WISKUNDE
GRAAD 11
NOG OEFENINGE
  
Omtrek, oppervlakte en volume : antwoorde
  
MATHEMATICS
GRADE 11
MORE EXERCISES
  
Perimeter, area and volume : answers
  
  
Antwoorde  / Answers  1.  
  

  1.1
  
      Lengte = L = 45 mm, breedte = B = 73 mm
  
      Oppervlakte = L X B
  
                          = 45 X 73 mm2
  
                          = 3 285 mm2                  [ V 1.1 ]
  
  1.2
  
      Basis = B = 56 cm, hoogte = H = 21 cm
  
      Oppervlakte = ½ X L X B
  
                          = ½ X 56 X 21 cm2
  
                          = 588 cm2                      [ V 1.2 ]
  
  1.3
  
      Basis = B = 45 m, hoogte = H = 22 m
  
      Oppervlakte = ½ X L X B
  
                          = ½ X 45 X 22 m2
  
                          = 495 m2                        [ V 1.3 ]
  
  1.4
  
      Sy = s = 72 cm
  
      Oppervlakte = s X s = s2
  
                          = 722 cm2
  
                          = 5 184 cm2                   [ V 1.4 ]
  
  1.5
  
      Basis = B = 3,8 m, hoogte = H = 1,8 m
  
      Oppervlakte = ½ X L X B
  
                          = ½ X 3.8 X 1,8 m2
  
                          = 3,42 m2                       [ V 1.5 ]
  
  1.6
  
      Basis = B = 56 m, hoogte = H = 46,4 m
  
      Oppervlakte = ½ X L X B
  
                          = ½ X 56 X 46,4 m2
  
                          = 1 299,2 m2                  [ V 1.6 ]
  
  1.7
  
      Radius = straal = r = 14,7 cm
  
      Oppervlakte = π X r2
  
                          = π X 14,72 cm2
  
                          = 678,867 cm2               [ V 1.7 ]
  
  1.8
  
      Middellyn = d = 3,14 m,  radius = r = 1,57 m
  
                                 π X d2
      Oppervlakte = ──────        =  π X r2
                                    4
  
                               π X 3,142
                          = ──────     =  π X 1,572  m2
                                   4
  
                          =  7,744  m2     =  7,744  m2
  
                                                                [ V 1.8 ]
  

  1.1
  
      Length = L = 45 mm, breadth = B = 73 mm
  
      Area = L X B
  
              = 45 X 73 mm2
  
              = 3 285 mm2                           [ Q 1.1 ]
  
  1.2
  
      Base = B = 56 cm, height = H = 21 cm
  
      Area = ½ X L X B
  
               = ½ X 56 X 21 cm2
  
               = 588 cm2                              [ Q 1.2 ]
  
  1.3
  
      Base = B = 45 m, height = H = 22 m
  
      Area = ½ X L X B
  
               = ½ X 45 X 22 m2
  
               = 495 m2                                [ Q 1.3 ]
  
  1.4
  
      Side = s = 72 cm
  
      Area = s X s = s2
  
               = 722  cm2
  
               = 5 184 cm2                           [ Q 1.4 ]
  
  1.5
  
      Base = B = 3,9 m, height = H = 1,8 m
  
      Area = ½ X L X B
  
               = ½ X 3,8 X 1,8 m2
  
               = 3,42 m2                                [ Q 1.5 ]
  
  1.6
  
      Base = B = 56 m, height = H = 46,4 m
  
      Area = ½ X L X B
  
               = ½ X 56 X 46,4 m2
  
               = 1 299,2 m2                          [ Q 1.6 ]
  
  1.7
  
      Radius = r = 14,7 cm
  
      Area = π X r2
  
               = π X 14,72 cm2
  
               = 678,867 cm2                       [ Q 1.7 ]
  
  1.8
  
      Diameter = d = 3,14 m,  radius = r = 1,57 m
  
                    π X d2
      Area = ──────        =  π X r2
                       4
  
                  π X 3,142
              = ──────     =  π X 1,572  m2
                        4
  
              =  7,744  m2     =  7,744  m2
  
                                                             [ Q 1.8 ]
  
  
  1.9
  
      Een reghoeksy is die basis en die ander een die hoogte. Bereken dus die ander reghoeksy deur
      Pythagoras se stellig te gebruik.
  
      One right-angled side is the base and the other the height. Calculate the length of the second
      right-angled side by using the theorem of Pythagoras.
  
