{"id":27455,"date":"2020-07-02T12:44:50","date_gmt":"2020-07-02T07:14:50","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=27455"},"modified":"2021-07-02T15:07:08","modified_gmt":"2021-07-02T09:37:08","slug":"kseeb-solutions-for-class-8-maths-chapter-6-additional-questions","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-8-maths-chapter-6-additional-questions\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions"},"content":{"rendered":"

Students can Download Maths Chapter 6 Theorems on Triangles Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions<\/h2>\n

Question 1.
\nFill up the blanks to make the following statements true
\na. Sum of the angles of a triangle in 180\u00b0<\/span>
\nb. An exterior angle of a triangle is equal to the sum of interior<\/span> opposite angles.
\nc. An exteior angle of a triangle is always more<\/span> than either of the interior opposite angles.
\nd. A triangle can not have more than one<\/span> right angles.
\ne. A triangle can not have more than one<\/span> obtuse angle.<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nChoose the correct answer from the given alternatives.
\na. In a triangle ABC, \u2220A = 80\u00b0 and AB = BAC then \u2220B is __________
\nA. 50\u00b0
\nB. 60\u00b0
\nC. 40\u00b0
\nD. 70\u00b0
\nSolution:
\nA. 50\u00b0<\/p>\n

b. In right-angled triangle, \u2220A is right angle and \u2220B = 35\u00b0 then \u2220C is _______
\nA. 65\u00b0
\nB. 55\u00b0
\nC. 75\u00b0
\nD. 45\u00b0
\nSolution:
\nB. 55\u00b0<\/p>\n

c. In \u2206ABC, \u2220B = \u2220C = 45\u00b0, then the triangle is ________
\nA. Right-angled
\nB. Acute angled
\nC. Obtuse angled
\nD. Equilateral triangle
\nSolution:
\nA. Right-angled<\/p>\n

d. In an equilateral triangle, each exterior angle is ________
\nA. 60\u00b0
\nB. 90\u00b0
\nC. 120\u00b0
\nD. 150\u00b0
\nSolution:
\nC. 120\u00b0<\/p>\n

e. Sum of the three exterior angles of a triangle is __________
\nA. two right angles
\nB. three right angles
\nC. one right angle
\nD. four right angles
\nSolution:
\nD. Four right angles<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nIn a triangle ABC, \u2220B = 70\u00b0 find \u2220A + \u2220C
\nSolution:
\n\u2220A + \u2220B + \u2220C = 180\u00b0 [Sum of the angles of a triangle is 180\u00b0]
\n\u2220A + 70 + \u2220C = 180\u00b0
\n\u2220A + \u2220C = 180\u00b0 – 70
\n\u2220A + \u2220C = 110\u00b0<\/p>\n

Question 4.
\nIn a triangle ABC, \u2220A =110\u00b0 and AB = AC find \u2220B and \u2220C
\nSolution:
\nAB = AC
\n\u2234 \u2220C = \u2220B [Base angles of an isosceles triangle]
\n\u2220A + \u2220B + \u2220C = 180\u00b0 [Sum of the angles of a triangle is 180\u00b0]
\n110\u00b0 + \u2220B + \u2220B = 180\u00b0
\n2\u2220B = 180 – 110
\n2\u2220B = 70\u00b0
\n\u2220B = \\(\\frac{70}{2}\\)
\n\u2220B = 35\u00b0
\n\u2220B = \u2220C = 35\u00b0<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nIf three angles of a triangle are in the ratio 2 : 3 : 5 determine three angles.
\nSolution:
\nLet the common ratio be x The three angles are 2x, 3x and 5x 2x + 3x + 5x = 180\u00b0 [Sum of the angles of triangle is 180\u00b0]
\n10x = 180\u00b0
\nx = \\(\\frac{180}{10}\\)
\n2x = 2 \u00d718 = 36\u00b0
\n3x = 3 \u00d7 18 = 54\u00b0
\n5x = 5 \u00d7 18 = 90\u00b0<\/p>\n

Question 6.
\nThe angles of a triangle are arranged in ascending order of magnitude. If the difference between consecuttive angles is 15\u00b0 find the three angles.
\nSolution:
\nLet the first angle be x then the second angle is x + 15 and third angle is x + 30.
\nx + x + 15 + x + 30 = 180\u00b0 [Sum of the angles]
\n3x + 45 = 180
\n3x = 180 – 45
\n3x = 135
\nx = \\(\\frac{135}{3}\\) = 45\u00b0
\nFirst angle = x = 45\u00b0
\nSecond angle = x + 15 = 45 + 15 = 60\u00b0
\nThird angle = x + 30 = 45 + 30 = 75\u00b0<\/p>\n

