WISKUNDE
GRAAD 11
NOG OEFENINGE
  
Omtrek, oppervlakte en volume
  
MATHEMATICS
GRADE 11
MORE EXERCISES
  
Perimeter, area and volume
  
  
Vraag  / Question  1  
  
  
      Bereken die oppervlakte van die volgende 2D
      2D figure:
  
      1.1  'n reghoek met lengte = 45 mm en
               breedte = 73 mm.                     [ A 1.1 ]
  
      1.2  'n reghoekige driehoek met die
               reghoeksye 56 cm en 21 cm.    [ A 1.2 ]
  
  
      1.3  'n driehoek met basis 45 m en
               hoogte 22 m.                            [ A 1.3 ]
  
      1.4  'n vierkant met sylengte 72 cm
                                                                [ A 1.4 ]
  
      1.5  'n gelykbenige driehoek met
             basis 3,8 m en hoogte 1,8 m.    [ A 1.5 ]
  
      1.6  'n gelyksydige driehoek met
             sy 56 m en hoogte 46,4 m.         [ A 1.6 ]
  
      1.7  'n sirkel met radius 14,7 cm.          [ A 1.7 ]
  
      1.8  'n sirkel met middellyn 3,14 m.      [ A 1.8 ]
  
      1.9  'n reghoekige driehoek met een
             reghoeksy = 1 cm en die
             skuinssy = 2,6 cm.
                                                                  [ A 1.9 ]
  
      Calculate the area of the following 2D figures :
  
  
      1.1  a rectangle with length = 45 mm and
             width = 73 mm.                          [ A 1.1 ]
  
      1.2  a right-angled triangle with the
             right-angled sides equal to 56 cm
             and 21 cm.                                [ A 1.2 ]
  
      1.3  a triangle with base 45 m and
             height 22 m.                                 [ A 1.3 ]
  
      1.4  a square with length 18 mm
                                                                  [ A 1.4 ]
  
      1.5  an isosceles triangle with base
             3,8 m and height 1,8 m.              [ A 1.5 ]
  
      1.6  an equlateral triangle with side
             56 m and height 46,4 m.               [ A 1.6 ]
  
      1.7  a circle with radius 14,7 cm.          [ A 1.7 ]
  
      1.8  a circle with diameter 3,14 m.       [ A 1.8 ]
  
      1.9  a right-angled triangle with one
             right-angked side equal to 1 cm and
             the hypotenuse equal to 2,6 cm.
                                                                  [ A 1.9 ]
  
  
Vraag  / Question  2  
  
  
  
        'n Muur is 3,25 m hoog en 6,60 m lank
  
         en is 2 stene breed. 'n Steen is 30 cm lank
  
         en 50 mm hoog. Bereken die aantal stene
  
         in die muur.
  
         Jy kan die volgende formule gebruik :
  
                                     oppervlakte van muur
         Aantal stene  =  ───────────────
                                     oppervlakte van steen
  
                                                                     [ A 2. ]
  
        A wall is 3,25 m high and 6,60 m long
  
         and is 2 bricks wide. A brick is 30 cm long
  
         and 50 mm high. Calculate the number of
  
         bricks in the wall.
  
         You may use the following formula :
  
                                            area of wall
         Number of bricks  =  ─────────
                                            area of brick
  
                                                                         [ A 2. ]
  
Vraag  / Question  3  
  
  

      Gebruik die volgende formules in die vraag :
  
      Reghoek :
  
          Opperlakte = lengte X breedte = L x B
  
          Omtrek = lengte + breedte + lengte +
  
                                                       breedte
  
          Omtrek = 2(L + B)
  
      Vierkant :
  
          Opperlakte = sy X sy = s x s = s2
  
          Omtrek = sy + sy + sy + sy
  
          Omtrek = 4 X s = 4s
  
      Driehoek :
  
          Opperlakte = ½ X basis X hoogte
  
          Opperlakte = ½ X B X H
  
          Omtrek = sy1 + sy2 + sy3
  
      Sirkel :
  
          Deursnit = Middellyn = D
  
          Straal = Radius = r
  
          Deursnit = 2 x straal; D = 2r
  
          Oppervlakte = π X straal2  =  πr2
  
                                     π X middellyn2
          Oppervlakte =  ───────────
                                               4
  
                                     π X D2
          Oppervlakte =  ──────
                                        4
  
          Omtrek = π X middellyn  =  πD
  
          Omtrek = π X middellyn  =  π X 2 X straal
  
          Omtrek = 2 X π X r    =  2πr
  
  
  
  
  

      Use the following formulae in this question :
  
      Rectangle :
  
          Area = length X breadth = L x B
  
          Perimeter = length + breadth + length +
  
                                                            breadth
  
          Perimeter = 2(L + B)
  