      Laat die skuinssy = a, een reghoeksy = b en die ander een = c  / 
      Let the hypotenuse be equal to a, one right-angled side = b and the other = c.
  
      a2  =  b2  +  c2
      2,62  =  12  +  c
      6,76  =  1  +  c
      6,76  -  1  =  c
      5,76  =  c2
      c  =  √5,76
      c  =  2,4
  
      Basis = Base = B = 1 cm, hoogte = height = H = 2,4 cm
  
      Oppervlakte / Area = ½ X L X B
                                     = ½ X 1 X 2,4 cm2
                                     = 1,2 cm2                                    [ Q 1.9 ]
  
  
Antwoord  / Answer  2.  
  


      Bereken die oppervlakte van die muur en die oppervlakte van 1 steen en gebruik dan
  
      die gegewe formule :
  
      Calculate the area of the wall and the area of a brick and then
  
      apply the given formula.
  
  
      Bereken al die oppervlaktes in m2  /  Calculate all areas in m2
  
      Muur is 6,60 m lank = L = length of wall en/and
  
      muur is 3,25 m hoog = B = height of wall.
  
      Oppervlakte / Area  =  L X B
  
                                      = 6,60 X 3,25 m2
  
                                      = 21,45 m2
  
      Steen is 30 cm lank = L = length of brick en/and 50 mm hoog = B = height of brick.
  
      Steen / Brick : L = 30 cm = (30 ÷ 100) m  =  0,3 m  ;    
  
      Steen / Brick : B  =  50 mm  =  (50 ÷ 1 000) m    =  0,05 m
  
      Oppervlakte van steen / Area of brick  =  L X B
  
                                                                   = 0,3 X 0,05 m2
  
                                                                   = 0,015 m2
  
                                                                     Oppervlakte van muur / Area of wall
      Aantal stene /  Number of bricks  =  ────────────────────────
                                                                     Oppervlakte van steen / Area of brick
  
                                                                 21,45 m2
                                                           =  ────────
                                                                 0,015 m2
  
                                                           =  1 430
  
       Maar die muur is 2 stene breed, dus is die totale aantal stene tweemaal 1 430,
       d.w.s. 2 860.
  
       But the wall is 2 bricks wide and therfore the total number of bricks is twice 1 430,
       i.e. 2 860
  
                                                                                                                                                          [ V/Q 2. ]qwe    
  
  
 OF / OR 

  
      Bereken al die oppervlaktes in cm2  /  Calculate all areas in cm2
  
      Muur is 6,60 m lank = L = length of wall = 6,60 X 100 cm = 660 cm en/and
  
      muur is 3,25 m hoog = B = height of wall = 3,25 X 100 cm = 325 cm.
  
      Oppervlakte / Area  =  L X B
  
                                      = 660 X 325 cm2
  
                                      = 214 500 cm2
  
      Steen is 30 cm lank = L = length of brick en/and
  
      steen is 50 mm hoog = B = height of brick = 50 ÷ 10 cm = 5 cm.
  
      Oppervlakte van steen / Area of brick  =  L X B
  
                                      = 30 X 5 cm2
  
                                      = 150 cm2
  
                                                                     Oppervlakte van muur / Area of wall
      Aantal stene /  Number of bricks  =  ────────────────────────
                                                                     Oppervlakte van steen / Area of brick
  
                                                                 214 500 cm2
                                                           =  ────────
                                                                 150 cm2
  
                                                           =  1 430
  
       Maar die muur is 2 stene breed, dus is die totale aantal stene tweemaal 1 430,
       d.w.s. 2 860.
  
       But the wall is 2 bricks wide and therfore the total number of bricks is twice 1 430,
       i.e. 2 860
  
                                                                                                                                                               [ V/Q 2. ] zxdrfv
  
  
 OF / OR 

  
       Bereken die oppervlakte van die muur in m2 en die opperlakte van die steen in cm2.
       Skakel dan die m2 om na cm2 of die cm2 om na m2
  
       Calculate the area of the wall in m2 and the area of the brick in cm2.
  
       Then convert m2 to cm2 or m2 to cm2.
  
      Oppervlakte / Area  =  L X B
  
                                      = 6,60 X 3,25 m2
  
                                      = 21,45 m2
  
      Oppervlakte van steen / Area of brick  =  L X B
  
                                      = 30 X 5 cm2
  
                                      = 150 cm2
  
       Skakel nou cm2 om na m2 / Convert cm2 to m2
  
                                  150
        150 cm2  =  ──────── m2    =  0,015  m2
                             100 X 100
  
       Bereken die aantal stene soos hierbo.  Calculate the number of bricks as in the calculations above.
  
