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== 2.3#h(0.6em)Solve a Formula for a Specific Variable

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Before you get started, take this readiness quiz.

Evaluate #math.equation(block: false, alt: "2 open parenthesis x plus 3 close parenthesis")[$2 ( x + 3 )$] when #math.equation(block: false, alt: "x equals 5 .")[$x = 5 .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "16")[$16$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The length of a rectangle is three less than the width. Let #emph[w] represent the width. Write an expression for the length of the rectangle. #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "w minus 3")[$w − 3$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Evaluate #math.equation(block: false, alt: "the fraction 1 over 2 b h")[$frac(1, 2) b h$] when #math.equation(block: false, alt: "b equals 14")[$b = 14$] and #math.equation(block: false, alt: "h equals 9 .")[$h = 9 .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "63")[$63$]
]
]


=== Solve a Formula for a Specific Variable

We have all probably worked with some geometric formulas in our study of mathematics. Formulas are used in so many fields, it is important to recognize formulas and be able to manipulate them easily.

It is often helpful to solve a formula for a specific variable. If you need to put a formula in a spreadsheet, it is not unusual to have to solve it for a specific variable first. We isolate that variable on one side of the equals sign with a coefficient of one and all other variables and constants are on the other side of the equal sign.

Geometric formulas often need to be solved for another variable, too. The formula #math.equation(block: false, alt: "V equals the fraction 1 over 3 π r squared h")[$V = frac(1, 3) π r^(2) h$] is used to find the #strong[volume] of a right circular cone when given the radius of the base and height. In the next example, we will solve this formula for the height.

#examplebox("Example 1")[][
Solve the formula #math.equation(block: false, alt: "V equals the fraction 1 over 3 π r squared h")[$V = frac(1, 3) π r^(2) h$] for #emph[h].

#solutionbox[
#figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([Write the formula.], [], [], [#figure(figph[The image displays the mathematical formula for the volume of a cone, which is V = (1/3)πr^2h, where V is the volume, r is the radius of the base, and h is the height of the cone.], alt: "The image displays the mathematical formula for the volume of a cone, which is V = (1/3)πr^2h, where V is the volume, r is the radius of the base, and h is the height of the cone.", caption: none)]),
  [Remove the fraction on the right.], [], [], [#figure(figph[A mathematical equation displays 3 multiplied by V (volume) on the left side, equaling 3 multiplied by (1/3)πr²h (volume of a cone formula) on the right side, highlighting the multiplication by 3.], alt: "A mathematical equation displays 3 multiplied by V (volume) on the left side, equaling 3 multiplied by (1/3)πr²h (volume of a cone formula) on the right side, highlighting the multiplication by 3.", caption: none)],
  [Simplify.], [], [], [#figure(figph[Mathematical equation showing a cone's volume formula: 3V = pi times r squared times h. V represents volume, r is the radius, and h is the height.], alt: "Mathematical equation showing a cone's volume formula: 3V = pi times r squared times h. V represents volume, r is the radius, and h is the height.", caption: none)],
  [Divide both sides by #math.equation(block: false, alt: "π r squared .")[$π r^(2) .$]], [], [], [#figure(figph[A mathematical equation shows h equals 3V divided by pi r squared, which is the formula for the height of a cone or pyramid given its volume and radius.], alt: "A mathematical equation shows h equals 3V divided by pi r squared, which is the formula for the height of a cone or pyramid given its volume and radius.", caption: none)],
))

We could now use this formula to find the height of a right circular cone when we know the volume and the radius of the base, by using the formula #math.equation(block: false, alt: "h equals the fraction 3 V over π r squared .")[$h = frac(3 V, π r^(2)) .$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Use the formula #math.equation(block: false, alt: "A equals the fraction 1 over 2 b h")[$A = frac(1, 2) b h$] to solve for #emph[b].

#solutionbox[
#math.equation(block: true, alt: "b equals the fraction 2 A over h")[$b = frac(2 A, h)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Use the formula #math.equation(block: false, alt: "A equals the fraction 1 over 2 b h")[$A = frac(1, 2) b h$] to solve for #emph[h].

#solutionbox[
#math.equation(block: true, alt: "h equals the fraction 2 A over b")[$h = frac(2 A, b)$]
]
]

In the sciences, we often need to change #strong[temperature] from Fahrenheit to Celsius or vice versa. If you travel in a foreign country, you may want to change the Celsius temperature to the more familiar Fahrenheit temperature.

#examplebox("Example 2")[][
Solve the formula #math.equation(block: false, alt: "C equals the fraction 5 over 9 open parenthesis F minus 32 close parenthesis")[$C = frac(5, 9) ( F − 32 )$] for #emph[F].

#solutionbox[
#figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([Write the formula.], [], [], [#figure(figph[A mathematical formula for converting temperature from Fahrenheit (F) to Celsius (C) is displayed: C = 5/9(F - 32).], alt: "A mathematical formula for converting temperature from Fahrenheit (F) to Celsius (C) is displayed: C = 5/9(F - 32).", caption: none)]),
  [Remove the fraction on the right.], [], [], [#figure(figph[An equation (9/5)C = (9/5) \* (5/9)(F - 32), showing a step in temperature conversion. Red numbers highlight fractions being multiplied on both sides.], alt: "An equation (9/5)C = (9/5) * (5/9)(F - 32), showing a step in temperature conversion. Red numbers highlight fractions being multiplied on both sides.", caption: none)],
  [Simplify.], [], [], [#figure(figph[A mathematical equation showing the conversion formula from Fahrenheit to Celsius, specifically (9/5)C = (F - 32). The equation is presented in a clear, typeset format against a white background.], alt: "A mathematical equation showing the conversion formula from Fahrenheit to Celsius, specifically (9/5)C = (F - 32). The equation is presented in a clear, typeset format against a white background.", caption: none)],
  [Add 32 to both sides.], [], [], [#figure(figph[A mathematical formula for converting Celsius to Fahrenheit, which reads as '9/5 C + 32 = F'.], alt: "A mathematical formula for converting Celsius to Fahrenheit, which reads as '9/5 C + 32 = F'.", caption: none)],
))

We can now use the formula #math.equation(block: false, alt: "F equals the fraction 9 over 5 C plus 32")[$F = frac(9, 5) C + 32$] to find the Fahrenheit temperature when we know the Celsius temperature.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve the formula #math.equation(block: false, alt: "F equals the fraction 9 over 5 C plus 32")[$F = frac(9, 5) C + 32$] for #emph[C].

#solutionbox[
#math.equation(block: true, alt: "C equals the fraction 5 over 9 open parenthesis F minus 32 close parenthesis")[$C = frac(5, 9) ( F − 32 )$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve the formula #math.equation(block: false, alt: "A equals the fraction 1 over 2 h open parenthesis b plus B close parenthesis")[$A = frac(1, 2) h ( b + B )$] for #emph[b].

