1. Calculations
On completion of this unit the Learner should be able to demonstrate an understanding of:
The formula used to calculate:
Areas
Volumes
Ratios
Volumes into weight
perimeters
Mean, Mode, Median and range

2. Calculations Areas of Rectangles, irregular shapes, circles.
Perimeter calculations.
Centreline method used to calculate volumes of concrete.
Volume for calculating quantities of concrete.
Converting volume into tonnage.
Mean = average (find the total and divide by the number youy have
Mode= most commonly occurring number
Median= put the numbers in order and find the middle number
Range= difference between highest and lowest number

3. Surface areas
Calculating areas of brickwork and blockwork we are in fact calculating the
surface area. Surface areas in construction are calculated to determine
quantities of materials required to cover that area. Examples of this are
bricks, blocks, floor tiles and block paving. To calculate for these
quantities the following procedures should be followed:
Calculate the total area to be covered ( m2)
Calculate the area of one unit (a brick,block,tile etc).
Divide the total area by the area one unit.
D) Add percentage to compensate for waste.

4. Calculate the following areas
Formula required to do this is: Area = L x H = M2
Calculate the area of a wall 7.5m long x 4.5m high with a window opening of
1.200m x 1.050m
Answer Formula L x H m x 4.5m = 33.75m2 – area of window
1.200m x 1.050m = 1.260m2
33.75m2 – 1.260m2 = 32.49m2

5. Example 1
Calculate the total of 0.100m x 0.200m paving block required to cover the drive area shown below, allow 5% for cutting/waste
8 m
4 m
Area = 8 x 4 = 32m2
Area of one paving block = 0.100m x 0.200m = 0.02m2
Total blocks required = 32m2 divided by 0.02m2
= 1600 blocks add 5% for cutting/waste
1600 x 105% = blocks
100
You will be expected to complete the following calculations.
Keep them simple and use the same method each time, ensuring neatness.
Complete your calculations using the same method each time.

6. Answer the following questions
Calculate the total 0.3m x 0.3m tiles required to cover the floor area shown below and allow 6% for waste.
6 m
4 m
Area = 6 x 4 = 24 m2
Area of one tile = 0.3m x 0.3m = 0.09m2
Total tiles required = 24m2 divided by 0.09m2
= tiles add 6% for cutting/waste
266.6 x 106% = tiles (round up to 283 tiles)
100

7. Area of one block = 0.225m x 0.450m = 0.10m2
Calculate the total 0.225m x 0.450m air-rated blocks are required to cover the wall area shown below and allow 8% for waste.
18 m
9 m
Area = 18 x 9 = 162 m2
Area of one block = 0.225m x 0.450m = 0.10m2
Total blocks required = 162m2 divided by 0.10m2 =1620 blocks
= 1620 blocks + 8% for cutting/waste
1620 x 108% = blocks
100

8. (A calculator can be used, but show all your workings.)
Calculate the area of insulation required for a cavity wall 4.725M long x 2.475M high in which there is a window 1.810M wide x 1.2M high, and allow 5% for wastage.
(A calculator can be used, but show all your workings.)
Area of wall m x 2.475m = m2
Area of window opening m x 1.2m = 2.172m2
Area of wall – area of window = m2 – 2.172m2 = 9.522m2
5% allowed for wastage = 9.522m2 x 105% = m2
100
Total area of insulation required = m2

9. Calculating the area of irregular shapes5m
Area = Length x Width
Length = average of two sides
8 + 5 divided by 2 = 6.5m
Length = 6.5m
width = 4m
6.5 x 4 = 26m2 4m 8m

10. Answer the following questions
6.5m
12m 8m 5m 7m 4m
Length = average of two sides
divided by 2 = 6.75m
Length = 6.75m
width = 5m
6.75 x 5 = 33.75m2
Length = average of two sides
divided by 2 = 8m
Length = 8m
width = 8m
8 x 8 = 64m2