      Square :
  
          Area = side X side = s x s = s2
  
          Perimeter = side + side + side + side
  
          Perimeter = 4 X s = 4s
  
      Triangle :
  
          Area = ½ X base X height
  
          Area = ½ X B X H
  
          Perimeter = side1 + side2 + side3
  
      Circle :
  
          Diameter = D
  
          Radius = r
  
          Diameter = 2 x radius; D = 2r
  
          Area = π X radius2  =  πr2
  
                          π X diameter2
          Area =  ───────────
                                 4
  
                         π X D2
          Area =  ──────
                             4
  
          Circumference = π X diameter  =  πD
  
          Circumference = π X diameter
  
                                   =  π X 2 X radius
  
          Circumference = 2 X π X r    =  2πr
  
  
  
      Bereken die
  
      3.1  sy en omtrek van 'n vierkant met
             oppervlakte van 552,25 m2
                                                                [ A 3.1 ]
  
      3.2  lengte en omtrek van 'n reghoek
             met breedte 3,45 m en 'n oppervlakte
             van 20,769 m2                           [ A 3.2 ]
  
      3.3  hoogte van 'n driehoek met
             basis 28,4 cm en 'n oppervlakte
             van 249,92 cm2                    [ A 3.3 ]
  
      3.4  basis van 'n driehoek met
             hoogte 1,26 m en 'n oppervlakte
             van 0,4095 m2                         [ A 3.4 ]
  
      3.5  deursnee en omtrek van 'n sirkel
             met oppervlakte 124,690 cm2
                                                             [ A 3.5 ]
  
      3.6  straal en omtrek van 'n sirkel met
             oppervlakte 3848,45 m2         [ A 3.6 ]
  
      Calculate the
  
      3.1  length of a side and the perimeter of
             a square having an area of 552,25 m2
                                                                    [ A 3.1 ]
  
      3.2  length and perimeter of a rectangle
             with breadth 3,45 m and an area
             of 20,769 m2                                  [ A 3.2 ]
  
      3.3  height of a triangle with base 28,4 cm
             and an area of 249,92 cm2
                                                                   [ A 3.3 ]
  
      3.4  base of a triangle with height
             1,26 m and an area of 0,4095 m2
                                                                  [ A 3.4 ]
  
      3.5  diameter and circumference of a circle
             with an area of 124,690 cm2       [ A 3.5 ]
  
  
      3.6  radius and circumference of a circle
             with an area of 3848,45 m2        [ A 3.6 ]
  
  
Vraag  / Question  4  
  

      Gebruik die volgende formules in die vraag :
  
      Lengte = L. breedte = B, hoogte = H
  
      Sy van kubus = s. Deursnit van silinder = d
  
  
      Straal = radius van silinder = r,
  
  
      Hoogte van silinder = h
  
  
      Reghoekige prisma :
  
          Buite-Opperlakte = BO
  
                    BO = 2(B + H)XL + 2XBXH
  
                    Volume = V = L X B X H = LBH
  
  
      Kubus :
  
          Buite-Opperlakte = BO
  
                    BO = 6 X s X s = 6s2
  
                    Volume = V = s X s X s = s3
  
  
      Silinder :
  
          Buite-Opperlakte = BO
  
                    BO = 2πrh + 2πr2
  
                    BO = 2πr(h + r)
  
                    Volume = V = πr2h
  
  
           BO  =  πdh + ½πd2
  
                  =  πd(h + ½d)
  
                         π d2 h
              V  =  ──────
                             4
  
  
  
  

      Use the following formulae in this question :
  
      Length = L. breadth = B, height = H
  
      Sideof cube = s. Diameter of cylinder = d
  
  
      Radius of cylinder = r,
  
  
      Height of cylinder = h
  
  
      Rectangular prism :
  
          Surface-Area = SA
  
                     SA = 2(B + H)XL + 2XBXH
  
                     Volume = V = L X B X H = LBH
  
  
      Cube :
  
          Surface-Area = SA
  
                    SA = 6 X s X s = 6s2
  
                    Volume = V = s X s X s = s3
  
  
      Cylinder :
  
          Surface-Area = SA
  
                    BO = 2πrh + 2πr2
  
                    BO = 2πr(h + r)
  
                    Volume = V = πr2h
  
  
           BO  =  πdh + ½πd2
  
                  =  πd(h + ½d)
  
                         π d2 h
              V  =  ──────
                             4
  
  
  
  
      Bereken die buite-oppervlakte en volume :
      van die volgende 3D-liggame:
  
      4.1  'n kubus met sy 34 mm.             [ A 4.1 ]
  
      4.2  'n regkoekige prisma met lengte 65 cm,
             breedte 40 cm en hoogte 24 cm.
                                                                [ A 4.2 ]
  