                                                                                                                                                                  [ VQ 2. ]
  
  
 OF / OR 

  
       Skakel nou m2 om na cm2 / Convert m2 to cm2
  
        21,45 m2  =  21,45 X (100 X 100) cm2    =  214 500  cm2
  
       Bereken die aantal stene soos hierbo.  Calculate the number of bricks as in the calculations above.
  
                                                                                                                                                                  [ VQ 2. ]
  
  
Antwoorde  / Answers  3.  
  

  3.1
  
       Oppervlakte van vierkant = sy2
  
                   552,25 = sy2
  
                            s = √552,25
  
                               =  23,5 m
  
       Vierkant se sy = 23,5 m
  
       Omtrek van vierkant = 4 X sy
  
                                        = 4 X 23,5 m
  
                                        = 94 m
  
       Omtrek van vierkant = 94 m                [ V 3.1 ]
  
  
  
  3.2
  
       Oppervlakte van reghoek = L X B
  
       L X B = Oppervlakte van reghoek
  
       L X 3,45 = 20,769
  
                           20,769
                  L  =  ─────    =  6,02
                             3,45
  
       Lengte van reghoek = 6,02 m
  
       Omtrek van reghoek = 2(L + B)
  
                                            = 2(6,02 + 3,45) m
  
                                            = 18,94 m       [ V 3.2 ]
  
  3.3
  
       Oppervlakte van driehoek = ½ BH
  
       ½ BH = Oppervlakte van driehoek
  
       ½ X 28,4 X H = 249,92
  
                                    2 X 249,92
                         H  =  ────────
                                        28,4
  
                           =  17,6
  
       Hoogte van driehoek = 17,6 cm
  
                                                                    [ V 3.3 ]
  
  3.4
  
       Oppervlakte van driehoek = ½ BH
  
       ½ BH = Oppervlakte van driehoek
  
       ½ X B X 1,26 = 0,4095
  
                                    2 X 0,4095
                         H  =  ────────
                                        1,26
  
                              =  0,65
  
       Basis van driehoek = 0,65 m
  
                                                                    [ V 3.4 ]
  
  3.5
  
       Oppervlakte van sirkel = π r2
  
       π r2 = Oppervlakte van sirkel
  
       π r2 = 124,690
  
                                    124,690
                         r2  =  ───────
                                        π
  
                              =  39,69005971
  
                         r  =  √39,69005971  =  6,300
  
       Radius van sirkel = 6,300 cm
  
       Deursnee van sirkel = D = 2r
  
                                        = 2 X 6,300 cm
  
                                        = 12,600 cm
  
       Omtrek van sirkel = 2πr
  
                                    = 2 X π X 6,300 cm
  
                                    = 39,584 cm
                                                                    [ V 3.5 ]
  
  
 OF 
  
  
                                     π X D2
          Oppervlakte =  ──────
                                        4
  
               π X D2
            ──────  =  124,690
                   4
  
                                   4 X 124,690
                      D2  =  ─────────
                                        π
  
                      D2  =  158,7602388
  
                      D   =  √158,7602388
  
                            =  12,600
  
       Deursnee van sirkel = 12,600 cm
  
       Omtrek van sirkel = πD
  
                                    = π X 12,600 cm
  
                                    =  39,584  cm
  
                                                                    [ V 3.5 ]
  
  3.6
  
       Oppervlakte van sirkel = π r2
  
       π r2 = Oppervlakte van sirkel
  
       π r2 = 3 848,45
  
                                    3 848,45
                         r2  =  ───────
                                        π
  
                              =  1 224,999681
  
                         r  =  √1 224,999681
  
                           =  34,99999545
  
       Radius van sirkel = 35,000 cm
  
       Omtrek van sirkel = 2πr
  
                                    = 2 X π X 35,000 cm
  
                                    = 219,911 cm
                                                                    [ V 3.6 ]
  
  
 OF 
  
  
                                     π X D2
          Oppervlakte =  ──────
                                        4
  
               π X D2
            ──────  =  3 848,45
                   4
  
                                  4 X 3 848,45
                      D2  =  ─────────
                                        π
  
                      D2  =  4 899,998726
  
                      D   =  √4 899,998726
  
                            =  69,9999909
  
       Deursnee van sirkel = 70,000 cm
  
       Straal van sirkel = D ÷ 2
  
                                  = 70,000 ÷ 2 cm
  
                                  = 35,000  cm
  
       Omtrek van sirkel = πD
  
                                    = π X 70,000 cm
  
                                    =  219,911  cm
  
                                                                    [ V 3.6 ]
  

  3.1
  
       Area of square = side2
  
                   552,25 = side2
  
                            s = √552,25
  
                               =  23,5 m
  
       Side of square = 23,5 m
  
       Perimeter of square = 4 X side
  
                                        = 4 X 23,5 m
  
                                        = 94 m
  
       Perimeter of square = 94 m                [ Q 3.1 ]
  
