Saturday X-tra
X-Sheet: 14
Revision of Grade 12 Space and Shape Part 3
Surface Area and Problem Solving with Space and Shape
Key Concepts
In this session we will focus on summarising what you need to know about:
- Surface Area
- Problem solving with Area, Perimeter, Volume and Surface Area
Terminology & definitions
- Area = the amount of space that the surface of a place or shape covers. Area
is expressed in square units, such as square metres.
- Surface Area – is the total area of all the faces of a 3D shape.
- Perimeter = the total length of the sides of a shape.
- Circumference = the distance around the edge of a circle
- Capacity – is the volume of a 3D container or shape
- Volume – is a measure of the amount that can be held or contained by
something (container).
- Capacity or volume is measured in units CUBED, e.g. cm3 or m3.
The capacity of fluids can be measured in millilitres, litres or kilolitres.
Please refer back to Space and Shape Part 1 and 2 for more details on Area,
Perimeter and Volume.
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Concept:
Surface Area
X-ample 1
An open top fish tank has the following dimensions:
Length = 2,5m
Breadth = 2m
Height = 1,5m
Determine the total surface area (in m2) of the glass used.
Where Surface Area of a rectangular prism = (l x b) + 2(l x h) + 2(b x h)
X-ample 2
Find the surface area of this cylinder if the
dimensions are:
Diameter = 10cm
Height = 20cm
Where Surface area of cylinder = 2πrh + 2πr2, using π = 3,14
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Concept: Problem solving with Area, Volume and Surface Area
X-ample 3
The athletics track shown below consists of two straight sides and two semicircles on
either end of the track.
The following formulae may be of use in order to answer the questions below:
Area of circle = π r2
Circumference of circle = 2 π r
Area of rectangle = l × b
r = radius l = length b = breadth, use π = 3,14
1. Determine the area of the field inside the track.
2. Determine the cost of covering 12 500 m2 of the track with grass, if grass
costs R83 per 2 m2.
3. The right bend marked A to B to C of the track is to be fenced with wire,
with supporting poles 2.85 m apart. Calculate how many poles will be
needed.
X-ample 4
A company manufactures electrical geysers out of steel in the following two
shapes:
Geyser 1:
radius = 0,4 metres, height = 1,2 metres
Geyser 2:
length = 80 centimetres
breadth = 80 centimetres
height = 120 centimetres
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1. Calculate the volume of Geyser 1 in m3.
Volume of cylinder = π x (radius)2 x height, using π = 3,14
2. The volume of Geyser 2 is 768 000 cm3.
If 1 000 cm3 = 1 litre, convert the volume of Geyser 2 to litres.
3. If 1 000 cm3 = 0, 22 gallon, how many gallons can Geyser 2 hold?
4. To prevent loss of heat, geysers are covered with an insulation material
pasted on all the outside surfaces. How many square metres of
insulation material will be needed to cover Geyser 1?
Surface area of cylinder = 2πrh + 2πr2, using π = 3,14
5. A 1 litre tin of glue used to paste the insulation material can cover a
surface area of 1,25 m2. Calculate the surface area that a 5 litre tin of
glue can cover.
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X-ample 5:
Thandi washes her dishes by hand three times daily in two identical cylindrical
basins. She uses one basin for washing the dishes and the other for rinsing the
dishes. Each basin has a radius of 30 cm and a depth of 40 cm, as shown in the
diagram below.
Thandi is considering buying a dishwasher that she will use to wash the dishes daily.
1. Calculate the volume of one cylindrical basin in cm3.
Volume of cylinder = π x (radius)2 x height, using π = 3,14
2. Thandi fills each basin to half its capacity whenever she washes or rinses the
dishes. Calculate how much water (in litres) she will use daily to wash and
rinse the dishes by hand. (1 000 cm3 = 1 litre)
3. A manufacturer of a dishwasher claims that their dishwasher uses nine
times less water in comparison to washing the same number of dishes by
hand.
a. How much water would this dishwasher use to wash Thandi's dishes
daily?
b. Is the claim of the manufacturer realistic? Justify your answer by giving a
reason(s).
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X-ample 6:
The diagram below shows the basic outlay of a rectangular garden. There is a
triangular sand pit in the top right corner and a circular pond in the middle of the
garden. The diagram is not drawn to scale.
