By the end of this chapter, you will be able to:
These skills will help you solve real-world problems confidently and ensure your work meets professional standards in your trade.
Carry out mensuration is fundamental in various professional fields across Kenya, from measuring land areas in agriculture to calculating volumes in construction and retail stock management. Accurate mensuration ensures correct resource allocation, cost estimation, and operational efficiency in organizations such as county hospitals, banks, and cooperatives. This chapter develops the ability to measure and calculate physical quantities using appropriate units and symbols, tailored for diverse professional applications.
Measurement units and their symbols are the universal language that professionals use to quantify physical quantities. Kenya, like many countries, primarily uses the International System of Units (SI), which standardizes measurements and facilitates clear communication across sectors. Understanding units of mass, distance, speed, temperature, and time is crucial for precise mensuration in all industries including healthcare, agriculture, hospitality, and public administration.
Mass is the quantity of matter contained in an object, independent of gravity, and is fundamental for inventory, dosing in healthcare, and material handling in retail and agriculture. The SI unit for mass is the kilogram (kg), with grams (g) and tonnes (t) as common subunits.
The basic formula relating mass, density, and volume is:
$$ m = \rho \times V $$
where \(m\) is mass in kilograms, \(\rho\) is density in kilograms per cubic metre (kg/m³), and \(V\) is volume in cubic metres (m³).
Example 1: A cooperative has a grain silo holding maize with a density of 720 kg/m³. If the silo volume is 5 m³, find the mass of maize stored.
Given: \(\rho = 720 \, \text{kg/m}^3\), \(V = 5 \, \text{m}^3\)
$$ m = \rho \times V $$$$ m = 720 \times 5 $$$$ m = 3600 \, \text{kg} $$ Answer: 3600 kg
Example 2: A hospital receives a shipment of medical supplies weighing 2.5 tonnes. Convert this mass to kilograms.
Given: \(m = 2.5 \, \text{t}\)
$$ 1 \, \text{t} = 1000 \, \text{kg} $$$$ m = 2.5 \times 1000 $$$$ m = 2500 \, \text{kg} $$ Answer: 2500 kg
Example 3: A retail business sells sugar in 500 g packets. How many kilograms are in 120 packets?
Given: packet mass = 500 g, number of packets = 120
Convert grams to kilograms:
$$ 500 \, \text{g} = \frac{500}{1000} = 0.5 \, \text{kg} $$
Total mass:
$$ m = 0.5 \times 120 $$$$ m = 60 \, \text{kg} $$ Answer: 60 kg
Example 4: A farm produces 15 tonnes of potatoes. If the density of potatoes is 700 kg/m³, find the volume occupied by the potatoes.
Given: \(m = 15 \, \text{t} = 15000 \, \text{kg}\), \(\rho = 700 \, \text{kg/m}^3\)
Rearranged formula for volume:
$$ V = \frac{m}{\rho} $$$$ V = \frac{15000}{700} $$$$ V \approx 21.43 \, \text{m}^3 $$ Answer: 21.43 m³
Example 5: A SACCO receives a metal container with a volume of 0.2 m³ and density of 7800 kg/m³. Calculate the mass of the container.
Given: \(V = 0.2 \, \text{m}^3\), \(\rho = 7800 \, \text{kg/m}^3\)
$$ m = \rho \times V $$$$ m = 7800 \times 0.2 $$$$ m = 1560 \, \text{kg} $$ Answer: 1560 kg
Distance measures the length between two points, essential in construction, land surveying, and transport logistics in Kenya’s counties and businesses. The SI unit for distance is the metre (m), with centimetres (cm) and kilometres (km) as common subunits.
The formula for distance when speed and time are known is:
$$ d = s \times t $$
where \(d\) is distance in metres or kilometres, \(s\) is speed, and \(t\) is time.
Example 1: A delivery truck from a retail business travels at 60 km/h for 2 hours. Calculate the distance covered.
Given: \(s = 60 \, \text{km/h}\), \(t = 2 \, \text{h}\)
$$ d = s \times t $$$$ d = 60 \times 2 $$$$ d = 120 \, \text{km} $$ Answer: 120 km
Example 2: A county official walks 5000 m to a meeting. Convert this distance to kilometres.
Given: \(d = 5000 \, \text{m}\)
$$ 1 \, \text{km} = 1000 \, \text{m} $$$$ d = \frac{5000}{1000} $$$$ d = 5 \, \text{km} $$ Answer: 5 km
Example 3: A university campus has a rectangular field 150 m long and 80 m wide. Find the perimeter of the field.