11. 10m 4m 3m 7m 8m 2m
Length = average of two sides
divided by 2 = 6m
Length = 6m
width = 7m
6 x 7 = 42m2
Length = average of two sides
8 + 4 divided by 2 = 6m
Length = 6m
width = 3m
6 x 3 = 18m2

12. Calculating the volume of concrete.
When taking off quantities of brickwork, excavation and concrete for a building, it is essential to calculate the perimeter of the structure.
The formula for this differs depending if the internal or external measurements are being used. Here is an alternative method to calculate the volume of concrete required. This is known as the centreline method
Using the external measurements
The mean perimeter = (2 x L) + ( 2 x W ) Less 4 x Thickness of foundations
Using the internal measurements
The mean perimeter = (2 x L) + ( 2 x W ) plus 4 x Thickness of foundations

13. Fig. 1 shows a foundation trench for a building.
The trench in fig 1 is 1.2m deep. Calculate the volume of material to be excavated. A
Formula when using
External dimensions is:
(2 x L) + ( 2 x W )
Less 4 x foundation width.
Volume = L X W x D = m3 C D
0.75m
8.0m B
10.00m
Materials such as concrete will have to be ordered on a weekly basis. When ordering concrete we will have to quote the volume required in this example we would order 29.7 cubic metres of concrete to complete the foundations.
Length 2 x 10 = 20
Width 2 x 8 = 16
36m (less 4 x 0.75m)
36m – 3m = 33m
33 x 0.75 x 1.2 = 29.7 m3 of concrete required for this foundation

14. Fig. 2 shows a foundation trench for a building.
Centre-line calculations
Fig. 2 shows a foundation trench for a building.
The trench in fig 2 is 1.5m deep. Calculate the volume of material to be excavated. A B
Formula when using
External dimensions is:
(2 x L) + ( 2 x W )
Less 4 x foundation width.
Volume = L X W x D = m3 C D
3.3m
0.60m A
16.5m
Length 2 x 16.5 = 33
Width 2 x 3.3 = 6.6
39.6m (less 4 x 0.60m)
39.6m – 2.4m = 37.2m
37.2 x 0.60 x 1.5 = 33.48m3 of concrete required for this foundation

15. Centre-line calculations
Depth of foundation 0.50m
0.65 4m
10m
Formula when using
Internal dimensions is:
(2 x L) + ( 2 x W )
plus 4 x foundation width.
Volume = L X W x D = m3
Length 2 x 10 = 20
Width 2 x 4 = 8
28m (plus 4 x 0.65m)
28m + 2.6m = 30.6m
30.6 x 0.65 x 0.50 = 9.945m3 of concrete required for this foundation.
If we calculated the amount of concrete required by the external measurements we would end up with too much concrete this is why we use centre line calculations.

16. Centre-line calculations
10m 3m
The trench is 0.65m wide and 0.50min depth. Calculate the volume of concrete required 7m 4m 6m 4m
The formula makes calculating volumes of irregular shaped foundations easier.
Length 2 x 10 = 20
Width 2 x 7 = 14
34m (less 4 x 0.65m) = 31.4m
34m – 2.6m = 31.4m
31.4 x 0.65 x 0.5 = m3 of concrete required for this foundation

17. Answer the following questions
Calculate the quantity of concrete required for this foundation.
N.B Width 0.60m depth 0.45m
18m 3m
12m
Length 2 x 18= 36
Width 2 x 3 = 6
42m (plus 4 x 0.60m)
42m + 2.4m = 44.4m
44.4 x 0.60 x 0.45 = m3 of concrete required for this foundation

18. Answer the following questions
Calculate (using the centre line method how many bricks would be required to build the one brick thick inspection chamber. (assume 120 bricks per square metre) depth of brickwork is 0.90m
0.7m
0.91m
12m
Length 2 x 0.91m = 1.82m
Width 2 x 0.7m = 1.4m
3.22m (plus 4 x 0.225m)
3.22m + 0.9m = 4.12m
L D B/M2
4.12 x 0.90 x 120 = bricks required to build the inspection chamber