  
  
  3.2
  
       Area of rectangle = L X B
  
       L X B = Area of rectangle
  
       L X 3,45 = 20,769
  
                           20,769
                  L  =  ─────    =  6,02
                             3,45
  
       Length of rectangle = 6,02 m
  
       Perimeter of rectangle = 2(L + B)
  
                                            = 2(6,02 + 3,45) m
  
                                            = 18,94 m          [ Q 3.2 ]
  
  3.3
  
       Area of triangle = ½ BH
  
       ½ BH = Area of triangle
  
       ½ X 28,4 X H = 249,92
  
                                    2 X 249,92
                         H  =  ────────
                                        28,4
  
                           =  17,6
  
       Height of triangle = 17,6 cm
  
                                                                       [ Q 3.3 ]
  
  3.4
  
       Area of triangle = ½ BH
  
       ½ BH = Area of triangle
  
       ½ X B X 1,26 = 0,4095
  
                                    2 X 0,4095
                         H  =  ────────
                                        1,26
  
                              =  0,65
  
       Base of triangle = 0,65 m
  
                                                                       [ Q 3.4 ]
  
  3.5
  
       Area of circle = π r2
  
       π r2 = Area of circle
  
       π r2 = 124,690
  
                                    124,690
                         r2  =  ───────
                                        π
  
                              =  39,69005971
  
                         r  =  √39,69005971  =  6,300
  
       Radius of circle = 6,300 cm
  
       Diameter of circle = D = 2r
  
                                    = 2 X 6,300 cm
  
                                    = 12,600 cm
  
       Circumference of circle = 2πr
  
                                             = 2 X π X 6,300 cm
  
                                             = 39,584 cm
                                                                    [ Q 3.5 ]
  
  
 OR 
  
  
                         π X D2
          Area =  ──────
                             4
  
               π X D2
            ──────  =  124,690
                   4
  
                                   4 X 124,690
                      D2  =  ─────────
                                        π
  
                      D2  =  158,7602388
  
                      D   =  √158,7602388
  
                            =  12,600
  
       Diameter of circle = 12,600 cm
  
       Circumference of circle = πD
  
                                             = π X 12,600 cm
  
                                             =  39,584  cm
  
                                                                    [ Q 3.5 ]
  
  3.6
  
       Area of circle = π r2
  
       π r2 = Area of circle
  
       π r2 = 3 848,45
  
                                    3 848,45
                         r2  =  ───────
                                        π
  
                              =  1 224,999681
  
                         r  =  √1 224,999681
  
                           =  34,99999545
  
       Radius of circle = 35,000 cm
  
       Circumference of circle = 2πr
  
                                             = 2 X π X 35,000 cm
  
                                             = 219,911 cm
                                                                    [ Q 3.6 ]
  
  
 OR 
  
  
                         π X D2
          Area =  ──────
                             4
  
               π X D2
            ──────  =  3 848,45
                   4
  
                                  4 X 3 848,45
                      D2  =  ─────────
                                        π
  
                      D2  =  4 899,998726
  
                      D   =  √4 899,998726
  
                            =  69,9999909
  
       Diameter of circle = 70,000 cm
  
       Radius of circle = D ÷ 2
  
                                 = 70,000 ÷ 2 cm
  
                                 = 35,000  cm
  
       Circumference of circle = πD
  
                                             = π X 70,000 cm
  
                                             =  219,911  cm
  
                                                                    [ Q 3.6 ]
  
  
Antwoorde  / Answers  4.  
  

  4.1
  
       BO van kubus = 6s2
  
                              = 342  mm2
  
                              =  1 156  mm2
  
       Volume van kubus = s3
                                    =  343  mm3
                                    =  39 304  mm3
  
                                                                    [ V 4.1 ]
  
  4.2
  
       BO van prisma = 2(B + H)L + 2BH
  
       BO  =  2(B + H)L + 2BH
              =  2(40 + 24) X 65 + 2 X 40 X 24  cm2
  
              =  10 240  cm2
  
       Volume van prisma = LBH
  
           V = LBH
  
              =  65 X 40 X 24  cm3
              =  62 400  cm3
                                                                    [ V 4.2 ]
  
  4.3
  
       lengte van silinder = hoogte = h
  
       h  =  450  mm  =  450  ÷  10  cm
  
           =  45  cm
  
       BO van silinder = πd(h + ½d)
  
       BO  =  πd(h + ½d)
              =  π X 6 X ( 45 + ½ X 6 )  cm  2
              =  904,779  cm  2
  
