Thermometer with Fahrenheit and Celsius unitsTo get an approximate conversion between the two scales, use these formulas:
- Fahrenheit to Celsius: Subtract 32 and halve the resulting number.
- Celsius to Fahrenheit: Double the number and add 32.
For an accurate conversion, use the following formulas:
- Fahrenheit to Celsius: (°F − 32) ÷ 1.8 = °C
- Celsius to Fahrenheit: (°C × 1.8) + 32 = °F
Imperial system of measurement
Canada is not totally metric. When we shop, the meat and vegetable prices are given in per pound as well as per kilogram. We still talk about how tall we are in feet and inches. Recipes still use cups and teaspoons.
The common imperial units of measurement are:
Quantity | Unit | Symbol |
length | foot | ft. |
volume | gallon | gal. |
weight | pound | lb. |
The following table contains some of the most common imperial conversions.
Length | Liquid Measure | Weight |
1 foot (ft.) = 12 inches (in.) | 1 pint (pt.) = 2 cups | 1 pound (lb.) = 16 ounces (oz.) |
1 yard (yd.) = 3 feet | 1 quart (qt.) = 2 pints | 1 ton = 2000 lb. |
1 mile (mi.) = 5280 feet | 1 gallon (gal.) = 4 quarts | |
When converting between units in the imperial system, use the familiar rule:
- When converting a larger unit to a smaller unit, multiply by the conversion factor.
- When converting a smaller unit to a larger unit, divide by the conversion factor.
Example 1
A two-by-four measures 5 feet 6 inches. How long is it in inches?
Solution
Since 1 foot = 12 inches, the conversion factor is 12, and since you are converting a larger unit to a smaller unit, multiply by 12.
5 feet = 5 × 12 = 60 inches
Now add the 6 inches to the 60 inches.
5 feet 6 inches = 60 + 6 = 66 inches or 5'6" = 66"
Note that "quotes" are sometimes used to represent feet and inches. For example, 8' means 8 ft., and 12" means 12 in..
Example 1
A block wall is 16 rows high and each block is 8" high. How high is the wall in feet and inches?
Solution
Each block is 8" times 16 levels, so multiply 8" times 16 to get the height in inches:
8" × 16 = 128"
Divide the inches by the conversion factor of 12 to get the number of feet.
128" ÷ 12 = 10.667'
Then convert the decimal portion of the foot measurement back into inches. This can be done a couple of ways:
or
Therefore, the total height of the wall is 10' 8".
Metric and imperial conversions
Most shops have conversion charts that give the conversions between the metric and imperial systems most useful to your trade. Specification guides will usually give you both metric and imperial measurements if appropriate.
Your calculator may also have the capacity to convert between the two systems. Read your calculator's instructions carefully; ask your instructor if you need help in figuring out how to use this feature.
U.S. measurements of capacity (pints, quarts, gallons) are different from imperial measurements of capacity. For example, a U.S. gallon is 32 ounces, or 3.785 L, while an imperial gallon is 40 ounces, or 4.546 L. There are some differences between U.S. and imperial measurements of mass, as well. Be sure that your conversion calculator or conversion chart uses the same measurement system you are working with.
The chart below gives the equivalents between commonly used metric and imperial measurements. The conversions that are not exact have been rounded off to the nearest thousandth.
Metric/U.S. imperial conversion factors
| Metric to imperial | Imperial to metric |
Length | 1 km = 0.621 mi. 1 m = 1.094 yd. 1 m = 3.28 ft. 1 cm = 0.394 in. 1 mm = 0.039 in. | 1 mi. = 1.61 km 1 yd. = 0.914 m 1 ft. = 0.305 m 1 in. = 2.54 cm 1 in. = 25.4 mm |
Mass | 1 kg = 2.20 lb. 1 g = 0.035 oz. 1 t (1000 kg) = 2204.6 lb. | 1 oz. = 28.4 g 1 lb. = 0.454 kg |
Capacity | 1 L = 2.11 pt. 1 L = 1.06 qt. | 1 pt. = 0.47 L 1 qt. = 0.95 L 1 U.S. gal. = 3.79 L 1 Imp. gal. = 4.55 L |
Examples of converting between metric and imperial measurements
It is not necessary to memorize the chart. You will be able to refer to a conversion chart for any calculations.
When you are using an approximate conversion factor, round your answers to the precision of your approximate factor, if necessary.
Converting units of length
Convert 38 mm to inches.
- Multiply mm by the metric to imperial conversion factor for millimetres to inches.
Convert 123 miles to kilometres.
- Multiply by the imperial to metric conversion factor for miles to kilometres.
Converting units of mass
Convert 12 pounds to kilograms.
- Multiply the imperial to metric conversion factor for pounds to kilograms.
Convert 15 grams to ounces.
- Multiply by the metric to imperial conversion factor for grams to ounces.
Converting units of capacity
Convert 8 U.S. gallons to litres.
- Multiply by the imperial to metric conversion factor for gallons to litres.
Convert 250 mL to pints.
- Convert mL to L.
250 mL = 0.25 L
- Multiply by the metric to imperial conversion factor for pints to litres.
Now complete the Learning Task Self-Test.
Self-Test 2
- Perform the following metric conversions.
- 2.36 m = mm
- 15 000 m = km
- 704 mm = m
- 1.2 m = mm
- 0.53 kg = g
- 25 mg = g
- 3 kg = t
- 0.01 L = mL
- 175 mL = cL
- 23.1 L = mL
- Using the chart below, perform the following metric/imperial conversions. Round your answers to the nearest thousandth if necessary.
- 24 mm = in.
- 18 in. = cm
- 30 kg = lb.
- 10 oz. = g
- 100 lb. = kg
- 10 qt. = L
- 40 km = mi.
- 59.7 L = qt.
- 3 gal. = L
- 5 t = lb.
| Metric to imperial (U.S.) | Imperial (U.S.) to metric |
Length | 1 km = 0.621 mi. 1 m = 1.094 yd. 1 m = 3.28 ft. 1 cm = 0.394 in. 1 mm = 0.039 in. | 1 mi. = 1.61 km 1 yd. = 0.914 m 1 ft. = 0.305 m 1 in. = 2.54 cm 1 in. = 25.4 mm |
Mass | 1 kg = 2.20 lb. 1 g = 0.035 oz. 1 t (1000 kg) = 2204.6 lb. | 1 oz. = 28.4 g 1 lb.= 0.454 kg |
Capacity | 1 L = 2.11 pt. 1 L = 1.06 qt. | 1 pt. = 0.47 L 1 qt. = 0.95 L 1 U.S. gal. = 3.79 L 1 Imp. gal. = 4.55 L |
- Perform the following metric conversions.
- 26.8 m = mm
- 6 mg = g
- 5 mL = cm3
- 0.68 kg = g
- 153 μm = mm
- 35.6 km = cm
- 0.015 Mg = kg
- 7 L = dL
- 1.3 cg = kg
- 3.29 × 102 m = m
- Using the chart below, perform the following metric/imperial conversions. Round your answers to the nearest thousandth, if necessary.
- 12 in. = cm
- 20 gal. = L.
- 420 g = oz.
- 3 ft. = cm
- 100 kg = lb.
- 100 m = ft.
- 8 mm = in.
- 82 mi. = km
- 1000 lb. = t
- 0.8 L = qt.
| Metric to imperial (U.S.) | Imperial (U.S.) to metric |
Length | 1 km = 0.621 mi. 1 m = 1.094 yd. 1 m = 3.28 ft. 1 cm = 0.394 in. 1 mm = 0.039 in. | 1 mi. = 1.61 km 1 yd. = 0.914 m 1 ft. = 0.305 m 1 in. = 2.54 cm 1 in. = 25.4 mm |
Mass | 1 kg = 2.20 lb. 1 g = 0.035 oz. 1 t (1000 kg) = 2204.6 lb. | 1 oz. = 28.4 g 1 lb.= 0.454 kg |
Capacity | 1 L = 2.11 pt. 1 L = 1.06 qt. | 1 pt. = 0.47 L 1 qt. = 0.95 L 1 U.S. gal. = 3.79 L 1 Imp. gal. = 4.55 L |
Learning Task 3
Solve problems involving ratio and proportion
Proportions are very useful in solving problems in everyday life. A proportion is made up of two equal fractions or ratios.
Many of the calculations you will have to do in your trade are likely to be of the kind you can solve by using ratio and proportion. For example, prices and quantities of materials, percentages of figures, forces required to move objects, and number of revolutions per minute can often be quickly calculated or estimated using ratio and proportion. By the same means, plans can be rescaled or gear ratios verified.
Key terms
Cross multiplication: Multiplying the numerator of the first ratio by the denominator of the second ratio, and the denominator of the first ratio by the numerator of the second ratio. If the two products are equal, the proportion is true. For example:
This is a true proportion.
Direct proportion: When one ratio is increased or decreased, it causes another to increase or decrease by the same factor, so the two remain equal.
Inverse proportion: The relation between quantities, the product of which is a constant. When one quantity increases, the other decreases by the same factor, and when one quantity decreases, the other increases by the same factor. For example, as speed increases, the time required to arrive at a destination decreases proportionately. Problems in this Competency will involve only direct proportions.
Inverse ratio: A ratio expressed with the terms inverted. For example, the inverse of the ratio 1:2 is 2:1.
Proportion: A statement of equal relationship between two ratios. The ratios can be expressed in one of three ways:
1 is to 2 as 8 is to 16
1:2 = 8:16
1/2 = 8/16
Ratio: A comparison between two quantities, or a way of stating the relationship between them using division. Ratio is expressed in one of four ways:
1 to 2
1:2
1÷2
½
The two numbers are known as the terms of the ratio.
Working with ratios
To write a ratio you must identify the quantities to be compared. For example, in a concrete mixture made with 10 parts sand and 2 parts cement, the relationship between the sand and the cement can be expressed as the ratio 10:2.
