WISKUNDE
GRAAD 12
NOG OEFENINGE
Omtrek, oppervlakte en volume : antwoorde
MATHEMATICS
GRADE 12
MORE EXERCISES
Perimeter, area and volume : answers
1.1 Oppervlakte = lengte X breedte
120
breedte = 120 mm = ──── cm
10
= 12 cm
Oppervlakte = lengte X breedte
= 20 X 12 cm
2
= 240 cm
2
[ V/Q 1.1 ]
1.2 Oppervlakte = ½ X basis X hoogte
= ½ X 45 X 31 cm
2
= 697,5 cm
2
[ V/Q 1.2 ]
1.3 Oppervlakte = ½ X basis X hoogte
= ½ X 30 X 5,5 m
2
= 82,5 m
2
[ V/Q 1.3 ]
1.4 Oppervlakte = lengte X hoogte
= 18 X 8 mm
2
= 144 mm
2
[ V/Q 1.4 ]
1.5 Oppervlakte = ½ X basis X hoogte
= ½ X 38 X 18 cm
2
= 342 cm
2
[ V/Q 1.5 ]
1.6 Oppervlakte = ½ X basis X hoogte
= ½ X 56 X 46,4 m
2
= 1 299,200 m
2
[ V/Q 1.6 ]
1.7 Oppervlakte = π X radius
2
= π X 23,4
2 cm
2
= 1 720,210 cm
2
[ V/Q 1.7 ]
1.8 radius = ½ middellyn
= ½ X 4,65 m = 2,325 m
Oppervlakte = π X radius
2
= π X 2,325
2 m
2
= 16,982 m
2
[ V/Q 1.8 ]
OF
π X middellyn
2
Oppervlakte = ──────────
4
π X 4,65
2
= ────────── m
2
4
= 16,982 m
2
[ V/Q 1.8 ]
1.1 Area = length X breadth
120
breadth = 120 mm = ──── cm
10
= 12 cm
Area = length X breadth
= 20 X 12 cm
2
= 240 cm
2
[ V/Q 1.1 ]
1.2 Area = ½ X base X height
= ½ X 45 X 31 cm
2
= 697,5 cm
2
[ V/Q 1.2 ]
1.3 Area = ½ X base X height
= ½ X 30 X 5,5 m
2
= 82,5 m
2
[ V/Q 1.3 ]
1.4 Area = length X height
= 18 X 8 mm
2
= 144 mm
2
[ V/Q 1.4 ]
1.5 Area = ½ X base X height
= ½ X 38 X 18 cm
2
= 342 cm
2
[ V/Q 1.5 ]
1.6 Area = ½ X base X height
= ½ X 56 X 46,4 m
2
= 1 299,200 m
2
[ V/Q 1.5 ]
1.7 Area = π X radius
2
= π X 23,4
2 cm
2
= 1 720,210 cm
2
[ V/Q 1.7 ]
1.8 radius = ½ diameter
= ½ X 4,65 m = 2,325 m
Area = π X radius
2
= π X 2,325
2 m
2
= 16,982 m
2
[ V/Q 1.8 ]
OR
π X diameter
2
Area = ──────────
4
π X 4,65
2
= ────────── m
2
4
= 16,982 m
2
[ V/Q 1.8 ]
Ons gebruik die volgende afkortings hier :
Lengte : L ; Breedte = B ; Wydte = W ;
hoogte : h; basis : b; sy : s
skuinssy : hs ; Buite-oppervlakte : BO;
Here we use the following abbreviations :
Length : L ; Breadth = B ; Width = W ;
height : h; base : b; side : s
hypotenuse : hs ; Surface area : SA;
2.1 Lengte / Length = L = 0,85 m = 0,85 X (10 X 10) cm
= 85 cm
185
Hoogte / Height = h = 185 mm = ─── cm = 18,5 cm
10
Volume = L X B X h
= 85 X 21 X 18,5 cm
3
= 33 022,5 cm
3
BO / SA = 2 X (L X B) + 2 X (L X h) + 2 X (B X h)
= 2 X (85 X 21) + 2 X (85 X 18,5) + 2 X (21 X 18,5) cm
2
= 3 570 + 3 145 + 777 cm
2
= 7 492 cm
2
OF / OR
BO / SA = (omtrek van boom / Perimeter of bottom) X h + 2 X (opperv. van kopstuk / area of end)
= 2(b + h) X L + 2 X (b X h)
= 2 X (21 + 18,5) X 85 + 2 X (21 X 18,5) cm
2
= 6 715 + 777 cm
2
= 7 492 cm
2
[ V/Q 2.1 ]
2.2
Volume = s X s X s = s
3
= 67
3 cm
3
= 300 703 cm
3
BO / SA = 6 X (oppervlakte van een sy / area of one side) = 6 X s
2
= 6 X 67
2
= 26 934 cm
2
[ V/Q 2.2 ]
2.3 Skuinssy / Hypotenuse = hs
: hs
2 = 15
2 + 18
2 = 549
hs = 23,431 cm
basis = reghoekige driehoek / base = right-angled triangle
