Page 67 - Focus SPM 2022 - Additional Mathematics
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Additional Mathematics  SPM   Chapter 1 Circular Measure

                  Example of HOTS                                        14
                             HOTS Question
                French drain using perforated circular pipes is a   The diagram below shows sector MON of a circle. The
                very effective rainwater management in solving   length of chord MN = 11 cm.
                landscaping water issues. The diagram below shows
                the cross section of the circular pipe used in the
                French drain. The pipe is partially filled with water       N    14 cm
                that accumulates as rainwater flows into it.                   11 cm
                                                                                     O
 1
 Minion Pro 10pt =  —                                                       M
 2
 ----------------------------------------
 Arial 8.5pt =   1                 O  5 cm                    Find the area, in cm , of the shaded segment.
                                                                              2
 2                                2 rad
 1
 Arial 8pt =
 2                                                            Solution
 ----------------------------------------  If the length of the circular pipe is 20 cm, calculate   For ΔMNO,
                the volume, in cm , of the water collected.
                              3
                                                                           2
                                                                                2
                                                                      2
 Cth KBAT_Optima 9pt =   80                                         11   = 14  + 14  – 2(14)(14)(cos ∠MON)
 1.25
 ----------------------------------------  Solution                      14  + 14  – 11 2
                                                                           2
                                                                               2
                Cross sectional area of the accumulated water  cos ∠MON =
 ----------------------------------------  =  r (q – sin q)                2(14)(14)
                  1 2
                  2                                                      = 0.6913
                  1
                =  (5) (2 – sin 2)
                     2
                                                                            –1
                  2                                              ∠MON  = cos (0.6913)
                         2
                = 13.63 cm
 1                                                                       = 0.8075 rad
                Volume of the accumulated water
                                                                          1
 XX               Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
                                                                   A
                                                                         =  r q
                                                                            2
                = Cross sectional area × Length of circular pipe    sector  2
 Penyelesaian   = 13.63 × 20
                                                                          1
 (a)  XX        = 272.6 cm 3                                             =   × 14  × 0.8075
                                                                               2
                                                                          2
                                                                         = 79.135 cm
                                                                                 2
                 Try this HOTS Question
                                                                          1
                 The circular pipes are widely used in sewage         s  =  (2r + p)
                                                                          2
                 treatment system. The diagram below shows the
                 sewage collected in the circular pipe.                   1
                                                                         =  [2(14) + 11]
                                                                          2
                                                                         = 19.5 cm
                              O  4 m
                             1.8 rad                              A triangle   =  s(s – p)(s – r) 2
                                      Sewage
         Form 5
                                                                         =  19.5(19.5 – 11)(19.5 – 14)
                                                                                               2
                 Calculate the volume, in m , of the sewage              = 70.81 cm
                                                                                 2
                                         3
                 collected  in  the  circular  pipe  with  the  length  of
                 1.5 km.                                      Area of shaded segment = A  sector  – A  triangle
                 Answer:  9 913.5 m 3                                            = 79.135 – 70.81
                                                                                 = 8.325 cm 2
                                                                 Try Question 8 in ‘Try This! 1.3’
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