Like fractions, ratios are normally reduced to their lowest terms. By dividing the terms of the sand and cement ratio by 2, you can express the ratio in lower terms, 5 to 1 (5:1 or 5÷1 or 5/1). In this way, 10:2 and 5:1 are said to be equivalent ratios.
In ratios it is important to be clear about which quantity is being compared to which. In the example above, the ratio of sand to cement is 5:1 (5/1) and the ratio of cement to sand is 1:5 (1/5). It is often helpful to set up the ratio in words before you express it in numbers.
Example
If a shaft turns 14 times per minute, what is the ratio of minutes to turns?
- Write a fractional ratio, using the word that precedes to or per in the question part of the problem as the fraction's numerator, and the word that follows to or per as the denominator. Note that minutes comes before to in the question part of the problem, even though it comes after per earlier in the problem.
- Substitute the quantities that correspond to the words:
Working with proportions
A ratio by itself is not especially useful in solving practical problems. However, two ratios that are equivalent can be written as a proportion. That proportion can then be used to solve certain kinds of problems.
Example
Two meshing gears are supposed to turn in the ratio of 1:20. You check the system while it is operating and find that the larger gear is rotating at 325 rpm and the smaller gear is rotating at 6500 rpm. Is the equipment operating correctly?
Solution
- Express the fractional ratio in words as in the example above.
- Write the given information as a fractional proportion, making sure that the terms are located to correspond to the relationship you showed in the previous step.
- Cross multiply to see if this is a true proportion.
(If it is a true proportion both products will be the same, indicating that the equipment is operating correctly.) - Both products are the same. Therefore, the equipment is operating correctly.
Solving word problems with ratio and proportion
You can use ratio and proportion to find an unknown quantity when the other three quantities are known.
Example
The ratio of the teeth on a driver gear to the teeth on a driven gear is 2:5. If the driver gear has 24 teeth, how many teeth does the driven gear have?
Solution
- Express the fractional ratio in words.
- Write the given information as a fractional proportion, making sure you use the correct relationships. Use the symbol N (or x) to identify the unknown quantity.
- Cross multiply.
- Divide both sides by the number that appears with N.
2N = 120
- In this case, divide both sides of the equation by 2.
- The driven gear has 60 teeth (N = 60).
N = 60
Similar triangles
Two triangles are said to be similar if their corresponding (or matching) angles are equal and the lengths of the corresponding sides are proportional. If we view triangles as having short, middle, and long side lengths, then corresponding sides are the same pair of sides on the respective triangles.
6 corresponds to 3 (short sides)
8 corresponds to 4 (middle sides)
10 corresponds to 5 (long sides)
4 corresponds to 3 (short sides)
8 corresponds to 6 (middle sides)
12 corresponds to 9 (long sides)
Example 1
The triangles are similar. Write a proportion and find x.
Solution
Create a ratio from the triangle containing the unknown side, x, such as,
Carefully note that this ratio, , compares the short side to the long side in the first triangle.
Set this ratio equal to the ratio of the other triangle. Then solve the proportion.
Example 2
Suppose you want to find the height of a tall tree that's too tall to measure directly. You can do this by comparing the length of its shadow cast by the sun to the length of the shadow of a known object.
Since, at any given time, the angle of the sun is the same for all objects in the immediate area, the triangles formed by objects and their shadows are similar.
Solution
In this case, the triangle formed by the tree and its shadow is similar to the triangle formed by the woman and her shadow. The height of the tree is represented by h.
To find h:
2.4 × h = 36 × 1.6
2.4h = 57.6
h = 24
Therefore, the tree height must be 24 m.
Scale drawings
Builders, designers, map-makers, etc., use scale drawings to represent real things. Proportions are used to relate the scale drawing to the actual object.
Scale drawings have exactly the same shape as the real object, but they are not always the same size. For example, the drawing of an insect would be larger than the actual insect, and the drawing of a room on a blueprint would be smaller than the actual room that will be built.
When working with scale drawings, you will need to identify the scale used. The scale is the ratio of the drawing measurement to the actual measurement.
When you are using an approximate conversion factor, round your answers to the precision of your approximate factor, if necessary.
Example 1
In a drawing of an insect with a scale of 10:1, a measurement of 10 cm on the drawing represents an actual length of 1 cm.
Find the actual length of the leg of an insect when the leg on the drawing measures 2 cm.
Solution
Set up a proportion to solve the problem. Let n be the actual length of the leg.
10 × n = 2 × 1
10n = 2
n = 0.2
The actual length of the leg is 0.2 cm or 2 mm.
Example 2
The scale of 1:40 is used for a drawing of a fridge. Find the actual width and height of the fridge.
Solution
On the drawing, the height of the fridge is 2.5 cm.
Let L be the actual height of the fridge.
L × 1 = 2.5 × 40
L = 100 cm
The actual height is 100 cm.
On the drawing, the width of the fridge is 2 cm.
so,
1 × w = 2 × 40
w = 80
The actual width is 80 cm.
Now complete the Learning Task Self-Test.
Self-Test 3
- Express the following ratios in lowest terms:
- 15 to 20 =
- 9 to 3 =
- 9:63 =
- 6:1 =
- 2.4:0.8 =
- Write the inverse of the following ratios:
- 5 to 7 =
- 4/5 =
- Find the missing term in these proportions:
- 35:5 = :2
- :3 = 28:12
- Of two gears that mesh, the smaller revolves at 84 rpm and the larger at 36 rpm. In lowest terms, what is the ratio of the speed of the smaller to the larger?
- 3:7
- 7:3
- 5:2
- 2:5
- Two oil tanks have capacities of 1989 L and 1105 L. In lowest terms, what is the ratio of the volume of the larger tank to the smaller?
- 3:1
- 1:2
- 9:5
- 6:11
- The primary winding of an electric transformer has 240 turns, and the secondary winding has 15 turns. The ratio of the primary winding to the secondary is 1:16.
- True
- False
Use proportion to solve the following problems:
- A cylindrical barrel contains 9 kL of gasoline when filled to a depth of 3 m. How many kL does the barrel contain when the level is 2 m?
- 6 kL
- 5 kL
- 4 kL
- 3 kL
- In the rear axle of a vehicle, the ratio of teeth on the ring gear to teeth on the pinion gear is 4 to 1. If there are 64 teeth on the ring gear, how many teeth are there on the pinion gear?
- 8
- 32
- 16
- 256
- The ratio of input power to output power of an amplifier is 1:3000. If the input power is 0.00806 W, the amplifier output power will be 24.18 W.
- True
- False
- George earns $18.50 an hour; his brother Mike earns $9.25 an hour. What is the ratio in lowest terms of George’s hourly wage to Mike’s?
- 1:2
- 2:1
- 5:10
- 10:5
- Basement windows in a house are 3000 mm wide by 900 mm high. What is the ratio in lowest terms of width to height?
- 3000:900
- 900:3000
- 3:10
- 10:3
- A starter gear has 12 teeth and a ring gear has 102 teeth. The ratio in lowest terms of the starter gear teeth to the ring gear teeth is 6:52.
- True
- False
- The ratio of the crankshaft pulley speed to the alternator pulley speed is 8:19. When the engine crankshaft speed is 736 rpm the speed of the alternator is 1748 rpm.
- True
- False
- A steel pipe has a uniform cross-section. The mass of 0.33 m of the pipe is 0.425 kg. What is the mass of a 3.6 m length of the pipe?
- 2.795 kg
- 4.636 kg
- 3.867 kg
- 5.385 kg
- You are told to construct a workbench for which the length is 4 1/2 times the width. If your bench is to be 1800 mm long, it must be made 450 mm wide.
- True
- False
- In a transformer, the turns ratio varies directly with the voltage ratio. In a power transformer in which the ratio of primary turns to secondary turns is 10:3, what is the secondary voltage when the primary voltage is 123 V?
- 41 V
- 36.9 V
- 410 V
- 1230 V
- In the following chart, A varies directly in proportion to B. Complete the chart.
- The slope of a roof is the ratio of unit rise to the unit of run. If the slope is 4:12 and the total run is 4.9 m, the total rise must be 1.633 m.
- True
- False
- In mixing concrete for a foundation slab, the ratio of gravel to cement is 7:2. If you estimate you need 12 m3 of gravel, how much cement powder will you need?
- 42 m3
- 3.8 m3
- 3.4 m3
- 4.2 m3
- A water pump discharges 50 L of water in 3.5 minutes. How long does it take to discharge 11 000 litres?
- 770 min.
- 63 min.
- 77 min.
- 550 min.
- If 600 kg of carbon steel alloy contains 4.80 kg of manganese, there would be 6.92 kg of manganese in 865 kg of the same steel alloy.
- True
- False
- The capacity of an oil reservoir is directly proportional to the depth of oil. If the reservoir holds 280 000 L of oil when filled to its maximum depth of 3.8 m, how much oil is there when the depth is 1.2 m?
- 23 333 L
- 37 255 L
- 66 398 L
- 88 421 L
- A 406 mm shaft has a uniform taper of 0.05 mm in 101.5 mm. What is the taper in the entire length of the shaft?
- 0.2 mm
- 12 mm
- 33 864 mm
- 824 810 mm
- The power of a gas engine increases with the area of the piston head. If an engine with a piston area of 8.3 in.2 develops 25.75 horsepower, to the nearest horsepower, how much horsepower will an engine have with a piston area of 25 in.2?
- 8 hp
- 77 hp
- 78 hp
- 92 hp
- If 3 L of paint covers a surface containing 82.5 m2, how much paint will be needed to cover 192.5 m2 of surface?
- 7 L
- 12 L
- 27.5 L
- 5293.75 L
Learning Task 4
Solve problems involving percent
Percent, or the symbol %, is very common in everyday life. For example, sales may be advertised as 25% or 50% off; sales tax is 7%; loans can be made with interest charges of anywhere from 1% to 20%. The word percent means per hundred. The word cent comes from the Latin word centum, which means hundred.