#solutionbox[
#math.equation(block: true, alt: "b equals the fraction 2 A minus B h over h")[$b = frac(2 A − B h, h)$]
]
]

The next example uses the formula for the #strong[surface area] of a right cylinder.

#examplebox("Example 3")[][
Solve the formula #math.equation(block: false, alt: "S equals 2 π r squared plus 2 π r h")[$S = 2 π r^(2) + 2 π r h$] for #emph[h].

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Write the formula.], [#figure(figph[A mathematical formula is displayed: S = 2πr² + 2πrh, representing the surface area of a cylinder with radius 'r' and height 'h'.], alt: "A mathematical formula is displayed: S = 2πr² + 2πrh, representing the surface area of a cylinder with radius 'r' and height 'h'.", caption: none)]),
  [Isolate the #emph[h] term by subtracting #math.equation(block: false, alt: "2 π r squared")[$2 π r^(2)$] from each side.], [#figure(figph[A mathematical equation is displayed, showing S minus two pi r squared equals two pi r squared minus two pi r squared plus two pi r h, written in black and red text on a white background.], alt: "A mathematical equation is displayed, showing S minus two pi r squared equals two pi r squared minus two pi r squared plus two pi r h, written in black and red text on a white background.", caption: none)],
  [Simplify.], [#figure(figph[Mathematical formula showing the relationship between total surface area (S), radius (r), height (h), and pi (π) for a cylinder: S - 2πr² = 2πrh.], alt: "Mathematical formula showing the relationship between total surface area (S), radius (r), height (h), and pi (π) for a cylinder: S - 2πr² = 2πrh.", caption: none)],
  [Solve for #emph[h] by dividing both sides by #math.equation(block: false, alt: "2 π r .")[$2 π r .$]], [#figure(figph[Mathematical equation: (S - 2πr^2) / (2πr) = (2πrh) / (2πr), an intermediate step in calculating the height of a cylinder from its surface area.], alt: "Mathematical equation: (S - 2πr^2) / (2πr) = (2πrh) / (2πr), an intermediate step in calculating the height of a cylinder from its surface area.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation showing h = (S - 2πr^2) / (2πr), likely representing the height (h) of a cylinder derived from its total surface area (S) and radius (r).], alt: "A mathematical equation showing h = (S - 2πr^2) / (2πr), likely representing the height (h) of a cylinder derived from its total surface area (S) and radius (r).", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve the formula #math.equation(block: false, alt: "A equals P plus P r t")[$A = P + P r t$] for #emph[t].

#solutionbox[
#math.equation(block: true, alt: "t equals the fraction A minus P over P r")[$t = frac(A − P, "P" r)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve the formula #math.equation(block: false, alt: "A equals P plus P r t")[$A = P + P r t$] for #emph[r].

#solutionbox[
#math.equation(block: true, alt: "r equals the fraction A minus P over P t")[$r = frac(A − P, "P" t)$]
]
]

Sometimes we might be given an equation that is solved for #emph[y] and need to solve it for #emph[x], or vice versa. In the following example, we’re given an equation with both #emph[x] and #emph[y] on the same side and we’ll solve it for #emph[y].

#examplebox("Example 4")[][
Solve the formula #math.equation(block: false, alt: "8 x plus 7 y equals 15")[$8 x + 7 y = 15$] for #emph[y].

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([We will isolate #emph[y] on one side of the equation.], [#figure(figph[A mathematical equation is displayed with a white background, reading '8x + 7y = 15'. The numbers and variables are in a dark gray font.], alt: "A mathematical equation is displayed with a white background, reading '8x + 7y = 15'. The numbers and variables are in a dark gray font.", caption: none)]),
  [Subtract #math.equation(block: false, alt: "8 x")[$8 x$] from both sides to isolate the term with #emph[y].], [#figure(figph[An equation is shown: 8x - 8x + 7y = 15 - 8x, demonstrating the step of subtracting 8x from both sides to isolate the 'y' term for simplification.], alt: "An equation is shown: 8x - 8x + 7y = 15 - 8x, demonstrating the step of subtracting 8x from both sides to isolate the 'y' term for simplification.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation is displayed, showing 7y = 15 - 8x. This linear equation relates variables 'y' and 'x' with constant terms.], alt: "A mathematical equation is displayed, showing 7y = 15 - 8x. This linear equation relates variables 'y' and 'x' with constant terms.", caption: none)],
  [Divide both sides by 7 to make the coefficient of #emph[y] one.], [#figure(figph[An equation showing 7y divided by 7 on the left, and (15 - 8x) divided by 7 on the right. The number 7 in the denominator is red on both sides.], alt: "An equation showing 7y divided by 7 on the left, and (15 - 8x) divided by 7 on the right. The number 7 in the denominator is red on both sides.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation shows y equals the fraction (15 minus 8x) all over 7. It is written as y = (15 - 8x) / 7.], alt: "A mathematical equation shows y equals the fraction (15 minus 8x) all over 7. It is written as y = (15 - 8x) / 7.", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve the formula #math.equation(block: false, alt: "4 x plus 7 y equals 9")[$4 x + 7 y = 9$] for #emph[y].

#solutionbox[
#math.equation(block: true, alt: "y equals the fraction 9 minus 4 x over 7")[$y = frac(9 − 4 x, 7)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve the formula #math.equation(block: false, alt: "5 x plus 8 y equals 1")[$5 x + 8 y = 1$] for #emph[y].

#solutionbox[
#math.equation(block: true, alt: "y equals the fraction 1 minus 5 x over 8")[$y = frac(1 − 5 x, 8)$]
]
]


=== Use Formulas to Solve Geometry Applications

In this objective we will use some common geometry formulas. We will adapt our problem solving strategy so that we can solve geometry applications. The geometry formula will name the variables and give us the equation to solve.

In addition, since these applications will all involve shapes of some sort, most people find it helpful to draw a figure and label it with the given information. We will include this in the first step of the problem solving strategy for geometry applications.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Solve geometry applications.]

+ #strong[Read] the problem and make sure all the words and ideas are understood.
+ #strong[Identify] what you are looking for.
+ #strong[Name] what we are looking for by choosing a variable to represent it. Draw the figure and label it with the given information.
+ #strong[Translate] into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
+ #strong[Solve] the equation using good algebra techniques.
+ #strong[Check] the answer in the problem and make sure it makes sense.
+ #strong[Answer] the question with a complete sentence.
]

When we solve geometry applications, we often have to use some of the properties of the figures. We will review those properties as needed.

The next example involves the #strong[area] of a triangle. The area of a triangle is one-half the base times the height. We can write this as #math.equation(block: false, alt: "A equals the fraction 1 over 2 b h ,")[$A = frac(1, 2) b h ,$] where #emph[b] = length of the base and #emph[h] = height.

