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Mathematics in Trades and Life

Megan Ossiander-Gobeille, Mark Fitch

Section 2.4 Proportions

Ratio problems presume that the ratio does not change. This makes sense in examples like conversion of units. For example 1 gallon is always 4 quarts. In contrast rates often change: your average speed may be 25 mph, but you must have driven slower and faster during that drive. For the ratios that do not change we can write equations and solve for properties. These are called (fixed) proportions.

Subsection 2.4.1 Proportion Examples

We can solve for values in ratio problems by setting up the equation (ratio equals ratio) and then multiplying and dividing as needed to solve. Often people remember this with the mnemonic device: cross multiplication which is also known as clearing the denominators.

Example 2.4.1. Proportion: Solving.

The Diamond DA-20 cruises at the rate of
\begin{equation*} \frac{110 \; \text{nautical miles}}{1 \; \text{hour}}\text{.} \end{equation*}
How long will it take to travel 236 nm?
Because this is a fixed ratio we can write
\begin{align*} \frac{110 \; \text{nm}}{\text{hour}} \amp = \frac{236 \; \text{nm}}{t \; \text{hours}}.\\ t \; \text{hours} \cdot \frac{110 \; \text{nm}}{1 \text{hour}} \amp = t \; \text{hours} \cdot \frac{236 \; \text{nm}}{t \; \text{hours}}.\\ t \cdot 110 \; \text{nm} \amp = 236 \; \text{nm}. \amp \; \text{Clearing the denominators.}\\ \frac{t \cdot 110 \; \text{nm}}{110 \; \text{nm}} \amp = \frac{236 \; \text{nm}}{110 \; \text{nm}}.\\ t \amp = \frac{236}{110}.\\ t \amp \approx 2.1 \; \text{hours}. \end{align*}

Checkpoint 2.4.2.

If the Diamond DA-20 climbs at the rate of
\begin{equation*} \frac{450 \; \text{feet}}{1.0 \; \text{minute}} \end{equation*}
how long will it take it to climb 2500 ft?
Answer.
\(5.5\)
Solution.
\begin{equation*} \begin{aligned} \frac{450 \; \text{ft}}{1.0 \; \text{minute}} \amp = \frac{2500 \; \text{ft}}{t \; \text{minutes}}.\\ t \; \text{minutes} \cdot \frac{450 \; \text{ft}}{1.0 \; \text{minutes}} \amp = t \; \text{minutes} \cdot \frac{2500 \; \text{ft}}{t \; \text{minutes}}.\\ t \cdot 450 \; \text{ft} \amp = 2500 \; \text{ft}.\\ \frac{t \cdot 450 \; \text{ft}}{450 \; \text{ft}} \amp = \frac{2500 \; \text{ft}}{450 \; \text{ft}}.\\ t \amp = \frac{2500}{450}\\ t \amp \approx 5.5 \text{ minutes}. \end{aligned} \end{equation*}

Example 2.4.3. Cheesecake Groceries: Double.

For a particular cheesecake recipe there is \(15\bar{0}\) g of eggs and \(150\bar{0}\) g of cream cheese. We will determine how many grams of eggs we need if we double the recipe. This means everything will be in ratio of 2/1. Note the 2 and 1 have infinite significant figures (they are exact numbers rather than approximated measurements).
\begin{align*} \frac{2}{1} & = \frac{E \; \text{g}}{15\bar{0} \; \text{g}}.\\ \frac{2}{1} \cdot 150 \; \text{g} & = \frac{E \; \text{g}}{15\bar{0} \; \text{g}} \cdot 150 \; \text{g}.\\ 2 \cdot 15\bar{0} \; \text{g} & = E \cdot 1. \amp \; \text{Eliminating the denominator.}\\ 30\bar{0} \; \text{g} & = E \text{ of eggs}. \end{align*}

Example 2.4.4. Cheesecake Groceries: Unknown.

For a particular cheesecake recipe there is \(15\bar{0}\) g of eggs and \(150\bar{0}\) g of cream cheese. If we have \(35\bar{0}\) g of egg how much cream cheese do we need? We know that the egg to cream cheese ratio must be 150/1500. We also notice this is 1/10.
\begin{align*} \frac{1}{10} & = \frac{35\bar{0} \; \text{g}}{C \; \text{g}}.\\ \frac{10}{1} & = \frac{C \; \text{g}}{35\bar{0} \; \text{g}}.\\ \frac{10}{1} \cdot 350 \; \text{g} & = \frac{C \; \text{g}}{350 \; \text{g}} \cdot 350 \; \text{g}.\\ 35\bar{0}0 & = C \; \text{g of cream cheese}. \end{align*}
Notice we flipped the ratio in the second step to make the arithmetic easier to follow. You can think of this as the addage “what you do to one side, you must also do to the other”.

Checkpoint 2.4.5.