19. Material which are ordered by weight such as hardcore (per tonne)
Material which are ordered by weight such as hardcore (per tonne). To calculate the quantity required we use the following formula. 1. calculate the volume as we do for concrete 2. Use the density method (volume x density) Example An area is to be filled with hardcore. The area is 8.5m long by 4.2m wide. The average depth is 0.15m Calculate the amount of hardcore required Volume = 8.5 x 4.2 x 0.15 = 5.36m3 (2 decimal places) Total amount = Volume x density = tonnage x 2.000kg/m = Tonnes (we would order 11 tonne)

20. Answer the following questions
An area is to be filled with hardcore. The area is 6.70 long by 5.50m wide,
with an average depth of concrete of 0.20m
Calculate the quantity of hardcore required. (Density = 2.000kg/m3)
Total cost of the hardcore at £18 per tonne
Remember Volume x density = tonnage
Volume = 6.70 x 5.50 x 0.20 = 7.37m3 ( 2 decimal places only)
7.37 x 2.000kg/m3 = tonne
we would order 15 tonne.
Cost 15 x £18 = £270.00

21. An area is to be filled with hardcore. The area is 8. 5m long by 4
An area is to be filled with hardcore. The area is 8.5m long by 4.2m wide. The average depth is 0.15m
Calculate the amount of hardcore required (Density = 2.000kg/m3)
Remember Volume x density = tonnage
Volume = 8.5 x 4.2 x 0.15 = 5.36m3 ( 2 decimal places only)
= 5.36 x 2.000kg/m3
= Tonnes
we would order = 11 Tonne.

22. An area is to be filled with concrete. The area is 10. 5m long by 6
An area is to be filled with concrete. The area is 10.5m long by 6.1m wide. The average depth is 0.55m
Calculate the amount of concrete required (Density = 2.400kg/m3)
Remember Volume x density = tonnage
Volume = 10.5 x 6.1 x 0.55 = 35.22m3 ( 2 decimal places only)
= x 2.400kg/m3
= Tonnes
we would order = 85 Tonne.

23. Ratios of materials
Material such as concrete as mixed together in ratios. A typical mix for concrete would be a 1:3:6 mix this means that
1 Portion of cement is required
3 portions of fine aggregate (sharp sand)
6 portions of course aggregate
To produce the required specified mix of concrete.
It is relatively easy to work out volume of concrete
Volume = (L x W x D).
From the volume we would then need to break this down into the proportions of material required to complete the concrete structure.

24. Example 1
Total portions is 1: 3: 6 = 10
Average density of concrete is 2400 kg/m3
(2.4 tonne of aggregate to produce a cubic meter of concrete)
Divide 2400 by 10 = 240kg therefore
1 portion of cement = 1 x 240 = 240kg
3 portions of fine aggregate = 3 x 240 = 720kg
6 portions of course aggregate = 6 x 240 = 1440kg
Are required to produce 1m3

25. Example 2
N.B Ratio of facings to commons (that is 3 facings to every 1 common)
Flemish bond is specified by the customer to construct a wall 23.6 long by 1.36 in height. Calculate the amount of facings and commons that would be required to complete this.
Area of wall 23.6 x 1.36 = m2
Total no of bricks required = x 120 = bricks
(120 bricks per m2 in any one brick wall)
Add together 3 and 1 together = 4
Therefore divide by 4 = (is equal to one portion)
Commons 1 x = bricks
Facings 3 x = bricks
check = total

26. Flemish bond is specified by the customer to construct a wall 15m long by 2m in height. Calculate the amount of facings and commons that would be required to complete this.
Area of wall 15m x 2m = 30m2
Total no of bricks required = 30 x 120 = 3600 bricks
(120 bricks per m2 in any one brick wall)
Add together 3 and 1 together = 4
Therefore divide 3600 by 4 = 900 (is equal to one portion)
Commons 1 x 900 = bricks
Facings 3 x 900 = 2700 bricks
check = 3600 total

