{"id":25249,"date":"2021-06-22T15:12:06","date_gmt":"2021-06-22T09:42:06","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=25249"},"modified":"2022-03-02T10:36:30","modified_gmt":"2022-03-02T05:06:30","slug":"ncert-solutions-for-class-9-maths-chapter-7-ex-7-2","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-9-maths-chapter-7-ex-7-2\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 7 Triangles Ex 7.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.2<\/h2>\n

Question 1.
\nIn an issosceies triangle ABC, with AB = AC, the bisectors of \u2220B and \u2220C intersect each other at O. Join A to O show that:
\n(i) OB = OC
\n(ii) AO bisects \u2220A
\nSolution:
\n(i) In \u2206ABC,
\nWe have given that
\nAB = AC
\n\u2234 \u2220B = \u2220C (Angle opposite to equal sides are equal)
\nor, \\(\\frac {1}{2}\\) \u2220B = \\(\\frac {1}{2}\\) \u2220C
\n\u2234 \u22201 = \u22202
\nNow in \u2220OBC,
\n\u22201 = \u22202
\n\u2234 OB = OC (Side opposite to equal angles are equal)<\/p>\n

\"NCERT<\/p>\n

(ii) In \u2206AOB and \u2206AOC
\nAB = AC
\nOB = OC
\n\u2220B = \u2220C
\n\u2234 \\(\\frac {1}{2}\\) \u2220B = \\(\\frac {1}{2}\\) \u2220B
\nor \u22203 = \u22204
\nTherefore, by S-A-S Congruency Condition,
\n\u2206AOB \u2245 \u2206AOC
\n\u2234 \u2220BAO = \u2220CAO (by C.P.C.T)
\n\u2234 AO is the bisector of \u2220A.<\/p>\n

Question 2.
\nIn \u2206ABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that \u2206ABC is an isosceles triangle in which AB = AC.
\n\"NCERT
\nSolution:
\nIn \u2206ABD and \u2206ACD
\nBD = CD (\u2235 AD bisects BC)
\n\u2220ADB = \u2220ADC (Each 90\u00b0)
\nAD = AD (Common)
\nBy S-A-S Congurency Condition,
\n\u2206ABD \u2245 \u2206ACD
\nTherefore, AB = AC (By C.P.C.T)
\n\u2234 ABC is an isosceles triangle.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nABC is an isosceles triangle in which altitudes BE and CF are drawn to side AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.
\n\"NCERT
\nSolution:
\nIn \u2206ABE and \u2206ACF,
\n\u2220A = \u2220A (Common)
\n\u2220AEB = \u2220AFC (each 90\u00b0)
\nAB = AC (Given)
\nBy, A-A-S Congruency Condition
\n\u2206ABE = \u2206ACP
\nTherefore, BE = CF (By C.P.C.T)<\/p>\n

Question 4.
\nABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that
\n(i) \u2206ABE \u2245 \u2206ACF
\n(ii) AB = AC, i.e. \u2206ABC is an isosceles triangle.
\n\"NCERT
\nSolution:
\n(i) \u2206ABE and \u2206ACF,
\nBE = CF (Given)
\n\u2220A = \u2220A (Common)
\n\u2220AEB = \u2220AFC (Each 90\u00b0)
\nBy A-A-S Congruency Condition
\n\u2206ABE = \u2206ACF<\/p>\n

(ii) Since \u2206ABE \u2245 \u2206ACF
\nSo, AB = AC (By C.P.C.T.)
\nor, \u2206ABC is an isosceles triangle.<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that \u2220ABD = \u2220ACD.
\n\"NCERT
\nSolution:
\nIn \u2206ABC,
\nAB = AC
\n\u2234 \u2220ABC = \u2220ACB …….(i)
\n(Angle opposite to equal sides of a \u2206 are equal)
\nAgain, In \u2206DBC
\nDB = DC
\n\u2220DBC = \u2220DCB ……(ii)
\n(Angle opposite to equal sides of a \u2206 are equal)
\nAdding equation (i) and (ii)
\n\u2220ABC + \u2220DBC = \u2220ACB + \u2220DCB
\nor \u2220ABD = \u2220ACD.<\/p>\n

