Perimeter Calculations
| Perimeter is the length around the outside of a figure. |
Normally we only look at 2-dimensional figures; however, the concept of total length can also apply to 3-dimensional figures which have a framework as well as any covering which will be attached to that framework.
Calculating the surface area of different 2D shapes
Calculating the surface area of different 2D and 3D shapes.
|
Name of Solids |
Figure |
Lateral or Curved Surface Area |
Total Surface Area |
Units |
|
Cuboid |
 |
2h(l+w) |
2(lw+wh+hl) |
l = Length w = Width h = Height |
|
Cube |
 |
4a2 |
6a2 |
a = Side |
|
Right Prism |
 |
Perimeter of base x height |
Lateral Surface Area + |
|
|
Right Circular Cylinder |
 |
2πrh |
2πr(r+h) |
r = Radius h = Height |
|
Right Pyramid |
 |
0.5(Perimeter of base x |
Lateral Surface Area |
|
|
Right Circular Cone |
 |
πrl |
πr(l+r) |
l = Slant h,=,Height r = Radius |
|
Sphere (Solid) |
 |
4πr2 |
4πr2 |
r = Radius |
|
Hemisphere (Solid) |
 |
2πr2 |
3πr2 |
r = Radius |
EXAMPLE:
Consider the metal archway in the following Figure:
Metal archway
Answer: Break the complicated figure down into more manageable pieces and then calculate the total.
Archway pieces
|
Total frame |
= 2 × (½ × π × diameter) + 4 × 1.6 m + 11 × 0.4 m = 2 × (0.5)(3.14)(1,2 m) + 6.4 m + 4.4 m = 3.768 m + 6.4 m + 4.4 m = 14.57 m |
|
Therefore, 3 lengths of steel bar will need to be purchased to create the archway. |
|
Wooden tile covering of metal archway
- The steel frame is going to be covered with wooden tiles that are 20 cm wide and overlap each other by 3 cm as in the diagram alongside. How many will be required to cover the outside of the frame?
Answer: The total outside length to be covered needs to be calculated first.
Wooden tile length for archway exterior
|
The total length of the front of the frame |
= (½ × π × diameter) + (2 × 1.6 m) = (0.5)(3.14)(1.2 m) + 3.2 m = 1.88 m + 3.2 m = 5.08 m |
Because they overlap, the width of the tile cannot be used as it is. The overlap of adjacent tiles needs to be considered to use the effective width of the tiles:
|
Effective Width |
= 20 cm – 3 cm = 17 cm |
|
Number of tiles needed |
= Total length ÷ Effective width = 508 cm ÷ 17 cm (Note that the units must be the same) = 29.88 tiles ≈ 30 tiles |
Area Calculations
Most complex areas can be broken down into simpler areas. Once a larger area is calculated, other calculations can be performed on it such as seeing how many of a smaller area can fit into it.
EXAMPLE:
Skateboarders often enjoy performing tricks on a special ramp called a halfpipe (see the following Figure).
- Use the Figure of a halfpipe to calculate the area of the curved skating surface of the halfpipe.
Answer: If a surface has the same width all along its length, it then forms a large rectangle, so we can calculate the length of the edge and use it as the length of the rectangle:
|
The total length of the edge |
= 2 × ¼ × π × d + 3.6 m = 2 × (0.25) × (3.142) × (5.48 m) + 3.6 m = 8.61 m + 3.6 m = 12.21 m |
||
|
Therefore, the total area of the ramp surface |
= length × breadth = 12,21 m × 13 m = 158,73 m2 |
Halfpipe
- A special board is used on the surface. The board is sold in rectangular pieces (width 1.5 m, length 3.0 m). Use the area of the board to estimate the number of boards needed to cover the skating surface of this halfpipe.
Answer: We can only estimate the number of boards as we would need to check how the boards fit together, but an estimate is useful when getting a rough idea of how many boards might be needed. We can estimate the number of boards by dividing the area of one board into the total area of the ramp:
|
Area of 1 board |
= length × breadth = 1.5 m × 3.0 m = 4,5 m2 |
|
Number of boards |
= Total area ÷ Area of 1 board = 158.73 m2 ÷ 4.5 m2 = 35.27 boards ≈ 36 boards needed |