27. Calculating diagonals
Remember the formula for working out diagonal dimensions is:
A2 + B2
Dimension C = ?
Dimension B = 8m
Being able to calculate the dimension of diagonals is a useful tool for most trades persons.
Bricklayers can use this to check of a corner has been set out correctly (square) for example during the setting out of a building onto foundations. Using a builders square and string lines is not a particularly accurate method.
A = 10 x 10 = 100
B = 8 x 8 = 64
= 164 square root of 164
√164 = 12.8m length of dimension C = 12.8 m
Dimension A = 10m

28. Calculating diagonals
Remember the formula for working out diagonal dimensions is:
A2 + B2
Dimension C = ?
Dimension B = 7m
Being able to calculate the dimension of diagonals is a useful tool for most trades persons.
Bricklayers can use this to check of a corner has been set out correctly (square) for example during the setting out of a building onto foundations. Using a builders square and string lines is not a particularly accurate method.
A = 12 x 12 = 144
B = 7 x 7 = 49
= 193 square root of 193
√193 = 13.89m length of dimension C = 13.89m
Dimension A = 12m

29. The diagram shows a section through a roof.
(Note rafter AB = AC2 + BC2) A
Rise of roof
2.100m
Calculation
AC2 = x = 4.41m2
BC2 = x = 6.25m2
AC2 + BC2 = = 10.66
√ 10.66
3.26m is the length of rafter AB B C
This is a useful formula as we can use it when setting out to check if a building is square or as in this case to find the length of materials.
2.500m
Half of total roof span

30. √ 58 = 7.61m is the length of the diagonal X
The diagram below shows a foundation of a building in the process of being set out. Calculate the length of X mathematically to ensure that the conservatory is square to the rear elevation of the building. A
(Note length of X = √ AC2 + BC2) X 7m
conservatory 3m
900 B C
Calculation
AC2 = 7m x 7m = 49m2
BC2= 3m x 3m = 9m2
AC2 + BC2 = = 58
√ 58 = 7.61m is the length of the diagonal X
The main building is always set-out first and the porch or other projecting part of the building is set-out from this.
Using this formula a bricklayer can check the position of the corner of the porch by calculating the length of the diagonal X.

31. Calculating areas of triangles
The formula required to do this is :-
Area = ½ Base dimension x Height = m2
Tradespersons need to be competent at calculating materials in order to submit accurate estimates. If we were to submit estimates by guessing quantities this would lead us to either not getting the job in the first place (overpriced estimate) or having to order extra materials which may lead to having no money left for wages. The most successful builders among us will be able to accurately estimate projects.
A triangular is a common shape that we construct. For example we would use this formula when calculating materials required to construct a gable end.
10m
Base 8 ÷ 2 = 4
4 x 10 = 40m2 8m

32. Calculate the areas of triangles below8m
12.3m
19.5m 7m
12m
37.7m
Now that we are able to work out the area it is simple to then multiply this number by the amount of bricks or blocks that is required to build a m2
We know that a half brick wall consists of 60 bricks per m2. Therefore if triangle A ) was a brick gable all we need do is multiply the area of the triangle by 60 this would provide us with the bricks required for construction of the gable.
28 x 60 = 1680 bricks A)
Base = 7 ÷ 2 = 3.5m
3.5 x 8 = 28m2 B)
Base = 12 ÷ 2 = 6m
6 x 12.3 = 73.8m2 C)
Base = 37.7 ÷ 2 = 18.85m
18.85 x 19.5 = 367.5m2

33. Calculating the area of a circle
The distance around a circle is called its circumference.
circumference
The distance across a circle is called its diameter.
diameter
The distance of half the diameter is called the radius.
Radius
For any size of circle, if you divide the circumference by its diameter, the value of 3.14 will be the answer. This is called (Pi).
There are many reasons that a bricklayer may need to work out areas of circles. You may be asked to tender for jobs that have circular windows within the structure or other circular features.
Depending on the information that we are supplied with will change the formula we use. The following examples will help you to complete the set questions.
R = 3in
Example 1. The radius of a circle is 3 in. What is its area ?
Area of circle = (Pi) x R x R
= 3.14 (Pi) x 3 x 3 = inch 2

