Solve Proportion and Similar Figure Applications

By the end of this section, you will be able to: solve proportions, solve applications using proportions, and solve similar figure applications.

Solve proportions

When two rational expressions are equal, the equation relating them is called a proportion.

Proportion. A proportion is an equation of the form

\tfrac{a}{b} = \tfrac{c}{d}

, where

b \neq 0

d \neq 0

. The proportion is read “

a

is to

b

, as

c

is to

d

.”

The equation $\tfrac{1}{2} = \tfrac{4}{8}$ is a proportion because the two fractions are equal. It is read “ $1$ is to $2$ as $4$ is to $8$ .”

Proportions are used in many applications to “scale up” quantities. Suppose a school principal wants to have $1$ teacher for $20$ students. To find the number of teachers needed for $60$ students, let $x$ be that number and set up a proportion, matching the units of the numerators and the units of the denominators:

\frac{1 \text{ teacher}}{20 \text{ students}} = \frac{x \text{ teachers}}{60 \text{ students}}

Since a proportion is an equation with rational expressions, we solve it the same way we solved rational equations — multiply both sides by the LCD to clear the fractions, then solve. Omitting the units until the last step:

\begin{array}{lrcl} & \tfrac{1}{20} &=& \tfrac{x}{60} \\[6pt] \text{Multiply both sides by the LCD, } 60. & \tfrac{1}{20} \cdot 60 &=& \tfrac{x}{60} \cdot 60 \\[6pt] \text{Simplify.} & 3 &=& x \end{array}

The principal needs $3$ teachers for $60$ students.

Example. Solve the proportion $\tfrac{x}{63} = \tfrac{4}{7}$ .

To isolate $x$ , multiply both sides by the LCD, $63$ , then simplify:

\begin{array}{lrcl} & \tfrac{x}{63} &=& \tfrac{4}{7} \\[6pt] \text{Multiply both sides by the LCD, } 63. & 63\!\left(\tfrac{x}{63}\right) &=& 63\!\left(\tfrac{4}{7}\right) \\[6pt] \text{Simplify.} & x &=& \tfrac{9 \cdot 7 \cdot 4}{7} \\[6pt] \text{Divide the common factors.} & x &=& 36 \end{array}

Check. Substitute $x = 36$ into the original proportion:

\frac{36}{63} \overset{?}{=} \frac{4}{7}, \qquad \frac{4 \cdot 9}{7 \cdot 9} \overset{?}{=} \frac{4}{7}, \qquad \frac{4}{7} = \frac{4}{7}\ \checkmark

Solve the proportion $\tfrac{n}{84} = \tfrac{11}{12}$ .

When we work with proportions, we exclude values that would make either denominator zero, just as we do for all rational expressions.

Example. Solve the proportion $\tfrac{144}{a} = \tfrac{9}{4}$ .

Multiply both sides by the LCD, $4a$ , remove the common factors, and solve:

\begin{array}{lrcl} & \tfrac{144}{a} &=& \tfrac{9}{4} \\[6pt] \text{Multiply both sides by the LCD.} & \tfrac{144}{a} \cdot 4a &=& \tfrac{9}{4} \cdot 4a \\[6pt] \text{Remove common factors.} & 4 \cdot 144 &=& a \cdot 9 \\[4pt] \text{Simplify.} & 576 &=& 9a \\[4pt] \text{Divide both sides by } 9. & 64 &=& a \end{array}

You can check that $\tfrac{144}{64} = \tfrac{9}{4}$ .

Solve the proportion $\tfrac{91}{b} = \tfrac{7}{5}$ .

When a variable appears in a sum inside a denominator, we clear the fraction the same way and then distribute.

Example. Solve the proportion $\tfrac{p+12}{9} = \tfrac{p-12}{6}$ .

The LCD of $9$ and $6$ is $18$ :

\begin{array}{lrcl} & \tfrac{p+12}{9} &=& \tfrac{p-12}{6} \\[6pt] \text{Multiply both sides by the LCD, } 18. & 18\!\left(\tfrac{p+12}{9}\right) &=& 18\!\left(\tfrac{p-12}{6}\right) \\[6pt] \text{Simplify.} & 2(p+12) &=& 3(p-12) \\[4pt] \text{Distribute.} & 2p + 24 &=& 3p - 36 \\[4pt] \text{Solve for } p. & 60 &=& p \end{array}

You can check that both sides equal $8$ when $p = 60$ .

Solve the proportion $\tfrac{v+30}{8} = \tfrac{v+66}{12}$ .

Solve applications using proportions

To solve applications with proportions, we follow our usual strategy for solving applications. But when we set up the proportion, we must make sure the units in the numerators match and the units in the denominators match.

Example. When pediatricians prescribe acetaminophen to children, they prescribe $5$ milliliters (ml) for every $25$ pounds of the child’s weight. If Zoe weighs $80$ pounds, how many milliliters will her doctor prescribe?

