{"id":2005,"date":"2020-09-08T04:27:56","date_gmt":"2020-09-07T22:57:56","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=2005"},"modified":"2021-07-02T15:03:45","modified_gmt":"2021-07-02T09:33:45","slug":"kseeb-solutions-class-9-maths-chapter-6-ex-6-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths-chapter-6-ex-6-2\/","title":{"rendered":"KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.2"},"content":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.2<\/strong> are part of KSEEB Solutions for Class 9 Maths<\/a>. Here we have given Karnataka Board Class 9 Maths Chapter 6 Constructions Exercise 6.2.<\/p>\n

Karnataka Board Class 9 Maths Chapter 6 Constructions Ex 6.2<\/h2>\n

Question 1.
\nConstruct a triangle ABC in which BC = 7 cm, \u2220B = 75\u00b0 and AB + AC = 13 cm,
\nSolution:
\nConstruction of a triangle ABC in which BC = 7 cm, \u2220B = 75\u00b0 and AB + AC = 13 cm
\n\"KSEEB
\nSteps of Construction :<\/p>\n

    \n
  1. Draw BC = 7 cm line segment.<\/li>\n
  2. Construct \u2220B = 75\u00b0 and produce it upto Bx.<\/li>\n
  3. With B as centre draw an arc BD = 13 cm such that it intersects Bx at D.<\/li>\n
  4. Construct PQ which is the perpendicular bisector of BD.<\/li>\n
  5. PQ intersects BD at A.<\/li>\n
  6. Join AC.<\/li>\n
  7. Now \u2206ABC, with measurement AB + AC = 13 cm, and \u2220B = 75\u00b0, BC = 7 cm is constructed.<\/li>\n<\/ol>\n

    Question 2.
    \nConstruct a triangle ABC in which BC = 8 cm, \u2220B = 45\u00b0 and AB – AC = 3.5 cm.
    \nSolution:
    \nConstructing a triangle ABC in which BC = 8 cm, \u2220B = 45\u00b0 and AB – AC = 3.5 cm.
    \n\"KSEEB
    \nSteps of Construction :<\/p>\n

      \n
    1. Draw a line segment BC = 8 cm.<\/li>\n
    2. Draw \u2220B = 45\u00b0 with the help of a protractor. Bx is joined.<\/li>\n
    3. Draw an arc BD = 3.5 cm.<\/li>\n
    4. When perpendicular bisector DC is drawn, that intersects Bx at A. AC is joined.<\/li>\n
    5. Now, DABC with BC = 8 cm, \u2220B = 45\u00b0 and AB – AC = 3.5 cm. is constructed.<\/li>\n<\/ol>\n

      Question 3.
      \nConstruct a triangle PQR in which QR = 6 cm, \u2220Q = 60\u00b0 and PR – PQ = 2 cm.
      \nSolution:
      \nConstructing a triangle PQR in which QR = 6 cm, \u2220Q = 60\u00b0 and PR – PQ = 2 cm.
      \n\"KSEEB
      \nSteps of Construction :<\/p>\n

        \n
      1. Draw a line segment QR = 6 cm.<\/li>\n
      2. With the help of a protractor draw a line segment Qx and \u2220Q = 60\u00b0.<\/li>\n
      3. Produce line segment xQ upto y. mark T such that QT = 2 cm. Join TR.<\/li>\n
      4. If the perpendicular bisector of TR is drawn, that meets line segment Qx at P. PR is joined.
        \nNow, \u2206PQR is constructed.<\/li>\n<\/ol>\n

        Question 4.
        \nConstruct a triangle XYZ in which \u2220Y = 30\u00b0, \u2220Z = 90\u00b0 and XY + YZ + ZX = 11 cm.
        \nSolution:
        \nConstructing a triangle XYZ in which \u2220Y = 30\u00b0, \u2220Z = 90\u00b0 and XY + YZ + ZX = 11 cm.
        \n\"KSEEB
        \nSteps of Construction :<\/p>\n

          \n
        1. Draw AB = 11 cm line segment with measurement XY + YZ + ZX.<\/li>\n
        2. With the help of protractor construct \u2220A = 30\u00b0, \u2220B = 90\u00b0.<\/li>\n
        3. Construct angular bisectors of \u2220CAB and \u2220DBA which meet at x. Ax, Bx are joined.<\/li>\n
        4. If perpendicular bisector AX is drawn, it intersects AB at Y and AB at Z.<\/li>\n
        5. By joining XY and XZ, \u2206XYZ is constructed.<\/li>\n<\/ol>\n

          Question 5.
          \nConstruct a right triangle whose base is 12 cm and the sum of its hypotenuse and the other side is 18 cm.
          \nSolution:
          \nConstructing a right-angled triangle ABC with base, AB =12 cm and the sum of hypotenuse BC and another side AC, AC + BC = 18 cm.
          \n\"KSEEB
          \nSteps of Construction :<\/p>\n

            \n
          1. Draw a line segment AB = 12 cm.<\/li>\n
          2. With the help of a protractor construct \u2220A = 90\u00b0 and produce upto AX.<\/li>\n
          3. Mark ‘D’ on AX such that AD = 18 cm.<\/li>\n
          4. Join BD.<\/li>\n
          5. Let the perpendicular bisector BD intersects AD at C. Join BC.<\/li>\n
          6. Now required triangle ABC is constructed with Base AB = 12cm, Side AC = – 5 cm, Hypotenuse BC = 13 cm and \u2220A = 90\u00b0.<\/li>\n<\/ol>\n

            We hope the KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.2 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 6 Constructions Exercise 6.2, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

            KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 6 Constructions Exercise 6.2. Karnataka Board Class 9 Maths Chapter 6 Constructions Ex 6.2 Question 1. Construct a triangle ABC in which BC = 7 …<\/p>\n","protected":false},"author":6,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2005"}],"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\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=2005"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2005\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=2005"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=2005"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=2005"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}