      4.3  'n silinder met lengte 450 mm en
             deursnit 6 cm.                            [ A 4.3 ]
  
      Calculate the surface area and the volume
      of the following 3D-bodies :
  
      4.1  a cube with side length 34 mm.    [ A 4.1 ]
  
      4.2  a rectangular prism with length 65 cm,
             breadth 40 cm and height 24 cm.
                                                                   [ A 4.2 ]
  
      4.3  a cylinder with length 450 mm adn
             diameter 6 cm.                              [ A 4.3 ]
  
  
Vraag  / Question  5  
  
  


      Die diagram hierbo stel 'n reghoekige
      blombedding omring met 'n 1 m breë
      voetpaadjie.
      Die lengte van die bedding is 9 m en sy
      breedte is 4 m. Bereken die
  
      5.1  oppervlakte van die blombedding.
                                                                  [ A 5.1 ]
  
      5.2  oppervlakte van die paadjie.        [ A 5.2 ]
  
      5.3  aantal stene benodig om die
             paadjie te plavei as 50 stene
             1 m2 bedek.                                [ A 5.3 ]
  
      5.4  koste om die paadjie te plavei as
             500 stene R1 130 kos.               [ A 5.4 ]
  

      The diagram above represents a rectangular
      flowerbed surrounded by a 1 m wide path.
  
      The length of the flowerbed is 9 m and its
      breadth is 4 m. Calculate the
  
      5.1  area of the flowerbed.                  [ A 5.1 ]
  
  
      5.2  area of the footpath.                     [ A 5.2 ]
  
      5.3  number of bricks needed to pave
             the path if 50 bricks cover 1 m2.
                                                                 [ A 5.3 ]
  
      5.4  cost to pave the path if the
             cost of 500 bricks is R1 130.       [ A 5.4 ]
  
  
Vraag  / Question  6  
  
  


      Diagram A hierbo stel 'n swembad
      omring deur 'n 2 m breë voetpaadjie voor.
      Die lengte van die swembad is 15 m en sy
      breedte is 5 m. Fig. B stel die sy-aansig van
      die bad voor, d.w.s. die diepte van die bad.
      Aan die vlakkant is die bad 1 m diep en
      aan die ander kant 2 m.Die vlakgedeelte is
      is 7 m lank en die diep gedeelte 5 m lank.
      Bereken die
  
      6.1  oppervlakte van die paadjie om
             die swembad.                              [ A 6.1 ]
  
      6.2  totale volume, in m3 en in ℓ,
             van die bad.                                [ A 6.2 ]
  
      6.3  volume water, in m3 en in ℓ, om die
             bad tot 100 mm van bo te vul.
                                                                [ A 6.3 ]
  

      Diagram A above represents a swimmingpool
      surrounded by a 2 m broad concrete path.
      The length of the pool is 15 m and its
      breadth is 5 m. Fig. B represents the side view
      ot the pool, i.e. the depth of the pool.
      The shallow end is 1 m deep and the deeper
      end is 2 m deep. The shallow part is 7 m
      long and the deep end 5 m.
      Calculate the
  
      6.1  area of the surrounding footpath.
                                                                  [ A 6.1 ]
  
      6.2  total volume, in m3 en in ℓ,
             of the pool.                                   [ A 6.2 ]
  
      6.3  volume water, in m3 en in ℓ, to fill
             the pool to 100 mm from the top.
                                                                 [ A 6.3 ]
  
  
Vraag  / Question  7  
  
  


      Die diagram hierbo stel 'n bedding
      omring deur 'n voetpaadjie voor.
      Die bedding bestaan uit 'n halfsirkel,
      'n reghoek en 'n gelykbenige driehoek.
  
      7.1  Verwys na die diagram om die
             volgende vrae te beantwoord :
      7.1.1  Hoe lank is die middellyn / deursnit
                van die groot halfsirkel?
                                                         [ A 7.1.1 ]
  
      7.1.2  Wat is die breedte van die
                driehoek in die bedding?
                                                         [ A 7.1.2 ]
  
      7.1.3  Wat is die lengte van die
                reghoek in die bedding?
                                                             [ A 7.1.3 ]
  
      7.1.4  Hoe breed is die voetpaadjie?
                                                             [ A 7.1.4 ]
  
      7.2  Gebruik Pythagoras se stelling en toon
             aan dat die skuinssye van die driehoek
             in die bedding gelyk is aan 3,536 m.
                                                             [ A 7.2 ]
  
      7.3  Bereken die totale oppervlakte van
             die bedding.                            [ A 7.3 ]
  
      7.4  Bereken die oppervlakte van
             die paadjie in m2.                    [ A 7.4 ]
  

      The diagram above represents a flower
      bed surrounded by a footpath.
      The bed consists of a semi-circle,
      a rectangle and an isosceles triangle.
  
