{"id":5035,"date":"2021-03-03T12:34:47","date_gmt":"2021-03-03T07:04:47","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=5035"},"modified":"2021-09-15T11:40:34","modified_gmt":"2021-09-15T06:10:34","slug":"ap-board-9th-class-maths-solutions-chapter-12-ex-12-2","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-9th-class-maths-solutions-chapter-12-ex-12-2\/","title":{"rendered":"AP Board 9th Class Maths Solutions Chapter 12 Circles Ex 12.2"},"content":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions<\/a> Chapter 12 Circles Ex 12.2 Textbook Questions and Answers.<\/p>\n

AP State Syllabus 9th Class Maths Solutions 12th Lesson Circles Exercise 12.2<\/h2>\n

\"AP<\/p>\n

Question 1.
\nIn the figure, if AB = CD and \u2220AOB = 90\u00b0 find \u2220COD.
\n\"AP
\nSolution:
\n‘O’ is the centre of the circle.
\nAB = CD (equal chords from the figure)
\n\u2234 \u2220AOB = \u2220COD
\n[ \u2235 equal chords make equal angles at the centre]
\n\u2234 \u2220COD = 90\u00b0 [ \u2235 \u2220AOB = 90\u00b0 given]<\/p>\n

\"AP<\/p>\n

Question 2.
\nIn the figure, PQ = RS and \u2220ORS = 48\u00b0.
\nFind \u2220OPQ and \u2220ROS.
\n\"AP
\nSolution:
\n‘O’ is the centre of the circle.
\nPQ = RS [given, equal chords]
\n\u2234\u2220POQ = \u2220ROS [ \u2235 equal chords make equal angles at the centre]
\n\u2234 In \u0394ROS
\n\u2220ORS + \u2220OSR + \u2220ROS = 180\u00b0
\n[angle sum property]
\n\u2234 48\u00b0 + 48\u00b0 + \u2220ROS = 180\u00b0
\n[ \u2235 OR = OS(radii); \u0394ORS is isosceles]
\n\u2234 \u2220ROS = 180\u00b0 – 96\u00b0 = 84\u00b0
\nAlso \u2220POQ = \u2220ROS = 84\u00b0
\n\u2234 \u2220OPQ = \u2220OQP
\n[\u2235 OP = OQ; radii]
\n= \\(\\frac { 1 }{ 2 }\\)[180\u00b0-84\u00b0] = 48\u00b0<\/p>\n

\"AP<\/p>\n

Question 3.
\nIn the figure, PR and QS are two diameters. Is PQ = RS ?
\n\"AP
\nSolution:
\n‘O’ is the centre of the circle.
\n[ \u2235 PR, QS are diameters]
\nOP = OR (\u2235 radii) .
\nOQ = OS (\u2235 radii)
\n\u2220POQ = \u2220ROS [vertically opp. angles]
\n\u2234 \u0394OPQ \u2245 \u0394ORS [SAS congruence]
\n\u2234 PQ = RS [CPCT]<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions Chapter 12 Circles Ex 12.2 Textbook Questions and Answers. AP State Syllabus 9th Class Maths Solutions 12th Lesson Circles Exercise 12.2 Question 1. In the figure, if AB = CD and \u2220AOB = 90\u00b0 find \u2220COD. Solution: ‘O’ is the centre of the circle. AB = CD …<\/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\/5035"}],"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=5035"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/5035\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=5035"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=5035"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=5035"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}