#figure(figph[The figure is a triangle with its height shown. Its base is b and its height is h. The formula for the area of the triangle is A is equal to one-half times b times h.], alt: "The figure is a triangle with its height shown. Its base is b and its height is h. The formula for the area of the triangle is A is equal to one-half times b times h.", caption: none)

#examplebox("Example 5")[][
The area of a triangular painting is 126 square inches. The base is 18 inches. What is the height?

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#strong[Step 1. Read] the problem.], []),
  [#strong[Step 2. Identify] what you are looking for.], [height of a triangle],
  [#strong[Step 3. Name.]], [],
  [Choose a variable to represent it.], [Let #math.equation(block: false, alt: "h equals")[$h =$] the height.],
  [Draw the figure and label it with the given information.], [Area = 126 sq. in.],
  [], [#figure(figph[A white equilateral triangle is depicted with a dashed line representing its height, labeled 'h', extending from the top vertex perpendicularly to the base. The base of the triangle is labeled '18 in'.], alt: "A white equilateral triangle is depicted with a dashed line representing its height, labeled 'h', extending from the top vertex perpendicularly to the base. The base of the triangle is labeled '18 in'.", caption: none)],
  [#strong[Step 4. Translate.]], [],
  [Write the appropriate formula.], [#math.equation(block: false, alt: "A equals the fraction 1 over 2 b h")[$#h(1.8em) A = frac(1, 2) b h$]],
  [Substitute in the given information.], [#math.equation(block: false, alt: "126 equals the fraction 1 over 2 times 18 times h")[$#h(1em) 126 = frac(1, 2) · 18 · h$]],
  [#strong[Step 5. Solve] the equation.], [#math.equation(block: false, alt: "126 equals 9 h")[$#h(1em) 126 = 9 h$]],
  [Divide both sides by 9.], [#math.equation(block: false, alt: "14 equals h")[$#h(1.5em) 14 = h$]],
  [#strong[Step 6. Check]. #linebreak()  #linebreak() #math.equation(block: false, alt: "A, equals, the fraction 1 over 2 b h; 126, equals ?, the fraction 1 over 2 times 18 times 14; 126, equals, 126 ✓")[$A & = & frac(1, 2) b h \ 126 & limits(=)^(?) & frac(1, 2) · 18 · 14 \ 126 & = & 126 ✓$]], [],
  [#strong[Step 7. Answer] the question.], [The height of the triangle is 14 inches.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The area of a triangular church window is 90 square meters. The base of the window is 15 meters. What is the window’s height?

#solutionbox[
The window’s height is 12 meters.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
A triangular tent door has an area of 15 square feet. The height is five feet. What is the length of the base?

#solutionbox[
The length of the base is 6 feet.
]
]

In the next example, we will work with a #strong[right triangle]. To solve for the measure of each angle, we need to use two triangle properties. In any triangle, the sum of the measures of the angles is #math.equation(block: false, alt: "180 ° .")[$180 "°" .$] We can write this as a formula: #math.equation(block: false, alt: "m ∠ A plus m ∠ B plus m ∠ C equals 180 .")[$m ∠ A + m ∠ B + m ∠ C = 180 .$] Also, since the triangle is a right triangle, we remember that a right triangle has one #math.equation(block: false, alt: "90 °")[$90 "°"$] angle.

Here, we will have to define one angle in terms of another. We will wait to draw the figure until we write expressions for all the angles we are looking for.

#examplebox("Example 6")[][
The measure of one angle of a right triangle is 40 degrees more than the measure of the smallest angle. Find the measures of all three angles.

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#strong[Step 1. Read] the problem.], []),
  [#strong[Step 2. Identify] what you are looking for.], [the measures of all three angles],
  [#strong[Step 3. Name.] Choose a variable to represent it.], [#math.equation(block: false, alt: "Let a, equals, 1 to the power st angle.; a plus 40, equals, 2 to the power nd angle; 90, equals, 3 to the power rd angle (the right angle)")[$"Let" #h(0.2em) a & = & 1^("st") #h(0.2em) "angle." \ a + 40 & = & 2^("nd") #h(0.2em) "angle" \ 90 & = & 3^("rd") #h(0.2em) "angle (the right angle)"$]],
  [Draw the figure and label it with the given information.], [#figure(figph[A right-angled triangle ABC is shown, with the right angle at C. Angle B is labeled 'a' and angle A is labeled 'a + 40'.], alt: "A right-angled triangle ABC is shown, with the right angle at C. Angle B is labeled 'a' and angle A is labeled 'a + 40'.", caption: none)],
  [#strong[Step 4. Translate.]], [],
  [Write the appropriate formula.], [#figure(figph[A mathematical equation displays the sum of the measures of angles A, B, and C as 180 degrees, represented as m∠A + m∠B + m∠C = 180, on a white background.], alt: "A mathematical equation displays the sum of the measures of angles A, B, and C as 180 degrees, represented as m∠A + m∠B + m∠C = 180, on a white background.", caption: none)],
  [Substitute into the formula.], [#figure(figph[A mathematical equation is displayed, showing 'a + (a + 40) + 90 = 180' in bold, dark gray text on a white background.], alt: "A mathematical equation is displayed, showing 'a + (a + 40) + 90 = 180' in bold, dark gray text on a white background.", caption: none)],
  [#strong[Step 5. Solve] the equation.], [#figure(figph[A math problem demonstrating the steps to find three angles. It begins with 2a + 130 = 180, solves for 'a' (25 for the first angle), then calculates the second angle (a+40=65), and lists the third angle as 90.], alt: "A math problem demonstrating the steps to find three angles. It begins with 2a + 130 = 180, solves for 'a' (25 for the first angle), then calculates the second angle (a+40=65), and lists the third angle as 90.", caption: none)],
  [#strong[Step 6. Check]. #linebreak()  #linebreak() #math.equation(block: false, alt: "25 plus 65 plus 90, equals ?, 180; 180, equals, 180 ✓")[$25 + 65 + 90 & limits(=)^(?) & 180 \ 180 & = & 180 ✓$]], [],
  [#strong[Step 7. Answer] the question.], [The three angles measure #math.equation(block: false, alt: "25 ° , 65 ° ,")[$25 "°" , 65 "°" ,$] and #math.equation(block: false, alt: "90 ° .")[$90 "°" .$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The measure of one angle of a right triangle is 50 more than the measure of the smallest angle. Find the measures of all three angles.

#solutionbox[
The measures of the angles are 20°, 70°, and 90°.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The measure of one angle of a right triangle is 30 more than the measure of the smallest angle. Find the measures of all three angles.

#solutionbox[
The measures of the angles are 30°, 60°, and 90°.
]
]

The next example uses another important geometry formula. The #strong[Pythagorean Theorem] tells how the lengths of the three sides of a right triangle relate to each other. Writing the formula in every exercise and saying it aloud as you write it may help you memorize the Pythagorean Theorem.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[The Pythagorean Theorem]

In any right triangle, where #emph[a] and #emph[b] are the lengths of the legs, and #emph[c] is the length of the hypotenuse, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse.

