{"id":1930,"date":"2020-09-08T12:25:16","date_gmt":"2020-09-08T06:55:16","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=1930"},"modified":"2021-07-02T15:03:37","modified_gmt":"2021-07-02T09:33:37","slug":"kseeb-solutions-class-9-maths-chapter-6-ex-6-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths-chapter-6-ex-6-1\/","title":{"rendered":"KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1"},"content":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1<\/strong> are part of KSEEB Solutions for Class 9 Maths<\/a>. Here we have given Karnataka Board Class 9 Maths Chapter 6 Constructions Exercise 6.1.<\/p>\n

Karnataka Board Class 9 Maths Chapter 6 Constructions Ex 6.1<\/h2>\n

Question 1.
\nConstruct an angle of 90\u00b0 at the initial point of a given ray and justify the construction.
\nSolution:
\nData: Constructing an angle of 90\u00b0 at the initial point of a given ray.
\n\"KSEEB
\nSteps of Construction:<\/p>\n

    \n
  1. Draw AB straight line.<\/li>\n
  2. Taking A as centre and any radius draw an arc and intersect AB at C.<\/li>\n
  3. Taking C as centre with the same radius draw an arc which intersects at D.<\/li>\n
  4. With the same radius taking D as centre, it intersects at E.<\/li>\n
  5. With centres E and D with the same radius draw two arcs which meet both at F.<\/li>\n
  6. If AF is joined, AB straight line is the line in point A with 90\u00b0, i.e., AF.
    \nIn the above construction \u2220DAB = 60\u00b0 and \u2220DAE = 60\u00b0.
    \nAngular bisector of \u2220DAE is AF.
    \n\u2234 \u2220DAF = \u2220EAF = 30\u00b0
    \n\u2234\u2220BAF = \u2220DAB + \u2220DAE = 60\u00b0+ 30\u00b0
    \n\u2234\u2220BAF = 90\u00b0.<\/li>\n<\/ol>\n

    Question 2.
    \nConstruct an angle of 45\u00b0 at the initial point of a given ray and justify the construction.
    \nSolution:
    \nData: To construct an angle of 45\u00b0 at the initial point of a given ray.
    \n\"KSEEB
    \nSteps of Construction :<\/p>\n

      \n
    1. Draw a straight line with any measurement.<\/li>\n
    2. Taking A as centre with any radius draw an arc, which meets AB at C.<\/li>\n
    3. With C centre, with the same radius draw an arc which meets at D.<\/li>\n
    4. With D centre, with the same radius, draw an arc which meets at D.<\/li>\n
    5. With centres E and D, with the previous radius draw two arcs which meet at F.<\/li>\n
    6. Now AF is joined. Now AP1AF at A. Hence \u2220BAF = 90\u00b0.<\/li>\n
    7. Now, AG which is the angular bisector of \u2220BAF is drawn.<\/li>\n
    8. \u2220BAG = 45\u00b0 is constructed.
      \n\u2220FAB = 90\u00b0
      \nAG is the angular bisector of \u2220FAB
      \n\u2234 \u2220FAG = \u2220GAB = 45\u00b0
      \n\u2234 \u2220GAB = 45\u00b0.<\/li>\n<\/ol>\n

      Question 3.
      \nConstruct the angles of the following measurements :
      \n(i) 30\u00b0
      \n(ii) \\(22 \\frac{1}{2}^{\\circ}\\)
      \n(iii) 15\u00b0.
      \nSolution:
      \n(i) To construct 30\u00b0 angle :
      \n\"KSEEB
      \nSteps of Construction :<\/p>\n

        \n
      1. Draw PQ straight line.<\/li>\n
      2. With P as centre with any radius draw \\(\\frac{1}{2}\\) arc- intersects PQ at A.<\/li>\n
      3. With A as centre with the same radius draw an arc which intersects at B. Join PB and produced.<\/li>\n
      4. With A and B centres, with radius more than half of AB draw two arcs which intersect at C. Join PC.<\/li>\n
      5. \u2220BPC = \u2220CPQ = 30\u00b0.<\/li>\n<\/ol>\n

        (ii) To draw an angle of \\(22 \\frac{1}{2}^{\\circ}\\)
        \nSolution:
        \n\"KSEEB
        \nSteps of Construction:<\/p>\n

          \n
        1. Draw a straight line AB.<\/li>\n
        2. With centre ‘A’, taking a convenient radius to draw an arc which intersect AB at P.<\/li>\n
        3. With P as centre with the same radius draw an arc at Q, with Q as centre with the same radius draw an arc which intersects at R.<\/li>\n
        4. With R and Q centres with the same radius draw two arcs which intersect at S. Join AS, \u2220BAS = 90\u00b0.<\/li>\n
        5. Now construct AT which is the angular bisector of \u2220BAS, and joined, \u2220TAB= 45\u00b0.<\/li>\n
        6. Now AU which is the angular bisector of \u2220TAB, AU is joined. Now, \u2220UAB = \\(22 \\frac{1}{2}^{\\circ}\\)<\/li>\n<\/ol>\n

          (iii) To construct an angle of 15\u00b0 :
          \nSolution:
          \n\"KSEEB
          \nSteps of Construction :<\/p>\n

            \n
          1. Draw a straight line AB.<\/li>\n
          2. With A as centre with a convenient radius draw an arc which intersect AB at C.<\/li>\n
          3. With C as centre, with the same radius, draw an arc which intersects at D.
            \nNow. \u2220DAB = 60\u00b0<\/li>\n
          4. Construct AE, the bisector of \u2220DAB. Join then, \u2220EAB = 30\u00b0.<\/li>\n
          5. Construct AF, the bisector of \u2220EAB. Join then, \u2220FAB = 15\u00b0.<\/li>\n<\/ol>\n