Applications for percentage in the trades include measuring efficiency of equipment, calculating proportions of mixtures, and calculating tolerances. Knowing how to calculate discounts and increases is important for equipment and material purchase, and essential if you plan to run your own business.
Percents (or percentages as they are more correctly called) are closely related to fractions and decimals.
7% means 7 per hundred
means 7 per hundred
0.07 means 7 per hundred
so
To change from percent notation to fraction or decimal notation, simply replace the percent sign with or 0.01.
Converting between decimals and percents
The symbol % does the work of two decimal places. Thus, 59% is equivalent to 0.59 and 0.172 is equivalent to 17.2%.
The "D - P Rule" is a simple way to remember which way to move the decimal point:
D comes before P in the alphabet, so it is always written first.
Note the use of zeros as placeholders in the above examples.
Examples of converting between decimals and percents
Example 1
Convert 7.25%, 136%, and 16¾% to decimals.
Solutions
- 7.25% = 7.25 × 0.01 = 0.0725 (move the decimal two places to the left)
- 136% = 136 × 0.01 = 1.36 (move the decimal two places to the left)
- 16¾% = 16.75 × 0.01 = 0.1675 (convert the fraction ¾ to the decimal 0.75, then move the decimal two places to the left)
Example 2
Convert 0.386, 2.45, and 0.008 to percent.
Solutions
- 0.386 = 0.386 × 100% = 38.6% (move the decimal two places to the right)
- 2.45 × 100% = 245% (move the decimal two places to the right)
- 0.008 = 0.008 × 100% = 0.8% (move the decimal two places to the right)
Converting between fractions and percents
Percent means per hundred. The percent symbol (%) following a number means that quantity out of a hundred. Since percent means per hundred, 100% of something means 100/100 or 1. Thus, 100% of 55 is 55. Seven percent (7%) means seven-hundredths, seven out of a hundred, or 7/100.
Percent to fractions
To convert a percentage to a fraction, you put the percentage over a denominator of 100 and reduce to lowest terms, if necessary. For example:
Not all conversions are this straightforward. For example:
Example 1
Convert 12%, , and 12.5% to fractions.
Solutions
Fractions to percent
Converting a fraction to a percent involves multiplying the numerator by 100 and dividing this product by the denominator.
Example 1
Convert to percents.
Solutions
Solve percent equations and word problems
You can use proportions to solve three types of percent problems.
The proportion can be used to find the:
- percent when the part and total are known
- part when the percent and total are known
- total when the part and percent are known
Total can also mean the whole amount or the final amount.
Any percent problem can be written using the following pattern:
____________ is ____________ % of ____________ .
The following examples demonstrate how to apply the two known pieces to solve any percent problem.
Example 1
If a mechanic has completed 6 out of the 30 hours required for a job, what percent of the job has been completed?
Solution
- Rephrase the question to fit the given pattern statement if necessary.
6 is what % of 30
- Substitute the following symbols in your pattern statement: "=" for "is"; "N" for "what"; "×" for "of."
6 = N% × 30
- Change the percent to a fraction with 100 as its denominator.
- Divide both sides of the equation by 30 to set up as a proportion. (See Learning Task 1 if you need to review this skill.)
- Solve as a proportion by multiplying each side by 100 to solve for N.
N = 20
- Insert your answer in the pattern statement.
6 is 20% of 30
Example 2
A certain alloy is 60% tin. How much tin is contained in 42 kilograms of this alloy?
Solution
- Rephrase to fit the given pattern statement.
what is 60% of 42
- Substitute the symbols as in the example above.
N = 60% × 42
- Change the percentage to a fraction.
- Divide both sides of the equation by 42 to set up as a proportion and solve.
- Insert your answer into the pattern statement.
25.2 is 60% of 42
Example 3
An article loses 17 mm in machining. This is 5% of its original size. What was the original size?
Solution
- Rephrase and substitute symbols.
17 = 5% × N
- Change the percentage to a fraction.
- Divide both sides of the equation by N to set up as a proportion and solve.
- Insert your answer into the pattern statement.
A simple way to remember the above procedure is to memorize this equation:
Once you have rephrased your questions to fit the pattern statement, put the number that follows of in the equation's OF location, the number that precedes the percent symbol in the % location, and the remaining number in the IS location.
Using a calculator's percent function to solve problems
The "%" button on your calculator can be used to find what percent one number is of another number. The following examples demonstrate the use of this function on your calculator.
Example 1
Out of 1757 motors, 7% were rejected by inspectors. How many motors were rejected?
Solution
- Rephrase to fit the pattern statement and substitute symbols.
N = 7% × 1757
- Enter the number that is to be multiplied by the known percentage.
1757
- Press "×" on the calculator, enter the known percentage and press "%."
× 7%
- Your answer will appear immediately.
123
Example 2
A certain ore yields 3.5% of iron. How many tons of ore are required to produce 150 tons of iron?
Solution
- Rephrase to fit the pattern statement and substitute symbols.
150 = 3.5% × N
- Divide both sides of the equation by 3.5%, and cancel to isolate the N.
- Enter 150, press "÷," enter 3.5, and press "%."
150 ÷ 3.5%
- Your answer will appear immediately.
4285.7142
Adding and subtracting percents
There are many cases where percents must be added or subtracted.
Example 1
The deductions to Frank's paycheque amounted to 36.2%. What percent of the paycheque did Frank take home?
Solution
The total paycheque must be represented as 100% of the paycheque.
100% − 36.2% = 63.8%
Example 2
Sulphuric acid by weight contains 2% hydrogen and 65% oxygen. The rest is sulphur. What percent of sulphuric acid is sulphur?
Solution
The total weight must be represented by 100%.
100% − (2% + 65%) = 100% − 67% = 33%
Common percent applications
Sales tax
Percent is widely used in sales tax computations. The provincial sales tax (PST) rate in British Columbia in 2015 is 7%. This means that the tax is 7% of the purchase price. Another sales tax, the goods and services tax (GST) is collected by the federal government and is 5% of the purchase price. The total price of an article is the price plus PST and GST.
Example
A refrigerator is priced at $699.95 plus PST and GST. How much is the PST, the GST, and the total price?
Solution
Find the PST:
Round to the nearest cent = $49.00
Now find the GST:
Round to the nearest cent = $35.00
The total price is $699.95 + $49.00 + $35.00 = $783.95
Another way to calculate the percentage is to look at it as multiplying your value by the decimal equivalent of the percentage, which can easily be calculated by moving the decimal two places to the left.
Simple interest
Whenever you open a savings account or buy a Canada Savings Bond, the bank or the government agrees to pay you interest for the use of your money. If you take out a loan with a bank, they expect you to pay them interest. The rate of the interest paid is always given
as a percent.
For example, if you buy a $100.00 Canada Savings Bond that pays 9% interest, then exactly one year later, the government will pay you $9.00 interest on your bond, because 9% of $100.00 = $9.00. Of course if you decide to cash in the bond before the year is out, then the interest will be less than $9.00.
To find simple interest, use the following formula:
I = Prt
where:
I = interest earned in dollars
P = principal amount that earns the interest in dollars
r = interest rate per year in percent
t = time in years
Example
Mary has a $500 GIC (guaranteed investment certificate) that earns interest at a rate of 8½% over three years. How much simple interest will it earn over three years?
Solution
Rewrite the percent number as a decimal:
8½% = 8.5% = 0.085
Write the simple interest formula using the information:
I = P × r × t
I = 500 × 0.085 × 3
Multiply and remember to write the answer in dollars and cents:
I = 127.5
The interest is $127.50.
Discounts and increases
The simplest way to figure out discounts and increases is to rephrase the problem before you begin. For example, if you are asked to calculate a 15% increase in cost, you would call the present cost 100%, add the 15% and call the new cost 115%. The following examples demonstrate how to calculate discounts and increases.
Example 1
If an article was priced at $189.50 and is now marked down 20%, what is its new price?
- Subtract the discount from 100%, rephrase to fit the pattern statement, and substitute symbols.
N = 80% × $189.50
- Solve.
$151.60
Example 2
Over 15 years, the cost of an item has increased 73%. The original cost was $58.47. What is the new cost?
- Add the increase to 100%, rephrase to fit the pattern statement and substitute symbols.
N = 173% × $58.47
- Solve.
$101.15
Here are some mental strategies for finding percentages:
- To find 50%, find half .
- To find 25%, find half and half again.
- To find 75%, find half and half again, and add them together .
- To find 10%, divide by 10
- To find 5%, find 10% and halve it.
- To find 20%, find 10% and double it.
Now complete the Learning Task Self-Test.
Self-Test 4
- Fill in the chart below with the required numbers. Provide all fractions in lowest terms.
| Fraction | Decimal | Percent |
a. | | 0.62 | |
b. | | | 8% |
c. | 2/9 | | |
d. | | 0.075 | |
e. | 14/5 | | |
f. | | | 95% |
- ________ is 6% of 63.
- 9 is ________% of 27.
- 84 is 375% of ________.
- 48 is 20% less than ________.
- 40% of 24 is what?
- 14.4
- 9.6
- 8.4
- 10.2
- What percent of 1.32 is 0.6?
- 46%
- 55%
- 45%
- 10%
- 3 is 15% of 22.
- True
- False
- 30 is 20% more than what?
- 25
- 24
- 22
- 20
- 102 is 15% less than what?
- 115
- 120
- 124
- 122.4
- 63 is what percent of 40, to the nearest percent?
- 63%
- 157%
- 158%
- 145%
- 12% of 30 is 3.6.
- True
- False
- 4.06 is 9% of 48.
- True
- False
- One part acid and four parts water are mixed as an electrolyte for a storage battery. What percent of the electrolyte is acid?
- 25%
- 20%
- 75%
- 80%
- The efficiency of a motor is 90%. If the motor delivers 13.5 horsepower, what is the input?