       Volume van silinder = V
  
                    π X D2 X h
          V  =  ────────
                           4
  
                    π X 62 X 45
              =  ────────
                           4
  
               =  1 272,345 cm3
                                                                    [ V 4.3 ]
  
  
 OF 

  
       lengte van silinder = hoogte = h
  
       h  =  450  mm  =  450  ÷  10  cm
  
           =  45  cm
  
       Straal van sirkel  =  r  =  ½D  =  3  cm
  
       BO van silinder = 2πr(r + h)
  
       BO  =  2πr(r + h)
              =  2 X π X 3 X ( 3 + 45)  cm  2
              =  904,779  cm  2
  
       Volume van silinder = V  =  πr2h
  
           V  =  πr2h
               =  π X 32 X 45  cm3
               =  1 272,345  cm3
                                                                    [ V 4.3 ]
  

  4.1
  
       SA of cube = 6s2
  
                          = 342  mm2
  
                          =  1 156  mm2
  
       Volume of cube = s3
                                 =  343  mm3
                                 =  39 304  mm3
  
                                                                    [ Q 4.1 ]
  
  4.2
  
       BO van prisma = 2(B + H)L + 2BH
  
       BO  =  2(B + H)L + 2BH
              =  2(40 + 24) X 65 + 2 X 40 X 24  cm2
  
              =  10 240  cm2
  
       Volume van prisma = LBH
  
           V = LBH
  
              =  65 X 40 X 24  cm3
              =  62 400  cm3
                                                                    [ Q 4.2 ]
  
  4.3
  
       lengte van silinder = hoogte = h
  
       h  =  450  mm  =  450  ÷  10  cm
  
           =  45  cm
  
       BO van silinder = πd(h + ½d)
  
       BO  =  πd(h + ½d)
              =  π X 6 X ( 45 + ½ X 6 )  cm  2
              =  904,779  cm  2
  
       Volume van silinder = V
  
                    π X D2 X h
          V  =  ────────
                           4
  
                    π X 62 X 45
              =  ────────
                           4
  
               =  1 272,345 cm3
                                                                    [ Q 4.3 ]
  
  
 OF 

  
       lengte van silinder = hoogte = h
  
       h  =  450  mm  =  450  ÷  10  cm
  
           =  45  cm
  
       Straal van sirkel  =  r  =  ½D  =  3  cm
  
       BO van silinder = 2πr(r + h)
  
       BO  =  2πr(r + h)
              =  2 X π X 3 X ( 3 + 45)  cm  2
              =  904,779  cm  2
  
       Volume van silinder = V  =  πr2h
  
           V  =  πr2h
               =  π X 32 X 45  cm3
               =  1 272,345  cm3
                                                                    [ Q 4.3 ]
  
  
Antwoorde  / Answers  5.  
  

  5.1
  
       Oppervlakte van bedding = LB
  
                                                =  9 X 4  m2
  
                                                =  36  m2
  
                                                               [ V 5.1 ]
  
  
  5.2
  
       Oppervlakte van paadjie = oppervlakte van
  
           2 lang dele + opperv. van 2 korter dele
  
                      =  2 X (11 X 1) + 2 X (4 X 1)  m2
  
                      =  30  m2                      [ V 5.2 ]
  
 
 OF 

       Oppervlakte van paadjie = totale
  
           oppervlakte − oppervlakte van bedding
  
                                  =  LB − LB
  
                                  =  11 X 6  −  9 X 4  m2
  
                                  =  66 − 36  m2
  
                                  =  30  m2              [ V 5.2 ]
  
  
  5.3
  
       Aantal stene =  Oppperv. van paadjie X 50
  
                            =  30 X 50
  
                            =  1 500                      [ V 5.3 ]
  
  
  5.4
  
                                        Aantal stene
       Aantal pakke  =  ────────────
                                    Aantal stene/pak
  
                                    1 500
                              =  ─────     = 3
                                      500
  
       Koste  =  R 1 130 X 3
  
                  =  R 3 390                           [ V 5.4 ]
  

  5.1
  
       Area of flowerbed = LB
  
                                    =  9 X 4  m2
  
                                    =  36  m2
  
                                                                  [ Q 5.1 ]
  
  
  5.2
  
       Area of path = area of 2 long parts + area
  
                                                 of two shorter parts
  
                            =  2 X (11 X 1) + 2 X (4 X 1) m2
  
                            =  30  m2                      [ Q 5.2 ]
  
 
 OR 

       Area of path = area of total area − area of
  
                                                       flower bed
  
                            =  LB − LB
  
                            =  11 X 6  −  9 X 4  m2
  
                            =  66 − 36  m2
  
                            =  30  m2                        [ Q 5.2 ]
  
  
  5.3
  
       Number of bricks =  Area of path X 50
  
                                    =  30 X 50
  
                                    =  1 500                 [ Q 5.3 ]
  
  
  5.4
  
                                            Number of bricks
       Number of packs  =  ───────────────
                                            Number of bricks / pack
  
                                          1 500
                                    =  ─────     = 3
                                            500
  
       Cost  =  R 1 130 X 3
  
                =  R 3 390                                 [ Q 5.4 ]
  
  
Antwoorde  / Answers  6.  
  

