{"id":992,"date":"2020-09-15T07:04:40","date_gmt":"2020-09-15T01:34:40","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=992"},"modified":"2021-07-02T15:01:51","modified_gmt":"2021-07-02T09:31:51","slug":"kseeb-sslc-class-10-maths-solutions-chapter-6-ex-6-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-6-ex-6-1\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 6 Constructions Ex 6.1"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 6 Constructions Ex 6.1<\/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 6.1.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 6 Constructions Exercise 6.1<\/h2>\n

In each of the following, give the justification of the construction also :<\/strong>
\nQuestion 1.
\nDraw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
\nSolution:
\nDivide a line segment of 7.6 cm length in the ratio 5 : 8 and measure.
\nm : n = 5 : 8
\nm + n = 5 + 8 = 13
\nAC : CB = 5 : 8.
\nby measurement: AC = 3cm, CB = 4.6cm.
\n\"KSEEB
\nSteps of Construction:<\/p>\n

    \n
  1. Draw any ray AB, such that AB = 7.6 cm.<\/li>\n
  2. Draw Ax ray at point A making acute angle.<\/li>\n
  3. Locate the points A1<\/sub>, A2<\/sub>, A3<\/sub>, ………., A13<\/sub> from A.<\/li>\n
  4. Join BA13<\/sub>. Draw BA13<\/sub> || A5<\/sub>C.
    \nNow, AC : CB = 5 : 8
    \nIf measured, AC = 3 cm, CB = 4.6 cm.<\/li>\n<\/ol>\n

    Question 2.
    \nConstruct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \\(\\frac{2}{3}\\) of the corresponding sides of the first triangle.
    \nSolution:
    \nConstruct a triangle ABC having sides 4 cm, 5 cm and 6 cm. For this we have to construct a triangle similar to it whose sides are \\(\\frac{2}{3}\\) of the corresponding sides of the first triangle.
    \n\"KSEEB
    \nSteps of Construction:<\/p>\n

      \n
    1. \u2206ABC is constructed, having AB = 4 cm, BC = 5 cm, CA = 6 cm.<\/li>\n
    2. Draw Bx ray such that it forms acute angle to BC which is adjacent to vertex A.<\/li>\n
    3. Locate 3 points on Bx such that BB1<\/sub> = B1<\/sub>B2<\/sub> = B2<\/sub>B3<\/sub>.<\/li>\n
    4. Join B3<\/sub>C. Draw a parallel line to B3<\/sub>C which intersect BC at C.<\/li>\n
    5. Draw a parallel line through C’ to CA and it intersects BA at A’.
      \nNow, A’BC’ is required triangle.<\/li>\n<\/ol>\n

      Question 3.
      \nConstruct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \\(\\frac{7}{5}\\) of the corresponding sides
      \nof the first triangle.
      \nSolution:
      \nConstruct an \u2206ABC having sides 5 cm, 6 cm and 7 cm. Then construct another triangle whose sides are \\(\\frac{7}{5}\\) of the corresponding sides of the first triangle.
      \n\"KSEEB
      \nSteps of Construction :<\/p>\n

        \n
      1. First construct a triangle ABC having sides AB = 5cm, BC = 6 cm, CA = 7 cm.<\/li>\n
      2. Draw a ray Bx such that it makes an acute angle at ‘B’.<\/li>\n
      3. Locate Points B1<\/sub>, B2, B3<\/sub>, B4<\/sub>, B5<\/sub>, B6<\/sub>, By points on Bx.<\/li>\n
      4. Join B5<\/sub>C. Draw B5<\/sub>C || B7<\/sub>C’, it intersects at ‘C’ which is produced BC line.<\/li>\n
      5. Draw AC || C’A’, it meets produced line BC at A’.<\/li>\n
      6. Now required \u2206A’BC’ is obtained whose sides are \\(\\frac{7}{5}\\) of the corresponding sides of the first triangle.<\/li>\n<\/ol>\n

        Question 4.
        \nConstruct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are \\(1 \\frac{1}{2}\\) times the corresponding sides of the isosceles triangle.
        \nSolution:
        \nConstruct an isosceles triangle with base 8 cm and altitude 4 cm and then another triangle whose sides are \\(1 \\frac{1}{2}\\) times the corresponding sides of the isosceles triangle.
        \n\"KSEEB
        \nSteps of Construction:<\/p>\n

