Page 14 - Mathematics Class - IX
P. 14

ACTIVITY 1.2





        OBJECTIVE
        To represent some irrational numbers on the number line.

        MATERIALS REQUIRED

              Squares of unit length                                 Coloured papers of two different colours
              Geometry box
        PRE-REQUISITE KNOWLEDGE

          1.  Concept of number line                             2.  Concept of Pythagoras theorem

        THEORY
          For all concepts, refer to Activity 1.1.


        PROCEDURE
          1.  Draw a  straight  line  X´OX as  shown in  Fig. (a)  and  mark  the  unit  length  as  1,  2,  3... and  –  1, –  2,
              – 3,... on either side of origin.
                                                              O
                 X´                                                                                        X
                        –3          –2           –1           0           1            2           3
                                                           Fig. (a)
          2.  Cut the square blocks from different coloured papers of same unit length as marked on the number line, as
              shown in Fig. (b).
                                                                      B







                                                     O                A
                                                           Fig. (b)
          3.  Fold the square block along the diagonal and place it on the number line as shown in Fig. (c).
                                                                               B







                                                                O               A
              X'                                                                                                   X
                      –3            –2            –1             0             1             2             3
                                                            Fig. (c)
          4.  The length of the diagonal OB represents  2  as ∆OAB is a right-angled triangle.

               So,                           OB =  OA  + AB     2
                                                 2
                                                         2
              ⇒                              OB =  1  + 1  2
                                                 2
                                                       2
              ⇒                               OB  =     2
          12
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