- 15 hp
- 14.85 hp
- 12.15 hp
- 15.28 hp
- A set of combination wrenches priced at $179.99 went on sale for 15% off. What is the sale price?
- $26.99
- $99.99
- $152.99
- $159.99
- A company has recently increased its workforce by 121/2%. There are now 18 employees in the company. How many workers were there before the increase?
- 12
- 14
- 16
- 18
- The operating spindle speed of a lathe spindle that rotates freely is 354 rpm. When 16% is lost through slippage and cutting pressure, the new speed is 297 rpm.
- True
- False
- A floor joist is allowed to vary 5% of its stated width of 184 mm. The maximum width would be what?
- 200 mm
- 188.4 mm
- 193.2 mm
- 197.9 mm
- What percent is wasted when 3.6 of every 120 sheets of metal are spoiled?
- 3%
- 0.3%
- 30%
- 300%
- The minimum idle speed for a particular engine is 630 rpm. This is 94% of the maximum idle speed of 668 rpm.
- True
- False
- A contractor wants to make a profit of 12% on a job. If the break-even cost is $2345, what should the total job price be, including the profit?
- $2626.00
- $2600.40
- $2600.00
- $2626.40
Learning Task 5
Solve problems involving powers, roots, and scientific notation
Key terms
Base: In an expression of a number to a power, the base is the number that is to be multiplied by itself. For example, in 83, the base is 8.
Cube: The result of a number being multiplied by itself three times. For example, the cube of 4 (4 cubed) is 43 = 4 × 4 × 4 = 64.
Cube root: The cube root of a number is the value that, when multiplied by itself three times, results in the number. The cube root of 125 is 5. Read as the "cube root of 125."
Exponent: A small number written above and to the right of a base. The exponent indicates how many times the base is to be multiplied by itself; also known as the degree of a power.
Index: The number written to the left and above the root sign; it indicates what root is to be taken. If the root sign has no index, it is understood that the square root is to be taken. The index 3 in indicates that the cube root of 8 is to be taken.
Power: The product that results from multiplying a base by itself the number of times indicated by the exponent. In the power 73, 7 is the base; 3 is the exponent. This power is read "seven to the third" or "seven cubed." It means 7 × 7 × 7.
Radical: An expression involving the root symbol .
Radicand: The number inside the root (or radical symbol). In the expression , 5 is called the radicand.
Root: The root of a number is the value that, when multiplied by itself the indicated number of times, results in the number.
Root sign: The symbol is the root sign.
Scientific notation: A shorter method of writing very large or very small numbers, using a power of 10 as a multiplier.
Square: The result of a number being multiplied by itself. For example, the square of 3 (3 squared, or 3 × 3) is 32 or 9.
Square root: The square root of a number is the value that, when multiplied by itself, results in the number. The square root of 9 is 3.
Important concepts
- Any number to the first power (having an exponent of 1) is equal to itself.
61 = 6 - Any number to the zero power (having an exponent of 0) equals one.
80 = 1 - Any number to a power between 0 and 1 will be less than the base.
160.5 = 4 - Any number to a power greater than 1 will be more than the base.
72 = 49
Powers
As you may recall, multiplying is a way of indicating how many times a number should be added. For example:
5 × 3 = 5 + 5 + 5
The power operation is a way of indicating how many times a number should be multiplied by the same number. For example:
53 = 5 × 5 × 5
The power operation is indicated by placing a small number, called an exponent, to the top right of some number, or base.
53 is called a power.
5 is called the base.
3 is called the exponent.
The power 53 is read "five to the third" or, more commonly, "five cubed." Powers with exponents of 2 are said to be "squared," while those with exponents of 3 are said to be "cubed." Below are some examples of how to read powers.
72 is read "seven squared."
73 is read "seven cubed."
94 is read "nine to the fourth."
128 is read "twelve to the eighth."
Notice that the exponent is not really a number—it is an instruction that says, "multiply the number below me by itself this many times." So 94 = 6561, not 36.
You can use exponents to find powers of integers, decimals, and fractions:
(–3)5 = (–3)(–3)(–3)(–3)(–3) = -243
(1.25)3 = (1.25)(1.25)(1.25) = 1.953125
You probably realize now why exponents are used; they save a lot of time writing down repeated multiplication.
Scientific calculators have a "raise to the power" key, usually labelled yx (sometimes xy or ax is used). To use this function, press the yx key in between the base number and the exponent, then press = .
Example 1
To find 35 = ? press: 3 5
Did you get 243?
Example 2
To find (1.25)3 = ? press: 1.25 3
Did you get 1.953125?
Example 3
To find (2.57)10 = ? press: 2.57 10
Did you get 12569.88294?
Depending on what kind of calculator you have, you may find that you cannot use a negative number as a base. For example, (–3)5 may result in an error message on the screen. If this is the case, you can raise the positive number to that power, and then make your answer positive or negative. Note that pairs of negative signs will cancel out to give positive answers:
(–3)(–3) = 9 (–3)(–3)(–3) = –27 (–3)(–3)(–3)(–3) = 81 so (–3)5 = –243
Something else to remember: when the base is negative, the negative sign is not part of the number that is raised to the power.
(–5)4 = (–5)(–5)(–5)(–5) = 625
–54 = –1 × 54 = –1 × 5 × 5 × 5 × 5 = –625
In the first example, the negative is part of the 5, but not part of the 4th power. Think of it as a negative one multiplied by 54, and remember that multiplying takes place after powers have been calculated.
Take a look at the following patterns:
24 = 2 × 2 × 2 × 2 = 16 104 = 10 × 10 × 10 × 10 = 10000
23 = 2 × 2 × 2 = 8 103 = 10 × 10 × 10 = 1000
22 = 2 × 2 = 4 102 = 10 × 10 = 100
21 = 2 101 = 10
20 = 1 100 = 1
Notice that each line is equal to one factor (2 or 10) divided into the line above it. It seems reasonable the 21 = 2 or 101 = 10. After all, the exponent gives the number of factors that are multiplied together. But why should 20 = 1 or 100 = 1? A zero exponent does look odd, but it has to follow the same pattern:
23 ÷ 2 = 22 = 4 103 ÷ 10 = 102 = 100
21 ÷ 2 = 20 = 1 101 ÷ 10 = 100 = 1
Note that your calculator can use zero as an exponent, except in the case 00, which is undefined. Otherwise, any base number raised to the 0 power equals 1.
Finally, here is a shortcut you can use on your calculator: The most common exponent used in algebra is 2. Your calculator will have a key labelled x2. You can use this key instead of the yx key, and save yourself some time:
13 or 13
Did you get 169?
Fractions can also be squared as follows:
To square the 2 and 3 separately:
2 is 4 and 3 is 9
The fraction squared can be rewritten as:
Application of powers
The following formulas all contain powers. They are examples from geometry, science, and finance.
Example 1: Area of a circle
A = πr2
where:
A = area
r = radius
π = 3.1415927
Find the area of a circle when r = 3 cm. (Round the answer to one decimal place.)
Solution
A = π (3)2
A = π (9)
A = 28.3 cm2
Example 2: Volume of a sphere
V = πr3
where:
V = volume
π = 3.1415927
r = radius
Find the volume of a sphere when r = 5 cm. (Round the answer to one decimal place.)
Solution
V = π(5)3
V = π(125)
V = 523.6 cm3
Example 3: Power required to overcome air resistance
Power required in kilowatts =
where:
S = speed in km/hr
A = frontal area in m2
23 400 is a constant for air resistance calculations
Find the power required to overcome air resistance when S = 40 km/hr and A = 5 m2. (Round the answer to one decimal place.)
Solution
Example 4: Gravitational force of Earth
where:
F = gravitational force in newtons
M = mass in kg
d = distance from Earth's centre in km
400 000 000 is a constant for Earth's gravity calculations
Find Earth's gravitational force on a 70 kg person when they are 6500 km from the centre of Earth. (Round the answer to one decimal place.)
Solution
Example 5: Compound interest
Compound interest is the interest calculated not only on the original principal, but also on all interests calculated in previous years.
A = P(1 + r)t
where:
A = total amount
P = principal amount
r = annual interest rate as a decimal
t = years
You have bought a Canada Savings Bond for $800 at an interest rate of 3.5%. In two years' time you cash it in. What is the total amount you will receive?
Solution
A = (800)(1 + .035)2
A = 800(1.035)2
A = 800(1.071225)
A = $856.98
Square roots
A root is the reverse of a power. Since 52 = 25, we say that 25 is the square of 5. We could reverse this statement, and say that 5 is the square root of 25. In symbols, this would be written:
The symbol means "find the number, that when squared, equals the number inside." Expressions involving the symbol are called radical expressions and the number inside it is called the radicand. The radicand in the expression is 25. The expression reads "the square root of 25."
Take a look at the following:
What about numbers that do not have whole number square roots? For example, what is the square root of 30?
Look for the button on your calculator. Use this button to find the above square roots. Did you get the right answers? Try the following, using your button.
= ____________ Did you get 1.414213562?
= ____________ Did you get 13.96424004?
= ____________ Did you get 0.223606797?
= ____________ Did you get 0.81649658?
= ____________ Did you get ERROR?
Apparently, the has no answer. This is because there is no number that can be multiplied by itself to give –4. Note that 2 × 2 = 4 and –2 × –2 = 4, and even though 2 × –2 = −4, 2 and −2 are not the same number.
Try entering Did you get −2? This is the method used to identify that the root is negative.
• = b because b × b = A.
• A must be a positive number.
• b can be a positive or negative number. It is assumed that b is positive unless a negative sign appears in front of the root symbol.
• The square root of a negative number is undefined.
So the is undefined. Since −5 × −5 = 25, we would have to ask for the negative root of 25: = –5.
