{"id":27602,"date":"2020-07-08T13:16:48","date_gmt":"2020-07-08T07:46:48","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=27602"},"modified":"2021-07-02T15:07:06","modified_gmt":"2021-07-02T09:37:06","slug":"kseeb-solutions-for-class-8-maths-chapter-11-additional-questions","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-8-maths-chapter-11-additional-questions\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 11 Congruency of Triangles Additional Questions"},"content":{"rendered":"

Students can Download Maths Chapter 11 Congruency of 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 11 Congruency of Triangles Additional Questions<\/h2>\n

Question 1.
\nFill in the blanks to make the statements true
\na. In right triangle the hypotenuse is the greatest<\/span> side
\nb. The sum of three altitudes of a triangle is less<\/span> than its perimeter.
\nc. The sum of any two sides of a triangle is greater<\/span> than the third side.
\nd. If two angles of a triangle are unequal, then the smaller<\/span> angle has the smaller side opposite to it.
\ne. If two sides of a triangle are unequal, then the larger side has larger<\/span> angle opposite to it.<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nJustify the following statements with reasons
\na. The sum of three sides of triangles is more than the sum of its altitudes
\nSolution:
\nIn \u2206 ABC Let AD, BE and CF be the altitudes The perpendicular is the shortest distance from a point to B the line
\n\"KSEEB
\n\u2234 AD < AB, BE < BC and CF < AC
\n\u2234 AD + BE + CF < AB T BC + AC or \u2234 AB + BC + AC > AD + BE + CF<\/p>\n

b. The sum of any two sides of a triangle is greater than twice the median drawn to the third side.
\nSolution:
\nIn \u2206 ABC AP is the median from A to BC
\n\"KSEEB
\nTo prove : AB + AC > 2APQ Construction : Produce AP to Q such that AP = PQ join CQ
\nProof: In \u2206 APB and \u2206 PQG
\nBP = PC [P is the midpoint of BC]
\nAP = PQ [construction]
\n\u2220APB = \u2220CPQ [V.O.A]
\n\u2234 \u2206 APB = \u2206 CPQ [SAS postulate]
\n\u2234 AB = CQ [corresponding sides]
\nIn \u2206 ACQ, AC + CQ > AQ [Sum of two sides is greater than the third side]
\nAC + AB > AP + PQ
\nAC + AB > 2 AP
\n[\u2234 AB = CQ and AP = PQ]<\/p>\n

\"KSEEB<\/p>\n

c. Difference of any two sides is less than the third side
\nSolution:
\nIn \u2206 ABC
\nTo prove (i) AC – AB < BC (ii) BC – AC < AB (iii) BC – AB < AC Construction Let AC > AB . On AC mark a point D such that AB = AD. Join BD
\nProof : AB = AD
\n\u2234\u22201 = \u22202 [Base angles of an isosceles triangle]
\nExtr \u22202 = \u22204 + \u22203
\n\u22202 > \u22204 [Exterior angle > each interior opp angle]
\nIn \u2206 ADB, AD is produced to C
\n\u2234 Extr \u22203 = \u22201 + \u2220A
\n\"KSEEB
\n\u22203 > \u22201 [Exterior angle > each interior angle]
\n\u22203 >\u22202 [\u2235 \u22201 = \u22202]
\n\u2234 \u22203 > \u22202 and \u22202 > \u22204 means \u22203 > \u22204
\n\u2234 BC > DC [side opposite to larger angle is larger].
\nBC > AC – AD [\u2235 DC = AC – AD]
\nBC >AC – AB [\u2235 AD = AB]
\nor
\nAC – AB < BC
\nSimilarly BC – AC < AB and BC – AB < AC<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nTwo triangles ABC and DBC have common base BC. Suppose AB = DC and \u2220ABC – \u2220BCD prove that AC = BD.
\nSolution:
\n\"KSEEB
\nIn \u2206 ABC and \u2206 DBC AB = DC (data) \u2220ABC = \u2220BCD [Data]
\nBC = BC (common side)
\n\u2234 \u2206 ABC \u2245 \u2206 DBC (SAS postulate)
\n\u2234 AC = BD (Corresponding sides)<\/p>\n

Question 4.
\nLet AB and CD be two line segments such that AD and BC intersect at O. Suppose AO = OC and BO = OD. Prove that AB = CD.
\nSolution:
\n\"KSEEB
\nIn \u2206 AOB and \u2206 COD
\nAO = OC [data]
\nBO = OD [data]
\n\u2220AOB = \u2220COD [vertically oppsoite angles]
\n\u2234 \u2206 AOB \u2245 \u2206 COD [SAS postulate]
\n\u2234 AB = CD [corresponding sides]<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nLet ABC be a triangle. Draw a triangle BDC externally on BC such that AB = BD and AC = CD. Prove that \u2206 ABC \u2245 \u2206 DBC
\nSolution:
\n\"KSEEB
\nIn \u2206 ABC and \u2206 DBC
\nAB = BD [data]
\nAC = CD [data]
\nBC = BC [common side]
\n\u2234 \u2206 ABC = ADBC [SSS postulate]<\/p>\n