20 m
12m
sandpit
pond
34,5m
The owner of the garden wants to erect a low wire fence around the pond as shown
below. He needs to calculate the length of fencing that he must buy. The following
needs to be taken into account:
•
•
•
The area of the entire garden is 1256 m2.
The pond takes up 25% of the garden.
The distance between the fence and the pond must be uniformly 20 cm, as
shown below.
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The following formulae may be of use in order to answer the questions below:
Circumference of a circle = 2×π ×r
Area of a circle = π × r2
Let π = 3,14
1.
Calculate the length of the radius (in meters) of the pond. Round your
answer to 1 decimal place.
2. The owner has 60 m of fencing. Is this enough fencing to fence the pond?
Motivate your answer with the use of calculations.
The sand pit (45 cm deep) is to be filled with sand. Sand can be bought as follows:
10 m3 (cubic meter) costs R125
5 m3 costs R 85
2 m3 costs R55
1 m3 costs R30
3. Determine, using the volume formula given below, whether an amount of R200
would be sufficient to cover the cost for the sand required to completely fill the
sandpit.
Volume of a triangular Prism = ½ × base × perpendicular height × depth
Base
depth
Perpendicular
Height
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X-ample 7:
In this question the following formulae can be used:
Area of rectangle = l × b
Total surface area of rectangular prism = 2 × [ l × b + l × h + h × b]
Volume of rectangular prism = l × b × h
Where l = length b = breadth h = height
The 'Choc Special' chocolate slabs are in the shape of a rectangular prism. The
dimensions of the chocolate slabs are shown in the diagram below.
15 cm
0,75cm
10cm
These chocolate slabs can be packed in Box A or Box B. The same number of slabs
can fit in each box.
24cm
15cm
20cm
15cm
10cm
1. Calculate the number of slabs that would fit in each box.
2. Determine the height of Box B.
3. Determine the total surface area of Box B.
4. If it costs R25 to make 4 slabs of chocolate, how much would it cost to fill box
A with slabs of chocolate?
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X-ercise :
Surface Area
Calculate the total outer surface area of a box with dimensions:
Length = 50 cm, breadth = 20 cm and height = 10 cm.
Use the formula: Total outer surface area = 2 (L x B + L x H + B x H)
where L = length, B = breadth and H = height
Problem Solving with Area, Volume and Surface Area
1. A bag of ready-mix cement makes 0,024 m3 of concrete.
The cost per bag of ready-mix cement is R150,00.
Volume = Length × Breadth × Height
A rectangular concrete pathway, 12 metres long, 0,6 metres wide and 0,1 metres
deep needs to be laid, as shown in the diagram below.
a.
b.
c.
d.
Calculate the volume of cement (in m3) required for the pathway.
How many bags of ready-mix cement will be required for the pathway?
Calculate the total cost of the ready mix cement required for the pathway.
A 5% discount is given for cash on delivery. A local supplier delivers the bags
of ready mix cement. Using your answer obtained in Question c, calculate the
discount amount that will be received.
2. Handyman Manufacturers were requested by a customer to make a table with
the top in the shape of a trapezium, as shown below:
The lengths of the parallel sides of the top of the table are (170 + A) cm and 130
cm. The perpendicular distance between the parallel sides of the top of the table
is 60 cm. The slant edges of the top of the table are each 72,1 cm long.
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170 cm
A
a. Calculate the length of A using Pythagoras. Round to the nearest whole
number.
b. Calculate the length of the longest parallel side.
c. Calculate the area of the top surface of the table.
Use the formula:
d. A decorative edging is to be glued around the edge of the table. Calculate the
length of the decorative edging needed in metres.
X-ercise Answers.
Surface Area
Total outer surface area
= 2 (L x B + L x H + B x H)
= 2(50x20 + 50x10 + 20x10) = 3400cm2
Problem Solving with Area, Volume and Surface Area
1. a. Volume = Length × Breadth × Height
= 12 x 0,6 x 0,1 = 0,72 m3
b. No Bags = 0,72 ÷ 0,024 = 30 bags
c. Cost = 30 x R150 = R4500
d. Discount = 5% of R4500 = R225
2. a. A2 = 72,12- 602 = 1598,41
A = 39,98… rounded to nearest whole number = 40cm
b. Length of the longest parallel side is then 170 + 40 = 210cm
c. Area of trapezium = ½ (60) x (210 + 130) = 10 200cm2
d. Perimeter = 210 + 72,1 + 130 + 72,1 = 484,2 cm = 4,842m
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