  6.1
  
       Oppervlakte van paadjie = oppervlakte van
  
        2 lang dele + oppervlakte van 2 korter dele
  
                      =  2 X (19 X 2) + 2 X (5 X 2)  m2
  
                      =  96  m2                        [ V 6.1 ]
  
 
 OF 

       Oppervlakte van paadjie =
  
           totale oppervlakte − oppervlakte van bad
  
                                =  LB − LB
  
                                =  19 X 9  −  15 X 5  m2
  
                                =  171 − 75  m2
  
                                =  96  m2             [ V 6.1 ]
  
  6.2
  
       Totale volume = TV = Total volume
  
       V(A), V(B) en/and V(C) = volume van
  
       elke deel/ of each part
  
       TV = V(A) + V(B) + V(C)
  
            = LBH + ½(a+b)H + LBH
  
            = 5X2X5 + ½(2+1)X3X5 + 7 X 1 X 5
  
            = 107,5  m3
  
            = 107,5 X 1 000  ℓ
  
            = 107 500  ℓ                            [ V/Q 6.2 ]
  
  
  
  
  
  
  
  
  
 
  OF / OR 

       Volume van liggaam = oppervlakte van
  
                                             basis X hoogte
  
       Gebruik die sy-aansig, Fig. B, se oppervlakte
  
       as die basis en die breedte van die bad as die
  
       hoogte. Kom ons gebruik V = volume van bad,
  
       A = sy-aansig se oppervlakte en
  
       H = die breedte van bad
  
       Gebruik V = AH as ons "formule"
  
       Gebruik die formula A = ½ X (a + b) X h
  
       om die oppervlakte van 'n trapesium, met
  
       a en b die ewewydige sye van die trapesium
  
       en h die loodregte afstand tussen die
  
       die ewewydig sye, te bereken. Die diagramme
  
       hieronder toon verskillende maniere om die
  
       oppervlakte van die sy-aansig te bereken.
  
       Gebruik die metode wat jy die beste verstaan.
  
  
       A(A), A(B) A(C) stel die oppervlaktes van
  
       figure A, B en C voor.  /
  
       A(A), A(B) A(C) represent the areas of
  
       figures A, B and C.
  
  
       A(bad/pool) = A(A) + A(B) + A(C)
  
                           = LB + ½BH + LB
  
                           = 2 X 5 + ½ X 3 X 1 + 10 X 1
  
                           = 21,5 m2
  
       Volume = AH
                    = 21,5 X 5  m3
                    = 107,5  m3
  
                    = 107,5 X 1 000  ℓ
  
                    = 107 500  ℓ                        [ V/Q 6.2 ]
  
  
  
  
 
  OF / OR 

  
       A(bad/pool) = A(A) + A(B) + A(C)
  
                           = LB + ½(a + b) X H + LB
  
                           = 2X5 + ½X(2+1)X3 + 7X1
  
                           = 21,5 m2
  
       Volume = AH
                    = 21,5 X 5  m3
                    = 107,5  m3
  
                    = 107,5 X 1 000  ℓ
  
                    = 107 500  ℓ                      [ V/Q 6.2 ]
  
  
  
  
  
  
  
  
 
  OF / OR 

  
       A(bad/pool) = A(A) + A(B) + A(C)
  
                           = LB + LB + ½BH
  
                           = 15 X 1 + 5 X 1 + ½ X 3 X 1
  
                           = 21,5 m2
  
       Volume = AH
                    = 21,5 X 5  m3
                    = 107,5  m3
  
                    = 107,5 X 1 000  ℓ
  
                    = 107 500  ℓ                        [ V/Q 6.2 ]
  
  
  
  
  
  
  
  
  
 
  OF / OR 

  
       A(bad/pool) = A(A) − A(B)
  
                           = LB − ½(a+b)H
  
                           = 15 X 2 − ½ X (7 + 10) X 1
  
                           = 21,5 m2
  
       Volume = AH
                    = 21,5 X 5  m3
                    = 107,5  m3
  
                    = 107,5 X 1 000  ℓ
  
                    = 107 500  ℓ                       [ V/Q 6.2 ]
  
