{"id":25322,"date":"2021-06-22T16:29:02","date_gmt":"2021-06-22T10:59:02","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=25322"},"modified":"2022-03-02T10:36:28","modified_gmt":"2022-03-02T05:06:28","slug":"ncert-solutions-for-class-9-maths-chapter-7-ex-7-4","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-9-maths-chapter-7-ex-7-4\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 7 Triangles Ex 7.4 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.4<\/h2>\n

Question 1.
\nShow that in a right angled triangle, the hypotenuse is the longest side.
\nSolution:
\n\"NCERT
\nIn right angled triangle ABC
\n\u2220A + \u2220B + \u2220C = 180\u00b0 (Angle sum property of A)
\nor, \u2220A + \u2220C + 90 = 180\u00b0 (\u2235 \u2220B = 90)
\nor, \u2220A + \u2220C = 90\u00b0
\n\u2234 \u2220B > \u2220A
\nTherefore, AC > BC …….(i)
\n(Side opposite to greater angle is greater)
\nAgain, \u2220B > \u2220C
\n\u2234 AC > AB ……(ii)
\n(Side opposite to greater angle is greater larger)
\nFrom equation (i) and (ii)
\nAC is longer then both AB and BC.
\nTherefore AC is longes side of \u2206ABC.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nIn Fig. 7.48 side AB and AC of \u2206ABC are extended to point P and Q respectively. Also, \u2220PBC < \u2220QCB. Show that AC > AB.
\n\"NCERT
\nSolution:
\nIn \u2206ABC, we have given,
\n\u2220PBC < \u2220QCB ……(i)
\nNow, \u2220PBC + \u2220ABC = 180\u00b0 …..(ii) (Linear pair)
\nAgain, \u2220QCB + \u2220ACB = 180\u00b0 ……(iii) (Linear pair)
\nFrom equation (ii) and (iii)
\n\u2220PBC + \u2220ABC = \u2220QCB + \u2220ACB
\nBut \u2220PBC < \u2220QBC Therefore, \u2220ABC > \u2220ACB
\n\u2234 AC > AB (Side opposite to greater angle is greater)<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nIn Fig. 7.49, \u2220B > \u2220A and \u2220C < \u2220D. Show thatn AD < BC.
\n\"NCERT
\nSolution:
\nIn \u2206BAO,
\n\u2220B < \u2220A
\n\u2234 OA < OB ……(i)
\n(Single opposite to greater angle is greater)
\nAgain, In \u2206CDO
\n\u2220C < \u2220D
\n\u2234 OD < OC ……(ii)
\n(Single opposite to greater angle is greater)
\nAdd equation (i) and (ii)
\nOA + OD < OB + OC
\nAD < BC<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nAB and CD are respectively the smallest and longest sides of quadrilateral ABCD (se Fig. 7.50). Show that \u2220A > \u2220C and \u2220B > \u2220D.
\n\"NCERT
\nSolution:
\nGiven ABCD is a quadrilateral in which AB is smallest and CD is longest side.
\nTo prove that:
\n(i) \u2220A > \u2220C
\n(ii) \u2220B > \u2220D
\nConstruction: Join A and C.
\nProof: In \u2206ABC
\nAB < BC (\u2235 AB is smallest side)
\n\u2234 \u2220BCA < \u2220BAC …..(i)
\n(\u2235 Angle opposite to larger side is greater)
\nAgain, In \u2206ACD,
\nAD < CD (\u2235 CD is largest side)
\n\u2234 \u2220ACD < \u2220ADC …….(ii)
\n(Angle opposite to larger side is greater)
\n\"NCERT
\nAdding equation (i) and (ii)
\n\u2220BCA + \u2220ACD < \u2220BAC + \u2220ADC
\n\u2220BCD < \u2220BAD
\nor, \u2220C < \u2220A Similarly by joining B and D we can prove that \u2220B > \u2220D<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nIn Fig. 7.51, PR > PQ and PS bisects \u2220QPR. Prove that \u2220PSR > \u2220PSQ.
\n\"NCERT
\nSolution:
\nIn \u2206PQR
\nPR > PQ
\n\u2220Q > \u2220R
\nAdding \u22201 both side,
\n\u2220Q + \u22201 > \u2220R + \u22201
\nor, \u2220Q + \u22201 > \u2220R + \u22202 (\u2235 \u22201 = \u22202)
\n\u2220PSR > \u2220PSQ
\n(\u2235 Exterior angle is equal to the sum of opposite interior angles)<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nShow that of all line segments drawn from a given point not on it, the perpendicular line segment is shortest.
\nSolution:
\nGiven: A straight line l and a point P not lying on l. PM \u22a5 l and N is any point on l other than M.
\n\"NCERT
\nTo prove that: PM < PN
\nProof: In \u2206PMN, we have
\n\u2220M = 90
\n\u2220N < 90
\n(\u2235 \u2220M = 90
\n\u21d2 \u2220MPN + \u2220PNM = 90
\n\u21d2 \u2220P + \u2220N = 90
\n\u21d2 \u2220N < 90)
\n\u21d2\u2220N< \u2220M
\n\u21d2 PM < PN (Side opposite to greater angle is larger)
\nHence, PM is shortest of all line segments from P to AB.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.4 Question 1. Show that in a right angled triangle, the hypotenuse is the longest side. Solution: In right angled triangle ABC …<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4<\/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 7.4 - 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-7-4\/\" \/>\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 7.4 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4 Questions and Answers are prepared by our highly skilled subject experts. 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