            Question 4.
            \nConstruct the following angles and verify by measuring them by a protractor :
            \n(i) 75\u00b0
            \n(ii) 105\u00b0
            \n(iii) 135\u00b0.
            \nSolution:
            \n(i) To construct angle 75\u00b0 :
            \n\"KSEEB
            \nSteps of Construction :<\/p>\n

              \n
            1. Draw AB with any measurement.<\/li>\n
            2. With A as centre, with convenient scale draw an arc AB which intersects at C.<\/li>\n
            3. With C as centre with the same radius draw an arc which intersects at D. \u2220DAB = 60\u00b0.<\/li>\n
            4. With D as centre with the same radius draw an arc which intersects at E.<\/li>\n
            5. With Centres E and D, by taking more than half of ED draw two arcs which meet at E. AF is joined.
              \nNow, \u2220FAB = 90\u00b0.<\/li>\n
            6. Draw AG bisector of \u2220EAD, AG joined. \u2220GAD = 15\u00b0.<\/li>\n
            7. \u2220GAD + \u2220DAB = 15 + 60 = 75\u00b0
              \n\u2234 \u2220GAB = 75\u00b0.<\/li>\n<\/ol>\n

              (ii) To construct angle 105\u00b0 :
              \nSolution:
              \n\"KSEEB
              \nSteps of Construction :<\/p>\n

                \n
              1. Draw PQ with any measurement.<\/li>\n
              2. With P as centre and with any radius, draw an arc which intersects PQ at A.<\/li>\n
              3. With A as centre, with the same radius draw an arc which intersect at B.<\/li>\n
              4. With B as centre with same radius draw another arc, it intersects at C.<\/li>\n
              5. With centres C and B if two arcs are drawn these two meet at D. Join PD. \u2220DPQ = 90\u00b0. Straight-line PD intersects the arc CB at E.<\/li>\n
              6. Now taking radius half of CE, if the line is drawn from C to E it intersects at F. FP is joined. Now \u2220FPD = 15\u00b0.<\/li>\n
              7. \u2220FPD + \u2220DPQ = 15 + 90 = 105\u00b0.
                \n\u2234 \u2220FPQ = 105\u00b0 is constructed.<\/li>\n<\/ol>\n

                (iii) To construct an angle of 135\u00b0
                \nSolution:
                \n\"KSEEB
                \nSteps of Construction :<\/p>\n

                  \n
                1. Draw a straight line AB.<\/li>\n
                2. With A as centre, a semicircle is drawn which meets AB at C and it intersects the produced BA line at G.<\/li>\n
                3. With C as centre with the same radius and centre CD, draw a radius of arc DE.<\/li>\n
                4. With centres E and D by taking a radius more than half draw two arcs which meet at F. Join AF. \u2220FAB = 90\u00b0. AF intersects ED at H.<\/li>\n
                5. With G and H centres by taking radius more than \\(\\frac{1}{2}\\) of the radius GH draw arcs which meet at I. AI is joined. Now, \u2220IAF = 45\u00b0.<\/li>\n
                6. \u2220IAF + \u2220FAB = 45\u00b0 + 90\u00b0 = 135\u00b0
                  \n\u2234 \u2220IAB = 135\u00b0 is constructed.<\/li>\n<\/ol>\n

                  Question 5.
                  \nConstruct an equilateral triangle, given its side, and justify the construction.
                  \nSolution:
                  \nData: Construct an equilateral triangle, given its side.
                  \n\"KSEEB
                  \nSteps of Construction :<\/p>\n

                    \n
                  1. Construct AB = 6 cm.<\/li>\n
                  2. With A as centre and convenient scale draw an arc, it meets AB at P.<\/li>\n
                  3. With P as centre, with same radius draw an arc, it intersects the first arc at Q, AQ is joined and produced. \u2220QAB = 60\u00b0.<\/li>\n
                  4. With A as centre and taking 6 cm radius draw an arc. It intersects at C which is the produced line of AQ.<\/li>\n
                  5. Next, BC is joined, ABC is an equilateral triangle.<\/li>\n<\/ol>\n

                    In \u2206ABC, \u2220A = \u2220B = \u2220C = 60\u00b0
                    \nAB = BC = CA = 6 cm.
                    \nIn \u2206ABC, AB = AC = 6 cm. \u2220A = 60\u00b0
                    \n\u2220B = 60\u00b0
                    \n\u2220A + \u2220B + \u2220C = 180\u00b0
                    \n60 + \u2220B + \u2220B =180 (\u2235 \u2220B = \u2220C)
                    \n60 + 2\u2220B = 180
                    \n2\u2220B = 180 – 60
                    \n2\u2220B = 120\u00b0
                    \n\u2234 \u2220B = \\(\\frac{120}{2}\\)
                    \n\u2234 \u2220B = 60\u00b0
                    \n\u2234\u2220B = \u2220C = 60\u00b0
                    \n\u2234\u2220A = \u2220B = \u2220C = 60\u00b0
                    \nAC = BC (Opposite sides of equal angles)
                    \nBut, AB = AC (known)
                    \n\u2234 AB = BC = CA = 6 cm.<\/p>\n

                    We hope the KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 6 Constructions Exercise 6.1, 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 6.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 6 Constructions Exercise 6.1. Karnataka Board Class 9 Maths Chapter 6 Constructions Ex 6.1 Question 1. Construct an angle of 90\u00b0 at the initial point …<\/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\/1930"}],"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=1930"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1930\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=1930"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=1930"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=1930"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}