1. Area & Volume Area is product of 2 units of length metre x metre = m2
Remember if the unit is squared so to is the conversion.
1 m2 = mm2 = mm2 or 1 x 106 mm2
It is important that you are able to find the areas and perimeters of standard shapes
Create a simple aid sheet for the following areas and perimeters in the following fashion:
eg1. Rectangle (sketch the shape)
Area = L x b
Perimeter = 2L + 2b
Breadth b
Length L
JR/2008

2. Area & Volume 2. Parallelogram
Use the format shown to create a sheet for the following shapes:
2. Parallelogram
3. Triangle ( please also research Hero’s formula & sine formula)
4. Trapezium
5. Circle
6. Sector of a circle
Useful refs :- Greer & Taylor – BTEC National Mathematics for Technicians
JR/2008

3. Area & Volume Volume or Capacity are similar with identical units.
Volume found from area x length of area gives units m2 x m = m3
Remember if the unit is cubed so to is the conversion.
1 m3 = mm3 = mm3 or 1 x 109 mm3
Liquid capacity is often measured in litres ( ℓ ). It must be noted that the litre is an approximate unit ( 1 litre = dm3 ) and should not be used for precise measurements.
In general terms 1 m3 = ℓ
To obtain volume it is usual to consider cross sectional area (csa) first ,
then multiply csa by length of csa.
JR/2008

4. Area & Volume
eg. find the volume of a 2m length of round bar of 20mm diameter.
Soln. First find csa of bar ( πr2 ) = π x 102 = mm2
Converting to m2 = ÷ = m2
( or more simply x 10-6 m2 )
Remember if the unit is cubed so to is the conversion.
Volume = csa x length of csa ( giving πr2 x h (height or length of area) )
Volume = x m2 x 2m = x 10-6 m3 ( m3 )
JR/2008

5. Area & Volume
eg2. find the volume (capacity in litres ) of a cylinder 88mm internal diameter (i/d) and 212mm depth.
Soln. First find area of cylinder ( πr2 ) = π x 442 = mm2 (1d.p.)
Volume = csa x length of csa
Volume = mm2 x 212 mm = mm3
Converting to m3 Remember if the unit is cubed so to is the conversion.
mm3 ÷ = m3
if 1000 litres = 1 m3 then m3 = x 1000 = litres
Remember – the litre is an approximate unit
JR/2008

6. Area & Volume One important property related to this work is DENSITY.
Density is taken as mass per unit volume, (mass ÷ volume)
Density has the symbol ρ (rho) and the unit kg/m3 or kg.m-3
The average density of pure water should be noted as 1000 kg m-3
Density knowledge can be put to good use in engineering eg.
Find the mass of steel needed to cast a solid ingot 400mm square x 2m long
given the density of steel to be 7600 kg m-3
Soln. Find the volume required = 0.4m x 0.4m x 2m = 0.32 m3
from density = mass/volume (ρ = m/v) then mass = density x volume
m = ρ x v giving mass m = 7600 kg m-3 x 0.32 m3 = kg (2.43 tonne)
JR/2008