WISKUNDE
GRAAD 11
NOG OEFENINGE
Omtrek, oppervlakte en volume
GRAAD 11
NOG OEFENINGE
Omtrek, oppervlakte en volume
MATHEMATICS
GRADE 11
MORE EXERCISES
Perimeter, area and volume
GRADE 11
MORE EXERCISES
Perimeter, area and volume
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 ]
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 ]
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 ]
'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. ]
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 ]
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 ]
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 (BO) 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 ]
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 (SA) and
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 ]
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 ]
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 ]
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
swimming pool 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 ]
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 the 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 ]
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 = ────
4
gebruik. [ 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 ]
'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 volume of the concrete.
[ A 9.1 ]
9.2 The concrete consists 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 ]