 cm. \nSolution: \n \nSteps of construction: \nStep-I: Draw the base BC = 7 cm and at point B make an angle say \u2220XBC = 75\u00b0. \nStep-II: Cut the line segment BD = AB + AC = 13 cm from die ray BX. \nStep-III: Join DC and draw perpendicular bisector of DC. Which intersect DB at A. \nStep-IV: Join AC. \nStep-V: Triangle ABC is the required triangle.<\/p>\n
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Question 2. \nConstruct a triangle ABC in which BC = 8 cm, \u2220B = 45\u00b0 and AB – AC = 3.5 cm. \nSolution: \n \nSteps of Construction: \nStep-I: Draw the base BC = 8 cm and make \u2220XBC = 45\u00b0. \nStep-II: Cut the Sine segment BD = AB – AC = 3.5 cm from ray BX. \nStep-III: Join DC and draw the perpendicular bisector of DC. \nStep-IV: Let the perpendicular bisector of DC intersect XB at A. \nStep-V: Join AC. \nStep-VI: Triangle ABC is the required triangle.<\/p>\n
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Question 3. \nConstruct a triangle PQR in which QR = 6 m, \u2220Q = 60\u00b0 and PR – PQ = 2 cm. \nSolution: \n \nSteps of Construction: \nStep-I: Draw the base QR = 6 cm and at point Q make an angle \u2220XQR = 60\u00b0. \nStep-II: Cutline segment QM = PR – PQ = 2 cm from the line XQ extended on the opposite side of line segment QR. \nStep-III: Join MC and draw the perpendicular bisector of MR which intersect XQ at P. \nStep-IV: Join PR. \nStep-V: Triangle PQR is the required triangle.<\/p>\n
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Question 4. \nConstruct a triangle XYZ in which \u2220Y = 30\u00b0, \u2220Z = 90\u00b0 and XY + YZ + ZX = 11 cm. \nSolution: \n \nSteps of Construction: \nStep-I: Draw the line segment PQ = 11 cm. (= XY + YZ + ZX) \nStep-II: At P construct an angle of 30\u00b0 and at Q, an angle of 90\u00b0. \nStep-III: Bisect these angles. Let the bisector of these angles intersect at a point X. \nStep-IV: Draw perpendicular bisectors of XP and XQ which intersect PQ at Y and Z respectively. \nStep-V: Join XY and XZ. \nStep-VI: Triangle XYZ is the required triangle.<\/p>\n
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Question 5. \nConstruct a right triangle whose base is 12 cm and the sum of its hypotenuse and other side is 18 cm. \nSolution: \n \nSteps of Construction: \nStep-I: Draw the base BC = 12 cm and at a point B makes an angle say \u2220XBC = 90\u00b0. \nStep-II: Cut the line segment BD = 18 cm (= AB + AC). \nStep-III: Join DC and make a \u2220DCY = \u2220BDC. \nStep-IV: CY intersect BX at A. \nStep-V: Join AC. \nStep-VI: Triangle ABC is the required triangle.<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 11 Constructions Exercise 11.2 Question 1. Construct a triangle ABC in which BC = 7 cm, \u2220B = 75\u00b0 and AB + AC = 13 …<\/p>\n
NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.2<\/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":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.2 - 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