Question 7.
\nThe sum of two angles of a triangle is equal to its third angle. Determine the measure of third angle.
\nSolution:
\nLet the sum of two angles be x and the third angle be y
\nx + y = 180\u00b0 [Sum of the angles of the triangle]
\ny + y = 180\u00b0 [\u2234 sum of two angles = third angle]
\n2y = 180\u00b0
\ny = \\(\\frac{180}{2}\\)
\ny = 90\u00b0
\n\u2234 The third angle is 90\u00b0<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nIn a triangle ABC if 2\u2220A = 3\u2220B = 6\u2220C determine \u2220A , \u2220B and \u2220C
\nSolution:
\n\"KSEEB
\n\u2220A + \u2220B + \u2220C = 180\u00b0 [Sum of the angles of a triangle]
\n\"KSEEB
\n\"KSEEB<\/p>\n

Question 9.
\nThe angles of a triangle are x – 40\u00b0, x – 20 and \\(\\frac{1}{2}\\) x + 15\u00b0 find the value of x
\nSolution:
\nx – 40 + x – 20 + \\(\\frac{1}{2}\\) x + 15 = 180\u00b0 [Sum of the angles of a triangle]
\nx + x + \\(\\frac{1}{2}\\) x – 60 + 15 = 1 80\u00b0
\nx + x + \\(\\frac{1}{2}\\) x – 45 = 180\u00b0
\n2x + \\(\\frac{1}{2}\\) x = 180 + 45
\n\\(\\frac{4 x+x}{2}\\) = 225
\n5x = 225 x 2
\n5x = 450
\nx = \\(\\frac{450}{5}\\)
\nx = 90\u00b0<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nIn \u2206ABC \u2220A – \u2220B = 15\u00b0 and \u2220B – \u2220C = 30\u00b0 find \u2206ABC \u2220A , \u2220B and \u2220C
\nSolution:
\n\u2220A – \u2220B = 15\u00b0
\n\u2234 \u2220A = 15 + \u2220B
\n\u2220B – \u2220C = 30
\n\u2234 \u2220B = 30 + \u2220C
\n\u2220A + \u2220B + \u2220C = 180\u00b0 [Sum of the angle of a triangle]
\n15 + \u2220B + \u2220B + \u2220C = 180\u00b0 (\u2234 \u2220A = 15 + \u2220B)
\n15 + 2\u2220B + \u2220C = 180\u00b0
\n15 + 2[30 + \u2220C + \u2220C = 180\u00b0 (\u2234 \u2220B = 30 + \u2220C)
\n15 + 60 + 2\u2220C] + \u2220C = 180\u00b0
\n75 + 3\u2220C = 180\u00b0
\n3\u2220C = 180 – 75
\n3\u2220C = 105\u00b0
\n\u2220C = \\(\\frac{105^{\\circ}}{3}\\)
\n\u2220C = 35\u00b0
\nNow \u2220B = 30 + \u2220C
\n\u2220B = 30 + 35
\n\u2220B = 65\u00b0
\n\u2220A = 15 + \u2220B
\n\u2220A = 15 + 65
\n\u2220A = 180\u00b0<\/p>\n

\"KSEEB<\/p>\n

Question 11.
\nThe sum of two angles of a triangle 80\u00b0 and their difference is 20\u00b0 find the angles of the triangle.
\nSolution:
\nLet the angles be x and y
\n\"KSEEB
\nx = \\(\\frac{100}{2}\\) ; x = 50\u00b0
\nx + y = 80\u00b0
\n50 + y = 80\u00b0
\ny = 80 – 50
\ny = 30\u00b0
\nLet the third angle be z
\nx + y + z = 180\u00b0 [Sum of the angles]
\n50 + 30 + z = 180\u00b0
\n80 + z = 180\u00b0
\nz = 180 – 80
\nz = 100\u00b0<\/p>\n

Question 12.
\nIn a triangle ABC, \u2220B = 60\u00b0, \u2220C = 80\u00b0. Suppose the bisectors of \u2220B and \u2220C meet at L find \u2220BLC.
\nSolution:
\n\"KSEEB
\nBL bisects \u2220B
\n\"KSEEB
\n\u2220LBC + \u2220BLC + \u2220LCB = 180\u00b0 (Sum of the angles of a triangle)
\n30 + \u2220BLC + 40\u00b0 = 180\u00b0
\n70 + \u2220BLC = 180\u00b0
\n\u2220BLC = 180 – 70
\n\u2220BLC = 110\u00b0<\/p>\n