34. Calculate of the areas of the circles below.
R = 10cm
Area of circle = (Pi) x R x R
(Pi) = 3.14 x 10 x 10 = 314cm 2
Area of circle = (Pi) x R x R
(Pi) = 3.14 x 4 x 4 = 50.24m 2
R = 4.0m

35. Example 2. The diameter of a circle is 8 cm. What is its area ?
Solution D = 8
R = D divided by 2 = 4cm
Area = (Pi) x 4 x 4 = cm2
D = 8 cm
Basic rule is to use the radius to calculate the area of a circle, therefore if a question only has the diameter size then dividing by 2 supplies us with the radius size.

36. Calculate of the areas of the circles below.
D = 15m
R = D divided by 2 = 7.5m
Area = (Pi) x 7.5 x 7.5 = m2
D = 15m
D = 300mm
R = D divided by 2 = 150mm
Area = (Pi) x 150 x 150 = 70650mm2
D = 300mm

37. The diameter of a circle is 12 in. What is its area ?
The radius of a circle is 9cm. What is its area ?
R= 9cm
Area of circle = (Pi) x R x R
3.14 x 9 x 9 = cm2
The diameter of a circle is 12 in. What is its area ?
D = 12 in
Area of circle = D divided by 2 = radius
(Pi) x R x R
3.14 x 6 x 6 = in2
The radius of a circular pool is 4ft. What is its area ?
R= 4ft
Area of circle = (Pi) x R x R
3.14 x 4 x 4 = ft2

38. Example 3. The area of a circle is 78.5m2. What is its radius ?
Solution Area divided by (Pi) = R x R
78.5m2 ÷ 3.14 = 25
25 = R x R
Therefore the square root of 25 will give us size of the radius.
We now need to find the square root of 25. Finding the square root of a number is the inverse operation of squaring that number. Remember, the square of a number is that number times itself. I.e the square root of 10 is 10 x 10 which = 100
We can do this with the aid of a calculator by simply pressing the
symbol after entering the square number.
So the answer for the sum above is :
25 = 5

39. The figure opposite shows a plan view of a timber floor.
The floor area is
64.28m2
86.13m2
100.28m2
158.14m2
R =3m 6m
Area of rectangle = L x H = m2
12 x 6 = 72m2 + area of circle
Area of circle = (Pi) x R x R
3.14 x 3 x 3 = m2
Answer = = m2
This is a typical exam question. They are trying to catch you out!
We need to divide the area of the circle in half as we have already calculated half of it as it is incorporated in the rectangle.
12m

40. Example 1
Calculate the total cost of the materials below from the prices given.
Allow 17.5% VAT
1500 Facing bricks are required to construct a £230/ 1000
calculate the cost of 1500 bricks
Calculate cost of bricks + vat
                   
1. Cost of bricks = 1500 x £230 = £345.00
1000
2. VAT= 345 x 117.5% = £ (cost of bricks including vat)
100

41. Example 2
Calculate the total cost of materials using the prices given, allowing
17.5% VAT
400 concrete blocks are required to construct a £6.50/ 10
calculate the cost of 400 blocks
Calculate cost of blocks + vat
100mm Concrete blocks £6.50/ 10
Cost of blocks = 400 x £6.50 = £260.00 10
VAT = 260 x 117.5% = £ (cost of blocks including vat)
100

42. Calculate the total cost of materials using the prices given, allowing
17.5% VAT
A) 2100 Facing Bricks £247/ 1000
Cost of bricks = 2100 x £247 = £518.70
1000
VAT x = £ (total cost of bricks including vat)
100

43. Calculate the total cost of materials using the prices given, allowing
17.5% VAT
B) mm Concrete Blocks £4.95/ 10
Cost of blocks = 650 x £4.95 = £321.75 10
VAT x = £ (total cost of blocks including vat)
100

