Lesson 8.7: Solve Proportion and Similar Figure Applications

New Concept

Master the power of proportions! You'll learn to solve equations of equal ratios, like $\dfrac{a}{b} = \dfrac{c}{d}$, and apply this to scale quantities and find unknown lengths in similar geometric figures, a key tool for real-world comparisons.

What’s next

Get ready to put this into action! You'll work through interactive examples of solving proportions, followed by challenge problems involving similar figures.

Property

A proportion is an equation of the form $\dfrac{a}{b} = \dfrac{c}{d}$, where $b \neq 0$, $d \neq 0$, $a \neq 0$. The proportion is read “$a$ is to $b$, as $c$ is to $d$.” Since a proportion is an equation with rational expressions, we solve it by multiplying both sides of the equation by the LCD to clear the fractions and then solve the resulting equation.

Examples

• To solve the proportion $\dfrac{x}{49} = \dfrac{3}{7}$, we multiply both sides by the LCD, 49. This gives $x = 49 \left( \dfrac{3}{7} \right)$, so $x = 21$.
• To solve $\dfrac{120}{a} = \dfrac{10}{3}$, we multiply by the LCD, $3a$. This gives $120 \cdot 3 = 10 \cdot a$, which simplifies to $360 = 10a$, so $a = 36$.
• To solve $\dfrac{y}{9} = \dfrac{7}{3}$, we multiply both sides by the LCD, 9. This gives $y = 9 \left( \dfrac{7}{3} \right)$, which simplifies to $y = 3 \cdot 7$, so $y = 21$.

Explanation

A proportion represents two equal ratios. Think of it as a balanced scale for fractions.

Property

To solve a proportion where the variable is part of an expression, such as $\dfrac{n}{n+14} = \dfrac{5}{7}$, the method remains the same.
Multiply both sides of the equation by the LCD, which in this case is $7(n+14)$.
This removes the denominators, resulting in a linear equation to solve: $7n = 5(n+14)$.
Use the distributive property and solve for the variable.

Examples

• To solve $\dfrac{x}{x+10} = \dfrac{3}{5}$, multiply by the LCD, $5(x+10)$, to get $5x = 3(x+10)$. Distributing gives $5x = 3x + 30$, so $2x = 30$, and $x=15$.
• To solve $\dfrac{y-8}{y} = \dfrac{7}{9}$, multiply by the LCD, $9y$. This gives $9(y-8) = 7y$. Distributing gives $9y - 72 = 7y$, so $2y = 72$, and $y=36$.
• To solve $\dfrac{p+10}{5} = \dfrac{p-6}{3}$, multiply by the LCD, 15. This gives $3(p+10) = 5(p-6)$. Distributing gives $3p+30 = 5p-30$, so $60=2p$, and $p=30$.

Explanation

Even when variables are in expressions like $(x+10)$, the core rule is unchanged. Treat the entire expression as one unit.

Property

Proportions are used to scale quantities in real-world applications. To solve, first identify the unknown and assign a variable. Then, translate the problem into a proportion, ensuring the units are consistent in the numerators and denominators (e.g., $\frac{\text{miles}}{\text{gallon}} = \frac{\text{miles}}{\text{gallon}}$). Finally, solve the proportion for the variable and state the answer in a complete sentence.

Examples

• A recipe for 4 servings needs 2 cups of flour. For 10 servings, how many cups are needed? Let $c$ be cups. $\dfrac{2 \text{ cups}}{4 \text{ servings}} = \dfrac{c \text{ cups}}{10 \text{ servings}} \implies 4c = 20 \implies c=5$. You need 5 cups.
• If 1 US dollar equals 0.92 Euros, how many Euros do you get for 500 US dollars? Let $e$ be Euros. $\dfrac{1 \text{ dollar}}{0.92 \text{ Euros}} = \dfrac{500 \text{ dollars}}{e \text{ Euros}} \implies e = 500(0.92) = 460$. You get 460 Euros.
• A car travels 150 miles on 5 gallons of gas. How many gallons are needed for a 450-mile trip? Let $g$ be gallons. $\dfrac{150 \text{ miles}}{5 \text{ gal}} = \dfrac{450 \text{ miles}}{g \text{ gal}} \implies 150g = 2250 \implies g=15$. You need 15 gallons.

Explanation

Proportions are perfect for real-life scaling, like adjusting recipes or converting currency.

Property

Two figures are similar if the measures of their corresponding angles are equal and their corresponding sides are in the same ratio. For triangles, if $\triangle ABC$ is similar to $\triangle XYZ$, this property holds true.

Corresponding angles are equal:
$\operatorname{m}\angle A = \operatorname{m}\angle X$
$\operatorname{m}\angle B = \operatorname{m}\angle Y$
$\operatorname{m}\angle C = \operatorname{m}\angle Z$

Corresponding sides are in the same ratio:

Property

The principle of similar figures is used to solve problems involving indirect measurement. Common applications include using shadows to find the height of tall objects or using a map's scale to find actual distances. The method involves setting up a proportion where the ratios of corresponding sides of the similar figures (e.g., triangles formed by objects and their shadows) are equal.

Examples

• A 5-foot-tall person casts a 4-foot shadow. A nearby flagpole casts a 20-foot shadow. How tall is the flagpole? Let $h$ be the height. $\dfrac{5 \text{ ft}}{4 \text{ ft}} = \dfrac{h \text{ ft}}{20 \text{ ft}} \implies 4h = 100 \implies h=25$. The pole is 25 feet tall.
• On a map, the distance between two cities is 3 inches. The map scale says 0.5 inches represents 40 miles. What is the actual distance? Let $d$ be the distance. $\dfrac{0.5 \text{ in}}{40 \text{ miles}} = \dfrac{3 \text{ in}}{d \text{ miles}} \implies 0.5d = 120 \implies d=240$ miles.
• A 10-foot street lamp casts a 15-foot shadow. A nearby fire hydrant casts a 3-foot shadow. What is the hydrant's height? Let $h$ be the height. $\dfrac{10 \text{ ft}}{15 \text{ ft}} = \dfrac{h \text{ ft}}{3 \text{ ft}} \implies 15h = 30 \implies h=2$. The hydrant is 2 feet tall.

Explanation

Can't measure a tall tree? Use its shadow! An object of known height and its shadow form a triangle.