#figure(figph[The figure is a right triangle with sides a and b, and a hypotenuse c. a squared plus b squared is equal to c squared. In a right triangle, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse.], alt: "The figure is a right triangle with sides a and b, and a hypotenuse c. a squared plus b squared is equal to c squared. In a right triangle, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse.", caption: none)
]

We will use the Pythagorean Theorem in the next example.

#examplebox("Example 7")[][
Use the Pythagorean Theorem to find the length of the other leg in

#figure(figph[This figure is a right triangle with one leg that is 12 units and a hypotenuse that is 13 units.], alt: "This figure is a right triangle with one leg that is 12 units and a hypotenuse that is 13 units.", caption: none)

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#strong[Step 1. Read] the problem.], []),
  [#strong[Step 2. Identify] what you are looking for.], [the length of the leg of the triangle],
  [#strong[Step 3. Name.]], [],
  [Choose a variable to represent it.], [Let #emph[a] = the leg of the triangle.],
  [Label side #emph[a].], [#figure(figph[A right-angled triangle with one leg labeled 'a', the other leg labeled '12', and the hypotenuse labeled '13'. A square symbol indicates the right angle.], alt: "A right-angled triangle with one leg labeled 'a', the other leg labeled '12', and the hypotenuse labeled '13'. A square symbol indicates the right angle.", caption: none)],
  [#strong[Step 4. Translate.]], [],
  [Write the appropriate formula. #linebreak() Substitute.], [#math.equation(block: false, alt: "a squared plus b squared, equals, c squared; a squared plus 12 squared, equals, 13 squared")[$a^(2) + b^(2) & = & c^(2) \ a^(2) + 12^(2) & = & 13^(2)$]],
  [#strong[Step 5. Solve] the equation. #linebreak() Isolate the variable term. #linebreak() Use the definition of square root. #linebreak() Simplify.], [#math.equation(block: false, alt: "a squared plus 144 to the power , equals, 169; a squared, equals, 25; a, equals, the square root of 25; a, equals, 5")[$a^(2) + 144^ & = & 169 \ a^(2) & = & 25 \ a & = & sqrt(25) \ a & = & 5$]],
  [#strong[Step 6. Check.] #linebreak() #figure(figph[A clear step-by-step verification of the Pythagorean theorem using the 5-12-13 triangle, demonstrating that 5 squared plus 12 squared equals 13 squared, resulting in 169 = 169.], alt: "A clear step-by-step verification of the Pythagorean theorem using the 5-12-13 triangle, demonstrating that 5 squared plus 12 squared equals 13 squared, resulting in 169 = 169.", caption: none)], [],
  [#strong[Step 7. Answer] the question.], [The length of the leg is 5.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Use the Pythagorean Theorem to find the length of the leg in the figure.

#figure(figph[The figure is a right triangle with legs that are b units and 15 units, and a hypotenuse that is 17 units.], alt: "The figure is a right triangle with legs that are b units and 15 units, and a hypotenuse that is 17 units.", caption: none)

#solutionbox[
The length of the leg is 8.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Use the Pythagorean Theorem to find the length of the leg in the figure.

#figure(figph[The figure is a right triangle with legs that are b units and 9 units, and a hypotenuse that is 15 units.], alt: "The figure is a right triangle with legs that are b units and 9 units, and a hypotenuse that is 15 units.", caption: none)

#solutionbox[
The length of the leg is 12.
]
]

The next example is about the #strong[perimeter] of a rectangle. Since the perimeter is just the distance around the rectangle, we find the sum of the lengths of its four sides—the sum of two lengths and two widths. We can write is as #math.equation(block: false, alt: "P equals 2 L plus 2 W")[$P = 2 L + 2 W$] where #emph[L] is the length and #math.equation(block: false, alt: "W")[$W$] is the width. To solve the example, we will need to define the length in terms of the width.

#examplebox("Example 8")[][
The length of a rectangle is six centimeters more than twice the width. The perimeter is 96 centimeters. Find the length and width.

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#strong[Step 1. Read] the problem.], []),
  [#strong[Step 2. Identify] what we are looking for.], [the length and the width],
  [#strong[Step 3. Name]. Choose a variable to represent the width. #linebreak() 
The length is six more than twice the width.], [Let #math.equation(block: false, alt: "w equals")[$#h(0.5em) w =$] width. #linebreak() #math.equation(block: false, alt: "2 w plus 6 equals")[$2 w + 6 =$] length #linebreak()  #linebreak() #figure(figph[A geometric illustration features a rectangle. The width of the rectangle is labeled as 'W', and its length is labeled with the expression '2W + 6', indicating the relationship between its dimensions.], alt: "A geometric illustration features a rectangle. The width of the rectangle is labeled as 'W', and its length is labeled with the expression '2W + 6', indicating the relationship between its dimensions.", caption: none)

 #linebreak()  #linebreak() #math.equation(block: false, alt: "P equals 96")[$P = 96$] cm],
  [#strong[Step 4. Translate.]], [],
  [Write the appropriate formula.], [#figure(figph[The formula P = 2L + 2W is displayed, representing the perimeter of a rectangle, where P is perimeter, L is length, and W is width. This fundamental equation is a cornerstone of geometry.], alt: "The formula P = 2L + 2W is displayed, representing the perimeter of a rectangle, where P is perimeter, L is length, and W is width. This fundamental equation is a cornerstone of geometry.", caption: none)],
  [Substitute in the given information.], [#figure(figph[A mathematical equation is displayed, showing 96 = 2(2W + 6) + 2W, which is an algebraic expression likely intended for solving for the variable W.], alt: "A mathematical equation is displayed, showing 96 = 2(2W + 6) + 2W, which is an algebraic expression likely intended for solving for the variable W.", caption: none)],
  [#strong[Step 5. Solve the equation.]], [#figure(figph[This image illustrates an algebraic solution for finding the width (W=14) and subsequent length (34 cm) of an object, starting from a perimeter equation and showing each calculation step.], alt: "This image illustrates an algebraic solution for finding the width (W=14) and subsequent length (34 cm) of an object, starting from a perimeter equation and showing each calculation step.", caption: none)],
  [#strong[Step 6.] Check. #linebreak()  #linebreak() #figure(figph[A rectangle is shown with a height of 14 cm and a width of 34 cm, indicating its dimensions.], alt: "A rectangle is shown with a height of 14 cm and a width of 34 cm, indicating its dimensions.", caption: none)

 #linebreak()  #linebreak() #math.equation(block: false, alt: "P, equals, 2 L plus 2 W; 96, equals ?, 2 times 34 plus 2 times 14; 96, equals, 96 ✓")[$P & = & 2 L + 2 W \ 96 & limits(=)^(?) & 2 · 34 + 2 · 14 \ 96 & = & 96 ✓$]], [],
  [#strong[Step 7. Answer the question.]], [The length is 34 cm and the width is 14 cm.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The length of a rectangle is seven more than twice the width. The perimeter is 110 inches. Find the length and width.

