{"id":4554,"date":"2021-03-01T16:30:29","date_gmt":"2021-03-01T11:00:29","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=4554"},"modified":"2021-09-15T11:40:36","modified_gmt":"2021-09-15T06:10:36","slug":"ap-board-9th-class-maths-solutions-chapter-7-ex-7-4","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-9th-class-maths-solutions-chapter-7-ex-7-4\/","title":{"rendered":"AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.4"},"content":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions<\/a> Chapter 7 Triangles Ex 7.4 Textbook Questions and Answers.<\/p>\n

AP State Syllabus 9th Class Maths Solutions 7th Lesson Triangles Exercise 7.1<\/h2>\n

Question 1.
\nShow that in a right angled triangle, the hypotenuse is the longest side.
\nSolution:
\n\"AP
\nLet a \u0394ABC be right angled at \u2220B.
\nThen \u2220A + \u2220C = 90\u00b0
\n(i.e.,) \u2220A and \u2220C are both acute.
\nNow, \u2220A < \u2220B \u21d2 BC < AC
\nAlso \u2220C < \u2220B \u21d2 AB < AC
\n\u2234 AC, the hypotenuse is the longest side.<\/p>\n

\"AP<\/p>\n

Question 2.
\nIn the given figure, sides AB and AC of \u0394ABC are extended “to points P and Q respectively. Also \u2220PBC < \u2220QCB. Show that AC > AB.
\n\"AP
\nSolution:
\nFrom the figure,
\n\u2220PBC = \u2220A + \u2220ACB
\n\u2220QCB = \u2220A + \u2220ABC
\nGiven that \u2220PBC < \u2220QCB
\n\u21d2\u2220A + \u2220ACB < \u2220A + \u2220ABC
\n\u21d2 \u2220ACB < \u2220ABC
\n\u21d2 AB < AC
\n\u21d2 AC > AB
\nHence proved.<\/p>\n

\"AP<\/p>\n

Question 3.
\nIn the given figure, \u2220B < \u2220A and \u2220C < \u2220D. Show that AD < BC.
\n\"AP
\nSolution:
\nGiven that \u2220B < \u2220A; \u2220C < \u2220D
\n\u2220B < \u2220A \u21d2 AO < OB [in \u0394AOB] ……………… (1)
\n\u2220C < \u2220D \u21d2 OD < OC [in \u0394COD]…… (2)
\nAdding (1) & (2)
\nAO + OD < OB + OC
\nAD < BC
\nHence proved.<\/p>\n

\"AP<\/p>\n

Question 4.
\nAB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that \u2220A >\u2220C and \u2220B > \u2220D.
\n\"AP
\nSolution:
\n\"AP
\nGiven that AB and CD are the smallest and longest sides of quadrilateral ABCD.
\nFrom the figure,
\nIn \u0394BCD
\n\u22201 > \u22202 [\u2235 DC > BC] ………………(1)
\nIn \u0394BDA
\n\u22204 > \u22203 [\u2235 AD > AB] ………….(2)
\nAdding (1) & (2)
\n\u22201 + \u22204 > \u22202 + \u22203
\n\u2220B > \u2220D
\nSimilarly,
\nIn \u0394ABC, \u22206 < \u22207 [ \u2235AB < BC] ……………….(3)
\nIn \u0394ACD
\n\u22205 < \u22208 …………. (4)
\nAdding (3) & (4)
\n\u22206 + \u22205 < \u22207 + \u22208
\n\u2220C < \u2220A \u21d2 \u2220A > \u2220C
\nHence proved.<\/p>\n

\"AP<\/p>\n

Question 5.
\nIn the given figure, PR > PQ and PS bisects \u2220QPR. Prove that
\n\u2220PSR > \u2220PSQ.
\n\"AP
\nSolution:
\nGiven that PR > PQ;
\n\u2220QPS =\u2220RPS
\nPR> PQ
\n\u2220Q > \u2220R
\nNow \u2220Q +\u2220QPS > \u2220R + \u2220RPS
\n\u21d2 180\u00b0 – (\u2220Q + \u2220QPS) < 180\u00b0 – (\u2220R + \u2220RPS)
\n\u21d2 \u2220PSQ < \u2220PSR \u21d2 \u2220PSR > \u2220PSQ
\nHence proved.<\/p>\n

Question 6.
\nIf two sides of a triangle measure 4 cm and 6 cm find all possible measurements (positive integers) of the third side. How many distinct triangles can be obtained ?
\nSolution:
\nGiven that two sides of a triangle are 4 cm and 6 cm.
\n\u2234 The measure of third side > Differ-ence between other two sides.
\nthird side > 6 – 4
\nthird side > 2
\nAlso the measure of third side < sum of other two sides
\nthird side <6 + 4 < 10
\n\u2234 2 < third side <10
\n\u2234 The measure of third side may be 3 cm, 4 cm, 5 cm, 6 cm, 7 cm, 8 cm, 9 cm
\n\u2234 Seven distinct triangles can be obtained.<\/p>\n

\"AP<\/p>\n

Question 7.
\nTry to construct a triangle with 5 cm, 8 cm and 1 cm. Is it possible or not ? Why ? Give your justification.
\nSolution:
\nAs the sum (6 cm) of two sides 5 cm and 1 cm is less than third side. It is not possible to construct a triangle with the given measures.<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.4 Textbook Questions and Answers. AP State Syllabus 9th Class Maths Solutions 7th Lesson Triangles Exercise 7.1 Question 1. Show that in a right angled triangle, the hypotenuse is the longest side. Solution: Let a \u0394ABC be right angled at \u2220B. Then \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\/4554"}],"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=4554"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/4554\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=4554"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=4554"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=4554"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}