{"id":4965,"date":"2021-03-03T09:41:11","date_gmt":"2021-03-03T04:11:11","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=4965"},"modified":"2021-09-15T11:40:35","modified_gmt":"2021-09-15T06:10:35","slug":"ap-board-9th-class-maths-solutions-chapter-11-ex-11-1","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-9th-class-maths-solutions-chapter-11-ex-11-1\/","title":{"rendered":"AP Board 9th Class Maths Solutions Chapter 11 Areas Ex 11.1"},"content":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions<\/a> Chapter 11 Areas Ex 11.1 Textbook Questions and Answers.<\/p>\n

AP State Syllabus 9th Class Maths Solutions 11th Lesson Areas Exercise 11.1<\/h2>\n

\"AP<\/p>\n

Question 1.
\nIn \u0394ABC, \u2220ABC = 90\u00b0; AD = DC; AB =12 cm, BC = 6.5 cm. Find the area of \u0394ADB
\n\"AP
\nSolution:
\n\u0394ADB = \\(\\frac { 1 }{ 2 }\\) \u0394ABC [ \u2235 AD is a median of \u0394ABC]
\n\\(\\frac { 1 }{ 2 }\\) = [ \\(\\frac { 1 }{ 2 }\\) AB x BC]
\n= \\(\\frac { 1 }{ 4 }\\) x 12 x 6.5
\n= 19.5 cm2<\/sup><\/p>\n

\"AP<\/p>\n

Question 2.
\nFind the area of a quadrilateral PQRS in which \u2220QPS = \u2220SQR = 90\u00b0, PQ = 12 cm, PS = 9 cm, QR = 8 cm and SR =17 cm.
\n[Hint: PQRS has two parts]
\n\"AP
\nSolution:
\nArea of \u0394QPS = \\(\\frac { 1 }{ 2 }\\) x base x height
\n= \\(\\frac { 1 }{ 2 }\\) x 9 x 12
\n= 54cm2<\/sup>
\nIn \u0394QPS
\nQS2<\/sup> = PQ2<\/sup> + PS2<\/sup>
\nQS = \\(\\begin{aligned}
\n\\sqrt{12^{2}+9^{2}} &=\\sqrt{144+81} \\\\
\n&=\\sqrt{225}=15
\n\\end{aligned}\\)
\nArea of \u0394QSR =\\(\\frac { 1 }{ 2 }\\) x base x height
\n= \\(\\frac { 1 }{ 2 }\\) x 15 x 8 = 60 cm2<\/sup>
\n\u2234 \u25a1PQRS = \u0394QPS + \u0394QSR
\n= 54 + 60= 114 cm2<\/sup><\/p>\n

\"AP<\/p>\n

Question 3.
\nFind the area of trapezium ABCD as given in the figure in which ADCE is a rectangle.
\n[Hint: ABCD has two parts]
\n\"AP
\nSolution:
\nArea of trapezium 1
\n= \\(\\frac { 1 }{ 2 }\\) (sum of parallel sides) x (distance between the parallel sides)
\n= \\(\\frac { 1 }{ 2 }\\) (a + b) h
\nFrom the figure, a = 3 + 3 = 6 cm
\nb = 3 cm
\n(\u2235 Opp. sides of rectangle)
\nh = 8 cm
\n\u2234 A = \\(\\frac { 1 }{ 2 }\\)(6 + 3)x8 = 36cm2<\/sup><\/p>\n

\"AP<\/p>\n

Question 4.
\nABCD is a parallelogram. The diago-nals AC and BD intersect each other at O. Prove that ar (\u0394AOD) = ar (\u0394BOQ. [Hint: Congruent figures have equal area]
\n\"AP
\nSolution:
\nGiven that \u25a1ABCD is a parallelogram.
\nDiagonals AC and BD meet at ‘O’.
\nIn \u0394AOD and \u0394BOC
\nAD = BC [ \u2235 Opp. sides of a ||gm]
\nAO = OC [ \u2235 diagonals bisect each
\nOD = OB other]
\n\u0394AOD = \u0394BOC [S.S.S. congruence]
\n\u2234 \u0394AOD = \u0394BOC (i.e., have equal area)<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions Chapter 11 Areas Ex 11.1 Textbook Questions and Answers. AP State Syllabus 9th Class Maths Solutions 11th Lesson Areas Exercise 11.1 Question 1. In \u0394ABC, \u2220ABC = 90\u00b0; AD = DC; AB =12 cm, BC = 6.5 cm. Find the area of \u0394ADB Solution: \u0394ADB = \u0394ABC …<\/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\/4965"}],"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=4965"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/4965\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=4965"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=4965"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=4965"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}