Given: length \(l = 150 \, \text{m}\), width \(w = 80 \, \text{m}\)
$$ P = 2(l + w) $$$$ P = 2(150 + 80) $$$$ P = 2(230) $$$$ P = 460 \, \text{m} $$ Answer: 460 m
Example 4: A cooperative member walks 3 km east, then 4 km north. Calculate the straight-line distance from the starting point to the final position.
Given: \(d_1 = 3 \, \text{km}\), \(d_2 = 4 \, \text{km}\)
Using Pythagoras theorem:
$$ d = \sqrt{d_1^2 + d_2^2} $$$$ d = \sqrt{3^2 + 4^2} $$$$ d = \sqrt{9 + 16} $$$$ d = \sqrt{25} $$$$ d = 5 \, \text{km} $$ Answer: 5 km
Example 5: A retail delivery vehicle travels 150 km at an average speed of 75 km/h. Find the time taken.
Given: \(d = 150 \, \text{km}\), \(s = 75 \, \text{km/h}\)
Rearranged formula for time:
$$ t = \frac{d}{s} $$$$ t = \frac{150}{75} $$$$ t = 2 \, \text{h} $$ Answer: 2 hours
Speed is the rate of change of distance with respect to time, vital for transport scheduling, logistics, and production planning in Kenyan sectors like county governments and retail. The SI unit is metres per second (m/s), but kilometres per hour (km/h) is more common for everyday use.
Speed is calculated as:
$$ s = \frac{d}{t} $$
where \(s\) is speed, \(d\) is distance, and \(t\) is time.
Example 1: A delivery bicycle covers 10 km in 30 minutes. Find the average speed in km/h.
Given: \(d = 10 \, \text{km}\), \(t = 30 \, \text{minutes} = 0.5 \, \text{h}\)
$$ s = \frac{d}{t} $$$$ s = \frac{10}{0.5} $$$$ s = 20 \, \text{km/h} $$ Answer: 20 km/h
Example 2: A farmer walks 2 km in 25 minutes. Calculate the speed in m/s.
Given: \(d = 2 \, \text{km} = 2000 \, \text{m}\), \(t = 25 \, \text{minutes} = 1500 \, \text{s}\)
$$ s = \frac{d}{t} $$$$ s = \frac{2000}{1500} $$$$ s = 1.33 \, \text{m/s} $$ Answer: 1.33 m/s
Example 3: A county government vehicle travels 180 km in 4 hours. Calculate average speed in km/h.
Given: \(d = 180 \, \text{km}\), \(t = 4 \, \text{h}\)
$$ s = \frac{d}{t} $$$$ s = \frac{180}{4} $$$$ s = 45 \, \text{km/h} $$ Answer: 45 km/h
Example 4: A taxi covers 15 km in 20 minutes. Find the speed in m/s.
Given: \(d = 15 \, \text{km} = 15000 \, \text{m}\), \(t = 20 \, \text{minutes} = 1200 \, \text{s}\)
$$ s = \frac{d}{t} $$$$ s = \frac{15000}{1200} $$$$ s = 12.5 \, \text{m/s} $$ Answer: 12.5 m/s
Example 5: A runner completes a 5 km race in 18 minutes. Calculate the speed in km/h.
Given: \(d = 5 \, \text{km}\), \(t = 18 \, \text{minutes} = 0.3 \, \text{h}\)
$$ s = \frac{d}{t} $$$$ s = \frac{5}{0.3} $$$$ s \approx 16.67 \, \text{km/h} $$ Answer: 16.67 km/h
Temperature measures the degree of hotness or coldness of a substance, critical in healthcare, food storage, and environmental monitoring in Kenya. The SI unit is the kelvin (K), but degrees Celsius (°C) is most commonly used.
Temperature conversions between Celsius and Fahrenheit use:
$$ F = \frac{9}{5}C + 32 $$
$$ C = \frac{5}{9}(F - 32) $$
where \(C\) is temperature in degrees Celsius and \(F\) in degrees Fahrenheit.
Example 1: Convert 25°C to Fahrenheit.
Given: \(C = 25\)
$$ F = \frac{9}{5} \times 25 + 32 $$$$ F = 45 + 32 $$$$ F = 77^\circ \text{F} $$ Answer: 77°F
Example 2: Convert 86°F to Celsius.
