{"id":28692,"date":"2021-07-08T18:19:04","date_gmt":"2021-07-08T12:49:04","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=28692"},"modified":"2022-03-02T10:30:13","modified_gmt":"2022-03-02T05:00:13","slug":"ncert-solutions-for-class-10-maths-chapter-11-ex-11-1","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-11-ex-11-1\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter 11 Constructions Ex 11.1 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 11 Constructions Exercise 11.1<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nDraw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.
\nSolution:
\nSteps of Construction:
\n\"NCERT<\/p>\n

    \n
  1. Draw a line segment AB = 7.6 cm.<\/li>\n
  2. Draw an acute angle BAX on base AB. Mark the ray as AX.<\/li>\n
  3. Locate 13 points A1<\/sub>, A2<\/sub>, A3<\/sub>, …… , A13<\/sub> on the ray AX so that AA1<\/sub> = A1<\/sub>A2<\/sub> = ……… = A12<\/sub>A13<\/sub><\/li>\n
  4. Join A13<\/sub> with B and at A5<\/sub> draw a line \u2225 to BA13<\/sub>, i.e. A5<\/sub>C. The line intersects AB at C.<\/li>\n<\/ol>\n

    On measure AC = 2.9 cm and BC = 4.7 cm.
    \nThen AC : CB = 5 : 8
    \nSince A3<\/sub>C is parallel to A13<\/sub>B therefore
    \n\"NCERT
    \nThis show that C divides AB in the ratio 5<\/p>\n

    \"NCERT<\/p>\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:
    \nSteps of Construction:
    \n\"NCERT
    \nFirst we draw line AB 6 cm and then cut an arc AC = 4 cm. After that cut an another are BC = 5 cm. Join A to C and B to C. We find a AABC.<\/p>\n

      \n
    1. Draw ray AX making an acute angle with AB.<\/li>\n
    2. Locate points A1<\/sub>, A2<\/sub>, A3<\/sub>, A4<\/sub>, A5<\/sub> on AX so that AA1<\/sub> = A1<\/sub>A3<\/sub>, = A2<\/sub>A3<\/sub><\/li>\n
    3. Join BA3<\/sub>.<\/li>\n
    4. Join A2<\/sub>B’ such that A2<\/sub>B’ || A3<\/sub>B and (\u2220AA3<\/sub>B = \u2220AA2<\/sub>B’)<\/li>\n
    5. Through B draw a ray B’C || BC’ and \u2220ABC – \u2220AB’C.<\/li>\n<\/ol>\n

      Hence ABC is the required triangle.<\/p>\n

      \"NCERT<\/p>\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 of the first triangle.
      \nSolution:
      \nSteps of Construction:
      \n\"NCERT<\/p>\n

        \n
      1. Draw a \u0394ABC with AB = 5 cm, BC = 7 cm and AC = 6 cm.<\/li>\n
      2. Draw an acute angle CBA below BC at point B.<\/li>\n
      3. Mark the ray BX as B1<\/sub>, B2<\/sub>, B3<\/sub>, B4<\/sub>, B5<\/sub>, B6<\/sub> and B7<\/sub> such that BB1<\/sub>= B1<\/sub>B2<\/sub> = B2<\/sub>B3<\/sub> = B3<\/sub>B4<\/sub> = B4<\/sub>B5<\/sub> = B5<\/sub>B6<\/sub> = B6<\/sub>B7<\/sub>.<\/li>\n
      4. Join B5<\/sub> to C.<\/li>\n
      5. Draw B7<\/sub>C’ parallel to B5<\/sub>C, where C’ is a point on extended line BC.<\/li>\n
      6. Draw A’C’ \u2225 AC, where A’ is a point on extended line BA.<\/li>\n<\/ol>\n

        \u2234 A’BC’ is the required triangle.<\/p>\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:
        \nSteps of Construction:
        \n\"<\/p>\n

          \n
        1. Draw base AB = 8 cm.<\/li>\n
        2. Draw perpendicular bisector of AB. Mark CD = 4 cm, on \u22a5 bisector where D is mid-point on AB.<\/li>\n
        3. Draw an acute angle BAX, below AB at point A.<\/li>\n
        4. Mark the ray AX with A1<\/sub>, A2<\/sub>, A3<\/sub> such that AA1<\/sub> =A1<\/sub>A2<\/sub> = A2<\/sub>A3
          \n<\/sub><\/li>\n
        5. Join A2<\/sub> to B. Draw A3<\/sub>B’ \u2225 A2<\/sub> B, where B’ is a point on extended line AB.<\/li>\n
        6. At B’, draw B’C’ 11 BC, where C’ is a point on extended line AC.<\/li>\n<\/ol>\n