Indexed radicals
Radicals can be used to represent other kinds of roots as well, by using an index number to indicate a cube (3rd) root, 4th root, or any higher root. For example:
The above statement would be read as: "the 5th root of 32 equals 2." Now examine the following:
since 2 × 2 × 2 × 2 × 2 = 32 or 25 = 32
since 33 = 27
since 48 = 65 536
since 92 = 81
No index number is used when it is a square root (index number = 2).
To find a root with a calculator, you need to use the or key. This function is usually above one of the other keys (often the yx key). This means that you will probably have to press the 2nd function key (sometimes labelled SHIFT or INV) first. Note also that some calculators that use direct algebraic logic (DAL) require a different sequence of key entries than older models (or Texas Instrument (TI) style calculators).
Key entries for TI-style calculator
To find = ? press 32 5 2
To find = ? press 216 3 6
Key entries for DAL-style calculator
To find = ? press 5 32 2
To find = ? press 3 216 6
Write down here the key entries that you would use on your own calculator: ______________ .
Note that the answer to is approximately 1.77828. Most roots will be irrational numbers, rather than whole numbers or fractions.
A root can also be written as a power, using a fractional exponent.
This means that a root can also be found using the yx key:
To find = ? press 32 1 5 2
To find = ? press 81 1 4 3
Scientific notation
Remember this sequence?
104 = 10 × 10 × 10 × 10 = 10000
103 = 10 × 10 × 10 = 1000
102 = 10 × 10 = 100
101 = 10
100 = 1
We can continue this sequence using negative exponents:
10-1 = = 0.1
10-2 = = 0.01
10-3 = = 0.001, and so on.
Since we use a base 10 counting system and a base 10 measurement system (SI, or the metric system), this sequence of powers of 10 is very useful.
Very large numbers can be expressed by multiplying a decimal number by a power of 10. For example:
15 trillion = 15 000 000 000 000 = 1.5 × 1013
260 billion = 260 000 000 000 = 2.6 × 1011
Note that this method of writing a number, or scientific notation, uses a decimal number between 1 and 10 followed by the "x 10" and the appropriate exponent of 10.
The exponent tells you the number of places and the direction that the decimal point would have to be moved.
Example 1
5.13 × 105 = 5.13 × 10 × 10 × 10 × 10 × 10 = 513 000
Notice that the decimal point moves five places to the right.
A negative exponent above the 10 does exactly the same thing for very small numbers. In this case, the negative exponent tells you how many places to move the decimal to the left.
Example 2
Scientific calculators usually have a "sci" mode, which will make all of your answers automatically appear in scientific notation. This function may also be used to convert decimals and whole numbers into scientific notation equivalents. Read your own calculator manual to see how this is done.
When entering a number in scientific notation into a calculator, you use the (enter exponent) key or .
For example, if you wanted to enter the mass of Earth (5.98 × 1021 tonnes), enter as follows:
5.98 21 or 5.98 21, depending on your calculator.
The or replaces the need to press "× 10".
Most calculators will show the exponent in smaller digits. (Some calculators even include the "10.") If your model does not, it will show a space between the decimal part of the number and the exponent.
By now, you should be comfortable dealing with exponents, roots. and scientific notation and be able to carry out these kinds of operations with your calculator.
Now complete the Learning Task Self-Test.
Self-Test 5
- In the power 52, 5 is the base and 2 is the exponent.
- True
- False
- (2/5)2 =
- 4/5
- 4/25
- 2/25
- 2/5
- 43 =
- 12
- 16
- 64
- 24
- 220 = 0
- True
- False
- 51 = 5
- True
- False
- 42 + 23 =
- 16
- 24
- 32
- 28
- (3/8)2 =
- 0.140
- 0.141
- 0.142
- 0.014
- 83 = 512
- True
- False
- 123 =
- 36
- 144
- 288
- 1728
- = 1.37
- True
- False
- =
- 2
- 2.33
- 2.25
- 2.67
- =
- 12
- 24
- 18
- 18.49
- = 6
- True
- False
- =
- 0.086
- 0.778
- 0.777
- 0.078
- = 1.1
- True
- False
Learning Task 6
Solve problems using equations and formulas
Using simple equations and formulas to solve problems is a useful skill for any trade. You can use this skill to convert temperatures between Fahrenheit and Celsius, to figure floor area for calculating how much flooring to buy, or to calculate the area of the top of the piston for determining displacement of an engine. Electricians make frequent use of Ohm's law and the Pythagorean theorem. Carpenters use the Pythagorean theorem (3:4:5 triangle) in connection with stair and roof construction. Mechanics use formulas to calculate compression ratios, hydraulic pressures, and torque, among other things.
Key terms
Addition/subtraction principle: The principle stating that if the same number is added or subtracted to both sides of an equation, the solution remains unchanged. For example, if x = a then x + 3 = a + 3.
Binomial: A polynomial consisting of two terms.
Brackets: Symbols, such as ( ) or [ ] or { }, used to enclose expressions.
Coefficient: The number part of an algebraic term; for example, –5 is the coefficient of the term –5x.
Common factor: A number and/or variable that divides evenly into each term of an expression.
Equation: A statement that says one expression is equal to another expression. For example, 2x + 3 = 9 is an equation.
Expression: Numbers, symbols, and operators (such as + and ×) grouped together that show the value of something. An expression is a group of terms separated by + or – signs; for example, 2x + 3 is an expression.
Factor a polynomial: The term used to describe writing a polynomial as a product.
Formula: An equation that contains two or more variables that represent unknowns. Formulas state how one quantity is related to other quantities; for example, P = 2L + 2W is a formula.
Like terms: Terms that have exactly the same variables including variable exponents; for example, 12x2 and 5x2 are like terms, but x3 and x2 are not like terms.
Monomial: A polynomial consisting of only one term.
Multiplication/division principle: The principle stating that if the same number is multiplied or divided to each side of an equation, the solution remains unchanged. For example, if x = a then 3x = 3a.
Polynomial: A many-termed algebraic expression; for example, x2 + 2x − 5 is a polynomial.
Sign: When it is used on its own by just saying "sign," it means negative or positive, such as "what sign is the number, positive or negative?"
Solution: Any variable replacement that makes an equation a true statement.
Term: A number or a number multiplied by one or more variables; for example, 5x2y is a term.
Trinomial: A polynomial consisting of three terms.
Variable: A symbol, usually a letter, used to represent a value in an equation. For example, the variable in the equation 2x + 3 = 9 is the letter x.
Solving equations
An equation is a statement that says one expression is equal to another expression. The equation
2x – 5 = 7
states that the expression 2x – 5 is equal to the term 7. The x in the equation 2x – 5 = 7 is called a variable. Initially the variable is an unknown.
To solve an equation means to find a number for the variable that makes the equation a true statement. For example, 6 is a solution to the equation 2x – 5 = 7, because when x = 6:
2x – 5 = 7
2(6) – 5 = 7
12 – 5 = 7
7 = 7
The statement 2(6) – 5 = 7 is a true statement. Note that there is no other number that is a solution to 2x – 5 = 7.
If x = 10, then 2(10) – 5 = 7 is a false statement and 10 is not a solution to the equation 2x – 5 = 7.
Some equations have no solution. For example, x = x + 1 has no solution. (No number can be equal to one more than itself.) There is no replacement for x that makes x = x + 1 a true statement.
Some equations have an infinite number of solutions. For example, 2x = x + x has an infinite number of solutions. Any replacement for x makes 2x = x + x a true statement. Try it.
The above kinds of equations are rare. Most of the equations we will deal with in this Learning Task have only one solution.
Multiplication and division principles
The equation
5x = 30
states that 5 times some number is 30. Since we know that 5 times 6 is 30, the solution to the equation, 5x = 30, is x = 6.
The equation
34x = 238
is not as simple to solve. Few of us know what number times 34 is 238. However, we can solve this equation algebraically using the division principle.
The division principle states that if we divide both sides of the equation by the same number, the solution to the equation will remain unchanged.
To solve the above equation means to find x or 1x. And, to find 1x, divide both sides of the equation by the coefficient of x term.
To solve 34x = 238 algebraically, write:
34x = 238
divide both sides by 34
1x = 7 note that
Study the following examples.
Example 1
Solve 5x = 36.
Solution
5x = 36
divide both sides by 5
x = 71/5 or x = 7.2
Check: 5(7.2) = 36, so the statement is true.
Example 2
.
Solution
–26 = 1x or x = –26
Check , so the statement is true.
Likewise, you can multiply both sides of an equation by the same number without affecting the solution of the equation.
Example 3
Use the multiplication principle to solve .
Solution
This is just the beginning. We will use various algebraic principles to solve more complicated equations. However, each time we will go through a series of steps that eventually leads to the statement 1x = the solution.
Addition and subtraction principles
Equations like
3x – 7 = 8
can be solved using two principles. If the same number is added to both sides of a true equation, the equation remains true.
Step 1
When solving an equation like
3x – 7 = 8,
the goal is to get the variable term isolated on one side of the equation and the number term on the other. If you add 7 to both sides of the above equation, you will have
3x – 7 + 7 = 8 + 7
and
3x = 15
The addition principle states that if the same number is added to both sides of a true equation, the equation remains true.
Step 2
To find x, now divide by 3, using the division principle.
x = 5
Another way of isolating the variable term is to move the number term to the other side and change its sign. For example:
3x – 7 = 8
3x = 8 + 7
When a term changes sides, it also changes signs.
Example 1
Solve 4x – 3 = 47.
Solution
To isolate the 4x term, move the –3 term to the other side and change its sign.
4x – 3 = 47
4x = 47 + 3
4x = 50
x = 12.5
Check this solution in the equation above.
Example 2
Solve 15 – 6x = 12,
Solution
Move the 15 to the other side and change its sign.
15 – 6x = 12
–6x = 12 – 15
–6x = –3
x = or 0.5
Check this solution in the equation above.
Example 3
Solve –4.5 = 0.2x – 7.5,
Solution
Move the –7.5 to the other side and change its sign.
