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Matematik Tambahan  Tingkatan 5  Bab 1  Sukatan Membulat

                  6.  Tentukan perimeter tembereng bagi setiap bulatan berpusat O yang berikut.  TP  4
                     Determine the perimeter of the segment of each of the following circles with centre O.

                                                                        (a)                                                 1
                       CONTOH
                                                                                         P
                                                                                            9.6 cm                         BAB
                                                                                              O
                                           O  6.2 cm  P                                Q   0.4  rad
                                       2 rad
                                         Q
                                                                                      0.4π × 360°
                                                                            0.4π rad  =
                       Penyelesaian:                                                      2π
                       Perentas  PQ dapat diperoleh dengan petua                    = 72°
                       kosinus, iaitu  PQ =   j  + j  − 2j kos q , dengan  q   Maka/Hence,
                                               2
                                            2
                                                    2
                       dalam darjah.                                        PQ  =  9.6  + 9.6  − 2(9.6)  kos72°
                                                                                           2
                                                                                                   2
                                                                                     2
                       The length of the chord PQ can be obtained by using the   Arc length PQ Bhd.

                                                                               = 11.28 cm
                       cosine rule, that is PQ =  j  + j  − 2j cos q , such that q   Panjang lengkok PQ = jq
                                             2
                                        Penerbitan Pelangi Sdn.
                                                     2
                                                2
                       is in degrees.                                                         = 9.6(0.4π)
                               2 × 360°
                       2 rad =         = 114.59°                                              = 12.06 cm
                                 2π                                         Perimeter = (12.06 + 11.28) cm
                       Maka/Hence,                                                   = 23.34 cm
                                               2
                       PQ  =  6.2  + 6.2  − 2(6.2)  kos 114.59°
                                 2
                                       2
                           = 10.43 cm
                       Panjang lengkok PQ = jq = 6.2(2)
                       Arc length PQ
                                              = 12.4 cm
                       Perimeter = (12.4 + 10.43) cm = 22.83 cm
                       (b)                                              (c)
                                            P
                                        8 cm                                           P  3 cm  4.5 rad
                                         O                                                    O
                                     30 cm    Q                                             Q

                           Panjang lengkok minor PQ = 2π(8) – 30
                           Minor arc length PQ      = 20.27 cm              Panjang lengkok minor PQ = 2π(3) – 4.5(3)
                                   20.27                                    Minor arc length PQ     = 5.35 cm
                           ∠POQ =                                           ∠POQ  = 2π − 4.5
                                     8                                            = 1.78 rad
                                 = 2.53 rad                                       = 102.17°
                                 = 145.14°                                  Maka, PQ =  3  + 3  − 2(3) kos102.17°
                                                                                         2
                                                                                             2
                                                                                                    2
                           Maka, PQ =    8  + 8 −2(8)  kos145.14°           Hence    = 4.67 cm
                                         2
                                             2
                                                   2
                                    = 15.27 cm                              Perimeter = 5.35 cm + 4.67 cm
                           Perimeter = 15.27 + 20.27                                 = 10.02 cm
                           Hence     = 35.54 cm



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         01 M&M Mate T Tg5.indd   7                                                                              09/12/2022   12:20 PM
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