Let $a$ be the milliliters of acetaminophen. Translate into a proportion, keeping ml in both numerators and pounds in both denominators:

\begin{array}{lrcl} \text{Translate.} & \tfrac{5}{25} &=& \tfrac{a}{80} \\[6pt] \text{Multiply both sides by the LCD, } 400. & 400\!\left(\tfrac{5}{25}\right) &=& 400\!\left(\tfrac{a}{80}\right) \\[6pt] \text{Remove common factors.} & 16 \cdot 5 &=& 5a \\[4pt] \text{Solve for } a. & 16 &=& a \end{array}

Since $80$ is about $3$ times $25$ , the medicine should be about $3$ times $5$ , so $16$ ml is reasonable. The pediatrician would prescribe $16$ ml of acetaminophen to Zoe.

Pediatricians prescribe $5$ milliliters (ml) of acetaminophen for every $25$ pounds of a child's weight. How many milliliters will the doctor prescribe for Emilia, who weighs $60$ pounds? Enter the number of milliliters.

Example. A $16$ -ounce iced caramel macchiato has $230$ calories. How many calories are there in a $24$ -ounce iced caramel macchiato?

Let $c$ be the calories in $24$ ounces. Translate into a proportion with calories in both numerators and ounces in both denominators, then solve:

\begin{array}{lrcl} \text{Translate.} & \tfrac{230}{16} &=& \tfrac{c}{24} \\[6pt] \text{Multiply both sides by the LCD, } 48. & 48\!\left(\tfrac{230}{16}\right) &=& 48\!\left(\tfrac{c}{24}\right) \\[6pt] \text{Simplify.} & 690 &=& 2c \\[4pt] \text{Solve for } c. & 345 &=& c \end{array}

Since $345$ calories for $24$ ounces is more than $230$ for $16$ ounces but not too much more, the answer is reasonable. There are $345$ calories in a $24$ -ounce iced caramel macchiato.

At a fast-food restaurant, a $22$ -ounce chocolate shake has $850$ calories. How many calories are in their $12$ -ounce chocolate shake? Round to the nearest whole number and enter the number of calories.

Example. Josiah went to Mexico for spring break and changed $325 into Mexican pesos. At that time the exchange rate had $1 US equal to $12.54$ Mexican pesos. How many pesos did he get?

Let $p$ be the number of Mexican pesos. Translate into a proportion with dollars in both numerators and pesos in both denominators, then solve:

\begin{array}{lrcl} \text{Translate.} & \tfrac{1}{12.54} &=& \tfrac{325}{p} \\[6pt] \text{Multiply both sides by the LCD, } 12.54p. & 12.54p\!\left(\tfrac{1}{12.54}\right) &=& 12.54p\!\left(\tfrac{325}{p}\right) \\[6pt] \text{Simplify.} & p &=& 4075.5 \end{array}

Since $100 would be about $1{,}254$ pesos and $325 is a little more than $3$ times that, the answer is reasonable. Josiah got $4075.5$ pesos for his trip.

Yurianna is going to Europe and wants to change $800 into Euros. At the current exchange rate, $1 US is equal to $0.738$ Euro. How many Euros will she have for her trip? Enter the number of Euros.

In this example we related the number of pesos to the number of dollars using a proportion. We could say the number of pesos is proportional to the number of dollars. If two quantities are related by a proportion, we say they are proportional.

Solve similar figure applications

When you shrink or enlarge a photo, figure out a distance on a map, or use a pattern to build a bookcase, you are working with similar figures. If two figures have exactly the same shape but different sizes, they are said to be similar. All their corresponding angles have the same measures and their corresponding sides are in the same ratio.

Similar figures. Two figures are similar if the measures of their corresponding angles are equal and their corresponding sides are in the same ratio.

For example, the two triangles below are similar. Each side of $\triangle ABC$ is $4$ times the length of the corresponding side of $\triangle XYZ$ :

\frac{16}{4} = \frac{20}{5} = \frac{12}{3} = 4

Property of Similar Triangles. If $\triangle ABC$ is similar to $\triangle XYZ$ , then their corresponding angle measures are equal and their corresponding sides are in the same ratio:

m\angle A = m\angle X, \quad m\angle B = m\angle Y, \quad m\angle C = m\angle Z

\frac{a}{x} = \frac{b}{y} = \frac{c}{z}

To solve applications with similar figures, we follow the same problem-solving strategy for geometry applications: read the problem and draw the figure, identify and name what we are looking for, translate into an equation using the proportional sides, solve, check, and answer in a complete sentence.

Example. $\triangle ABC$ is similar to $\triangle XYZ$ . The lengths of two sides of each triangle are given. Find the lengths of the third sides.