          \u2234 \u2206AB’C’ is the required triangle.<\/p>\n

          \"NCERT<\/p>\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:
          \nSteps of Construction:
          \n\"NCERT<\/p>\n

            \n
          1. Draw a line segment BC = 6 cm and at point B draw an \u2220ABC = 60\u00b0.<\/li>\n
          2. Cut AB = 5 cm. Join AC. We obtain a \u0394ABC.<\/li>\n
          3. Draw a ray BX making an acute angle with BC on the side opposite to the vertex A.<\/li>\n
          4. Locate 4 points A1<\/sub>, A2<\/sub>, A3<\/sub> and A4<\/sub> on the ray BX so that BA1<\/sub> = A1<\/sub>A2<\/sub> = A2<\/sub>A3<\/sub> = A3<\/sub>A4<\/sub>.<\/li>\n
          5. Join A4<\/sub> to C.<\/li>\n
          6. At A3<\/sub>, draw A3<\/sub>C’ \u2225 A4<\/sub>C, where C’ is a point on the line segment BC.<\/li>\n
          7. At C’, draw C’A’ \u2225 CA, where A’ is a point on the line segment BA.<\/li>\n<\/ol>\n

            \u2234 \u2206A’BC’ is the required triangle.<\/p>\n

            \"NCERT<\/p>\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:
            \nIn \u2206ABC, \u2220A + \u2220B + \u2220C = 180\u00b0
            \n\u21d2 105\u00b0 + 45\u00b0 + \u2220C = 180\u00b0
            \n\u21d2 150\u00b0 + \u2220C = 180\u00b0
            \n\u21d2 \u2220C = 30\u00b0
            \nSteps of Construction:
            \n\"NCERT<\/p>\n

              \n
            1. Draw a line segment BC = 7 cm. At point B, draw an \u2220B = 45\u00b0 and at point C, draw an \u2220C = 30\u00b0 and get \u0394ABC.<\/li>\n
            2. Draw an acute \u2220CBX on the base BC at point B. Mark the ray BX with B1<\/sub>, B2<\/sub>, B3<\/sub>, B4<\/sub>, such that BB1<\/sub> = B1<\/sub>B2<\/sub> = B2<\/sub>B3<\/sub> = B3<\/sub>B4<\/sub><\/li>\n
            3. Join B3<\/sub> to C.<\/li>\n
            4. Draw B4<\/sub>C’ \u2225 B3<\/sub>C, where C’ is point on extended line segment BC.<\/li>\n
            5. At C’, draw C’A’ \u2225 AC, where A’ is a point on extended line segment BA.<\/li>\n<\/ol>\n

              \u2234\u00a0\u2206A’BC’ is the required triangle.<\/p>\n

              \"NCERT<\/p>\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 }{ 3F }\\) times the corresponding sides of the given triangle.
              \nSolution:
              \nSteps of Construction:
              \n\"NCERT<\/p>\n

                \n
              1. Draw a right angled \u2206ABC with AB = 4 cm, AC = 3 cm and \u2220A = 90\u00b0.<\/li>\n
              2. Make an acute angle BAX on the base AB at point A.<\/li>\n
              3. Mark the ray AX with A1<\/sub>, A2<\/sub>, A3<\/sub>, A4<\/sub>, A5<\/sub> such that AA1<\/sub> = A1<\/sub>A2<\/sub> = A2<\/sub>A3<\/sub> = A3<\/sub>A4<\/sub> = A4<\/sub>A5<\/sub>.<\/li>\n
              4. Join A3<\/sub>B. At A5<\/sub>, draw A5<\/sub>B’ \u2225 A3<\/sub>B, where B’ is a point on extended line segment AB.<\/li>\n
              5. At B’, draw B’C’ \u2225 BC, where C’ is a point on extended line segment AC.<\/li>\n<\/ol>\n

                \u2234 \u2206AB’C’ is the required triangle.<\/p>\n","protected":false},"excerpt":{"rendered":"

                These NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 11 Constructions Exercise 11.1 Question 1. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts. …<\/p>\n

                NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[2],"tags":[],"yoast_head":"\nNCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 - MCQ Questions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-11-ex-11-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 Questions and Answers are prepared by our highly skilled subject experts. 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