–4.5 = 0.2x – 7.5
7.5 – 4.5 = 0.2x
3 = 0.2x
15 = x
Check this solution in the equation above.
Many-termed equations
We have already seen that when a term changes sides, it also changes signs. We can extend this idea to changing variable terms from one side of an equation to another side.
Example 1
Solve 5x = 16 – 3x
Solution
The strategy is to get all the x-terms on the same side and have the number term on the other side.
Move the –3x to the other side and change its sign.
5x = 16 – 3x
3x + 5x = 16
8x = 16
x = 2
Example 2
Solve 14x – 8 = 9x + 10
Solution
Here, move the x-terms to one side and the number terms to the other.
14x – 8 = 9x + 10
14x – 9x = 10 + 8
5x = 18
x = or 3.6
Example 3
The length of a rectangle is 4 cm more than the width of the rectangle. If the perimeter of the rectangle is 100 cm, what are the dimensions of the rectangle?
Solution
Draw a diagram of the rectangle and label its sides. Note that if the width is x cm, the length is 4 more cm, or x + 4.
The equation is
x + x + 4 + x + x + 4 = 100
4x + 8 = 100
4x = 100 – 8
4x = 92
x = 23
The width is x or 23 cm.
The length is x + 4 or 23 + 4 = 27 cm.
Note that the question asked for the dimensions of the rectangle, so the answer must give both the length and width.
Equations with brackets
There is another type of equation that involves parentheses (or brackets). For example, the equation
3 + 2(x – 5) = 29
has brackets. Before you can solve this equation, you must simplify it to remove the brackets. Then you can use the principles of addition and subtraction to solve it.
Example 1
Remove brackets from 2(3 + 8x).
Solution
Multiply both the 3 and the 8x by 2.
2(3 + 8x)
= 2(3) + 2(8x)
= 6 + 16x
Example 2
Remove brackets from –(x – 4).
Solution
There doesn't appear to be a number in front of the brackets, but there is. It is –1. Multiply both x or 1x and 4 by –1.
–(x – 4)
= –1(x –4)
= –1(1x) – (–1)(4)
= –1x – (–4)
= –x + 4
Note in the above example that a negative sign in front of the brackets has the effect of changing the inside terms to their opposites. The shortcut is:
–(x – 4) = –x + 4
Example 3
Simplify the expression 2 – 5 (2x – 7). Recall the order of operations, BEDMAS.
Solution
Remove the brackets first, but do not subtract 2 – 5. Multiply both 2x and 7 by –5. Be careful, and notice that –(–35) means +35. Add 2 + 35.
2 – 5 (2x – 7)
= 2 – 10x – (–35)
= 2 – 10x + 35
= –10x + 37
Formulas
Formulas are equations that contain two or more variables that represent unknown values. If you know the values of all but one of the variables, you can substitute these values into the formula and solve the formula for the unknown variable.
Example 1
You can use the formula A = LW to find the area (A) of a rectangle if you know the length (L) and the width (W). Find the area of a rectangle with a length of 8 cm and a width of 5 cm.
Solution
A = LW
A = 8 cm × 5 cm
A = 40 cm2
Example 2
Use the formula P = 2L + 2W to find the perimeter of the above rectangle.
Solution
P = 2L + 2W
P = 2(8 cm) + 2(5 cm)
P = 16 cm + 10 cm
P = 26 cm
Example 3
The formula is used to find Celsius temperature given Fahrenheit temperature. Find the Celsius temperature that corresponds to 104° Fahrenheit.
Solution
C = 40°, so 104°F is equal to 40°C.
Formula manipulation (transposing)
It is often useful to rearrange a formula to solve for an unknown variable. To do this, isolate the required letter on one side of the equals sign while all other symbols are on the opposite side. Remember that formulas are equations. Use the same principles in solving that you use for any other equations.
Example 1
A = P + I is a formula that gives the amount of money (A) an investor collects on a principal (P) when the interest is (I).
Suppose you know the amount collected (A) and the interest (I). You wish to calculate the principal (P).
Solution
To solve the formula for P, you need to get P alone on one side.
Solve for P
P + I = A
Subtract I from each side
P + I – I = A – I
P = A – I
Example 2
A = LW is the formula that gives the area (A) of a rectangle with length (L) and width (W).
Solution
Solve for W
A = LW
Divide both sides by L
Example 3
A = bh finds the measure of the area (A) of a triangle with base (b) and height (h).
Solution
Solve for b.
Multiply both sides by 2
Divide both sides by h
Example 4
is a means of finding the temperature in degrees Fahrenheit (F) when the temperature in degrees Celsius (C) is known.
Solution
Solve for C
Subtract 32 from both sides
Multiply through by 5
Divide through by 9
Example 5
A = πr2 is the formula for the area (A) of a circle where r is the radius and π is a constant.
Solution
Solve for r
πr2 = A
Divide both sides by r
Take the square root of both sides
Some hints for solving a formula:
1. To solve a formula for a given variable you must isolate the variable on one side of the equals sign.
2. To isolate a variable, remove all other terms from the same side. Do this by performing the inverse operation:
• division is the inverse of multiplication.
• addition is the inverse of subtraction.
• square rooting is the inverse of squaring (x2).
3. Remember that formulas are equations. Whatever operations you perform on one side of the equals sign must also be performed on the other side.
Now complete the Learning Task Self-Test.
Self-Test 6
- 11 = n – 3
- n = 7
- n = 14
- n = 21
- n = 10
- 27 = 8(n + 3)
- n = 2
- n = 3/4
- n = 3/8
- n = 3/16
- x2 + 4 = 20
- 4
- 8
- 16
- 24
- C = π42
- C = 133.648
- C = 131.946
- C = 131.947
- C = 131.946
- 2[13 + (8 – 5)] = x
- x = 52
- x = 32
- x = 23
- x = 43
- N + 2(3 – 1) = 8
- N = 2
- N = 0
- N = 4
- N = 6
- 3N = 123
- N = 14
- N = 41
- N = 36
- N = 63
- 15 = 29 – (2x)
- x = 7
- x = 5
- x = 14
- x = –7
- Use the formula A = LW, where
A = area
L = length
W = width
Solve for the missing dimensions.
RECTANGLE | LENGTH | WIDTH | AREA |
A | 250 cm | 150 cm | ______cm2 |
B | 7.5 cm | ______cm | 18 cm2 |
C | ______cm | 475 cm | 570 000 cm2 |
D | 28 cm | ______cm | 644 cm2 |
- Use the formula P = EI, where
P = power in watts (w)
E = force in volts (v)
I = current in amperes (A)
Calculate the power in watts for an electric clothes dryer circuit that has 24 A and 220 V.
- P = 9.16 W
- P = 1200 W
- P = 2640 W
- P = 5280 W
- Use the formula for calculating the taper of a shaft t = , where
t = taper in centimetres per metre (cm/m)
D = diameter of the larger end
d = diameter of the smaller end
L = length in metres
Determine the taper of a shaft that has end diameters of 5.3 cm and 3.1 cm and a length of 1.3 metres.
- t = 1.7 cm/m
- t = 1.4 cm/m
- t = 1.2 cm/m
- t = 1.1 cm/m
- Use the formula F = PA, where
F = force in newtons (N)
P = pressure in pascals (Pa)
A = area in square metres (m2)
Calculate the pressure in the hydraulic system illustrated below:
- P = 80 Pa
- P = 8000 Pa
- P = 20 000 Pa
- P = 200 000 Pa
- Use the conversion formula F = (C + 32), where
F = degrees Farhenheit
C = degrees Celsius
Convert 175°C to degrees Farhenheit.
- F = 337°F
- F = 372.6°F
- F = 378.8°F
- F = 388.8°F
- Use the formula A = πr2, where
A = area of a circle
r = radius of the circle
Determine the radius in centimetres of a circle with an area of 255 cm2.
- 9 cm
- 18 cm
- 44 cm
- 81.169 cm
Learning Task 7
Solve problems involving perimeters, areas, and volumes
Working with pipe and conduit requires knowledge of the area and volume of circles and cylinders, as does many aspects of engine work in the mechanics trades. Virtually all parts of any trade require working with these dimensions. The precise dimensions required for detail and layout work in all trades are equally dependent on careful calculation.
Key terms
Altitude: Also referred to as height. The perpendicular distance between the base of a triangle or other figure and its uppermost side or point.
Area: The amount of surface enclosed by a figure.
Base: The lower side of a triangle or other figure.
Circumference: The distance around a circle.
Composite: Something made up of several simpler parts.
Diameter: The line segment that joins two points on a circle and passes through the centre of the circle.
Formula: A shortcut method of finding an unknown numerical quantity when other quantities are known; for example, a formula to find the area of a rectangle:
area = length × width, or A = L × W
Height: See Altitude.
Perimeter: The distance around a figure.
Perpendicular: A term used to mean "at right angles." Two lines are perpendicular when the angle between them is a right angle, or has a measure of 90°.
Pi (π): A Greek letter, which stands for the irrational number that begins 3.14159265...
Quadrilateral: A four-sided figure.
Radius: The line segment from the centre of a circle to any point on the circle. The length of a radius is one-half the length of a diameter.
Volume: The amount of space occupied by a solid.
Figures
Perimeter
The perimeter of a figure is the total distance around the figure.
Example
Find the perimeter of the quadrilateral shown. (A quadrilateral is a four-sided figure.)
Solution
Measure and label each side of the figure. Add the four sides.
P = 4 cm + 2 cm + 6 cm + 3.5 cm = 15.5 cm
(The letter P is the symbol for perimeter.)
Perimeter formulas
Finding the perimeter of a figure by measuring all of its sides can sometimes be time consuming. For certain figures, such as squares, rectangles, and circles, you can use perimeter formulas that shortcut the measuring process. For example, the perimeter of a square can be found by measuring one side and multiplying by four. The formula is:
Perimeter = 4 × side
P = 4s
(Recall that "4s" means "4 times s")
Study the following.