Volume = oppervlakte van basis X loodregte hoogte / Volume = area of base X perpendicular height
= ( ½ X b X h ) X L
= ( ½ X 15 X 18 ) X 34 cm
3
= 4 590 cm
3
BO / SA = omtrek van driehoek X L + 2 X oppervlakte van driehoek /
= perimeter of triangle X L + 2 X area of triangle
BO / SA = (23,431 + 15 + 18) X 34 + 2 X ( ½ X 15 X 18) cm
2
= 1 918,654 + 270 cm
2
= 2 188,654 cm
2
[ V/Q 2.3 ]
Totale oppervlakte = Oppervlakte van twee driehoeke + oppervlakte van een reghoek
Total area = area of two triangles + area of one rectangle
Hoogte van die twee driehoeke = Totale lengte − lengte van reghoek. /
Height of the two triangles = Total length − length of rectangle.
Hoogte van een driehoek / Height of one triangle = ½ ( 6 − 3 ) = 1,5 m
Totale oppervlakte / Total area = 2 X ( ½ X b X h ) + L X b
= 2 X ( ½ X 3 X 1,5 ) + 3 X 3 m
2
= 13,5 m
2
[ V/Q 3.1 ]
Totale oppervlakte = Oppervlakte van twee halwe sirkels + oppervlakte van een reghoek
Total area = area of two semi-circles + area of one rectangle
Total area = area of two semi-circles + area of one rectangle
Totale opervlakte / Total area opp. van 1 sirkel / area of 1 circle + opp. van reghoek / area of rectangle
Deursnit van sirkel / Diameter of circle = D = 2 m
Radius van sirkel / Radius of circle = ½ X D = ½ X 2 = 1 m
Totale opervlakte / Total area
= π r
2 + LB
= π X 1
2 + 3 X 3 m
2
= 12,142 m
2
OF / OR
π X D
2
Totale opervlakte / Total area
= ────── + LB
4
π X 2
2
= ────── + 3 X 3 m
2
4
= 12,142 m
2
[ V/Q 3.2 ]
Totale omtrek Total perimeter
= Beweeg een maal op die buitelyne van die figuur om die figuur. /
Total perimeter
= Move once around the figure following the exterior lines.
Totale omtrek / Total perimeter
= Omtrek van twee halwe sirkels + twee 4 m sye van reghoek
Totale omtrek / Total perimeter
= Perimeter of two semi-circles + two 4 m sides of the rectangle.
Totale omtrek / Total perimeter
= ½ X π X D + 4 + ½ X π X D + 4
= ½ X π X 2 + 4 + ½ X π X 2 + 4 m
= 14,283 m
[ V/Q 3.2 ]
Totale oppervlakte = Oppervlakte van 'n driehoek + oppervlakte van 'n reghoek
Total area = area of a triangle + area of a rectangle
Totale oppervlakte / Total area = ½ X b X h + L X b
= ½ X 20 X 5,5 + 20 X 14 m
2
= 335 m
2
[ V/Q 3.3 ]
3.4 Totale oppervlakte = Oppervlakte van twee reghoekige driehoeke
Total area = area of two right-angled triangles
Ons moet die lengte van sy AC van ΔADC bereken. AC is ook die skuinssy van ΔABC.
We have to calculate the length of side AC of ΔADC. AC is also the hypotenuse of ΔABC.
In ΔABC
: AC
2 = AB
2 + BC
2 . . . Pythagoras
= 200
2 + 150
2
= 62 500
∴ AC = √62500 = 250
Totale oppervlakte / Total area = ½ X b X h + ½ X b X h
= ½ X 200 X 150 + ½ X 600 X 250 m
2
= 90 000 m
2
[ V/Q 3.4 ]
Omtrek / Perimeter = AB + BC + CD + DA
Ons moet CD se lengte bepaal. CD is die skuinssy van ΔACD /
We have to calculate the length of side CD of ΔADC. CD is also the hypotenuse of ΔACD.