#solutionbox[
The length is 39 inches and the width is 16 inches.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The width of a rectangle is eight yards less than twice the length. The perimeter is 86 yards. Find the length and width.

#solutionbox[
The length is 17 yards and the width is 26 yards.
]
]

The next example is about the #strong[perimeter] of a triangle. Since the perimeter is just the distance around the triangle, we find the sum of the lengths of its three sides. We can write this as #math.equation(block: false, alt: "P equals a plus b plus c ,")[$P = a + b + c ,$] where #emph[a], #emph[b], and #emph[c] are the lengths of the sides.

#examplebox("Example 9")[][
One side of a triangle is three inches more than the first side. The third side is two inches more than twice the first. The perimeter is 29 inches. Find the length of the three sides of the triangle.

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#strong[Step 1. Read] the problem.], []),
  [#strong[Step 2. Identify] what we are looking for.], [the lengths of the three sides of a triangle],
  [#strong[Step 3. Name]. Choose a variable to #linebreak() represent the length of the first side.], [#math.equation(block: false, alt: "Let x, equals, length of 1 to the power st side.; x plus 3, equals, length of 2 to the power nd side; 2 x plus 2, equals, length of 3 to the power rd side")[$"Let" #h(0.2em) x & = & "length of" #h(0.2em) 1^("st") #h(0.2em) "side." \ x + 3 & = & "length of" #h(0.2em) 2^("nd") #h(0.2em) "side" \ 2 x + 2 & = & "length of" #h(0.2em) 3^("rd") #h(0.2em) "side"$] #linebreak() #figure(figph[A triangle with side lengths x, x+3, and 2x+2, with the text 'Perimeter is 29 in.' displayed next to it.], alt: "A triangle with side lengths x, x+3, and 2x+2, with the text 'Perimeter is 29 in.' displayed next to it.", caption: none)],
  [#strong[Step 4. Translate.] #linebreak() Write the appropriate formula. #linebreak() Substitute in the given information.], [#figure(figph[A mathematical formula is shown, P = a + b + c, typically representing the perimeter of a shape or a sum of three variables.], alt: "A mathematical formula is shown, P = a + b + c, typically representing the perimeter of a shape or a sum of three variables.", caption: none)

 #linebreak() #figure(figph[A mathematical equation is displayed, showing 29 equal to the sum of x, (x + 3), and (2x + 2).], alt: "A mathematical equation is displayed, showing 29 equal to the sum of x, (x + 3), and (2x + 2).", caption: none)],
  [#strong[Step 5. Solve] the equation.], [#figure(figph[Algebraic problem solution: solving 29 = 4x + 5 for x, then calculating side lengths x+3 and 2x+2 by substituting the value of x.], alt: "Algebraic problem solution: solving 29 = 4x + 5 for x, then calculating side lengths x+3 and 2x+2 by substituting the value of x.", caption: none)],
  [#strong[Step 6. Check.] #linebreak() #figure(figph[A scalene triangle with side lengths labeled as 6, 9, and 14.], alt: "A scalene triangle with side lengths labeled as 6, 9, and 14.", caption: none)

 #linebreak() #math.equation(block: false, alt: "29 equals ? 6 plus 9 plus 14")[$#h(2em) 29 limits(=)^(?) 6 + 9 + 14$] #linebreak() #math.equation(block: false, alt: "29 equals 29 ✓")[$#h(2em) 29 = 29 ✓$]], [],
  [#strong[Step 7. Answer] the question.], [The lengths of the sides of the triangle #linebreak() are 6, 9, and 14 inches.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
One side of a triangle is seven inches more than the first side. The third side is four inches less than three times the first. The perimeter is 28 inches. Find the length of the three sides of the triangle.

#solutionbox[
The lengths of the sides of the triangle are 5, 11 and 12 inches.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
One side of a triangle is three feet less than the first side. The third side is five feet less than twice the first. The perimeter is 20 feet. Find the length of the three sides of the triangle.

#solutionbox[
The lengths of the sides of the triangle are 4, 7 and 9 feet.
]
]

#examplebox("Example 10")[][
The perimeter of a rectangular soccer field is 360 feet. The length is 40 feet more than the width. Find the length and width.

#figure(figph[The figure is an illustration of rectangular soccer field.], alt: "The figure is an illustration of rectangular soccer field.", caption: none)

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#strong[Step 1. Read] the problem.], []),
  [#strong[Step 2. Identify] what we are looking for.], [the length and width of the soccer field],
  [#strong[Step 3. Name.] Choose a variable to represent it. #linebreak() The length is 40 feet more than the width. #linebreak() Draw the figure and label it with the #linebreak() given information.], [Let #emph[w] = width. #linebreak() #math.equation(block: false, alt: "w plus 40 equals")[$w + 40 =$] length #linebreak() #figure(figph[A soccer field diagram illustrates a perimeter problem. The field has a perimeter of 360 feet, with one side labeled 'W' and the other 'W + 40', setting up an algebraic equation.], alt: "A soccer field diagram illustrates a perimeter problem. The field has a perimeter of 360 feet, with one side labeled 'W' and the other 'W + 40', setting up an algebraic equation.", caption: none)],
  [#strong[Step 4. Translate]. #linebreak() Write the appropriate formula and #linebreak() substitute.], [#figure(figph[The mathematical formula P = 2L + 2W, which represents the perimeter of a rectangle where P is the perimeter, L is the length, and W is the width, is displayed on a white background.], alt: "The mathematical formula P = 2L + 2W, which represents the perimeter of a rectangle where P is the perimeter, L is the length, and W is the width, is displayed on a white background.", caption: none)

 #linebreak() #figure(figph[A mathematical equation is displayed, stating 360 equals 2 multiplied by the quantity of w plus 40, plus 2 multiplied by w. The equation is '360 = 2(w + 40) + 2w'.], alt: "A mathematical equation is displayed, stating 360 equals 2 multiplied by the quantity of w plus 40, plus 2 multiplied by w. The equation is '360 = 2(w + 40) + 2w'.", caption: none)],
  [#strong[Step 5. Solve] the equation.], [#figure(figph[Calculation showing the steps to determine the width (70) and length (110) of a field, derived from the equation 360 = 2w + 80 + 2w.], alt: "Calculation showing the steps to determine the width (70) and length (110) of a field, derived from the equation 360 = 2w + 80 + 2w.", caption: none)],
  [#strong[Step 6. Check.] #linebreak()  #linebreak() #math.equation(block: false, alt: "P, equals, 2 L plus 2 W; 360, equals ?, 2 open parenthesis 110 close parenthesis plus 2 open parenthesis 70 close parenthesis; 360, equals, 360 ✓")[$P & = & 2 L + 2 W \ 360 & limits(=)^(?) & 2 ( 110 ) + 2 ( 70 ) \ 360 & = & 360 ✓$]], [],
  [#strong[Step 7. Answer] the question.], [The length of the soccer field is 110 feet #linebreak() and the width is 70 feet.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The perimeter of a rectangular swimming pool is 200 feet. The length is 40 feet more than the width. Find the length and width.