\"KSEEB<\/p>\n

Question 13.
\nIn a triangle, each of the smaller angles is half of the largest angle. Find the angles.
\nSolution:
\nLet the smallest angle be x then the largest angle is 2x.
\nx + x + 2x = 180\u00b0 [Sum of the angles of triangle]
\n4x = 180\u00b0
\nx = \\(\\frac{180}{4}\\)
\nx = 45\u00b0
\n2x = 2 \u00d7 45\u00b0 = 90\u00b0
\n\u2234 The angles are 45\u00b0, 45\u00b0 and 90\u00b0<\/p>\n

Question 14.
\nIn a triangle each of the bigger angle is twice the third angle find the angles.
\nSolution:
\nLet the third angle be x. then the bigger angles is 2x 2x + 2x + x = 180\u00b0 [Sum of the angles of triangle]
\n5x = 180\u00b0
\nx = \\(\\frac{180}{5}\\)
\nx = 36\u00b0
\n2x = 2 \u00d7 36\u00b0 = 72\u00b0
\n\u2234 The angles are 72\u00b0, 72\u00b0 and 36\u00b0<\/p>\n

Question 15.
\nIn a triangle ABC, \u2220B = 50\u00b0 and \u2220A = 60\u00b0. Suppose BC is extended D. Find \u2220ACD
\nSolution
\n\u2220ACD = \u2220 ABC + \u2220BAC [Exterior angle = sum of interior opposite angles]
\n\"KSEEB
\n\u2220ACD = 50 + 60
\n\u2220ACD = 110\u00b0<\/p>\n

Question 16.
\nIn an isosceles triangle, the vertex angle is twice the sum of the base angles. Find the angles of the triangle.
\nSolution:
\n\"KSEEB
\nIn \u2206ABC, AB = BC
\n\u2234 \u2220B = \u2220C
\n\u2220Base angles of an isosceles triangle]
\nGiven \u2220A = 2(\u2220B + \u2220C)
\n\u2220A +\u2220B + \u2220C = 180\u00b0
\n[Sum of the angles of a triangle]
\n2(\u2220B + \u2220C) + \u2220B + \u2220C = 180\u00b0
\n[Substituted for \u2220A ]
\n2\u2220B + 2\u2220C + \u2220B + \u2220C = 180\u00b0
\n2\u2220B + 2\u2220B + \u2220B + \u2220B = 180\u00b0 [\u2234 \u2220B = \u2220C]
\n6\u2220B = 180\u00b0 \u2220B = \\(\\frac{180}{6}\\) = 30\u00b0
\n\u2220B = 30\u00b0, \u2220C = 30\u00b0 and \u2220A = 2(\u2220B + \u2220C)
\n= 2 (30 + 30) = 2 (60) = 120\u00b0<\/p>\n

\"KSEEB<\/p>\n

Question 17.
\nFind the sum of all the angles at the five vertices of the adjoining star.
\n\"KSEEB
\nSolution:
\nSince all the sides of the star are equal to each other the Afes AGL, CDB, EFD, GHF and KHL are isosceles triangles. Let each base angle equal to ‘x’.
\nIn the figure, BDFHL is a pentagon & its sides are extended in order. \u2234 The sum of exterior angles = 360\u00b0
\n\u2220DBC + \u2220FDE + \u2220GFH + \u2220KHL + \u2220BLA = 360\u00b0
\nx + x + x + x + x = 360\u00b0
\n5x = 360\u00b0
\nx = \\(\\frac{360^{\\circ}}{5}\\) = 72\u00b0
\nIn \u2206ABL,
\n\u2220A + \u2220ABL + \u2220ALB = 180\u00b0
\n\u2220A + x + x = 180\u00b0
\n\u2220A + 72 + 72 = 180\u00b0
\n\u2220A + 144 = 180\u00b0
\n\u2220A = 180\u00b0 – 144 \u2220A = 36\u00b0
\n\u2220C, \u2220E, \u2220G and \u2220K = 36\u00b0
\n\u2234 Sum of all the angles at the five vertices = 36\u00b0 \u00d7 5 = 180\u00b0<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 6 Theorems on Triangles Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions Question 1. Fill up the …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27455"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=27455"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27455\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=27455"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=27455"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=27455"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}