Question 6.
\n\u2206ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that \u2220BCD is a right angle.
\n\"NCERT
\nSolution:
\nIn \u2206ABC,
\nAB = AC
\n\u2220ABC = \u2220ACB ……(i)
\n(Angles opposite to equal sides of a \u2206ABC)
\nNow, In \u2206ACD,
\nAC = AD
\n\u2220ACD = \u2220ADC ……(ii)
\n(Angles opposite to equal sides of \u2206ACD)
\nNow, \u2220BAC + \u2220CAD = 180\u00b0 ……(iii) (Linear pair)
\nAlso, \u2220CAD = \u2220ABC + \u2220ACB (Exterior angle of \u2206ABC)
\n\u2220CAD = 2\u2220ACB …….(iv)
\nFrom equation (i),
\n\u2220ABC = \u2220ACB
\nand \u2220BAC = \u2220ACD + \u2220ADC (Exterior angle of \u2206ADC)
\n\u2220BAC = 2\u2220ACD …….(v)
\n(From equation (ii), \u2220ACD = \u2220ADC)
\nFrom equation (iii) we have,
\n\u2220BAC + \u2220CAD = 180\u00b0
\n\u21d2 2\u2220ACD + 2\u2220ACB = 180\u00b0 (From equation (iv) and (v))
\n\u21d2 2(\u2220ACD + \u2220ACB) = 180\u00b0
\n\u21d2 2(\u2220BCD) = 180\u00b0
\n\u21d2 \u2220BCD = 90\u00b0<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nABC is a right-angled triangle in which \u2220A = 90\u00b0 and AB = AC. Find \u2220B and \u2220C.
\n\"NCERT
\nSolution:
\nIn \u2206ABC,
\nAB = AC
\n\u2234 \u2220B = \u2220C (Angle opposite to equal sides of \u2206ABC)
\nNow, \u2220A + \u2220B + \u2220C = 180\u00b0 (Angle sum property)
\nor, 90\u00b0 + \u2220B + \u2220B = 180\u00b0 (\u2235 \u2220A = 90\u00b0 and \u2220B = \u2220C)
\nor, 2\u2220B = 90\u00b0
\n\u2234 \u2220B = 45\u00b0
\nTherefore, \u2220B = \u2220C = 45\u00b0.<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nShow that the angles of equilateral triangles are 60\u00b0 each.
\n\"NCERT
\nSolution:
\nIn \u2206ABC,
\nAB = BC = AC (Because ABC is an equilateral triangle)
\n\u2234 \u2220C = \u2220A = \u2220B (Angle opposite to equal sides of \u2206ABC)
\nNow, \u2220A + \u2220B + \u2220C = 180\u00b0 (Angle sum property of \u2206)
\nor, \u2220A + \u2220A + \u2220A = 180\u00b0 (\u2235 \u2220A = \u2220B = \u2220C)
\nor, 3\u2220A = 180
\nTherefore, \u2220A = 60\u00b0
\nAgain, \u2220A = \u2220B = \u2220C
\n\u2234 \u2220A = \u2220B = \u2220C = 60\u00b0
\nTherefore, each angle of an equilateral \u2206 is 60\u00b0.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.2 Question 1. In an issosceies triangle ABC, with AB = AC, the bisectors of \u2220B and \u2220C intersect each other at …<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2<\/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 7 Triangles Ex 7.2 - MCQ Questions<\/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:\/\/mcq-questions.com\/ncert-solutions-for-class-9-maths-chapter-7-ex-7-2\/\" \/>\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 7 Triangles Ex 7.2 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 Questions and Answers are prepared by our highly skilled subject experts. 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