By the end of this chapter, you will be able to:
Mastering these skills will help you solve real-world engineering problems with confidence and precision.
Mensuration is fundamental in biomedical engineering for accurately determining dimensions, areas, and volumes of biological structures, medical devices, and components. Precision in these measurements supports design, diagnostics, and treatment planning in hospitals, research labs, and manufacturing of prosthetics and implants. This chapter focuses on applying mensuration concepts to solve practical biomedical problems using plane geometry.
Mensuration is the measurement of geometric quantities: lengths, perimeters, areas and volumes. A biomedical engineering technician uses it to size electrode pads and dressings, to find the capacity of reagent reservoirs and fluid containers, to choose tubing and cable lengths, and to check that a replacement part will fit before it is ordered. Correct results depend on two habits: choosing the right formula for the shape, and keeping every quantity in consistent units.
Almost every engineering task starts with a measurement that must be turned into a number that can be used: how much material, how much fluid, how much surface. Mensuration gives the formulas that link measured dimensions to area and volume. Errors in these calculations lead to parts that do not fit, containers that overflow or doses that are wrong, so the working is always written out and the units are stated.
Typical applications include:
Example 1: A circular wound has a diameter of 8 cm. Find the area of dressing needed to cover it exactly.
Radius = 8 ÷ 2 = 4 cm
Area = π × r² = 3.1416 × 4²
= 3.1416 × 16
Answer: 50.27 cm²
Example 2: A cylindrical reagent reservoir has a radius of 4 cm and a height of 10 cm. Find its volume in cm³ and in litres.
Volume = π × r² × h = 3.1416 × 4² × 10
= 3.1416 × 16 × 10
= 502.65 cm³
Since 1 000 cm³ = 1 litre, 502.65 cm³ = 0.503 litres.
Answer: 502.65 cm³, which is 0.503 litres.
Example 3: A catheter must reach 30 cm inside a vein and protrude 5 cm outside the skin. Find the total length needed.
Total length = 30 + 5
Answer: 35 cm
Example 4: A workshop floor measures 6 m by 4.5 m and is to be covered with square tiles of side 0.5 m. How many tiles are needed before any allowance for cutting?
Floor area = 6 × 4.5 = 27 m²
Area of one tile = 0.5 × 0.5 = 0.25 m²
Number of tiles = 27 ÷ 0.25 = 108
Answer: 108 tiles
A measurement is a number together with a unit, and the unit must be stated every time. Length is measured in units of length, area in units of length squared, and volume in units of length cubed. Capacity, the volume of a fluid, is often given in litres. Two systems of units are met in practice: the metric system (SI), which is used for the answers in these notes, and the imperial system, which still appears on some imported equipment, pipe fittings and manuals.
The SI base unit of length is the metre (m). Area is measured in square metres (m²) and volume in cubic metres (m³). The litre (L) is accepted for use with SI units for capacity, where 1 L = 1 000 cm³ = 0.001 m³ and 1 mL = 1 cm³. Other units are made from the base unit by adding a prefix that stands for a power of ten.
| Prefix | Symbol | Factor | Example |
|---|---|---|---|
| mega | M | 1 000 000 (10⁶) | 1 MΩ = 1 000 000 Ω |
| kilo | k | 1 000 (10³) | 1 km = 1 000 m |
| centi | c | 0.01 (10⁻²) | 1 cm = 0.01 m |
| milli | m | 0.001 (10⁻³) | 1 mm = 0.001 m |
| micro | µ | 0.000 001 (10⁻⁶) | 1 µm = 0.000 001 m |
Because area and volume involve two and three lengths, their conversion factors are the length factor squared and cubed. Since 1 m = 100 cm, 1 m² = 100 × 100 = 10 000 cm², and 1 m³ = 100 × 100 × 100 = 1 000 000 cm³.
Example 1: Convert 250 mm to metres.