  
  
  
  
  
  
  
  6.3
  
       Volume om swembad te vul = TV − "oop spasie"
  
       Bereken die volume van die "oop spasie"
  
       deur die totale oppervlakte van die bad te
  
       vermenigvuldig met 100 mm
  
                                                      100
       V(oop spasie)  = 15 X 5 X ─────  m3
                                                     1000
  
                               = 7,5  m3
  
                               = 7,5 X 1 000   = 7 500  ℓ
  
       Volume water om bad te vul
  
                             =  asTV − V(oop spasie)
  
       V(water)  = 107,5 − 7,5  m3
  
                      = 100  m3
  
                      = 100 000  ℓ                           [ V 6.3 ]
  

  6.1
  
       Area of path = area of 2 long parts + area
  
                                          of two shorter parts
  
                            =  2 X (19 X 2) + 2 X (5 X 2)  m2
  
                            =  96  m2                         [ Q 6.1 ]
  
 
 OR 

       Area of path = area of total area − area of
  
                                                       flower bed
  
                            =  LB − LB
  
                           =  19 X 9  −  15 X 5  m2
  
                            =  171 − 75  m2
  
                            =  96  m2                       [ Q 6.1 ]
  
  6.2
  
  

  
  
  
 
  OR 

       Volume of a body = area of base X height
  
       Use the side view of the pool, Fig. B,
  
       as the base and the breadth of the
  
       pool as the height.
  
       Let us use V = volume of pool, A = area of
  
       side view and H = the breadth of the pool
  
       Let us use V = AH as our "formula"
  
       Let's Use the formula A = ½ X (a + b) X h
  
       to calculate the area of a trapezium with a
  
       and b the parallel sides of the trapezium
  
       and h the prependicular distance
  
       between the parallel sides.
  
       The diagrams below show different ways to
  
       calculate the area of the side view
  
       of the pool.
  
       Use the method that you understand best.
  
  
  

  
  
  
  
  
 
  OF / OR 

  

  
  
 
  OF / OR 

  

  
  
 
  OF / OR 

  

  
  
  6.3
  
       Volume to fill the pool = TV − "open space"
  
       Calculate the volume of the "unfilled space"
  
       by multiplying the total area of the pool
  
       by 100 mm
  
                                                        100
       V(open space)  = 15 X 5 X ─────  m3
                                                       1000
  
                               = 7,5  m3
  
                               = 7,5 X 1 000   = 7 500  ℓ
  
       Volume water to fill the pool
  
                                         =  TV − V(open space)
       V(water)  = 107,5 − 7,5  m3
  
                       = 100  m3
  
                       = 100 000  ℓ                         [ V 6.3 ]
  
  
Antwoorde  / Answers  7.  
  

  7.1.1    Diameter = 7 m            [ V 7.1 ]
  
  7.1.2    Breadth = 5 m              [ V 7.1 ]
  
  7.1.3    Length = 8 m                [ V 7.1 ]
  
  7.1.4    Width = 1 m                  [ V 7.1 ]
  
  
  
  
  7.2   
           s2  =  b2  +  h2      . . . Pythagoras 
  
                =  2,52  +  2,52
  
                =  12,5
  
            s  =  √12,5
  
                =  3,536
  
           Skuinssye = 3,536 m
  
           Hypotenuses = 3,536 m
  
                                                                [ V/Q 7.2 ]
  
  

  7.1.1    Diameter = 7 m            [ Q 7.1 ]
  
  7.1.2    Breadth = 5 m              [ Q 7.1 ]
  
  7.1.3    Length = 8 m                [ Q 7.1 ]
  
  7.1.4    Width = 1 m                  [ Q 7.1 ]
  
  
  
  

  
  
  7.3                πD2
           TA  =  ────  +  LB  +  ½BH
                        8
  
                       π X 72
           TA  =  ────  +  8 X 7  +  ½ X 7 X 3,5  m2
                          8
  
                 =  19,242  +  56  +  12,25  m2
  
                 =  87,492  m2                                                             [ V/Q 7.3 ]
  
  
  
  7.4    Oppervlakte van paadjie  =  Totale oppervlakte (TA)  −  oppervlakte van bedding
  
           Area of path  =  Total area (TA)  −  area of flowerbed
  
                                                                                                     πD2
           Oppervlakte van paadjie / Area of path   =  87,492  −  [ ────  +  LB  +  ½BH    ]  m2
                                                                                                      8
                                                                                                     π X 52
                                                                            =  87,492  −  [ ────  +  8 X 5  +  ½ X 5 X 2,5   ]  m2
                                                                                                      8
  
                                                                            =  87,492  −  [ 9,817  +  40  +  6,25 ]  m2
  
                                                                            =  87,492  −  56,067  m2
  
                                                                            =  31,425  m2                                                   [ V/Q 7.4 ]
  
  
Antwoorde  / Answers  8.  
  