44. Calculate the total cost of materials using the prices given, allowing
17.5% VAT
C) 3.5 Tonne Building Sand £18/ Tonne
Cost of sand = 3.5 x £18 = £63.00
VAT x = £ (total cost of sand including vat)
100

45. Calculate the total cost of materials using the prices given, allowing
17.5% VAT
D) 250 Wall Ties £17.00/ 100
Cost of wall ties = 250 x £17.00 = £42.50
100
VAT x = £ (total cost of wall ties including vat)

46. Assessment questions Question 1
Calculate the total number of bricks required to build the ½ B wall shown, allowing 5% waste.
Main area – door and window
1) 12.00m x 3.00m = 36m2
Door = 2.00m x 1.00m = 2m2
Window = 2.00m x 1.50m = 3m2
36m2 – 5m2 = 31m2
2) 31 x 60 = 1860 bricks x 105 = 1953 bricks including waste
100

47. Question 2
The uppermost part of a gable wall forms the shape of a triangle.
Calculate the total number of bricks required to build the ½ B gable wall shown, add 3 ½% for waste.
Area of a triangle = half base x height
9 divided by 2 = 4.5m x 2.0m = 9 m2
9 x 60 = 540 bricks % waste
540 x = bricks (round up to 559)
100

48. Question 3
An area is to be filled with hardcore. The area is 6.70m long by 5.50m wide. The average depth of hardcore is 0.20m.
1) Calculate the amount of hardcore required. (Density = 2.000kg/m3)
2) Total cost of the hardcore at £30 per tonne, including VAT at 17 ½ %
Area to be filled with hardcore =
6.70 x 5.50 x 0.20 = 7.37m3
Volume x density = tonnage 7.37m3 x = tonnes
(round up to 15 tonnes) 15 x £30 = £450.00
450 x 17.5 = 78.75
100
total cost = = £528.75

49. Question 4
An area is to be filled with hardcore. The area is 7.00 long by 5.50m wide, with an average depth of 0.20m
1. Calculate the quantity of hardcore required. (Density = 2000kg/m3)
2. 2.  Total cost of the hardcore at £27 per tonne
Area to be filled with hardcore = 7.00m x 5.50m x 0.20m = 7.7m3
Volume x density = tonnage 7.7m3 x = 15.4 tonnes
(round up to 16 tonne)
16 x £27= £432

50. Question 5
Calculate the area of the circle which has a radius of 9m.
Area of circle = (Pi) x R x R = m2
3.14 x 9 x 9 = m2

51. Question 6
1750 Facing bricks are required to construct a £320/ 1000
1.       calculate the cost of 1750 bricks
2.      Calculate cost of bricks % vat
1750 x 320 = £560
1000
560 x = £658
100

52. Formula when using
External dimensions is:
(2 x L) + ( 2 x W )
Less 4 x foundation width.
Volume = L X W x D = m3
Question 7
Calculate the concrete required for the foundation shown below, given that the trench width is 0.4m and the concrete is to be 0.5m deep.
N.B all measurements are external
6 x 2 = 12m
4 x 2 = 8m
Less 4 x width of foundation 4 x 0.4m = 1.6m
12m + 8m = 20m – 1.6m = 18.4m
Volume x 0.4 x 0.5 = 3.68m3

53. Question 8
Calculate the area of the circle which has a diameter of 29m.
Diameter divided by 2 = radius
Area of circle = (pi) x R x R
3.14 x 14.5 x 14.5 = m3
Question 9
Add the following dimensions together 1240mm + 1.2m + 2m +1537mm
1.240
1.200
2.000
1.537
5.977m or 5977mm

54. Question 10
An area which has a length of 16m and width of 9m requires tiling.
1) Calculate the number of 0.100m x 0.200m tiles required to cover the
area.
2) Add 13% for waste
Area to be tiled = 16m x 9m = 144m2
Area of one tile = 0.100m x = 0.02m2
144m2 divided by area of one tile = 144 divided by 0.02 = 7200
Add 13% for waste 7200 x 113 = 8136 tiles required
100