1 mm = 0.001 m
250 × 0.001 = 0.25
Answer: 0.25 m
Example 2: Convert 1 500 cm² to m².
1 m² = 10 000 cm²
1 500 ÷ 10 000 = 0.15
Answer: 0.15 m²
The imperial system uses the inch, foot, yard and mile for length. Each has an exact metric value, so a measurement in imperial units can always be converted to metric without loss of accuracy.
| Quantity | Imperial unit | Metric equivalent |
|---|---|---|
| Length | 1 inch (in) | 25.4 mm |
| Length | 1 foot (ft) = 12 in | 0.3048 m |
| Length | 1 yard (yd) = 3 ft | 0.9144 m |
| Length | 1 mile = 1 760 yd | 1.609344 km |
| Area | 1 square inch (in²) | 645.16 mm² |
| Area | 1 square foot (ft²) | 0.092903 m² |
| Volume | 1 cubic inch (in³) | 16.387 cm³ |
| Capacity | 1 UK gallon | 4.54609 L |
| Capacity | 1 US gallon | 3.78541 L |
The UK gallon and the US gallon are different sizes, so a gallon figure on a manual must be checked for which one is meant. Pipe and fitting sizes are often quoted in inches, for example a half-inch pipe, and the inch is widely met on imported fittings and tools.
Example 1: A fitting is marked 3.5 in long. Give its length in millimetres.
Length = 3.5 × 25.4
Answer: 88.9 mm
Example 2: A pipe run is 12 ft long. Give its length in metres.
Length = 12 × 0.3048
Answer: 3.6576 m, or 3.66 m to 3 significant figures.
To convert a measurement from one unit to another, multiply or divide by the conversion factor so that the unit that is not wanted cancels. For area and volume the factor is squared or cubed. All the quantities in one calculation are first brought into the same unit, and the answer is then converted if needed.
Steps for converting units:
Example 1: Convert 5 litres to cubic centimetres.
1 L = 1 000 cm³
5 × 1 000 = 5 000
Answer: 5 000 cm³
Example 2: A rectangular steel plate is 30 in by 18 in. Find its area in m².
30 in = 30 × 0.0254 = 0.762 m
18 in = 18 × 0.0254 = 0.4572 m
Area = 0.762 × 0.4572 = 0.34839 m²
Check using area units: 30 × 18 = 540 in², and 540 × 0.00064516 = 0.34839 m².
Answer: 0.348 m² (to 3 significant figures).
Example 3: A water tank holds 20 litres. Express the capacity in UK gallons and in m³.
In UK gallons: 20 ÷ 4.54609 = 4.399
In m³: 20 ÷ 1 000 = 0.02
Answer: 4.40 UK gallons (to 3 significant figures), or 0.02 m³.
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Create a free accountThis chapter introduced the fundamental concepts of mensuration, emphasizing its importance in engineering and practical applications, and covered the different units of measurement including SI and Imperial units along with their conversions. It then explored plane geometry by detailing how to calculate the area and perimeter of basic shapes such as squares, rectangles, parallelograms, triangles, trapezoids, circles, and ellipses, as well as techniques for handling composite figures through decomposition. The study of solid geometry followed, focusing on the surface area and volume of various solids including prisms, cylinders, pyramids, cones, spheres, and spheroids, with methods to analyze composite solids by breaking them down into simpler components. The chapter also addressed the calculations related to frustums of cones and pyramids, covering both surface area and volume. Finally, it explained the mensuration of sectors and segments of a circle, including how to determine the area of a sector, the length of an arc, and the area of a segment, highlighting their engineering relevance. Throughout, the chapter provided a comprehensive foundation for solving mensuration problems in an engineering context.
Calculate the area and perimeter of a rectangular biomedical sensor platform measuring 12 cm by 8 cm. (2 marks)
A circular catheter has an external diameter of 10 mm. Calculate its circumference and cross-sectional area. (3 marks)
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