Maths Quest A Year 11 for Queensland
WorkSHEET 7.2
Chapter 7 Basics of construction
Basics of construction
WorkSHEET 7.2
Name: ___________________________
Questions 1 – 6 refer to information available from the floor plan below.
1
Determine the actual size of the house.
1
The scale is 1 : 150.
Length measurement of house on plan = 12 cm
 Actual length  12 cm  150
 1800 cm
 18 m
Width of house on plan  8 cm
 Actual width  8 cm  150
 1200 cm
 12 m
Maths Quest A Year 11 for Queensland
2
Chapter 7 Basics of construction
WorkSHEET 7.2
If the house was pegged out correctly so that it
was rectangular in shape with the corner angles
being right angles, what would its diagonal
length be?
Using Pythagoras’ Theorem.
hypot. 2  base 2  ht 2
diagonal 2  18 2  12 2
 324  144
 468
So, diagnonal  468
 21.6 m
Diagonal length  21.6 m
2
Maths Quest A Year 11 for Queensland
3
Chapter 7 Basics of construction
WorkSHEET 7.2
If 6 m lengths of trench mesh were laid in the
footings around the perimeter of the house,
how many lengths would be required for two
layers of mesh?
The mesh must overlap at the corners and
where two sheets meet, the overlap must be
500 mm. Note that the mesh lengths can be cut
if necessary.
Remember to allow for 500 mm overlap.
The length of the plan would require
3 lengths (17 m) + 1.5 m.
So, both lengths would require
6 lengths + 3 metres
i.e. 6 12 lengths
The width of the plan would require
2 lengths (11.5 m) + 1 metre
So, both widths would require
4 lengths + 2 metres
Total required
= (6 lengths + 3 m) + (4 lengths + 2 m)
= 10 lengths + 5 m
So, number of lengths required for 1 layer
= 11
 Two layers would require
11  2 = 22 lengths
3
Maths Quest A Year 11 for Queensland
4
Chapter 7 Basics of construction
WorkSHEET 7.2
What would be the cost of concrete for the
footing around the perimeter of the house, if
they are 350 mm wide and 200 mm deep, and
concrete costs $130/m3?
Outer area of floor = 18 m  12 m
= 216 m2
Inner area of floor
= (18 – 2  0.35) m  (12 – 2  0.35) m
= 17.3 m  11.3 m
= 195.49 m2
 Surface area of footings
= outer area – inner area
= 216 m2 – 195.49 m2
= 20.51 m2
Volume of footings
= surface area  depth
= 20.51 m2  0.2 m
= 4.102 m3
 Cost of concrete = 4.102 m3  $130/m3
= $533.26
5
Determine the cost of concrete for the slab of
the house (100 mm thick) at $130/m3.
Volume of concrete = L  W  H
= 18 m  12 m  0.1 m
= 21.6 m3
Cost of concrete
= 21.6 m3  $130/m3
= $2808
4
Maths Quest A Year 11 for Queensland
6
Chapter 7 Basics of construction
WorkSHEET 7.2
The wall separating the lounge from bedroom 3
requires bracing. The height of the wall is
2.4 m and its length is 4.2 m.
Would it be acceptable to put a brace
diagonally across the wall from a floor corner
to a ceiling corner?
Explain your answer.
For a brace to be acceptable, it must lie within
an angle range of 37º to 53º with the horizontal.
tan  
2 .4
4 .2
 2 .4 

 4 .2 
  tan 1 
 29.7 
This brace is not acceptable as it lies outside
the range 37º to 53º.
Questions 7 – 10 refer to information available from the front elevation shown below.
5
Maths Quest A Year 11 for Queensland
7
Chapter 7 Basics of construction
WorkSHEET 7.2
The main section of the roof is in the shape of
a gable roof with a pitch angle of 28º.
Express this as a pitch ratio.
Pitch ratio is the tangent ratio in the form 1 : x.
1
tan 28  
x
x tan 28  1
1
x
tan 28 
 1 .9
So, pitch ratio is 1 : 1.9.
8
Use the scale indicated on the plan to determine Front length measures 10.8 cm.
the length of the front of the house.
Scale is 1 : 100.
 Length of front of house
 10.8 cm  100
 10.8 m
9
If the side of the house (from front to rear) is
8 m, determine the area of the roof. (Assume
that the roof does not overhang the walls in any
direction.)
The roof consists of 2 rectangles
x metres by 10.8 metres
adj.
hypot.
4

x
cos 28 
x cos 28  4
x
4
cos 28
 4.5 m
Area of roof  10.8 m  4.5 m  2 m
 97.2 m 2
6
Maths Quest A Year 11 for Queensland
10
Chapter 7 Basics of construction
The brick wall beneath the front of the house
approximates the shape of a triangle. Use the
scale of the plan, together with the estimated
requirement of 50 bricks per square metre, to
determine the number of bricks required for
this wall.
WorkSHEET 7.2
7
Measure length ' h' on plan  9.1 cm
So, actual length of ' h'  9.1 cm  100
 9.1 m
Measured length ' b' on plan  1.3 cm
So, actual length of ' b'  1.3 cm  100
 1.3 m
b h
2
1.3  9.1

2
Area of bricked region 
 5.915 m 2
Brick requiremen t  50/m 2
 Number of bricks required for this region
 5.915  50
 295.75
 approximat ely 300