{"id":1044,"date":"2020-09-15T08:41:52","date_gmt":"2020-09-15T03:11:52","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=1044"},"modified":"2021-07-02T15:01:51","modified_gmt":"2021-07-02T09:31:51","slug":"kseeb-sslc-class-10-maths-solutions-chapter-6-ex-6-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-6-ex-6-2\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 6 Constructions Ex 6.2"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 6 Constructions Ex 6.2<\/strong> are part of KSEEB SSLC Class 10 Maths Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 6 Constructions Exercise 6.2.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 6 Constructions Exercise 6.2<\/h2>\n

In each of the following, give also the justification of the construction :
\n<\/strong>Question 1.
\nDraw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
\nSolution:
\nr = 6 cm,, d = 10 cm., t = ?
\n\"KSEEB
\nTangent, PQ = PR = 8 cm.
\nSteps of Construction :<\/p>\n

    \n
  1. A circle with radius 6 cm is drawn taking O’ as centre.<\/li>\n
  2. Point P is marked at 10 cm apart from centre of circle.<\/li>\n
  3. With the half of compass mark M which is the mid-point of OP.<\/li>\n
  4. A circle is drawn at M taking radius MO or MP which intersect the given circle at Q and R.<\/li>\n
  5. Now join PQ from P, and Join PR, required tangents are obtained. If measured it is PQ = PR = 8 cm.<\/li>\n<\/ol>\n

    Question 2.
    \nConstruct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also, verify the measurement by actual calculation.
    \nSolution:
    \nSteps of Construction:<\/p>\n

      \n
    1. With ‘O’ as centre two circles are constructed with radii 4 cm and 6 cm.<\/li>\n
    2. Draw OM radius for small circle and drawn perpendicular at M which meets the big circle at P and Q.<\/li>\n
    3. Now PQ is the tangent drawn for a small circle.
      \n\"KSEEB<\/li>\n<\/ol>\n

      Tangent PMQ = 9 cm
      \nVerification : In \u22a5\u2206OMP, \u2220M = 90\u00b0.
      \n\u2234 OM2<\/sup> + PM2<\/sup> = OP2<\/sup>
      \n(4)2<\/sup> + PM2<\/sup> = (6)2<\/sup>
      \nPM2<\/sup> = (6)2<\/sup> – (4)2<\/sup> = 36 – 16
      \nPM2<\/sup> = 20
      \n\u2234 PM = \\(\\sqrt{20}\\) = 4.5 cm.
      \nSimilarly,
      \nMQ = 4.5 cm.
      \n\u2234 PQ = PM + MQ = 4.5 + 4.5
      \n\u2234 PQ = 9 cm.<\/p>\n

      Question 3.
      \nDraw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
      \nSolution:
      \nDraw Pw, Px tangents to circle with centre ‘O’ at P.
      \nDraw Qy, Qz tangents to circle with centre \u2018O\u201d at Q.
      \n\"KSEEB
      \nTantent, OP = OQ = 7 cm.
      \nSteps of Construction :<\/p>\n

        \n
      1. Construct a circle with radius 3 cm with centre O’. Draw the diameter AB and on both sides mark P and Q such that OP = OQ = 7 cm.<\/li>\n
      2. Bisect PO, Mark point M and draw circle of radius MO and centre M. It intersects given circle at w and x.
        \nIf joined Pw and Px, required tangents are constructed.<\/li>\n
      3. Similarly, bisect OQ, mark point N. Draw a circle of radius NO or NQ with centre N, it meets given circle at y and z points. Join Qy and Qz. These are the required tangents.<\/li>\n<\/ol>\n

        Question 4.
        \nDraw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60\u00b0.
        \nSolution:
        \nr = 5 cm.
        \nAngle between tangents is 60\u00b0.
        \n\"KSEEB
        \nSteps of Construction :<\/p>\n

          \n
        1. A circle is drawn with centre ‘O’ and radius 5 cm. Draw OA and OB radii and \u2220AOB = 120\u00b0<\/li>\n
        2. Tangents are drawn at A and B and these intersect at P.<\/li>\n
        3. By measurement, \u2220APB = 60\u00b0.
          \nAngle between tangents PA and PB is 60\u00b0.
          \n\u2235 \u2220AOB and \u2220APB Supplementary angles.<\/li>\n<\/ol>\n

          Question 5.
          \nDraw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
          \nSolution:
          \n\"KSEEB
          \nSteps of Construction :<\/p>\n

            \n
          1. Draw a line segment AB = 8 cm. Draw two circles with radii 4 cm and 3 cm at A and B.<\/li>\n
          2. To construct tangent from centre A to Centre B, it intersects at ‘O’ which is the midpoint of AB.
            \nDraw circle with radius OA, OB at Centre ‘O’, it intersects circle with centre A at C and D and with centre ‘B’ at E and F.<\/li>\n
          3. Now join E to A centre and F, required AE and AF tangents are constructed.<\/li>\n
          4. Similarly ‘B’ centre is joined to C and D, BC and BF tangents are constructed.<\/li>\n
          5. AE and AF are tangents drawn from centre ‘A’ to centre ‘B’.
            \nTangents drawn from centre B to centre A are BC and BD.<\/li>\n<\/ol>\n

            Question 6.
            \nLet ABC be a right triangle in which AB= 6 cm, BC = 8 cm and \u2220B = 90\u00b0. BD is perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
            \nSolution:
            \nAB and AR are tangents drawn for circle with centre P from external point A.
            \nSteps of Construction :<\/p>\n

              \n
            1. \u22a5\u2206 ABC is constructed with BC = 8 cm. (base), \u2220B = 90\u00b0, AB = 6 cm.<\/li>\n
            2. Draw BD \u22a5 AC. , Draw perpendicular bisectors of DC and BC, they intersect at P.<\/li>\n
            3. A circle is drawn with centre P through A, B, D, and C.
              \n\"KSEEB
              \nFor this circle AB and AR tangents are constructed from external point A.
              \nTangent, AB = AR = 6 cm.<\/li>\n<\/ol>\n

              Question 7.
              \nDraw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
              \nSolution:
              \n\"KSEEB
              \nSteps of Construction:<\/p>\n

                \n
              1. Draw a circle with the help of a circular bangle kept on paper.<\/li>\n
              2. To draw AB and CD chords for this circle which has no centre. Their perpendicular bisectors meet at ‘O’. Now ‘O’ is the centre of the circle.<\/li>\n
              3. Mark point A on the outside of the circle from centre. Join OA. R is the midpoint of OA.<\/li>\n
              4. A circle is drawn with radius RO or RA which intersects the first circle at P and Q. Now join P from A and from A to Q, required two tangents AP and AQ are constructed.
                \nTangent, AP = Tangent, AQ.<\/li>\n<\/ol>\n

                We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 6 Constructions Ex 6.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions 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 SSLC Class 10 Maths Solutions Chapter 6 Constructions Ex 6.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 6 Constructions Exercise 6.2. Karnataka SSLC Class 10 Maths Solutions Chapter 6 Constructions Exercise 6.2 In each of the following, give also the justification …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1044"}],"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\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=1044"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1044\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=1044"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=1044"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=1044"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}