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Matematik Tambahan  Tingkatan 5  Bab 1  Sukatan Membulat
                       Sudut KBATKBAT                                                                       KBAT

                                                                                                            EKSTRA
                                                                                                                            1


                  Rajah di bawah menunjukkan sebuah bulatan           (a)  Cari sudut SOP dalam radian.                    BAB
                  berpusat O dan berjejari 10 cm. PQ dan SR adalah dua      Find the angle of SOP in radians.
                  perentas yang selari dengan PQ = 10 cm dan ∠SOR =   (b)  Tentukan luas rantau berlorek itu.
                  120°.                                                  Determine the area of the shaded region.
                  The diagram below shows a circle with centre O and radius   (c)  Tunjukkan bahawa perimeter rantau berlorek ialah

                                                                           
                  10 cm. PQ and SR are two parallel chords such that PQ = 10   10 1 +  3 +  π   cm.
                  cm and ∠SOR = 120°.  KBAT   Menganalisis                          3
                                                                         Show  that the  perimeter  of the  shaded  region  is
                                                                           
                                       P        Q                        10 1 +  3 +  π   cm.
                                     S           R                                  3
                                           O






                             2
                                  2
                                          2
                  (a)  PQ  = 10  + 10  − 2(10)  kos POQ                  Luas tembereng SR / Area of segment SR
                        2
                                                                                        1
                                       2
                                  2
                                10 + 10  − 10 2                          =  1  (10) 2  2π   −  (10)  sin 120°
                                                                                              2
                       kos POQ  =                                          2       3    2
                                   2(100)
                       ∠POQ  = 60°                                       = 61.42 cm 2
                                120°−60°                                 Maka, luas berlorek
                       ∠SOP  =     1 Penerbitan Pelangi Sdn. Bhd.
                                                                         Hence, the shaded area

                                   2
                                     π                                   = 61.42 – 9.06
                              = 30° =   rad                              = 52.36 cm 2
                                     6
                                          10 cm                       (c)  Panjang lengkok SP = QR
                                       P        Q                        Arc length SP = QR
                                     S           R
                                                                            π
                                           120                           10    =  5π
                                             10 cm                          6     3
                                           O
                                                                           2
                                                                                              2
                                                                                      2
                                                                         SR  = 10  + 10  − 2(10)  kos 120°
                                                                                 2
                                                                         SR  = 10 3
                                                                                           5π
                  (b)  Luas tembereng PQ /Area of segment PQ             Perimeter = 10 + 2    + 10 3
                       1      π                                                            3
                     =   (10) 2    −  (10)  sin 60°                                      π
                                         2
                                                                                       
                       2      3    2                                               = 10 1+   +  3 
                     = 9.06 cm 2                                                           3

                                                                                                              Jawapan
                                                                                         Enrolment key        Bab 1
                                                                                        M#maTT5#m


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         01 M&M Mate T Tg5.indd   21                                                                             09/12/2022   12:21 PM
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