For a particular cheesecake recipe there is \(15\bar{0}\) g of eggs and \(150\bar{0}\) g of cream cheese. Suppose you have \(35\bar{0}\) g of eggs and \(340\bar{0}\) g of cream cheese. How much of the egg and cream cheese can you use?
Egg:
Cream cheese:
Answer 1.
\(340\)
Answer 2.
\(3400\)
Solution.
First we can figure out whether the egg or the cream cheese is the limiting ingredient. The egg to cream cheese must be in a
\begin{equation*} \frac{15\bar{0}}{150\bar{0}}=\frac{1.00}{10.0} \end{equation*}
ratio. We can calculate the ratio of ingredients we have
\begin{equation*} \frac{35\bar{0} \; \text{g}}{340\bar{0} \; \text{g}} \approx 0.103 \end{equation*}
Because this is more than \(0.103 \gt 0.1=1/10,\) we have more egg than we can use, because the numerator (egg) is bigger. Thus we will set up the proportion using all of the cream cheese.
We can setup the ratio multiple ways. The first is using the egg to cheese ratio.
\begin{equation*} \begin{aligned} \frac{15\bar{0}}{150\bar{0}} \amp = \frac{E \; \text{g}}{340\bar{0} \; \text{g}}.\\ \frac{1.00}{10.0} \amp = \frac{E \; \text{g}}{340\bar{0} \; \text{g}}.\\ \frac{1.00}{10.0} \cdot 340\bar{0} \; \text{g} \amp = \frac{E \; \text{g}}{340\bar{0} \; \text{ g}} \cdot 340\bar{0} \; \text{g}.\\ 34\bar{0}.0 \amp = E \; \text{g}. \end{aligned} \end{equation*}
We can use 340 of the 350 grams of egg.
Another way is using the recipe to scaled up ratio.
\begin{equation*} \begin{aligned} \frac{15\bar{0}}{E} \amp = \frac{150\bar{0}}{340\bar{0}}.\\ \frac{15\bar{0}}{E}E \amp = \frac{150\bar{0}}{340\bar{0}}E.\\ 15\bar{0} \amp = \frac{150\bar{0}}{340\bar{0}}E.\\ \frac{340\bar{0}}{150\bar{0}} \cdot 15\bar{0} \amp = \frac{340\bar{0}}{150\bar{0}} \cdot \frac{150\bar{0}}{340\bar{0}}E.\\ \frac{340\bar{0} \cdot 15\bar{0}}{150\bar{0}} \amp = E.\\ 34\bar{0} \amp = E. \end{aligned} \end{equation*}
We can use 340 of the 350 grams of egg.

Subsection 2.4.2 Similar polygons

Definition 2.4.6. Similar Triangles.

Two triangles are similar if and only if corresponding angles are congruent.
Congruent in this case means the same length. Because triangles are similar corresponding side lengths are proprotional.

Instructions.

Notice that the the two triangles have the same angles. Calculate ratios of corresponding sides (in this case correspond means between the same two angles). Notice that every ratio you calculate is the same.
Figure 2.4.7. Similar Triangles
We can use the proportionaliy of similar triangle sides lengths to calculate the lengths using the same technique as Example 2.4.1.

Example 2.4.8.

Suppose triangle A has angles \(30^\circ, 60^\circ, 90^\circ\) with side lengths \(1.000, 1.732, 2.000\text{.}\) If triangle B also has angles \(30^\circ, 60^\circ, 90^\circ\) it is similar. Suppose the smallest side length is \(2.000\text{.}\) Then we know.
\begin{align*} \frac{1.000}{2.000} & = \frac{1.732}{s}.\\ \frac{1.000}{2.000} \cdot 2.000 s & = \frac{1.732}{s} \cdot 2.000 s. \amp \text{ Clearing the denominators.}\\ s & = 1.732(2.000).\\ s & = 3.464. \end{align*}
We can calculate the length of the third side in the same way.
\begin{align*} \frac{1.000}{2.000} & = \frac{2.000}{s}.\\ \frac{1.000}{2.000} \cdot 2.000s & = \frac{2.000}{s} \cdot 2.000s.\\ s & = 4.000 \end{align*}

Checkpoint 2.4.9.

Suppose triangle A has angles \(40^\circ, 60^\circ, 80^\circ\) and side lengths 1.000, 1.347, 2.000. If triangle B has the same angle measures and the shortest side is length 2.500, what are the other two side lengths?
Corresponds to 1.347:
Corresponds to 2.000:
Answer 1.
\(3.368\)
Answer 2.
\(5\)
Solution.
We setup the proportions for the other two sides.
\begin{equation*} \begin{aligned} \frac{1.000}{2.500} \amp = \frac{1.347}{s_2}.\\ s_2 \amp = 3.368.\\ \frac{1.000}{2.500} \amp = \frac{2.000}{s_3}.\\ s_3 \amp = 5.000. \end{aligned} \end{equation*}
Shapes other than triangles can be similar. For example there are similar rectangles and similar pentagons. To be similar they must have the same number of sides, corresponding angles must be the same, and corresponding sides must be in the same ratio. Note that just having the same angles is insufficient: any two rectangles have all the same angles (right angles) but not every pair is similar.
One place where similar shapes (beyond triangles) is used is scale drawing and scale models. If you ever built a model of a car or a plane or some such there was most likely a scale given. For example they may be 1/32 scale. This means that one inch on the model is 32 inches on the actual object.

Exercises 2.4.3 Exercises

1. Proportion (no context).

2. Proportion (no context).

3. Proportion (no context).

4. Proportion (no context).

5. Proportion Application.

6. Proportion Application.

7. Proportion Application.

8. Proportion Application.

9. Proportion Application.

10. Proportion Application.

11. Solution.

12. Similar Triangles.

13. Similar Triangles.

14. Similar Triangles.

15. Similar Triangles.

16. Similar Triangles.

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