Circles
The diameter of a circle is the distance from one side of the circle through the centre to the other side. The radius is the distance from the centre of the circle to the edge of the circle. The radius is one half the diameter or:
d = 2r
Perimeter = pi × diameter, or
Perimeter = 2 × pi × radius
P = πd or P = 2πr
The number pi, or π, is impossible to state exactly. A good approximation for π is 3.1416. The number π can only be approximated. It is one of many irrational numbers. Most scientific calculators have a π button. π is the Greek letter pronounced "pi." π = 3.14159265...
The perimeter of a circle is also referred to as the circumference of the circle.
Example 1
Find the circumference of the circle. Round to one decimal place.
Solution
Measure the diameter of the circle. Use the formula:
P = πd
P = 3.1416 × 5.2
P = 16.33632 cm ≈ 16.3 cm
The symbol ≈ means "approximately equal to."
Example 2
Find the perimeter of the parallelogram shown.
Solution
Measure two adjacent sides. Use the formula:
P = 2(a + b)
P = 2 (3.3 cm + 5.1 cm)
P = 2(8.4 cm)
P = 16.8 cm
Example 3
Find the perimeter of the semicircle shown. Round to one decimal place.
Solution
Measure the diameter. Find the perimeter of the half circle.
P = πd ÷ 2
P = 3.1416 × 5.8 cm ÷ 2
P = 18.22128 cm ÷ 2
P = 9.11064 cm
Now add the diameter length and round the answer to one decimal place.
9.11064 + 5.8 cm = 14.91064 ≈ 14.9 cm
Area
The area of a figure is the amount of surface enclosed by the figure.
The area of a figure can be determined by counting the number of squares (and parts of squares) that are needed to cover the particular surface. For example, in the rectangle shown below, 13.5 square centimetres (count them) are needed to cover the surface of the rectangle.
We say that the area of this rectangle is 13.5 cm2.
The common metric units of area are km2, m2, and cm2.
Area formulas
Counting squares and parts of squares is a fairly tedious way of determining the area of a figure. Fortunately, there are simple area formulas for the more common geometrical figures.
Note: s2 means s × s and s2 is read as "s squared."
Knowing the area of a rectangle enables us to find the area of parallelograms, trapezoids, and triangles. Notice how the parallelogram below has been transformed into a rectangle with exactly the same area. The corner triangle can be "cut off" and shifted to the far side of the parallelogram.
To find the area of a parallelogram you need only two measurements: the base and the altitude. The altitude or height is the perpendicular distance between the lower and upper parallel sides of the parallelogram.
Consider the trapezoid below. This time, imagine that you trace out the original trapezoid, flip it upside down, and slide it over to the far end. Now you have created a parallelogram that has twice the area of the original trapezoid. Or, the trapezoid is one-half the area of the larger parallelogram.
In the triangle below, you can again trace another identical triangle and flip it about one of the triangle's sides to, once again, obtain a parallelogram with an area twice that of the original triangle. So, the area of the triangle is one-half the area of the parallelogram.
Now, consider the circle below. Imagine slicing it into thin pieces called sectors. Now form a row of sectors. The thinner the slices or sectors are, the more this row will form the figure of a rectangle with a width equal to the radius of the circle and a length equal to one-half the circumference (or perimeter) of the circle.
Examples using area formulas
Study the following examples.
Example 1
Find the area of the trapezoid.
Solution
Measure the length of the upper base, lower base, and altitude. Label these lengths.
Apply the trapezoid area formula.
A = ½h(a + b)
A = ½(2.5 cm)(4.3 cm + 6 cm)
A = 0.5(2.5 cm)(10.3 cm)
A = 12.875 cm2 ≈ 12.9 cm2
Example 2
Find the area of the triangle.
Solution
Simply measure and label the base and altitude. (Note that the altitude must make a 90-degree angle with the base, which in this case is extended.)
Use the area formula.
A = ½bh
A = ½(4.2 cm)(3.6 cm)
A = 0.5(15.12 cm2)
A = 7.56 cm2 ≈ 7.6 cm2
Example 3
Find the area of a circle with a diameter of 12 m.
Solution
The area formula for the circle involves only the radius. To find the radius,
r = 12 m ÷ 2 = 6 m
To find the area,
A = πr2 = 3.1416(6 m)2
A = 3.1416(36 m2)
A = 113.0976 m2 ≈ 113.1 m2
Areas of irregular figures
An irregular figure can be thought of as a combination of familiar figures. To determine the area of an irregular figure, either divide the figure into several familiar areas and find their sum or subtract one area from another. Study the examples below.
Example 1
Find the area of the quadrilateral.
Solution
Step 1: Divide the quadrilateral into two triangles, A1 and A2, as shown. Use a 90 degree right angle to construct an altitude for the upper triangle. Note that one side of the lower right triangle is also an altitude.
Step 2: Measure and label the bases and their altitudes.
These are the only measurements needed.
Step 3: Now determine the area of both triangles, A1 and A2, and then find their sum, A1 + A2.
A1 = ½b1h1
A1 = ½(5.5 cm)(2.8 cm)
A1 = 0.5(15.4 cm2) = 7.7 cm2
A2 = ½b2h2
A2 = ½(5 cm)(2.3 cm)
A2 = 0.5(11.5 cm) = 5.75 cm2
A1 + A2 = 7.7 cm2 + 5.75 cm2 = 13.45 cm2 ≈ 13.5 cm2
As seen in the example above, there are three important steps in finding the area of an irregular figure:
Step 1: Divide the figure into several familiar areas.
Step 2: Make only those measurements necessary to find the familiar areas.
Step 3: Calculate the areas and add or subtract as necessary.
Example 2
Find the area of the square washer with a hole.
s = 24 cm
Solution
Step 1: Let A1 be the area of the square and A2 be the area of the circle. Subtract A1 – A2 to determine the area of the washer.
Step 2: The necessary measurements are given.
Step 3: Determine A1 and A2 and then calculate A1 – A2 to find the area.
A1 = s2
A1 = (24 cm)2
A1 = 576 cm2
A2 = πr2
A2 = 3.1416(6 cm)2 ≈ 113.0976 cm2
A1 – A2 = 576 cm2 – 113.0976 cm2 = 462.9024 cm2 ≈ 462.9 cm2
Example 3
Find the area of the figure shown.
Solution
Step 1: Divide the figure into the semicircle A1 and the rectangle A2.
Step 2: Measure only the radius of A1 and the length and width of A2.
Step 3: Calculate the areas.
A1 = ½πr2
A1 = ½(3.1416)(1.6 cm)2
A1 = ½(3.1416)(2.56 cm2)
A1 = 4.021248 cm2
A2 = LW = 4.0 cm (3.2 cm)
A2 = 12.8 cm2
A1 + A2 = 4.021248 cm2 + 12.8 cm2 = 16.821248 cm2 ≈ 16.8 cm2
Volume
The perimeter of an object is simply the total distance around that object. You can measure that distance with a ruler (a flexible ruler in the case of curved distance), and the units you use are mm, cm, m, or km.
Area measures the amount of surface taken up by a figure. You can measure a figure's area by counting the number of squares it takes to cover the figure. Fortunately, the area formulas can be used in place of actually counting squares. Area is a two-dimensional idea, hence the use of "2" in the area units of measurement: cm2, m2, and km2.
Volume measures the amount of three-dimensional space an object takes up. You can measure the volume of an object, or solid, by counting the number of cubes that fit inside that object. Can you determine how many cubes there are inside the rectangular prism below?
Did you "count" 36 cubes? Did you also notice that in the prism, L = 4, W = 3, and H = 3, and LWH = 4(3)(3) = 36?
The basic units of volume measurement are the:
- cubic centimetre (cm3)
- cubic kilometre (km3)
- cubic metre (m3)
Fortunately, you do not have to go through the tedious activity of counting the number of cubes (and parts of cubes) that fit inside a regular solid object. Once again, you can apply some fairly simple volume formulas to determine the volume of a regular solid.
Although we will not spend a lot of time rigorously developing volume formulas for the following solids, you will probably notice a common idea running through each formula. The idea is that the volume of a regular solid can be determined by multiplying the base area of the solid by the solid's altitude or height. For example:
Formulas for determining the volume of different solid shapes are as follows:
Cube
Rectangular prism
Triangular prism
Cylinder
Pyramid
V = 1/3b2h or V = b2h ÷ 3
where h is the vertical height of the pyramid
Cone
V = 1/3πr2h or πr2h ÷ 3
where h is the vertical height
Sphere
V = 4/3πr3 or V = 4πr3 ÷ 3
Example 1
Find the volume of the rectangular prism.
Solution
The measurements are given. Use the formula.
V = LWH
V = (30 cm)(12 cm)(2 cm)
V = 720 cm3
Example 2
Find the volume of a pipe whose radius is 0.5 m and height is 16 m.
Solution
A pipe is a cylinder. Sketch and label it. Apply the formula.
V = πr2h
V = 3.1416 (0.5 m)2(16 m)
V = 3.1416 (0.25 m2)(16 m)
V = 12.5664 m3
Example 3
Find the volume of the cone.
Solution
Note that the diameter of the base is given. Its radius is one-half this.
Apply the formula.
V = πr2h ÷ 3
V = (3.1416)(20 cm)2(80 cm) ÷ 3
V = 33 510.4 cm3 ≈ 33 500 cm3
Example 4
Find the volume of a basketball with a radius of 20 cm.
Solution
Apply the volume formula.
V = 4πr3 ÷ 3
V = 4(3.1416)(20 cm)3 ÷ 3
V = 33 510.4 cm3 ≈ 33 500 cm3
Volume of irregular solid figures
An irregular solid is a combination of or a portion of some regular solids. The solid figures below are irregular solids. Study the examples.