Given: \(F = 86\)
$$ C = \frac{5}{9} (86 - 32) $$$$ C = \frac{5}{9} \times 54 $$$$ C = 30^\circ \text{C} $$ Answer: 30°C
Example 3: A pharmacy stores vaccines at 4°C. Convert this temperature to Fahrenheit.
Given: \(C = 4\)
$$ F = \frac{9}{5} \times 4 + 32 $$$$ F = 7.2 + 32 $$$$ F = 39.2^\circ \text{F} $$ Answer: 39.2°F
Example 4: A weather report shows 95°F. Convert to Celsius.
Given: \(F = 95\)
$$ C = \frac{5}{9}(95 - 32) $$$$ C = \frac{5}{9} \times 63 $$$$ C = 35^\circ \text{C} $$ Answer: 35°C
Example 5: A cold storage unit is set to -10°C. Find the equivalent Fahrenheit temperature.
Given: \(C = -10\)
$$ F = \frac{9}{5} \times (-10) + 32 $$$$ F = -18 + 32 $$$$ F = 14^\circ \text{F} $$ Answer: 14°F
Time is the continuous progression of existence and events, essential for scheduling, production, and monitoring activities in all Kenyan professional fields. The SI unit of time is the second (s), with minutes (min), hours (h), and days (d) commonly used.
Time calculations often involve conversions and arithmetic operations.
Example 1: Convert 3 hours 45 minutes to minutes.
Given: \(3 \, \text{h} = 3 \times 60 = 180 \, \text{min}\), plus 45 minutes
$$ t = 180 + 45 = 225 \, \text{min} $$ Answer: 225 minutes
Example 2: Convert 150 minutes to hours and minutes.
Given: \(t = 150 \, \text{min}\)
Divide by 60:
$$ \text{hours} = \frac{150}{60} = 2.5 \, \text{h} $$
Separate integer and decimal:
$$ 2 \, \text{h} + 0.5 \times 60 = 2 \, \text{h} + 30 \, \text{min} $$ Answer: 2 hours 30 minutes
Example 3: A bank opens at 8:30 a.m. and closes at 5:15 p.m. Calculate the number of hours it operates.
Convert times to minutes past midnight:
8:30 a.m. = 8 × 60 + 30 = 510 min
5:15 p.m. = 17 × 60 + 15 = 1035 min
Calculate duration:
$$ t = 1035 - 510 = 525 \, \text{min} $$
Convert to hours and minutes:
$$ \text{hours} = \frac{525}{60} = 8.75 \, \text{h} $$
$$ 8 \, \text{h} + 0.75 \times 60 = 8 \, \text{h} + 45 \, \text{min} $$ Answer: 8 hours 45 minutes
Example 4: Convert 2.75 hours to hours and minutes.
Given: \(t = 2.75 \, \text{h}\)
Separate integer and decimal:
$$ 2 \, \text{h} + 0.75 \times 60 = 2 \, \text{h} + 45 \, \text{min} $$ Answer: 2 hours 45 minutes
Example 5: A meeting starts at 9:15 a.m. and lasts for 2 hours 30 minutes. At what time does it end?
Start time in minutes:
$$ 9 \times 60 + 15 = 555 \, \text{min} $$
Duration:
$$ 2 \times 60 + 30 = 150 \, \text{min} $$
End time:
$$ 555 + 150 = 705 \, \text{min} $$
Convert back to hours and minutes:
$$ \frac{705}{60} = 11.75 \, \text{h} $$
$$ 11 \, \text{h} + 0.75 \times 60 = 11 \, \text{h} + 45 \, \text{min} $$ Answer: 11:45 a.m.
Create a free account to open more of this chapter.
Free: practical guides, quick cards, workplace scenarios and more.
Create a free accountThis chapter covered the fundamental units and symbols used in measurement, focusing on mass, distance, speed, temperature, and time as essential quantities in applied mathematics. It distinguished between imperial and metric systems, emphasizing the importance of accurate conversions between these units for practical applications. The concept of perimeter was explored with calculations for regular shapes, providing the basis for understanding boundary measurements. Area calculations for regular geometric figures were then examined, highlighting methods to determine surface extents in two dimensions. The chapter concluded with volume measurement techniques for regular solids, enabling the computation of three-dimensional space. Throughout, emphasis was placed on applying formulas and conversions accurately to solve mensuration problems in civil engineering contexts. This foundation prepares students to handle various measurement tasks essential for design, construction, and quality control.
At the start of this chapter we promised you would be able to:
Tick each one you can genuinely do.
So, are you there yet?
You're competent when you can confidently do 50% or more of what this chapter promised.
Sign in to record how you're doing.