Page 28 - Hybrid PBD 2022 Tg 5 - Matematik Tambahan
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Matematik Tambahan   Tingkatan 5  Bab 1  Sukatan Membulat
                  8.  Tentukan luas sektor berlorek bagi setiap bulatan yang berikut. Beri jawapan betul kepada dua tempat
                     perpuluhan.
                     Determine the area of the shaded sector for each of the following circles. Give your answer correct to two decimal places.    SP 1.3.1     TP3

                                                                       (a)
                                                                                               4.8 cm
                                      A      O     B                                           O     θ
                                                                                               __
                                                                                               6π
                                                                                                5
                                                 C
                                                                              
                      Diberi AOB ialah diameter dengan panjangnya         q =  2π –   6π 2  rad
                      ialah 7.4 cm, dan /AOC = 2/BOC.                         4π    5
                      Given AOB is the diameter of length 7.4 cm and        =     rad
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                      /AOC = 2/BOC.                                            5
                                                                              1 2
                      /AOC =   2π  rad                                    L  =  j q
                               3                                              2
                                                                              1
                                                                                      2
                          1 2
                      L  =  j q                                             =   (4.8) 2  4π
                          2                                                   2        5
                          1
                                  2
                        =    (3.7) 2  2π   = 14.34 cm 2                     = 28.95 cm 2
                          2        3
                      (b)                                              (c)
                                                                                        20 cm
                                          3.8 cm
                                           O   θ    7.5 cm
                                                                                                 θ  3.85 rad
                                                                                                    O
                         q =  7.5                                         q  = (2π – 3.85)
                             3.8                                            = 2.43 rad
                           = 1.97 rad                                     Jejari / Radius =   20   = 8.23 cm
                                                                                        2.43
                         L  =  j q                                            1 2
                             1 2
                             2                                            L  =  j q
                                                                              2
                             1
                           =   (3.8) (1.97)                                   1
                                   2
                                                                                     2
                             2                                              =   (8.23) (2.43)
                                     2
                           = 14.22 cm                                         2
                                                                            = 82.30 cm
                                                                                      2
                  9.  Tentukan jejari bagi sektor bulatan berlorek yang berikut. Beri jawapan betul kepada dua tempat perpuluhan.
                     Determine the radius of the following shaded sectors of  circles. Give your answer correct to two decimal places.     SP 1.3.1     TP4
                                                                  (a)  Diberi luas sektor ialah 49 cm 2
                      Diberi luas sektor yang berlorek ialah                                   2
                      .8 cm 2                                        Given the area of the sector is 49 cm .
                      Given the area of the shaded sector is 23.8 cm . 2
                           A                                                             1.63 rad
                              1.78 rad
                               O
                                     B
                                                                        49  =  1.63
                                               Tip Penting              πj 2   2π
                                                luas sektor  q                49 × 2
                                                                          2
                       23.8   =  1.78      Guna   luas bulatan   =    2π     j  =   1.63
                       πj 2    2π          Use   area of sector   =   q    j = 7.75 cm
                                     2        area of circle  2p
                           2
                          j  = 23.8 ×
                                    1.78
                           j = 5.17 cm






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