In ΔACD
: CD
2 = AD
2 + BC
2 . . . Pythagoras
= 600
2 + 62 500
2
= 422 500
∴ CD = √422 500 = 650
Omtrek / Perimeter = AB + BC + CD + DA
= 200 + 150 + 650 + 600 m
= 1 600 m
[ V/Q 3.4 ]
Totale oppervlakte = Oppervlakte van 'n halfsirkel + oppervlakte van 'n reghoekige driehoek /
Total area = area of a semi-circle + area of a right-angled triangle.
Ons moet die lengte van sy AB van ΔABC bereken. AB is 'n reghoeksy van ΔABC.
We have to calculate the length of side AB of ΔABC. AB is a right-angled side of ΔABC.
In ΔABC
: BC
2 = AB
2 + AC
2 . . . Pythagoras
13
2 = AB
2 + 5
2
169 = AB
2 + 25
AB
2 = 169 − 25
= 144
∴ AC = √144 = 12
∴ D = 12 of / or r = 6
π X D
2
Totale opervlakte / Total area
= ────── + ½ X b X h
4
π X 12
2
= ½ X ────── + ½ X 5 X 12 m
2
4
= 86,549 m
2
[ V/Q 3.5 ]
OF / OR
Totale oppervlakte / Total area = ½ X π X r
2 + ½ X b X h
= ½ X π X 6
2 + ½ X 12 X 5 m
2
= 86,549 m
2
[ V/Q 3.5 ]
Omtrek / Perimeter = omtrek van halwe sirkel / perimeter of a semi-circle + BC + CA
Omtrek / Perimeter = ½ X π X D + BC + CA
= ½ X π X 12 + 13 + 5 m
= 3,850 m
[ V/Q 3.5 ]
4.1 Oppervlakte van grond / Area of soil = Oppervlakte van reghoek / Area of a rectangle = LB
Oppervlakte van grond / Area of soil = 15 X 20 m
2
= 300 m
2
[ V/Q 4.1 ]
4.2 Nee, die plastic is te naby aan die grond langs die kante en daardie oppervlakte kan nie bewerk word nie.
No, the plastic is too close to the ground along the sides and that area can not be used.
[ V/Q 4.2 ]
4.3 Volume lug / Volume of air = Volume van 'n halwe silinder / Volume of a half cylinder
π D
2
= ½ ──── X L
4
π 15
2
= ½ ──── X 20 m
3
4
= 1 767,146 m
3
[ V/Q 4.3 ]
5 000
5.1 Volume van tenk / Volume of tank = ───── | 1000 liter = 1 m
3
1 000
= 5 m
3
[ V/Q 5.1 ]
5.2 Reghoekige prisma / Rectangular prism
:
Volume = LBh en / and V = 5 m
3
L = 2 X B | lengte is tweemaal breedte / length is double its breadth
∴ 2B X B X 2 = 5
5
B
2 = ───
4
B = 1,118 en dus / and therefore L = 2,236
Dus afmetings is : / Dimensions are
: L = 2,236 m , B = 1,118 m en / and h = 2 m
[ V/Q 5.2 ]
Kubus / Cube
: Volume = s
3
∴ s
3 = 5
3
s = √ 5 = 1,710
Dus afmetings is : / Dimensions are
: sy / side = 1,170 m
[ V/Q 5.2 ]
π X D
2 X h
Silinder / Cylinder
: Volume = ───────
4
π X D
2 X h
─────── = Volume
4
π X D
2 X 2
─────── = 5
4
π X D
2 = 10
D
2 = 3,183098862
D = √3,183098862 = 1,784
Dus afmetings is : / Dimensions are
: D = 1,784 m en / and h = 2 m
[ V/Q 5.2 ]
1,65
Lengte van 'n derde van die rak / Length of one third of the shelf = L = ──── = 0,55 m = 55 cm
3
Rakwydte / Width of shelf = W = 25 cm ; Hoogte tussen rakke / Height between shelves =RSH = 23 cm
Pilchards
:
D = deursnit van blik / diameter of tin = 7,5 cm
H = hoogte van blik / height of tin = 110 mm = 11 cm
L 55
Aantal blikke in een ry / Number of tins in one row = ─── = ─── = 7,333
D 7,5
∴ Daar kan 7 blikke in 1 ry gepsk word. / 7 tins can be palced in one row.