#solutionbox[
The length of the swimming pool is 70 feet and the width is 30 feet.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The length of a rectangular garden is 30 yards more than the width. The perimeter is 300 yards. Find the length and width.

#solutionbox[
The length of the garden is 90 yards and the width is 60 yards.
]
]

Applications of these geometric properties can be found in many everyday situations as shown in the next example.

#examplebox("Example 11")[][
Kelvin is building a gazebo and wants to brace each corner by placing a 10” piece of wood diagonally as shown.

#figure(figph[The figure is an illustration of a gazebo whose corner forms a right triangle with a 10 inch piece of wood that is placed diagonally to brace it.], alt: "The figure is an illustration of a gazebo whose corner forms a right triangle with a 10 inch piece of wood that is placed diagonally to brace it.", caption: none)

How far from the corner should he fasten the wood if wants the distances from the corner to be equal? Approximate to the nearest tenth of an inch.

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#strong[Step 1. Read] the problem.], []),
  [#strong[Step 2. Identify] what we are looking for.], [the distance from the corner that the #linebreak() bracket should be attached],
  [#strong[Step 3. Name.] Choose a variable to represent it. #linebreak() Draw the figure and label it with the given #linebreak() information.], [Let #math.equation(block: false, alt: "x equals")[$x =$] the distance from the corner. #linebreak() #figure(figph[A right-angled triangle with two equal sides labeled 'x' and a hypotenuse of 10. This diagram is used to illustrate the Pythagorean theorem or an isosceles right triangle.], alt: "A right-angled triangle with two equal sides labeled 'x' and a hypotenuse of 10. This diagram is used to illustrate the Pythagorean theorem or an isosceles right triangle.", caption: none)],
  [#strong[Step 4. Translate.] #linebreak() Write the appropriate formula and substitute.], [#linebreak() #math.equation(block: false, alt: "a squared plus b squared equals c squared")[$a^(2) + b^(2) = c^(2)$] #linebreak() #math.equation(block: false, alt: "x squared plus x squared equals 10 squared")[$x^(2) + x^(2) = 10^(2)$]],
  [#strong[Step 5. Solve] the equation. #linebreak() Isolate the variable. #linebreak() Use the definition of square root. #linebreak() Simplify. Approximate to the nearest tenth.], [#math.equation(block: false, alt: "2 x squared, equals, 100; x squared, equals, 50; x, equals, the square root of 50; x, approximately equals, 7.1")[$2 x^(2) & = & 100 \ x^(2) & = & 50 \ x & = & sqrt(50) \ x & ≈ & 7.1$]],
  [#strong[Step 6. Check.] #linebreak() #math.equation(block: false, alt: "a squared plus b squared, equals, c squared; open parenthesis 7.1 close parenthesis squared plus open parenthesis 7.1 close parenthesis squared, approximately equals, 10 squared Yes.")[$a^(2) + b^(2) & = & c^(2) \ attach(( 7.1 ), t: 2) + attach(( 7.1 ), t: 2) & ≈ & 10^(2) #h(1em) "Yes."$]], [],
  [#strong[Step 7. Answer] the question.], [Kelvin should fasten each piece of wood #linebreak() approximately 7.1” from the corner.],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
John puts the base of a 13-foot ladder five feet from the wall of his house as shown in the figure. How far up the wall does the ladder reach?

#figure(figph[The figure is an illustration that shows a ladder placed against the wall of a house. The ladder forms a right triangle with the side of the house. The ladder is 13 feet long and the base of the ladder is 5 feet from the wall of the house.], alt: "The figure is an illustration that shows a ladder placed against the wall of a house. The ladder forms a right triangle with the side of the house. The ladder is 13 feet long and the base of the ladder is 5 feet from the wall of the house.", caption: none)

#solutionbox[
The ladder reaches 12 feet.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Randy wants to attach a 17-foot string of lights to the top of the 15 foot mast of his sailboat, as shown in the figure. How far from the base of the mast should he attach the end of the light string?

#figure(figph[The figure is an illustration of a sailboat that has a 15 foot mast. A string of lights that are 17 feet long are placed diagonally from the top of the mast.], alt: "The figure is an illustration of a sailboat that has a 15 foot mast. A string of lights that are 17 feet long are placed diagonally from the top of the mast.", caption: none)

#solutionbox[
He should attach the lights 8 feet from the base of the mast.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Access this online resource for additional instruction and practice with solving for a variable in literal equations.

- #link("https://openstax.org/l/37literalequat")[Solving Literal Equations]
]


=== Key Concepts

- #strong[How To Solve Geometry Applications]
+ #strong[Read] the problem and make sure all the words and ideas are understood.
+ #strong[Identify] what you are looking for.
+ #strong[Name] what you are looking for by choosing a variable to represent it. Draw the figure and label it with the given information.
+ #strong[Translate] into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
+ #strong[Solve] the equation using good algebra techniques.
+ #strong[Check] the answer in the problem and make sure it makes sense.
+ #strong[Answer] the question with a complete sentence.
- #strong[The Pythagorean Theorem]
- In any right triangle, where #emph[a] and #emph[b] are the lengths of the legs, and #emph[c] is the length of the hypotenuse, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse. #linebreak() 
#figure(figph[The figure is a right triangle with sides a and b, and a hypotenuse c with the formula, a squared plus b squared is equal to c squared.], alt: "The figure is a right triangle with sides a and b, and a hypotenuse c with the formula, a squared plus b squared is equal to c squared.", caption: none)


==== Practice Makes Perfect

#strong[Solve a Formula for a Specific Variable]

In the following exercises, solve the given formula for the specified variable.

Solve the formula #math.equation(block: false, alt: "C equals π d")[$C = π d$] for #emph[d].

#solutionbox[
#math.equation(block: true, alt: "d equals the fraction C over π")[$d = frac(C, π)$]
]

Solve the formula #math.equation(block: false, alt: "C equals π d")[$C = π d$] for #math.equation(block: false, alt: "π .")[$π .$]

Solve the formula #math.equation(block: false, alt: "V equals L W H")[$V = L W H$] for #emph[L].

#solutionbox[
#math.equation(block: true, alt: "L equals the fraction V over W H")[$L = frac(V, W H)$]
]

Solve the formula #math.equation(block: false, alt: "V equals L W H")[$V = L W H$] for #emph[H].

Solve the formula #math.equation(block: false, alt: "A equals the fraction 1 over 2 b h")[$A = frac(1, 2) b h$] for #emph[b].

#solutionbox[
#math.equation(block: true, alt: "b equals the fraction 2 A over h")[$b = frac(2 A, h)$]
]

Solve the formula #math.equation(block: false, alt: "A equals the fraction 1 over 2 b h")[$A = frac(1, 2) b h$] for #emph[h].