          \n
        1. Construct \u2206ABC having base AB = 8 cm, altitude AD = 4 cm.<\/li>\n
        2. \u2220BAx acute angle is constructed at A. From ‘A’ locate A1<\/sub>, A2<\/sub>, A3<\/sub>.<\/li>\n
        3. Join A2<\/sub>B. Draw A2<\/sub>B || A3<\/sub>B’. B is marked on line AB produced.<\/li>\n
        4. Draw AB || B’C\u2019. C’ is marked on AC produced and draw AC’.<\/li>\n
        5. Now \u2206AB’C’ is constructed which is similar to \u2206ABC and corresponding sides,
          \ni.e., \\(1 \\frac{1}{2}=\\frac{3}{2}\\)
          \n\u2206ABC ||| \u2206AB’C’<\/li>\n<\/ol>\n

          Question 5.
          \nDraw a triangle ABC with side BC = 6 cm, AB = 5 cm, and \u2220ABC = 60\u00b0. Then construct a triangle whose sides are \\(\\frac{3}{4}\\) of the corresponding sides of the triangle ABC.
          \nSolution:
          \nConstruct a triangle ABC with sides BC = 6 cm, AB = 5 cm and \u2220ABC = 60\u00b0. Then construct a triangle whose sides are \\(\\frac{3}{4}\\) of the corresponding sides of the triangle ABC.
          \n\"KSEEB
          \nSteps of Construction:<\/p>\n

            \n
          1. Construct a AABC having BC = 6 cm, AB = 5 cm, and \u2220ABC = 60\u00b0.<\/li>\n
          2. Draw a ray Bx to form acute angle at B and mark BB1<\/sub>, B2<\/sub>, B3<\/sub>, at equal intervals of distance.<\/li>\n
          3. Join B4<\/sub>C. Draw B4<\/sub>C || B3<\/sub>C’.<\/li>\n
          4. Draw AC || A’C’.<\/li>\n
          5. Now \u2206ABC’ is equal to \u2206ABC.
            \n\u2234 \u2206A’BC’ ||| \u2206ABC
            \n3 : 4<\/li>\n<\/ol>\n

            Question 6.
            \nDraw a triangle ABC with side BC = 7 cm, \u2220B = 45\u00b0, \u2220A = 105\u00b0. Then, construct a triangle whose sides are \\(\\frac{4}{3}\\) times the corresponding sides of \u2206ABC
            \nSolution:
            \nDraw a triangle ABC with side BC = cm, \u2220B = 45\u00b0, \u2220A = 105\u00b0. The construct a triangle whose sides are \\(\\frac{4}{3}\\) times the corresponding sides of \u2206ABC.
            \n\"KSEEB
            \nSteps of Construction :<\/p>\n

              \n
            1. Construct a triangle ABC having BC = 7 cm, \u2220A = 45\u00b0 and \u2220B = 105\u00b0.<\/li>\n
            2. Draw Bx ray at B to form acute angle.<\/li>\n
            3. Mark points B1<\/sub>, B2<\/sub>, B3<\/sub>, B4<\/sub> from B. Draw B4<\/sub>C\u2019.<\/li>\n
            4. Draw B4<\/sub>C || B3<\/sub>C’. Draw AC || A’C’.<\/li>\n
            5. Now, \u2206A’BC’ ||| \u2206ABC is constructed.
              \n\u2206A’BC’ : \u2206ABC<\/li>\n<\/ol>\n

              Question 7.
              \nDraw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \\(\\frac{5}{3}\\) times the corresponding sides of the given triangle.
              \nSolution:
              \nDraw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \\(\\frac{5}{3}\\) times the corresponding sides of the given triangle.
              \n\"KSEEB
              \nSteps of Construction:<\/p>\n

                \n
              1. Construct Right angled triangle ABC having sides BC = 4 cm, BA = 3 cm, and \u2220B = 90\u00b0.<\/li>\n
              2. Draw a ray Bx to form acute angle at B.<\/li>\n
              3. Mark points B1<\/sub>, B2<\/sub>, B3<\/sub>, B4<\/sub> from B at equal intervals.<\/li>\n
              4. Join B4<\/sub>C. Draw B4<\/sub>C || B3<\/sub>C’.<\/li>\n
              5. Draw AC ‘|| C’A\u2019, then \u2206A’BC’ is constructed.<\/li>\n
              6. Now, \u2206A’BC’ ||| \u2206ABC.
                \n\u2234 \u2206A’BC’ : \u2206ABC = 5 : 3.<\/li>\n<\/ol>\n

                We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 6 Constructions Ex 6.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions 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 SSLC Class 10 Maths Solutions Chapter 6 Constructions Ex 6.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 6 Constructions Exercise 6.1. Karnataka SSLC Class 10 Maths Solutions Chapter 6 Constructions Exercise 6.1 In each of the following, give the justification of …<\/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\/992"}],"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=992"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/992\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=992"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=992"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=992"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}