      7.1  Refer to the diagram to answer the
             following questions :
      7.1.1  What is the length of the diameter
                of the larger semi-circle?
                                                         [ A 7.1.1 ]
  
      7.1.2  What is the breadth of the
                triangle in the bed?
                                                         [ A 7.1.2 ]
  
      7.1.3  What is the length of the
                rectangle in the bed?
                                                         [ A 7.1.3 ]
  
      7.1.4  How wide is the footpath?
                                                         [ A 7.1.4 ]
  
      7.2  Use Pythagoras's theorem and show
             that the hypotenuse of teh triangle in
             the bed is equal to 3,536 m.
                                                             [ A 7.2 ]
  
      7.3  Calculate the total area of
             the bed.                            [ A 7.3 ]
  
      7.4  Calculate the area of the
             path in m2.                    [ A 7.4 ]
  
  
Vraag  / Question  8  
  


      Thabo giet sement stene wat vir plaveiwerk
      gebruik word. Hy giet sirkelvormige,
      vierkantige en heksagonale stene.Al die stene
      het dieselfde oppervlakte en is 60 mm dik.
      Hy gebruik 'n sement-sand mengsel wat in
      die verhouding sement:sand = 1:3 vermeng
      word. 50 stene van dieselfde vorm word
      gelyktydig gegiet.
  
      8.1  Die sirlelvormige steen het 'n middellyn
              van 380 mm. Bereken die oppervlakte
              van die steen in cm2. Jy kan die formule :
                                                         πD2
              oppervlakte van sirkel  =  ────    gebruik
                                                           4
                                                                      [ A 8.1 ]
      8.2  Bereken die sylengte van die
              vierkantige steen.                            [ A 8.2 ]
      8.3  Bereken die volume van 'n
              heksagonale steen.                         [ A 8.3 ]
      8.4  Bereken die volume van die sement
             en van die sand wat benodig word om
             50 stene te giet.                               [ A 8.4 ]

      Thabo manufactures cement bricks used for
      paving. The forms are circular, square
      and hexagonal.All the bricks have the same
      area and they are 60 mm thick.He uses a
      cement-sand mixture, which is mixed in
      the proportion sement : sand = 1 : 3.
      50 bricks of a shape are made simultaneously.
  
  
      8.1  The circular brick has a diameter
              of 380 mm. Calculate the area of the
              brick in cm2. You may use the formula :
                                          πD2
              area of circle  =  ────
                                            4                  [ A 8.1 ]
  
      8.2  Calculate the side length of the
              square brick.                              [ A 8.2 ]
      8.3  Calculate the volume of a
              hexagonal brick.                        [ A 8.3 ]
      8.4  Calculate the volume of cement and
              of sand needed to make 50 bricks.
                                                                [ A 8.4 ]
  
  
Vraag  / Question  9  
  


      'n Man wil 'n beton paadjie om sy huis gooi.
      Die oppervlakte van die paadjie is 80 m2.
      Die paadjie moet 150 mm dik wees.
  
      9.1  Bereken die volume van die beton.
                                                                      [ A 9.1 ]
  
      9.2  Die beton bestaan uit 'n mengsel van
             sement, sand en klip, vermeng in die
             verhouding sement : sand : klip = 1 : 2 : 4.
             Bereken die volume van elke bestanddeel.
                                                                      [ A 9.2 ]
  
      9.3  Die volume van een sakkie sement is 25 dm3.
             Bereken die aantal sakkies sement benodig.
                                                                      [ A 9.3 ]
  
  
      9.4  Bereken die koste van elke komponent
             van die beton en dan ook die totale koste.
             Die prys van sement is R98,00 per sakkie.
             Sand kos R450 per m3 en klip kos
             R645 per m3.
                                                                    [ A 9.4 ]

      A man wants to cast a concrete path around
      his house. The area of the path is 80 m2
      and it has to be 150 mm thick.
  
      9.1  Calculate the vo;ume of the concrete.
                                                                [ A 9.1 ]
  
      9.2  The concrete exists of a mixture of
             cement, sand and stone, mixed in the
             proportion cement : sand : stone = 1 : 2 : 4.
             Calculate the volume of each component.
                                                                [ A 9.2 ]
  
      9.3  The volume of a pocket of cement is
             25 dm3. Calculate the number of
             pockets of cement necessary.
                                                                [ A 9.3 ]
  
      9.4  Calculate the cost of each component
             of the concrete as well as the total cost.
             Cement costs R98,00 per pocket.
             Sand costs R450 per m3 and stone
             costs R645 per m3.
                                                                [ A 9.4 ]
  
  
  
  
  
  
  
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