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Friday, January 25, 2013




                    AIRIL AHMAD
Friday, January 25, 2013




POLYGON
       By
 Airil Bin Ahmad


                                       AIRIL AHMAD
Friday, January 25, 2013




Polygon is a plane figure bounded by
straight lines




                                              AIRIL AHMAD
Friday, January 25, 2013




2. There are 2 types of polygon:
   a. Irregular Polygon
     A polygon in which not all the sides
     are of equal length or not all the
     interior angles are of equal sides.
         1200


   300          300



                                                AIRIL AHMAD
Friday, January 25, 2013




b. Regular Polygon
  A polygon in which all the sides
  are of equal length and all the
  interior angles are of equal size.




                                             AIRIL AHMAD
Friday, January 25, 2013




Interior Angle

a             b


                  Angles a, b, c and d
    d     c       are interior angles.


                                               AIRIL AHMAD
Friday, January 25, 2013


         3.The name of polygon :

No.of side     Name of figure   No.of triangle      Sum of interior angles
   (n)                             n -2               (n -2) x 180
    3        Triangle             3–2=       1     (3 -2) x 180 =      1800

    4        Quadilateral         4–2=       2     (4 -2) x 180 =      3600

    5        Pentagon             5–2=       3     (5 -2) x 180 =      5400
    6        Hexagon                         4                         7200
    7        Heptagon                        5                         9000
    8        Octagon                         6                         10800
    9        Nonagon                         7                         12600
    10       Decagon                         8                 14400

                                                                       AIRIL AHMAD
Friday, January 25, 2013




a0 + b0 + c0 = 1800
                      c0



                 a0        b0


                                                    AIRIL AHMAD
Friday, January 25, 2013




a0 + b0 + c0 = 1800                           0
x0 + y0 + z0 = 1800

                  c0
a0
                        z0
     b0         y0
          x0

                                           AIRIL AHMAD
Friday, January 25, 2013




a0 + b0 + c0 = 1800                            0
q0 + r0 + s0 = 1800
x0 + y0 + z0 = 1800

       b r00   y0
a0                       z0
 c0                    x0
      q0            s0



                                            AIRIL AHMAD
Friday, January 25, 2013




MATHEMATICS G
                           AIRIL AHMAD
Friday, January 25, 2013




Exercise:
1. Find the sum of the interior angle of each of the
  following regular polygons:


 a. 12-sides                  b. 26-sides
 c. 22-sides                  d. 36-sides




                                                        AIRIL AHMAD
Friday, January 25, 2013




                    AIRIL AHMAD
Friday, January 25, 2013



        Exercise:       2. Find the value of x:




x0 +       +        +        +       = 540        1200


       x + 455    = 540               x0                        1000
                                                                   0




                 x = 5400- 455
                 x = 850                                 1100
                                           1250




                                                                AIRIL AHMAD
Friday, January 25, 2013



        3. Find the value of x:   4. Find the value of x:



              1250                              1390



1180                      1020     x0




       1290          x0                 1130           1310




                                                              AIRIL AHMAD
Friday, January 25, 2013



 5. Find the value of x:   6. Find the value of x:


                                   1020   1090


1250              1020


                            x0                   2390



       650   x0

                                   1130   1010




                                                        AIRIL AHMAD
Friday, January 25, 2013



   7. Find the value of x:    8. Find the value of x:


             x0                        1030       2x0



1250                   1020
                              2x0                       2390




       950        x0
                                       1150     1030




                                                        AIRIL AHMAD
Friday, January 25, 2013




                    AIRIL AHMAD
Friday, January 25, 2013




SUM OF EXTERIOR
    ANGLES
          By
    Airil Bin Ahmad



                                          AIRIL AHMAD
Friday, January 25, 2013




    e

                f


            Angles e, f, g and h
h           are exterior angles.

        g
                                        AIRIL AHMAD
Friday, January 25, 2013




        c
            b

d                             Exterior
                              Angle


            a
    e
                                    AIRIL AHMAD
Friday, January 25, 2013




        c       a+b+c+d+e=3600
            b
d


            a
    e
                                       AIRIL AHMAD
Friday, January 25, 2013




SUM OF EXTERIOR
  ANGLES = 360 0




                               AIRIL AHMAD
Friday, January 25, 2013




Solving Problem Example 1:         Find the value of x:


  +    +     +       +   =   360      600


x + 28500= 3600
  - 285                                                  800


      x = 360    0            x0


      x = 750
                                                        700

                                    750


                                                       AIRIL AHMAD
Friday, January 25, 2013




Solving Problem Example 2:         Find the value of x:


  +    +     +       +   =   360      650


x + 27100= 3600
  - 271                                                  800


      x = 360    0            x0


      x = 890
                                                        560

                                    700


                                                       AIRIL AHMAD
Friday, January 25, 2013



  Example 3:       Find the value of x:



                                           65 0
 +      +      +        +     = 360 0             1150

                                                               800
x + 271 00 = 360 0
  - 271
                              x0


        x =        360 0
                                        1100             1240 560
        X = 89      0
                                         700



                                                             AIRIL AHMAD
Friday, January 25, 2013




EXERCISE
   PG 28
   EX 2.2C
   NO 1, 2 AND 3



                                        AIRIL AHMAD
Friday, January 25, 2013




NO. OF SIDES
        By
  Airil Bin Ahmad


                                        AIRIL AHMAD
Friday, January 25, 2013




 No.of sides(n)=
360 ÷ exterior angle

                                  AIRIL AHMAD
Friday, January 25, 2013



Example 1:       Find the no. of sides if :


 1 a. Exterior angle = 600


    n = 3600 ÷ 600

       = 6      = Hexagon




                                                         AIRIL AHMAD
Friday, January 25, 2013



Example 2:        Find the no. of sides if :


 1b. Interior angle = 1400
    Exterior angle = 1800 – 140=40


    n = 3600 ÷ 400

        = 9      = Nonagon



                                                          AIRIL AHMAD
Friday, January 25, 2013



      Exercise:
      Find the no. of sides in each of the regular polygons.

1.                              2.




                 450                               360


 3.                              4.




                  400                               200


                                                              AIRIL AHMAD
Friday, January 25, 2013



      Exercise:
      Find the no. of sides in each of the regular polygons.

5.                                6.




          1600                               1440


 7.                               8.




           1700                              1080


                                                              AIRIL AHMAD
Friday, January 25, 2013




SOLVING PROBLEM
  OF POLYGONS
          By
    Airil Bin Ahmad


                                          AIRIL AHMAD
Friday, January 25, 2013


     EXERCISE:

1.                2.       X0

             X0

 Y0                                  Y0




                                           AIRIL AHMAD
Friday, January 25, 2013


     EXERCISE:
3.                4.