Solve the formula #linebreak() #math.equation(block: false, alt: "A equals the fraction 1 over 2 d sub 1 d sub 2")[$A = frac(1, 2) d_(1) d_(2)$] for #math.equation(block: false, alt: "d sub 1 .")[$d_(1) .$]

#solutionbox[
#math.equation(block: true, alt: "d sub 1 equals the fraction 2 A over d sub 2")[$d_(1) = frac(2 A, d_(2))$]
]

Solve the formula #linebreak() #math.equation(block: false, alt: "A equals the fraction 1 over 2 d sub 1 d sub 2")[$A = frac(1, 2) d_(1) d_(2)$] for #math.equation(block: false, alt: "d sub 2 .")[$d_(2) .$]

Solve the formula #linebreak() #math.equation(block: false, alt: "A equals the fraction 1 over 2 h open parenthesis b sub 1 plus b sub 2 close parenthesis")[$A = frac(1, 2) h ( b_(1) + b_(2) )$] for #math.equation(block: false, alt: "b sub 1 .")[$b_(1) .$]

#solutionbox[
#math.equation(block: true, alt: "b sub 1 equals the fraction 2 A over h minus b sub 2")[$b_(1) = frac(2 A, h) − b_(2)$]
]

Solve the formula #linebreak() #math.equation(block: false, alt: "A equals the fraction 1 over 2 h open parenthesis b sub 1 plus b sub 2 close parenthesis")[$A = frac(1, 2) h ( b_(1) + b_(2) )$] for #math.equation(block: false, alt: "b sub 2 .")[$b_(2) .$]

Solve the formula #linebreak() #math.equation(block: false, alt: "h equals 54 t plus the fraction 1 over 2 a t squared")[$h = 54 t + frac(1, 2) a t^(2)$] for #emph[a].

#solutionbox[
#math.equation(block: true, alt: "a equals the fraction 2 h minus 108 t over t squared")[$a = frac(2 h − 108 t, t^(2))$]
]

Solve the formula #linebreak() #math.equation(block: false, alt: "h equals 48 t plus the fraction 1 over 2 a t squared")[$h = 48 t + frac(1, 2) a t^(2)$] for #emph[a].

Solve #math.equation(block: false, alt: "180 equals a plus b plus c")[$180 = a + b + c$] for #emph[a].

#solutionbox[
#math.equation(block: true, alt: "a equals 180 minus b minus c")[$a = 180 − b − c$]
]

Solve #math.equation(block: false, alt: "180 equals a plus b plus c")[$180 = a + b + c$] for #emph[c].

Solve the formula #linebreak() #math.equation(block: false, alt: "A equals the fraction 1 over 2 p l plus B")[$A = frac(1, 2) p l + B$] for #emph[p].

#solutionbox[
#math.equation(block: true, alt: "p equals the fraction 2 A minus 2 B over l")[$p = frac(2 A − 2 B, l)$]
]

Solve the formula #linebreak() #math.equation(block: false, alt: "A equals the fraction 1 over 2 p l plus B")[$A = frac(1, 2) p l + B$] for #emph[l].

Solve the formula #linebreak() #math.equation(block: false, alt: "P equals 2 L plus 2 W")[$P = 2 L + 2 W$] for #emph[L].

#solutionbox[
#math.equation(block: true, alt: "L equals the fraction P minus 2 W over 2")[$L = frac(P − 2 W, 2)$]
]

Solve the formula #linebreak() #math.equation(block: false, alt: "P equals 2 L plus 2 W")[$P = 2 L + 2 W$] for #emph[W].

In the following exercises, solve for the formula for #emph[y].

Solve the formula #linebreak() #math.equation(block: false, alt: "8 x plus y equals 15")[$8 x + y = 15$] for #emph[y].

#solutionbox[
#math.equation(block: true, alt: "y equals 15 minus 8 x")[$y = 15 − 8 x$]
]

Solve the formula #linebreak() #math.equation(block: false, alt: "9 x plus y equals 13")[$9 x + y = 13$] for #emph[y].

Solve the formula #linebreak() #math.equation(block: false, alt: "−4 x plus y equals −6")[$−4 x + y = −6$] for #emph[y].

#solutionbox[
#math.equation(block: true, alt: "y equals −6 plus 4 x")[$y = −6 + 4 x$]
]

Solve the formula #linebreak() #math.equation(block: false, alt: "−5 x plus y equals −1")[$−5 x + y = −1$] for #emph[y].

Solve the formula #linebreak() #math.equation(block: false, alt: "x minus y equals −4")[$x − y = −4$] for #emph[y].

#solutionbox[
#math.equation(block: true, alt: "y equals 4 plus x")[$y = 4 + x$]
]

Solve the formula #linebreak() #math.equation(block: false, alt: "x minus y equals −3")[$x − y = −3$] for #emph[y].

Solve the formula #linebreak() #math.equation(block: false, alt: "4 x plus 3 y equals 7")[$4 x + 3 y = 7$] for #emph[y].

#solutionbox[
#math.equation(block: true, alt: "y equals the fraction 7 minus 4 x over 3")[$y = frac(7 − 4 x, 3)$]
]

Solve the formula #linebreak() #math.equation(block: false, alt: "3 x plus 2 y equals 11")[$3 x + 2 y = 11$] for #emph[y].

Solve the formula #linebreak() #math.equation(block: false, alt: "2 x plus 3 y equals 12")[$2 x + 3 y = 12$] for #emph[y].

#solutionbox[
#math.equation(block: true, alt: "y equals the fraction 12 minus 2 x over 3")[$y = frac(12 − 2 x, 3)$]
]

Solve the formula #linebreak() #math.equation(block: false, alt: "5 x plus 2 y equals 10")[$5 x + 2 y = 10$] for #emph[y].

Solve the formula #linebreak() #math.equation(block: false, alt: "3 x minus 2 y equals 18")[$3 x − 2 y = 18$] for #emph[y].

#solutionbox[
#math.equation(block: true, alt: "y equals the fraction 18 minus 3 x over −2")[$y = frac(18 − 3 x, −2)$]
]

Solve the formula #linebreak() #math.equation(block: false, alt: "4 x minus 3 y equals 12")[$4 x − 3 y = 12$] for #emph[y].

#strong[Use Formulas to Solve Geometry Applications]

In the following exercises, solve using a geometry formula.

A triangular flag has area 0.75 square feet and height 1.5 foot. What is its base?

#solutionbox[
1 foot
]

A triangular window has area 24 square feet and height six feet. What is its base?

What is the base of a triangle with area 207 square inches and height 18 inches?

#solutionbox[
23 inches
]

What is the height of a triangle with area 893 square inches and base 38 inches?

The two smaller angles of a right triangle have equal measures. Find the measures of all three angles.

