{"id":4553,"date":"2021-03-01T15:45:49","date_gmt":"2021-03-01T10:15:49","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=4553"},"modified":"2021-09-15T11:40:36","modified_gmt":"2021-09-15T06:10:36","slug":"ap-board-9th-class-maths-solutions-chapter-7-ex-7-3","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-9th-class-maths-solutions-chapter-7-ex-7-3\/","title":{"rendered":"AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.3"},"content":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions<\/a> Chapter 7 Triangles Ex 7.3 Textbook Questions and Answers.<\/p>\n

AP State Syllabus 9th Class Maths Solutions 7th Lesson Triangles Exercise 7.3<\/h2>\n

\"AP<\/p>\n

Question 1.
\nAD is an altitude of an isosceles triangle ABC in which AB = AC. Show that, (i) AD bisects BC (ii) AD bisects \u2220A.
\nSolution:
\n\"AP
\nGiven that in \u0394ABC, AB = AC
\nand AD \u22a5 BC
\ni) Now in \u0394ABD and \u0394ACD
\nAB = AC (given)
\n\u2220ADB = ADC (given AD \u22a5 BC)
\nAD = AD (common)
\n\u2234 \u0394ABD \u2245 \u0394ACD (\u2235 RHS congruence)
\n\u21d2 BD = CD (CPCT)
\n\u21d2 AD, bisects BC.<\/p>\n

ii) Also \u2220BAD = \u2220CAD
\n(CPCT of \u0394ABD \u2245 \u0394ACD )
\n\u2234 AD bisects \u2220A.<\/p>\n

\"AP<\/p>\n

Question 2.
\nTwo sides AB, BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of \u0394PQR (see figure). Show that:
\n(i) \u0394ABM \u2245 \u0394PQN
\nii) \u0394ABC \u2245 \u0394PQR
\n\"AP
\nSolution:
\nGiven that
\nAB = PQ
\nAM = PN
\ni) Now in \u0394ABM and \u0394PQN
\nAB = PQ (given)
\nAM = PN (given)
\nBM = QN (\u2235 BC = QR \u21d2 \\(\\frac { 1 }{ 2 }\\)BC = \\(\\frac { 1 }{ 2 }\\)QR \u21d2 BM = QN)
\n\u2234 \u0394ABM \u2245 \u0394PQN
\n(\u2235 SSS congruence)<\/p>\n

ii) In \u0394ABC and \u0394PQR
\nAB = PQ (given)
\nBC = QR (given)
\n\u2220ABC = \u2220PQN [\u2235 CPCT of \u0394ABM and \u0394PQN from (i)]
\n\u2234 \u0394ABC \u2245 \u0394PQR
\n(\u2235 SAS congruence)<\/p>\n

\"AP<\/p>\n

Question 3.
\nBE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
\nSolution:
\n\"AP
\nIn \u0394ABC altitude BE and CF are equal.
\nNow in \u0394BCE and \u0394CBF
\n\u2220BEC = \u2220CFB (\u2235 given 90\u00b0)
\nBC = BC (common; hypotenuse)
\nCF = BE (given)
\n\u2234 \u0394BEC \u2245 \u0394CBF
\n\u21d2 \u2220EBC = \u2220FCB (\u2235 CPCT)
\nBut these are also the interior angles opposite to sides AC and AB of \u0394ABC.
\n\u21d2 AC = AB
\nHence proved.<\/p>\n

\"AP<\/p>\n

Question 4.
\n\u0394ABC is an isosceles triangle in which AB = AC. Show that \u2220B = \u2220C.
\n(Hint : Draw AP \u22a5 BQ (Using RHS congruence rule)
\nSolution:
\n\"AP
\nGiven the \u0394ABC is an isosceles triangle and AB = AC
\nLet D be the mid point of BC; Join A, D.
\nNow in \u0394ABD and \u0394ACD
\nAB = AC (given)
\nBD = DC (construction)
\nAD = AD (common)
\n\u2234 \u0394ABD \u2245 \u0394ACD (\u2235 SSS congruence)
\n\u21d2 \u2220B = \u2220C [\u2235 CPCT]<\/p>\n

Question 5.
\n\u0394ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that \u2220BCD is a right angle.
\n\"AP
\nSolution:
\n\"AP
\nGiven that in \u0394DBC; AB = AC; AD = AB
\nIn \u0394ABC
\n\u2220ABC + \u2220ACB = \u2220DAC …………… (1)
\n[\u2235 exterior angle]
\nIn \u0394ACD
\n\u2220ADC + \u2220ACD = \u2220BAC ………………(2)
\nAdding (1) & (2)
\n\u2220DAC + \u2220BAC = 2 \u2220ACB + 2\u2220ACD
\n[\u2235 \u2220ABC = \u2220ACB
\n\u2220ADC = \u2220ACD]
\n180\u00b0 = 2 [\u2220ACB + \u2220ACD]
\n180\u00b0 = 2[\u2220BCD]
\n\u2234 \u2220BCD = \\(\\frac{180^{\\circ}}{2}\\) = 90\u00b0
\n(or)
\nFrom the figure
\n\u22202 = x + x = 2x
\n\u22201 = y + y = 2y
\n\u22201 + \u22202 = 2x + 2y
\n180\u00b0 = 2 = (x + y)
\n\u2234 x + y\u00a0 = \\(\\frac{180^{\\circ}}{2}\\) = 90\u00b0
\nHence proved.<\/p>\n

\"AP<\/p>\n

Question 6.
\nABC is a right angled triangle in which \u2220A = 90\u00b0 and AB = AC, Show that \u2220B = \u2220C.
\nSolution:
\n\"AP
\nGiven \u0394ABC; AB – AC
\nJoin the mid point D of BC to A.
\nNow in \u0394ADC and \u0394ADB
\nAD = AD (common)
\nAC = AB (giyen)
\nDC = DB (construction)
\n\u21d2 \u0394ADC \u2245 \u0394ADB
\n\u21d2 \u2220C = \u2220B (CPCT)<\/p>\n

\"AP<\/p>\n

Question 7.
\nShow that the angles of an equilateral triangle are 60\u00b0 each.
\nSolution:
\n\"AP
\nGiven \u0394ABC is an equilateral triangle
\nAB = BC = CA
\n\u2220A = \u2220B (\u2235 angles opposite to equal sides)
\n\u2220B = \u2220C (\u2235 angles opposite to equal sides)
\n\u21d2 \u2220A = \u2220B = \u2220C = x say
\nAlso \u2220A+\u2220B + \u2220C =180\u00b0
\n\u21d2 x + x + x = 180\u00b0
\n3x = 180\u00b0
\n\u21d2 x = \\(\\frac{180}{3}\\) = 60\u00b0
\nHence proved.<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.3 Textbook Questions and Answers. AP State Syllabus 9th Class Maths Solutions 7th Lesson Triangles Exercise 7.3 Question 1. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that, (i) AD bisects BC (ii) AD bisects \u2220A. …<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/4553"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=4553"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/4553\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=4553"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=4553"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=4553"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}