{"id":27479,"date":"2021-06-30T17:18:23","date_gmt":"2021-06-30T11:48:23","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=27479"},"modified":"2021-06-30T17:18:23","modified_gmt":"2021-06-30T11:48:23","slug":"ncert-solutions-for-class-7-maths-chapter-6-ex-6-3","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-7-maths-chapter-6-ex-6-3\/","title":{"rendered":"NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.3"},"content":{"rendered":"

These NCERT Solutions for Class 7 Maths<\/a> Chapter 6 The Triangles and Its Properties Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Exercise 6.3<\/h2>\n

Question 1.
\nFind the value of the unknown x in the following diagrams:
\n\"NCERT
\n\"NCERT
\nAnswer:
\n(i) x + 50\u00b0 + 60\u00b0= 180\u00b0
\n(angle sum property of a triangle)
\nx + 110\u00b0 = 180\u00b0
\nx = 180\u00b0- 110\u00b0
\n= 70\u00b0
\n\u2220x = 70\u00b0<\/p>\n

(ii) x + 90\u00b0 + 30\u00b0= 180\u00b0
\n(angle sum property of a triangle)
\nx + 120\u00b0 = 180\u00b0
\nx = 180\u00b0 – 120\u00b0
\n= 60\u00b0
\n\u2220x = 60\u00b0<\/p>\n

(iii) 30\u00b0 + 110\u00b0 + x = 180\u00b0
\n(angle sum property of a triangle)
\n140\u00b0+ x = 180\u00b0
\nx = 180\u00b0 – 140\u00b0
\n= 40\u00b0
\n\u2220x =40\u00b0<\/p>\n

(iv) x + x + 50\u00b0= 180\u00b0
\n(angle sum property of a triangle)
\n2x + 50\u00b0 = 180\u00b0
\n2x = 180\u00b0 – 50\u00b0
\n2x = 130\u00b0
\nx = \\(\\frac{130^{\\circ}}{2}\\) = 65\u00b0
\n\u2220x = 65\u00b0<\/p>\n

(v) x + x + x = 180\u00b0
\n(angle sum property of a triangle)
\n3x = 180\u00b0
\nx = \\(\\frac{180^{\\circ}}{3}\\)
\nx = 60\u00b0
\n\u2220x = 60\u00b0<\/p>\n

(vi) x + 2x + 90\u00b0 = 180\u00b0
\n(angle sum property of a triangle)
\n3x + 90\u00b0 = 180\u00b0
\n3x = 180\u00b0 – 90\u00b0
\n= 90\u00b0
\nx = \\(\\frac{90^{\\circ}}{3}\\)
\n= 30\u00b0
\n\u2234 \u2220x = 30\u00b0<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nFind the values of the unknown x and y in the following diagrams:
\n\"NCERT<\/p>\n

Answer:
\n(i) \u2220y + 120\u00b0= 180\u00b0
\n(linear pair of angles)
\n\u2220y = 180\u00b0 – 120\u00b0
\n= 60\u00b0
\n\u2220x + \u2220y + 50\u00b0 = 180\u00b0
\n(using the angle sum property of a triangle)
\n\u2220x + 60\u00b0 + 50\u00b0 = 180\u00b0
\n\u2220x + 110\u00b0 = 180\u00b0
\n\u2220x = 180\u00b0- 110\u00b0
\n= 70\u00b0
\nThus, \u2220x = 70\u00b0 and \u2220y = 60\u00b0<\/p>\n

(ii) \u2220y = 80\u00b0
\n(vertically opposite angles are equal)
\n\u2220x + \u2220y + \u222050\u00b0 = 180\u00b0
\n(using the angle sum property of a triangle)
\n\u2220x + 80\u00b0 + 50\u00b0= 180\u00b0
\n\u2220x + 130\u00b0 = 180\u00b0
\n\u2220x = 180\u00b0 – 130\u00b0
\n= 50\u00b0
\nThus, \u2220x = 50\u00b0 and \u2220y = 80\u00b0<\/p>\n

(iii) 50\u00b0 + 60\u00b0 + \u2220y = 180\u00b0
\n(using the angle sum property of a triangle)
\n110\u00b0 + \u2220y – 180\u00b0
\n\u2220y = 180\u00b0- 110\u00b0
\n= 70\u00b0
\n\u2220x and \u2220y form a linear pair.
\n\u2220x + \u2220y = 180\u00b0
\n\u2220x + 70\u00b0 = 180\u00b0
\n\u2220x = 180\u00b0 – 70\u00b0
\n= 110\u00b0
\nThus, \u2220x =110\u00b0 and \u2220y = 70\u00b0<\/p>\n

(iv) \u2220x = 60\u00b0
\n(vertically opposite angles)
\n\u2220x+ \u2220y + 30\u00b0 = 180\u00b0
\n(angle sum property of a triangle)
\n60\u00b0 + \u2220y + 30\u00b0 = 180\u00b0
\n\u2220y + 90\u00b0 = 180\u00b0
\n\u2220y = 180\u00b0 – 90\u00b0
\n\u2220y = 90\u00b0
\nThus, \u2220x = 60\u00b0 and \u2220y = 90\u00b0<\/p>\n

(v) \u2220y = 90\u00b0
\n(vertically opposite angles are equal)
\n\u2220x + \u2220x + \u2220y = 180\u00b0
\n(angle sum property of triangle)
\n2\u2220x + 90\u00b0 = 180\u00b0
\n2x = 180\u00b0 – 90\u00b0
\n2x = 90\u00b0
\nx = \\(\\frac{90^{\\circ}}{2}\\) = 45\u00b0
\nThus, \u2220y = 90\u00b0,\u2220x = 45\u00b0<\/p>\n

(vi) One angle of the triangle = y
\nEach of the other two angles is equal to their vertically opposite angle \u2018x\u2019
\n\u2220x + \u2220x + \u2220y = 180\u00b0
\n(angle sum property of a triangle)
\n2x + y = 180\u00b0
\n2x + x = 180\u00b0
\n3x = 180\u00b0
\nx = \\(\\frac{180^{\\circ}}{3}\\)
\nx = 60\u00b0
\ny = x
\n(Vertically opposite angles) y = 60\u00b0
\nThus, x = 60\u00b0 and y = 60\u00b0<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Exercise 6.3 Question 1. Find the value of the unknown x in the following diagrams: …<\/p>\n

NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.3<\/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":[9],"tags":[],"yoast_head":"\nNCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.3 - 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-7-maths-chapter-6-ex-6-3\/\" \/>\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 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.3 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts. 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