             Y0        Y0




       X0
                            X0




                                           AIRIL AHMAD
Friday, January 25, 2013




5. In the regular polygon, find x0 and y0   6. In the regular polygon, find x0 and y0


                                                             y0




         x0

                                 y0
                                                    x0


                                                                               AIRIL AHMAD

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NOTE MATH PMR POLYGON

  • 1. Friday, January 25, 2013 AIRIL AHMAD
  • 2. Friday, January 25, 2013 POLYGON By Airil Bin Ahmad AIRIL AHMAD
  • 3. Friday, January 25, 2013 Polygon is a plane figure bounded by straight lines AIRIL AHMAD
  • 4. Friday, January 25, 2013 2. There are 2 types of polygon: a. Irregular Polygon A polygon in which not all the sides are of equal length or not all the interior angles are of equal sides. 1200 300 300 AIRIL AHMAD
  • 5. Friday, January 25, 2013 b. Regular Polygon A polygon in which all the sides are of equal length and all the interior angles are of equal size. AIRIL AHMAD
  • 6. Friday, January 25, 2013 Interior Angle a b Angles a, b, c and d d c are interior angles. AIRIL AHMAD
  • 7. Friday, January 25, 2013 3.The name of polygon : No.of side Name of figure No.of triangle Sum of interior angles (n) n -2 (n -2) x 180 3 Triangle 3–2= 1 (3 -2) x 180 = 1800 4 Quadilateral 4–2= 2 (4 -2) x 180 = 3600 5 Pentagon 5–2= 3 (5 -2) x 180 = 5400 6 Hexagon 4 7200 7 Heptagon 5 9000 8 Octagon 6 10800 9 Nonagon 7 12600 10 Decagon 8 14400 AIRIL AHMAD
  • 8. Friday, January 25, 2013 a0 + b0 + c0 = 1800 c0 a0 b0 AIRIL AHMAD
  • 9. Friday, January 25, 2013 a0 + b0 + c0 = 1800 0 x0 + y0 + z0 = 1800 c0 a0 z0 b0 y0 x0 AIRIL AHMAD
  • 10. Friday, January 25, 2013 a0 + b0 + c0 = 1800 0 q0 + r0 + s0 = 1800 x0 + y0 + z0 = 1800 b r00 y0 a0 z0 c0 x0 q0 s0 AIRIL AHMAD
  • 11. Friday, January 25, 2013 MATHEMATICS G AIRIL AHMAD
  • 12. Friday, January 25, 2013 Exercise: 1. Find the sum of the interior angle of each of the following regular polygons: a. 12-sides b. 26-sides c. 22-sides d. 36-sides AIRIL AHMAD
  • 13. Friday, January 25, 2013 AIRIL AHMAD
  • 14. Friday, January 25, 2013 Exercise: 2. Find the value of x: x0 + + + + = 540 1200 x + 455 = 540 x0 1000 0 x = 5400- 455 x = 850 1100 1250 AIRIL AHMAD
  • 15. Friday, January 25, 2013 3. Find the value of x: 4. Find the value of x: 1250 1390 1180 1020 x0 1290 x0 1130 1310 AIRIL AHMAD
  • 16. Friday, January 25, 2013 5. Find the value of x: 6. Find the value of x: 1020 1090 1250 1020 x0 2390 650 x0 1130 1010 AIRIL AHMAD
  • 17. Friday, January 25, 2013 7. Find the value of x: 8. Find the value of x: x0 1030 2x0 1250 1020 2x0 2390 950 x0 1150 1030 AIRIL AHMAD
  • 18. Friday, January 25, 2013 AIRIL AHMAD
  • 19. Friday, January 25, 2013 SUM OF EXTERIOR ANGLES By Airil Bin Ahmad AIRIL AHMAD
  • 20. Friday, January 25, 2013 e f Angles e, f, g and h h are exterior angles. g AIRIL AHMAD
  • 21. Friday, January 25, 2013 c b d Exterior Angle a e AIRIL AHMAD
  • 22. Friday, January 25, 2013 c a+b+c+d+e=3600 b d a e AIRIL AHMAD
  • 23. Friday, January 25, 2013 SUM OF EXTERIOR ANGLES = 360 0 AIRIL AHMAD
  • 24. Friday, January 25, 2013 Solving Problem Example 1: Find the value of x: + + + + = 360 600 x + 28500= 3600 - 285 800 x = 360 0 x0 x = 750 700 750 AIRIL AHMAD
  • 25. Friday, January 25, 2013 Solving Problem Example 2: Find the value of x: + + + + = 360 650 x + 27100= 3600 - 271 800 x = 360 0 x0 x = 890 560 700 AIRIL AHMAD
  • 26. Friday, January 25, 2013 Example 3: Find the value of x: 65 0 + + + + = 360 0 1150 800 x + 271 00 = 360 0 - 271 x0 x = 360 0 1100 1240 560 X = 89 0 700 AIRIL AHMAD
  • 27. Friday, January 25, 2013 EXERCISE  PG 28  EX 2.2C  NO 1, 2 AND 3 AIRIL AHMAD
  • 28. Friday, January 25, 2013 NO. OF SIDES By Airil Bin Ahmad AIRIL AHMAD
  • 29. Friday, January 25, 2013 No.of sides(n)= 360 ÷ exterior angle AIRIL AHMAD
  • 30. Friday, January 25, 2013 Example 1: Find the no. of sides if : 1 a. Exterior angle = 600 n = 3600 ÷ 600 = 6 = Hexagon AIRIL AHMAD
  • 31. Friday, January 25, 2013 Example 2: Find the no. of sides if : 1b. Interior angle = 1400 Exterior angle = 1800 – 140=40 n = 3600 ÷ 400 = 9 = Nonagon AIRIL AHMAD
  • 32. Friday, January 25, 2013 Exercise: Find the no. of sides in each of the regular polygons. 1. 2. 450 360 3. 4. 400 200 AIRIL AHMAD
  • 33. Friday, January 25, 2013 Exercise: Find the no. of sides in each of the regular polygons. 5. 6. 1600 1440 7. 8. 1700 1080 AIRIL AHMAD
  • 34. Friday, January 25, 2013 SOLVING PROBLEM OF POLYGONS By Airil Bin Ahmad AIRIL AHMAD
  • 35. Friday, January 25, 2013 EXERCISE: 1. 2. X0 X0 Y0 Y0 AIRIL AHMAD
  • 36. Friday, January 25, 2013 EXERCISE: 3. 4. Y0 Y0 X0 X0 AIRIL AHMAD
  • 37. Friday, January 25, 2013 5. In the regular polygon, find x0 and y0 6. In the regular polygon, find x0 and y0 y0 x0 y0 x0 AIRIL AHMAD