Question 6.
\nLet ABCD be a square and let points P on AB and Q on DC be such that DP = AQ. Prove that BP = CQ.
\nSolution:
\n\"KSEEB
\nIn \u2206 APD and \u2206 AQD \u2220PAD = \u2220QDA [= 90\u00b0]
\nAD = AD
\n[common side]
\nDP = AQ [data]
\n\u2206 APD \u2245 \u2206 AQD [RHS postulate]
\n\u2234 AP = DQ [corresponding sides]
\n\u2234 AB = DC [siodes of a square]
\nAB – AP = DC – DQ
\nBP = CQ [Axiom 3]<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nIn a triangle ABC, AB = AC. Suppose P is a point on AB and Q is a point on AC such that AP = AQ. Prove that \u2206 APC \u2245 \u2206 AQB.
\nSolution:
\n\"KSEEB
\nIn \u2206 APC and \u2206 AQB AP = AQ [data]
\nAC = AB [data]
\n\u2220PAC – \u2220BAQ [common angle]
\n\u2234 \u2206 APC \u2245\u00a0\u2206 AQB [SAS postulate]<\/p>\n

Question 8.
\nIn an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.
\nSolution:
\n\"KSEEB
\nIn \u2206 ABC, Let AB = BC A
\n\u2220B = \u2220C [\u2234 base angles of isosceles triangle]
\n\u2220A = 2[\u2220B + \u2220C] [data] B C
\n\u2220A = 2[\u2220B + \u2220B] [\u2234 \u2220B = \u2220C]
\n\u2220A = 2[2\u2220B]
\n\u2220A = 4\u2220B
\nIn \u2206 ABC, \u2220A + \u2220B + \u2220C = 180\u00b0 [Sum of the angles of the triangle]
\n4\u2220B + \u2220B = \u2220B = 180\u00b0 [substituting for \u2220A = 4\u2220B and |C = [B]
\n6\u2220B = 180\u00b0 \u2220B = \\(\\frac{180}{6}\\) = 30\u00b0
\n\u2220B = 30\u00b0 \u2220C = 30\u00b0
\n\u2220A = 4\u2220B = 4 x 30 = 120\u00b0<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nIf the bisector of the vertical angle of a triangle bisects the base, show that the triangle is isosceles.
\nSolution:
\n\"KSEEB
\nData : In \u2206 ABC, AD bisects \u2220A and BD = DC
\nTo prove : AB = AC Construction : produce AD to P such that AD = DP and B joint PC Proof : In
\nADP and CDP
\nAD = DP [construction]
\nBD = DC [data]
\n\u2220APB = \u2220CDP [V.O.A]
\n\u2234 \u2206 ABD \u2245 \u2206 CDP [SAS postulate]
\n\u2234 AB = CP [corresponding sides]
\n\u2220BAD = \u2220DPC [corresponding angles] But \u2220BAD = \u2220CAD [AD bisects \u2220A ]
\n\u2234 \u2220CAD = \u2220DPC
\n\u2234 CP = AC
\n[sides opposite to equal angles] But CP = AB [proved]
\n\u2234 AB = AC<\/p>\n

Question 10.
\nSuppose ABC is an isosceles triangle with AB = AC side BA is produced to D such that BA = AD. Prove that \u2220BCD is a right angle .
\nSolution:
\n\"KSEEB
\nLet \u2220ABC = x
\nIn \u2206 ABC, AB = AC [data]
\n\u2234 \u2220ABC = \u2220ACB = x [Base angles of isosceles triangle]
\nIn \u2206 ADC, Let \u2220ADC = y
\nAB = AC = AD [data]
\n\u2220ACD = \u2220ADC = y
\n[Base angles of isosceles triangle]
\nIn \u2206 BCD, \u2220DBC +\u2220BCD + \u2220BDC = 180\u00b0
\n[Sum of the angles of a triangle]
\n\u2220DBC + \u2220ACD + \u2220ACB + \u2220BDC = 180\u00b0
\nx + y + x + y = 180\u00b0
\n2x + 2y = 180\u00b0
\n2(x + y) = 180\u00b0
\nx + y = \\(\\frac{180}{2}\\) =90\u00b0
\n\u2220BCD = 90\u00b0 [ \u2235 x + y = \u2220BCD<\/p>\n

\"KSEEB<\/p>\n

Question 11.
\nLet AB, CD be two line segements such that AB || CD and AD || BC. Let E be the midpoint of BC and let DE extended meet AB in F. Prove that AB = BF.
\nSolution:
\nIn \u2206 BEF and \u2206 DEC
\n\"KSEEB
\nBE = EC [data]
\n\u2220BEF – \u2220DEC [vertically opposite angles]
\n\u2220EBF = \u2220ECF [Alternate angles]
\n\u2234 \u2206 BEF \u2245 \u2206 DEC [ASA postulate]
\n\u2234 BF = DC [corresponding sides]
\nBut AB = DC [data]
\n\u2234 AB = BF<\/p>\n

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

Students can Download Maths Chapter 11 Congruency of 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 11 Congruency of Triangles Additional Questions Question 1. Fill in 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\/27602"}],"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=27602"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27602\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=27602"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=27602"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=27602"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}