Page 11 - PBD Plus Matematik Tambahan T5 (EG)
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Matematik Tambahan   Tingkatan 5  Bab 1  Sukatan Membulat

                 (c)               11.7 cm       (d)                              (e)  Diberi AB ialah diameter.
                                                                  13.1 cm  2 θ       Given that AB is a diameter.
                                    θ
                                                                 O
                                   O
                                                                                         A        2 θ  B
                         8.9 cm
                                                                                                O
                                                               26.8 cm
                    2πj = 11.7 + 2(8.9)                         26.8
                                                                                          π
                       j = 4.695 cm                    2π – 2q  =   13.1             2q =   × 3
                      q =   11.7                 	   	     q	= 2.12 rad                   5
                          4.695                                                      q =  3π  rad
                 	  	  	= 2.492 rad                                                      10
                                                                                       = 0.942 rad


           SP 1.2.2  Menentukan perimeter tembereng suatu bulatan.

             6.  Tentukan perimeter tembereng yang berlorek bagi setiap bulatan berpusat O yang berikut.  TP 4
               Determine the perimeter of the shaded segment of each of the following circles with centre O.
                     Contoh
                                  Tip Penting                     (a)
                    P                                                                P          Q
                                  Perentas PQ dapat diperolehi dengan                    1.3 rad
                                  petua kosinus, iaitu                                    O  7.5 cm
                                  PQ = √
                                       2
                                          2
                                             2
                                       j  + j  – 2j  kos q, dengan q
                       0.4π rad   dalam darjah.
                      9.6 cm      The chord PQ can be obtained by using the
                    Q                                                              180°
                                  cosine rule, that is               1.3 rad = 1.3 ×    = 74.48°
                                  PQ = √                                           π
                                            2
                                       2
                                       j  + j  – 2j  cos q, such that q is in
                                         2
                                  degrees.
                                                                                 7.5  + 7.5  – 2(7.5)  kos 74.48°
                                                                     Maka, PQ = √
                                                                                          2
                                                                                                  2
                                                                                    2
                                 180°                                Hence    = 9.08 cm
                 0.4p rad = 0.4p ×    = 72°
                                  π Penerbitan Pelangi Sdn. Bhd.
                                                                     Panjang lengkok PQ = jq = 7.5(1.3)
                 Maka, PQ = √                                       Arc length of PQ
                             9.6  + 9.6  – 2(9.6)  kos 72°
                                2
                                             2
                                      2
                 Hence    = 11.29 cm                                                        = 9.75 cm
                 Panjang lengkok PQ  = jq = 9.6(0.4π)                Perimeter tembereng berlorek  = 9.08 + 9.75
                 Arc length of PQ   = 12.06 cm                       Perimeter of the shaded segment   = 18.83 cm
                 Perimeter tembereng berlorek  = 12.06 + 11.29
                 Perimeter of the shaded segment   = 23.35 cm
                                                                  (c)                     P
                 (b)                         P
                                                                                     2.8 cm
                                       7.2 cm
                                                                                       O       5.25 cm
                                           O
                                        4.2 rad
                                               Q
                                                                                           Q
                    Panjang lengkok minor PQ                         Panjang lengkok major PQ
                    Minor arc length of PQ                                           Major arc length of PQ
                    = (2π – 4.2) × 7.2                               = 2π(2.8) – 5.25 = 12.34 cm.
                    = 15 cm                                          ∠POQ =   5.25
                                                                               2.8
                    /POQ = 2π – 4.2 = 2.08 rad                              = 1.875 rad
                                    = 119.36°                                        180
                    Maka, PQ = √                                          = 1.875 ×   p
                                7.2  + 7.2  – 2(7.2)  kos 119.36°
                                   2
                                         2
                                                2
                    Hence    = 12.43 cm                                     = 107.43°
                                                                                 2.8  + 2.8  – 
                                                                                          2
                                                                                                  2
                                                                                    2
                    Perimeter tembereng berlorek  = 15 + 12.43       Maka, PQ = √ 2(2.8)  kos 107.43°
                    Perimeter of the shaded segment   = 27.43 cm     Hence    = 4.51 cm
                                                                     Perimeter tembereng berlorek  = 12.34 + 4.51
                                                                     Perimeter of the shaded segment   = 16.85 cm
                Cuba jawab   Praktis SPM 1, K1: S6
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