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    O-Level Additional Math

    Scheduled Pinned Locked Moved Secondary Schools - Academic Support
    809 Posts 301 Posters 489.6k Views 1 Watching
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    • T Offline
      tianzhu
      last edited by

      Hi schellen and tankee


      I believe Guan Hui is a tutor.I think he is trying to start a dedicated thread similar to that of mathsguru in the primary section.

      Maybe the moderator can consider to give him a sticky thread for him to help to solve Secondary Maths problems for members. This is a win-win situation, members can get help from a professional tutor and GH can gain more exposure about his tutoring service.Let’s us take a small step first and monitor the responses, questions for all secondary levels can be lumped together.But members asking questions should state the level of their question.

      Perhaps, the thread’s heading can be
      Tutor Guan Hui ------ Ask me your Secondary Maths Questions

      As for GH, I hope you are mentally prepared for the long journey.It takes a lot of stamina to do what mathsguru is doing.

      Best wishes

      1 Reply Last reply Reply Quote 0
      • G Offline
        Guan Hui
        last edited by

        Thx tianzhu it is exactly what i intended. 😄 😄 and thankyou for your wishes 😄

        1 Reply Last reply Reply Quote 0
        • S Offline
          schellen
          last edited by

          Then, go ahead and edit this thread's title, Guan Hui...and good luck! 🙂

          1 Reply Last reply Reply Quote 0
          • T Offline
            tianzhu
            last edited by

            Tutor Guanhui- Ask Your Secondary Questions Here!


            Hi GH

            What are the subjects?

            Best wishes

            1 Reply Last reply Reply Quote 0
            • G Offline
              Guan Hui
              last edited by

              oppps... haha thanks for reminding 😄

              1 Reply Last reply Reply Quote 0
              • T Offline
                tweet
                last edited by

                Hi Guan Hui


                Pls help with these 2 questions:

                1) Given that A is the point (0,2), B is the point (9,0) and C is a point on the line y=2x + 2 such that AB = BC, find the coordinates of C. If ABCD forms a parallelogram, find the coordinates of point D.
                Answers: C(2,6) and D(-7,8 )

                2) Find the area of triangle ABC whose coordinates are A(3,3), B(-1,0) and C (5,-3) Hence, or otherwise, calculate the distance from C to AB.
                Answer: 15;6

                TIA

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                • G Offline
                  Guan Hui
                  last edited by

                  zzz… typing this for the 3rd time… haha here we go


                  let c be (x,y)
                  distance of ab=sqrt(9^2+2^2)
                  =sqrt(85)
                  distance of bc=sqrt((9-x)^2+y^2)
                  since ab=bc
                  sqrt((9-x)^2+y^2)=sqrt(85)
                  (9-x)^2+y^2=85
                  y^2=85-81+18x-x^2
                  y^2=4+18x-x^2------------------1
                  y=2x+2
                  y^2=(2x+2)^2
                  y^2=4x^2+8x+4------------------2
                  4+18x-x^2=4x^2+8x+4
                  5x^2-10x=0
                  x=0 (rejected cause this value is for a)
                  or x=2
                  put x=2 into y=2x+2
                  y=6
                  c(2,6)

                  since abcd is parallelogram, ab ll cd and bc ll ad
                  gradient ab=(2-0)/(0-9)
                  =-2/9
                  y =-2/9x +c
                  sub c in,
                  6=-4/9+c
                  c=58/9
                  equation of cd: y=-2/9x+58/9

                  gradient of bc=(0-6)/(9-2)
                  =-6/7
                  equation of line ad: y= -6/7x +2

                  -2/9x+58/9=-6/7x+2
                  -14x+406=-54x+126
                  40x=-280
                  x=-7
                  put x=-7 into y=-6/7x+2
                  y=8

                  d(-7,8 )

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                  • G Offline
                    Guan Hui
                    last edited by

                    ab=sqrt((3+1)^2+3^2)

                    =sqrt(25)
                    bc=sqrt((-1-5)^2+(0+3)^2)
                    =sqrt(45)
                    ac=sqrt((3-5)^2+(3+3)^2)
                    =sqrt(40)
                    let the angle at point c be x
                    cosx=( ac^2+bc^2-ab^2 )/(2 acbc)
                    =60/2 sqrt(1800)
                    =sqrt(0.5)
                    x= 45 degree
                    sin 45 =sqrt(0.5)
                    area = 0.5 bc
                    ac sin 45
                    =15

                    since ab= sqrt 25= 5
                    area of triangle:
                    0.5 x 5x h =15
                    h=6

                    **note
                    sqrt (1800)= sqrt (3600
                    0.5
                    = 60 sqrt (0.5)

                    1/2sqrt(0.5)=sqrt (0.5)

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                    • T Offline
                      tweet
                      last edited by

                      Hi Guan Hui


                      Thanks for your help.Will show the worked solution to my son.

                      Rgds

                      1 Reply Last reply Reply Quote 0
                      • E Offline
                        emerald
                        last edited by

                        Hi, need your help to solve the following using Commutative Laws, Associative Laws &/or Distributive Laws (Easy method):


                        (54.2 x 1.8 + 0.2 x 25.86)


                        TIA.

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