  8.1                                             πD2
              oppervlakte / area  =  ────
                                                     4
  
                                                   π X (380 ÷ 10)2
                                             =  ───────────  cm2
                                                             4
  
                                             =  1 134  cm2                                                                                        [ V/Q 8.1 ]
  
  
  8.2    Oppervlakte van vierkant / Area of square  =  sy2  =  side2  =  s2
  
                                                                                     s2  =  1 134
  
                                                                                      s  =  √1 134
  
                                                                                          =  33,675  cm                                           [ V/Q 8.2 ]
  
  
  8.3    Volume  =  oppervlakte X hoogte / area X height    =  AH
                                                                                                                                                                                                                                                               V  =  1 134 X (60 ÷ 10)  cm3
                                                                                           =  6 804  cm3                                          [ V/Q 8.3 ]
  
  8.4    Verhouding / Proportion  :  /  sement / cement : sand  =  1 : 3
  
           Totale volume / Total volume  =  50  X  6 804  =  340 200  cm3
  
  Totale volume   Volume sement   Volume sand
  Total volume   Volume cement   Volume sand
        4 (1+3)             1             3
            1             1/4             3/4
     340 200   1/4 X 340 200   3/4 X 340 200
     340 200         85 050         255 150
  
  
            Volume sement / cement  =  85 050  cm3 6    en / and  volume sand  =  255 150  cm3        [ V/Q 8.4 ]
  
  
  OF / OR  

  
            Volume sement / cement  =  1/4 X totale volume / total volume
                                                     =  1/4 X 340 200  cm3
                                                     =  85 050  cm3
  
            Volume sand  =  3/4 X totale volume / total volume
                                   =  3/4 X 340 200  cm3
                                   =  255 150  cm3                                                                                              [ V/Q 8.4 ]
  
  
Antwoorde  / Answers  9.  
  

  9.1    Volume  =  oppervlakte van paadie / area of path  X  dikte / thickness
  
                                         150
                         =  80 X ─────   m3    =  12  m3                                                                                [ V/Q 8.1 ]
                                       1 000
  
  9.2    Totaal / Total   Sement / Cement   :   Sand   :  klip / stone
  
                      7                       1                       2               4
  
                                               1                       2               4
                      1                     ───                 ───         ───
                                               7                       7               7
  
                                        1                       2                      4
                      12           ─── X 12         ─── X 12        ─── X 12
                                        7                       7                      7
  
                      12                 1,714                3,429          6,857
  
                      Volume (sement / cement)  =  1,714 m3;            Volume (sand)  =  3,429 m3;  en / and
  
                      Volume (klip / stone)  =  6,857 m3;                                                                                  [ V/Q 9. ]
  

                                 25
  9.3    25 dm3  =  ─────    =   0,025 m3 
                              1 000
  
                                                                                                         1,714
           Aantal sakkies sement / Number of pockets of cement  =  ─────    =   68,56  =  69              [ V/Q 9. ]                                                                                                          0,025
  
  
 OF / OR 
 
           0,025 m3 is die volume van 1 sakkie sement    |    0,025 m3 is the volume of 1 pocket of cement
  
                                                       1                                                                                 1
            1 m3 is die volume van ───── sakkies sement    |     1 m3 is the volume of ───── pockets of cement
                                                   0,025                                                                           0,025
  
                                          d.i  40  sakkies                                                              i.e.  40  pockets
  
              1,714 m3 is die volume van 40 X 1,714  sakkies  |   12 m3 is the volume of 1,714 X 40  pockets
  
                                                          = 68,56 = 69  sakkies  |                                    = 68,56 = 69 pockets
  
            Aantal sakkies sement  =  69  |    Number of pockets of cement  =  69                                   [ V/Q 9. ]
  

  9.4    Koste van sement  /  Cost of cement  =  R 69 X 98,00 =  R 6 762,00 
  
           Koste van sand  /  Cost of sand  =  R 3,429 X 450,00 =  R 1 543,05 
  
           Koste van klip  /  Cost of stone  =  R 6,857 X 645,00 =  R 4 422,77 
  
  
           Totale koste van materiaal  /  Total cost of material  =  R 12 727,82                                         [ V/Q 9. ]