W 25
Aantal rye op die rak / Number of rows on the shelf = ─── = ─── = 3,333
D 7,5
∴ Daar kan 3 rye op die rak gepak word. / 3 rows can be packed on the shelf.
RSH 23 cm
Aantal blikke op mekaar / Number of tins on top of each other = ─── = ──── = 2,091
H 11 cm
∴ Daar kan 2 blikke op mekaar gepak word. / 2 tins can be placed on top of one another.
Aantal blikke in een laag = 7 blikke in 'n ry X 3 rye = 21
Number of tins in one layer = 7 tine per row X 3 rows = 21
Total aantal blikke op die rak = 21 blikke per laag X 2 lae = 42
Total number of tins on the shelf = 21 tins per layer X 2 layers = 42
Tuna
:
d = deursnit van blik / diameter of tin = 2r = 2 X 43 mm = 86 mm = 8,6 cm
h = hoogte van blik / height of tin = 37 mm = 3,7 cm
L 55
Aantal blikke in een ry / Number of tins in one row = ─── = ─── = 6,395
D 8,6
∴ Daar kan 6 blikke in 1 ry gepsk word. / 6 tins can be palced in one row.
W 25
Aantal rye op die rak / Number of rows on the shelf = ─── = ─── = 2,907
D 8,6
∴ Daar kan 2 rye op die rak gepak word. / 2 rows can be packed on the shelf.
RSH 23 cm
Aantal blikke op mekaar / Number of tins on top of each other = ─── = ──── = 6,216
H 3,7 cm
∴ Daar kan 6 blikke op mekaar gepak word. / 6 tins can be placed on top of one another.
Aantal blikke in een laag = 6 blikke in 'n ry X 2 rye = 12
Number of tins in one layer = 6 tine per row X 2 rows = 12
Total aantal blikke op die rak = 12 blikke per laag X 6 lae = 72
Total number of tins on the shelf = 12 tins per layer X 6 layers = 72
Jack kan 42 blikkies Pilchards, 72 blikkies Tune in Water en 72 blikkies Tuna in Oil op die rak pak.
Jack can place 42 tins of Pilchards, 72 tins of Tuna in Water and 72 tins of Tuna in Oil on the shelf
[ V/Q 6. ]
Antwoorde / Answers 7.
Lengte van bakkie / Length of bakkie = L = 3 X 100 = 300 cm
Breedte van bakkie / Breadth of bakkie = B = 1,75 X 100 = 175 cm
Diepte/hoogte van bakkie / Deapth/Height of bakkie = H = 400 ÷ 10 = 40 cm
600
Lengte van sakkie / Length of pocket = l = ──── = 60 cm
10
320
Breedte van sakkie / Breadth of pocket = l = ──── = 32 cm
10
Diepte/hoogte van sakkie / Deapth/Height of pocket = h = 130 ÷ 10 = 13 cm
300
7.1
Aantal sakkies / Number of pockets = ──── = 5
[ V/Q 7. ]
60
175
7.2
Aantal sakkies / Number of pockets = ──── = 5,469 = 5
[ V/Q 7. ]
32
 23;
[ V/Q 7. ]
7.3
Aantal sakkies in een laag / Number of pockets in one layer = 5 X 5 = 25
[ V/Q 7. ]
175
7.4
Aantal sakkies agter mekaar / Number of pockets across bakkie = ──── = 2,917 = 2
60
300
Aantal sakkies oor lengte / Number of pockets lengthwise = ──── = 9,375 = 9
32
Aantal sakkies in een laag / Number of pockets in one layer = 2 X 9 = 18
[ V/Q 7. ]
40
7.5
Aantal sakkies op mekaar / Number of layers = ──── = 3,077 = 3
13
Aantal lae / Number of layers = 3
Maksimum aantal sakkies = aantal per laag X aantal lae = 25 X 3 = 75
Maximum number of pockets = number per layer X number of layers = 25 X 3 = 75
[ V/Q 7. ]
7.6
Massa van 75 sakkies / Mass of 75 pockets = 75 X 50 = 3 750 kg
Maksimum vrag / Maximum load = 2 000 kg = 40 sakkies / pockets
Nee, die vrag is te swaar. / No the load is too great. 56
[ V/Q 7. ]
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