Let $a$ be the length of the third side of $\triangle ABC$ and $y$ the length of the third side of $\triangle XYZ$ . Since the triangles are similar, the corresponding sides are proportional. The side $AB = 4$ corresponds to the side $XY = 3$ , so $\tfrac{AB}{XY} = \tfrac{4}{3}$ . We write equations using $\tfrac{AB}{XY}$ to find each unknown side:

\begin{array}{lrcl} \text{To find } a. & \tfrac{4}{3} &=& \tfrac{a}{4.5} \\[6pt] \text{Solve.} & 3a &=& 4(4.5) \\[4pt] & a &=& 6 \\[6pt] \text{To find } y. & \tfrac{4}{3} &=& \tfrac{3.2}{y} \\[6pt] \text{Solve.} & 4y &=& 3(3.2) \\[4pt] & y &=& 2.4 \end{array}

Checking, $4(4.5) = 6(3)$ gives $18 = 18\ \checkmark$ and $4(2.4) = 3.2(3)$ gives $9.6 = 9.6\ \checkmark$ . The third side of $\triangle ABC$ is $6$ and the third side of $\triangle XYZ$ is $2.4$ .

$\triangle ABC$ is similar to $\triangle XYZ$ . In the smaller triangle $AB = 17$ and the unknown side $BC = a$ ; in the larger triangle the corresponding sides are $XY = 25.5$ and $YZ = 12$ . Using $\tfrac{17}{25.5} = \tfrac{a}{12}$ , find the length of side $a$ .

The next example shows how similar triangles are used with maps.

Example. On a map, San Francisco, Las Vegas, and Los Angeles form a triangle. The map distance from Los Angeles to Las Vegas is $1$ inch and from Los Angeles to San Francisco is $1.3$ inches. The actual distance from Los Angeles to Las Vegas is $270$ miles. Find the actual distance from Los Angeles to San Francisco.

Let $x$ be the distance from Los Angeles to San Francisco. The map triangle and the actual triangle are similar, so the corresponding sides are proportional. Translate with miles in both numerators and inches in both denominators, then solve:

\begin{array}{lrcl} \text{Translate.} & \tfrac{x \text{ miles}}{1.3 \text{ inches}} &=& \tfrac{270 \text{ miles}}{1 \text{ inch}} \\[6pt] \text{Solve.} & 1.3\!\left(\tfrac{x}{1.3}\right) &=& 1.3\!\left(\tfrac{270}{1}\right) \\[6pt] & x &=& 351 \end{array}

On the map, the distance from Los Angeles to San Francisco is more than the distance from Los Angeles to Las Vegas. Since $351$ is more than $270$ , the answer makes sense. The distance from Los Angeles to San Francisco is $351$ miles.

On a map, Seattle, Portland, and Boise form a triangle. The map distance from Seattle to Boise is $4$ inches and from Seattle to Portland is $1.5$ inches. If the actual distance from Seattle to Boise is $400$ miles, find the distance from Seattle to Portland. Enter the number of miles.

We can also use similar figures to find heights that we cannot directly measure.

Example. Tyler is $6$ feet tall. Late one afternoon his shadow was $8$ feet long. At the same time, the shadow of a tree was $24$ feet long. Find the height of the tree.

Tyler and his shadow form a triangle similar to the one formed by the tree and its shadow. Let $h$ be the height of the tree. The small triangle is similar to the large triangle, so the corresponding sides are proportional:

\begin{array}{lrcl} \text{Translate.} & \tfrac{h}{24} &=& \tfrac{6}{8} \\[6pt] \text{Solve.} & 24\!\left(\tfrac{h}{24}\right) &=& 24\!\left(\tfrac{6}{8}\right) \\[6pt] \text{Simplify.} & h &=& 18 \end{array}

Tyler’s height is less than his shadow’s length, so it makes sense that the tree’s height is less than the length of its shadow. The tree is $18$ feet tall.

A telephone pole casts a shadow that is $50$ feet long. Nearby, an $8$ -foot tall traffic sign casts a shadow that is $10$ feet long. How tall is the telephone pole? Enter the height in feet.

Key terms

proportion — an equation of the form $\tfrac{a}{b} = \tfrac{c}{d}$ (with $b \neq 0$ and $d \neq 0$ ) stating that two ratios are equal; read “ $a$ is to $b$ as $c$ is to $d$ .” proportional — two quantities are proportional when they are related by a proportion. similar figures — two figures with the same shape but possibly different sizes, so their corresponding angles are equal and their corresponding sides are in the same ratio.

This section is adapted from Elementary Algebra 2e, Section 8.7: Solve Proportion and Similar Figure Applications by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: condensed the worked examples into aligned step tables and prose, recast the “How To” procedure as a proportion strategy paragraph, recreated the similar-triangle and map figures with the accessible <Figure /> component; omitted the Be Prepared quiz, Self Check checklist, media links, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.

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