APPLIED GEOMETRY
MATHEMATICAL APPLICATIONS ESSENTIALS STAGE 2
Area calculations for irregular shapes
Consider the irregular shape below:
d0
d1
d2
w
w
The irregularly shaped area is divided into an even number of strips of equal width w. Three measurements of length,
equally spaced apart, are taken do, d1, d2. This divides the shape into two strips of equal width.
We can achieve a better approximation by dividing the area into four strips:
d0
d1
d2
d3
The number
of strips must
be even.
d4
w
w
w
w
Simpson’s rule is used twice: once to approximate the area of first two strips and then to approximate the area in the next
two strips.
Simpson’s rule works best when the area is divided into an even number of strips.
Area of first two strips:
Area of next two strips:
A1+2 ≈ w (d0 + 4d1 + d2)
3
A3+4 ≈ w (d2 + 4d3 + d4)
3
Note that the two formulae in this example can be combined into a single formula:
A1+2+3+4 ≈ w (d0 + 4d1 + d2) + w (d2+ 4d3 + d4)
3
3
≈ w (d0 + 4d1 + d2 + d2 + 4d3 + d4)
3
≈ w (d0 + 4d1 + 2d2 + 4d3 + d4)
3
EXAMPLE 1
The diagram represents a mound in the back corner of the local football oval. Calculate the approximate cross-sectional
area of this mound.
3m
5m
4m
3m
4m
6m
4m
1.5 m
4m
Solution
A ≈ w (d0 + 4d1 + 2d2 + 4d3 + d4)
3
A ≈ 4 (3 + (4 × 5) + (2 × 3) + (4 × 6) + 1.5)
3
A ≈ 72.67 m2
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APPLIED GEOMETRY
MATHEMATICAL APPLICATIONS
7m
12.3 m
(1) What volume of water is in the dam before irrigating begins?
6.5 m
14.7 m
12 m
The potato grower uses a large dam to irrigate his paddock. It fills with
water during the day and is then pumped to the irrigation system at night.
Before irrigating begins, the dam has an average depth of 3.85 metres.
When irrigating finishes, the average depth is 1.25 metres.
12 m
EXAMPLE 2
15 m
12 m
14.6 m
13.5 m
12 m
13.7 m
(2) How much water (in cubic metres) is pumped during the night?
(3) Calculate the rate of flow of the water (in kL per minute) if it takes 7 hours to pump the volume of water calculated in
part (2).
(4) State two assumptions that are being made about the volume and flow of water.
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APPLIED GEOMETRY
MATHEMATICAL APPLICATIONS ESSENTIALS STAGE 2
Solution
The different ways that the area of the dam can be calculated using Simpson’s Rule are demonstrated below:
AL ≈ 12 (7 + (4 × 12.3) + (2 × 14.6) + (4 × 13.7) + 0)
3
AL ≈ 560.8 m2
AR ≈ 12 (6.5 + (4 × 14.7) + (2 × 15) + (4 × 13.5) + 0)
3
AR ≈ 597.2 m2
Area of left side
Area of right side
Total area:
left side + right side
Total area ≈ 560.8 + 597.2 ≈ 1158 m2
Awhole ≈ 12 (13.5 + (4 × 27) + (2 × 29.6) + (4 × 27.2) + 0)
3
Awhole ≈ 1158 m2
Awhole ≈ 12 (0 + (4 × 27.2) + (2 × 29.6) + (4 × 27) + 13.5)
3
Awhole ≈ 1158 m2
Area of whole dam
(from top to bottom)
Area of whole dam
(from bottom to top)
The volume of water is 1158 m2 × 3.85 m = 4458.3 m3
(2) How much water (in cubic metres) is pumped during the night?
Volume of water after pumping is 1158 m2 × 1.25 m = 1447.5 m3
Volume of water pumped = 4458.3 m3 – 1447.5 m3 = 3010.8 m3
(3) Calculate the rate of flow of the water (in kL per minute) if it takes 7 hours to pump the volume of water calculated in
part (2).
1 m3 holds 1 kilolitre of water 3010.8 m3 = 3010.8 kL
7 hours = 420 minutes
Rate of flow = 3010.8 ≈ 7.2 kL/min
420
(4) State two assumptions that are being made about the volume and flow of water.
The flow of water is constant during the entire 7 hours.
The depth of the water in the dam is the same across the entire dam.
EXAMPLE 3
The water and environment department is responsible for ensuring high water quality in reservoirs. The diagram below
(not to scale) shows a cross section of a small rural reservoir:
A
B
40 m
40 m
37 m
21 m
40 m
40 m
6m
Scale 1:500
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MATHEMATICAL APPLICATIONS
APPLIED GEOMETRY
(1) (a) Using the scale, show that the drawn length of AB on the scaled plan would be 32 cm.
(b) Calculate the area of the cross section of the reservoir. Round your answer to the nearest square metre.
(2) (a) The surface area of the water covers 63 hectares. Convert this to square metres.
(b) The capacity of the reservoir is 6 370 megalitres. Using this information and the information from part (a)
calculate the average depth of the reservoir.
(3) Water is pumped from the River Murray pipeline at a rate of 1 400 megalitres per week into this reservoir.
If the reservoir is empty at the end of summer, how many days will it take to fill the reservoir?
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APPLIED GEOMETRY
MATHEMATICAL APPLICATIONS ESSENTIALS STAGE 2
(4) Reservoir water is disinfected before it becomes suitable for drinking. The disinfectant, Chloramine, is added to the
water at the rate of 4 milligrams per litre.
(a) How many kg of Chloramine is required to disinfect 1 megalitre?
(b) If Chloramine comes in 25 kg drums, how many drums would be required to disinfect 6 000 megalitres?
(5) Discuss one limitation of the measurements and answers to calculations found in parts (a) to (d) above.
Solution
(1) (a) Using the scale, show that the drawn length of AB on the scaled plan would be 32 cm.
× by 32
1 : 500
1 cm : 500 cm
1 cm : 5 m
? cm : 160 m
32 cm : 160 m
AB = 160 m
160 m = 16 000 cm
16 000 = 32 cm
500
OR
(b) Calculate the area of the cross section of the reservoir. Round your answer to the nearest square metre.
Using Simpson’s Rule
A ≈ 40 (0 + (4 × 37) + (2 × 21) + (4 × 6) + 0)
3
Note the 0 values.
A ≈ 2853 m2
A
B
40 m
37 m
40 m
21 m
Using the trapezoidal rule
40 m
40 m
6m
Scale 1:500
A≈ 40 ( 1 × 0 + 37 + 21 + 6 + 1 × 0)
2
2
A ≈ 2560 m2
(2) (a) The surface area of the water covers 63 hectares. Convert this to square metres.
63 × 10 000 = 630 000 m2
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