Example 1
Find the volume of the solid.
Solution
Divide the solid into the cube, V1, and the triangular prism, V2. Find V1 and V2.
V1 = s3
V1 = (5 cm)3
V1 = 125 cm3
V2 = ½bah
V2 = ½(8 cm)(5 cm)(5 cm)
V2 = 100 cm3
The total volume is
V1 + V2 = 125 cm3 + 100 cm3 = 225 cm3
Example 2
Find the volume of the support plate.
Solution
Consider the plate to be a portion of a cone. The volume of the large cone is:
V2 = πr2h ÷ 3
V2 = 3.1416(8 cm)2(20 cm) ÷ 3
V2 = 1340.4 cm3
The volume of the smaller "missing" cone is:
V1 = πr2h ÷ 3
V1 = 3.1416(6 cm)2(20 cm – 5 cm) ÷ 3
V1 = 565.488 cm3
The volume of the support plate is:
V2 – V1 = 1340.4 cm3 – 565.5 cm3 = 774.9 cm3 ≈ 775 cm3
Solving formulas with other variables
When you use formulas in practical applications, you may know the numbers for perimeter, area, or volume but be missing some other quantity. The examples below demonstrate how to solve formulas in such cases.
- The area of a room is 36 m2. If the room's width is 4.5 m, what is its length?
- Insert the known quantities into the appropriate formula. A = LW
36 = L(4.5)
- Divide both sides of the equation by 4.5 to isolate the variable L.
8 = L
- Solve for L in metres.
L = 8 m
- A cylinder 15 cm tall holds 3000 cm3 of liquid. What is the diameter of the cylinder?
- Insert the known quantities into the appropriate formula. V = πr2h
3000 = 3.1416 (r2)15
- Divide both sides of the equation by 47.124 (3.1416 × 15) to isolate the variable r2. Round the result to the precision of the approximate value of π.
- Take the square root of each side of the equation.
63.6618 = r2
- Because diameter is two times the radius, multiply the radius 7.9788 by 2 to find the diameter D in centimetres.
Now complete the Learning Task Self-Test.
Self-Test 7
- A construction site 27 m by 76 m must be fenced in before excavation starts. How many metres of fencing are required to enclose it?
- 2052 m
- 103 m
- 206 m
- 309 m
- A contractor is deciding which floor covering to use in the room shown below. Cork floor tiles 0.25 m by 0.25 m cost $1.10 each, and carpeting costs $16.95 a square metre. Floor tiles would be cheaper than carpet.
- True
- False
- A rectangular panel measuring 36" x 18" has two holes 6" in diameter to receive stereo speakers and a rectangular opening 8" x 10" for a rear window defroster. What is the surface area of the panel after the holes have been cut to the nearest square inch?
- 511 sq. in.
- 540 sq. in.
- 560 sq. in.
- 584 sq. in.
- The cylindrical water tank servicing a construction site must be erected on a wooden cradle and fastened by three metal straps around the outside. If the tank is 44" in diameter, what is the total length of metal required for the straps, to the nearest 1/16"?
- 1381/4"
- 2761/2"
- 4529/16"
- 41411/16"
- A pipe 35 cm in diameter must be inserted into a storage tank. What is the area of the hole that must be cut in the side of the tank if you leave 0.3 cm clearance on the outside of the pipe?
- 111.84 cm2
- 995.38 cm2
- 223.68 cm2
- 1029.22 cm2
- A four cylinder engine has a bore (diameter) of 7.5 cm and a stroke (height) of 1.1 cm. What is the maximum displacement (volume) of each cylinder to the nearest cubic centimetre?
- 49 cm3
- 26 cm3
- 52 cm3
- 50 cm3
- A solvent tank measures 135 cm x 50 cm x 45 cm. To the nearest litre, how many litres of solvent will it take to fill it three-quarters full? 1 cm3 equals 1 mL.
- 228 L
- 304 L
- 508 L
- 104 L
- Find the area of the washer in the figure below with an inside diameter of 12 mm and an outside diameter of 22 mm.
- 85.6 mm2
- 31.4 mm2
- 212.08 mm2
- 267.04 mm2
- There are two rectangular containers. The dimensions of container A are 10 m × 6 m × 5 m. The dimensions of container B are 7 m × 4 m × 8 m. Container A holds more.
- True
- False
- A circle and a triangle with a base of 9 in. both have an area of 28.2744 in.2. The diameter of the circle is greater than the altitude of the triangle.
- True
- False
- Find the area of the metal plate shown shaded below.
- 26 460 cm2
- 32 694 cm2
- 44 725 cm2
- 15 298 cm2
- What is the area of a triangle with base of 21.5" and an altitude (or height) of 8"?
- 172"2
- 104"2
- 86"2
- 44"2
- A truck has a cylindrical air pressure reservoir for its air brake system with an outside diameter of 39 cm and an outside length of 84 cm. What is the internal volume of the reservoir in cubic centimetres if the cylinder walls are 0.5 cm thick?
- 95 266 cm3
- 88 463 cm3
- 100 346 cm3
- 122 863 cm3
- A rectangular steel bar has a cross sectional area of 468 mm2. The bar is to be reformed into a triangular-shaped bar with the same cross-sectional area. If the height of the triangle is 36 mm, what is the length of the base?
- 22 mm
- 26 mm
- 28 mm
- 32 mm
- Calculate the cost of all moulding needed to go around the window in the figure below, if the cost per metre is $6.50. Round your answer to the nearest cent.
- $12.70
- $25.40
- $15.55
- $10.85
- How much ceiling trim is required for the perimeter of the room shown in the figure below?
- 12.4 m
- 18.5 m
- 24.8 m
- 32.5 m
- A DC generator contains four brush assemblies, with each assembly containing six brushes. The surface of each brush is 0.065 m by 0.025 m. Calculate in square metres the total surface area of all the brushes.
- 0.028 m2
- 0.039 m2
- 0.0016 m2
- 0.0098 m2
- A cleaning tank for small parts measures 15 cm by 35 cm by 12 cm deep. The tank has a fill line 4 cm from the top. How many litres of cleaning fluid will the tank hold if filled to the fill line? (1 cm3 = 1 mL)
- 6.3 L
- 4.2 L
- 3.8 L
- 5.8 L
- A sheet metal worker is building a heating duct in which the rectangular vent is 15 cm high. This vent must have the same area as the rectangular opening, which is 20 cm by 30 cm. Calculate the length of the vent.
- 20 cm
- 28 cm
- 36 cm
- 40 cm
Learning Task 8
Solve problems involving geometric shapes
When we think of geometry, we usually think about angles, circles, triangles, and squares. Geometry is simply the study of points, lines, planes, and space. As long as 4000 years ago, the Egyptians used geometrical ideas to survey land that was flooded yearly by the Nile River. They were also concerned about the orientation of their temples and had to accurately determine north and south, east and west lines. The angle between these lines had to be 90 degrees (a right angle). The Egyptians had known for some time that a 12-cubit-long rope marked out in 3-, 4-, and 5-cubit lengths would create a triangle with a 90-degree angle.
This method of creating a 90-degree angle was also known to the Chinese 3000 years ago, but it was the Egyptians who first attempted to classify and extend such geometrical ideas. Around 600 BC, the Greek Thales of Miletus visited Egypt, and when he returned to Greece, he taught geometry. Thales is credited with being the first person to use deductive logic to determine new geometrical facts.
It was another Greek, Pythagoras (born around 580 BC), who forever changed the nature of mathematics. Unlike Thales, who worked only with concrete numbers and facts, Pythagoras treated geometry in a purely abstract manner. Finally, around 300 BC, Euclid, another Greek mathematician, wrote what is probably the most famous textbook of all time, The Elements.
In his book, Euclid stated five "truths" or postulates about points and lines. He then went on to prove hundreds of other truths or theorems about points and lines using deductive logic. It is exactly this technique, of beginning with a few postulates and then deriving many other theorems through deductive logic, that has been at the heart of mathematics for the last 2300 years.
Geometry is still relevant today. Almost every machine or handmade construction involves using geometry; walls must be straight, fences must be level, shapes have to be fitted properly. Whenever triangles, rectangles, and circles are constructed, geometry is being used.
In this Learning Task we will study plane Euclidean geometry, the most useful of the geometries that deal with flat surfaces.
Key terms
Term | Definition | Symbol | Figure |
Circle | A closed figure where all of the points on the figure are the same distance from a fixed point, called the centre. | no symbol | |
Intersecting lines | Lines that do intersect and have one point in common | no symbol | |
Line | A set of points extending endlessly in both directions; has length but no thickness | or m | |
Parallel lines | Lines in the same plane that do not intersect | / | | m or m | | / ↔→ ↔→ | |
Parallelogram | A quadrilateral with opposite sides parallel and equal in length | no symbol | |
Plane | A flat surface that extends in all directions; it has length and width but not thickness. (Think of a tabletop extending endlessly in all directions.) | no symbol | |
Point | A location that assumes a position but that does not extend through space to also assume dimension, magnitude, or direction | A | •A |
Polygon | A closed figure made up of three or more straight line segments | no symbol | |
Quadrilateral | A four-sided plane figure. (Quad means “four” and lateral means “side.”) | no symbol | |
Ray | A part of a line with one endpoint that extends endlessly in one direction. (Note that the starting point is always given first.) | → AB | AB
|
Rectangle | A parallelogram with four right angles | no symbol | |
Rhombus | A parallelogram with four equal sides; its opposite sides are parallel and its opposite angles are equal. | no symbol | |
Right angle | An angle that measures 90 degrees |
| |
Segment | A part of a line consisting of two endpoints and all the points in between | | |
Square | A rhombus with four right angles | no symbol | |
Trapezoid | A quadrilateral with one and only one pair of opposite sides parallel | no symbol | |
Angles
An angle is formed whenever two rays have a common endpoint or vertex. In Figure 1, rays
and have a common endpoint, B.