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>\nNCERT 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 \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
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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 \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
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Question 3. \nIn Fig. 7.49, \u2220B > \u2220A and \u2220C < \u2220D. Show thatn AD < BC. \n \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
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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 \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 \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
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Question 5. \nIn Fig. 7.51, PR > PQ and PS bisects \u2220QPR. Prove that \u2220PSR > \u2220PSQ. \n \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
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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 \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 \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n