#solutionbox[
#math.equation(block: true, alt: "45 ° , 45 ° , 90 °")[$45 "°" , 45 "°" , 90 "°"$]
]

The measure of the smallest angle of a right triangle is #math.equation(block: false, alt: "20 °")[$20 "°"$] less than the measure of the next larger angle. Find the measures of all three angles.

The angles in a triangle are such that one angle is twice the smallest angle, while the third angle is three times as large as the smallest angle. Find the measures of all three angles.

#solutionbox[
#math.equation(block: true, alt: "30 ° , 60 ° , 90 °")[$30 "°" , 60 "°" , 90 "°"$]
]

The angles in a triangle are such that one angle is 20 more than the smallest angle, while the third angle is three times as large as the smallest angle. Find the measures of all three angles.

In the following exercises, use the Pythagorean Theorem to find the length of the hypotenuse.

#figure(figph[The figure is a right triangle with sides 9 units and 12 units.], alt: "The figure is a right triangle with sides 9 units and 12 units.", caption: none)

#solutionbox[
15
]

#figure(figph[The figure is a right triangle with sides 16 units and 12 units.], alt: "The figure is a right triangle with sides 16 units and 12 units.", caption: none)

#figure(figph[The figure is a right triangle with sides 15 units and 20 units.], alt: "The figure is a right triangle with sides 15 units and 20 units.", caption: none)

#solutionbox[
25
]

#figure(figph[The figure is a right triangle with sides 5 units and 12 units.], alt: "The figure is a right triangle with sides 5 units and 12 units.", caption: none)

In the following exercises, use the Pythagorean Theorem to find the length of the leg. Round to the nearest tenth if necessary.

#figure(figph[The figure is a right triangle with sides 6 units and 10 units.], alt: "The figure is a right triangle with sides 6 units and 10 units.", caption: none)

#solutionbox[
8
]

#figure(figph[The figure is a right triangle with a side that is 9 units and a hypotenuse that is 13 units.], alt: "The figure is a right triangle with a side that is 9 units and a hypotenuse that is 13 units.", caption: none)

#figure(figph[The figure is a right triangle with a side that is 5 units and a hypotenuse that is 13 units.], alt: "The figure is a right triangle with a side that is 5 units and a hypotenuse that is 13 units.", caption: none)

#solutionbox[
12
]

#figure(figph[The figure is a right triangle with a side that is 16 units and a hypotenuse that is 20 units.], alt: "The figure is a right triangle with a side that is 16 units and a hypotenuse that is 20 units.", caption: none)

#figure(figph[The figure is a right triangle with a side that is 8 units and a hypotenuse that is 13 units.], alt: "The figure is a right triangle with a side that is 8 units and a hypotenuse that is 13 units.", caption: none)

#solutionbox[
#math.equation(block: true, alt: "10.2")[$10.2$]
]

#figure(figph[The figure is a right triangle with sides that are both 6 units.], alt: "The figure is a right triangle with sides that are both 6 units.", caption: none)

#figure(figph[The figure is a right triangle with sides that are 5 units and 11 units.], alt: "The figure is a right triangle with sides that are 5 units and 11 units.", caption: none)

#solutionbox[
#math.equation(block: true, alt: "9.8")[$9.8$]
]

#figure(figph[The figure is a right triangle with sides that are 5 units and 7 units.], alt: "The figure is a right triangle with sides that are 5 units and 7 units.", caption: none)

In the following exercises, solve using a geometry formula.

The width of a rectangle is seven meters less than the length. The perimeter is 58 meters. Find the length and width.

#solutionbox[
18 meters, 11 meters
]

The length of a rectangle is eight feet more than the width. The perimeter is 60 feet. Find the length and width.

The width of the rectangle is 0.7 meters less than the length. The perimeter of a rectangle is 52.6 meters. Find the dimensions of the rectangle.

#solutionbox[
#math.equation(block: false, alt: "13.5")[$13.5$] m, #math.equation(block: false, alt: "12.8")[$12.8$] m
]

The length of the rectangle is 1.1 meters less than the width. The perimeter of a rectangle is 49.4 meters. Find the dimensions of the rectangle.

The perimeter of a rectangle of 150 feet. The length of the rectangle is twice the width. Find the length and width of the rectangle.

#solutionbox[
25 ft, 50 ft
]

The length of the rectangle is three times the width. The perimeter of a rectangle is 72 feet. Find the length and width of the rectangle.

The length of the rectangle is three meters less than twice the width. The perimeter of a rectangle is 36 meters. Find the dimensions of the rectangle.

#solutionbox[
#emph[l] = 11 m, #emph[w] = 7 m
]

The length of a rectangle is five inches more than twice the width. The perimeter is 34 inches. Find the length and width.

The perimeter of a triangle is 39 feet. One side of the triangle is one foot longer than the second side. The third side is two feet longer than the second side. Find the length of each side.

#solutionbox[
12 ft, 13 ft, 14 ft
]

The perimeter of a triangle is 35 feet. One side of the triangle is five feet longer than the second side. The third side is three feet longer than the second side. Find the length of each side.

One side of a triangle is twice the smallest side. The third side is five feet more than the shortest side. The perimeter is 17 feet. Find the lengths of all three sides.

#solutionbox[
3 ft, 6 ft, 8 ft
]

One side of a triangle is three times the smallest side. The third side is three feet more than the shortest side. The perimeter is 13 feet. Find the lengths of all three sides.

The perimeter of a rectangular field is 560 yards. The length is 40 yards more than the width. Find the length and width of the field.

#solutionbox[
120 yd, 160 yd
]

The perimeter of a rectangular atrium is 160 feet. The length is 16 feet more than the width. Find the length and width of the atrium.

A rectangular parking lot has perimeter 250 feet. The length is five feet more than twice the width. Find the length and width of the parking lot.

#solutionbox[
40 ft, 85 ft
]

A rectangular rug has perimeter 240 inches. The length is 12 inches more than twice the width. Find the length and width of the rug.

In the following exercises, solve. Approximate answers to the nearest tenth, if necessary.

A 13-foot string of lights will be attached to the top of a 12-foot pole for a holiday display as shown. How far from the base of the pole should the end of the string of lights be anchored?

#figure(figph[The figure is an illustration that shows a 13 foot string of lights attached diagonally to the top of a 12 foot pole.], alt: "The figure is an illustration that shows a 13 foot string of lights attached diagonally to the top of a 12 foot pole.", caption: none)

#solutionbox[
5 feet
]

am wants to put a banner across her garage door diagonally, as shown, to congratulate her son for his college graduation. The garage door is 12 feet high and 16 feet wide. Approximately how long should the banner be to fit the garage door?

#figure(figph[The figure is an illustration of a banner positioned diagonally across a garage door that is 12 feet high and 16 feet wide.], alt: "The figure is an illustration of a banner positioned diagonally across a garage door that is 12 feet high and 16 feet wide.", caption: none)

Chi is planning to put a diagonal path of paving stones through her flower garden as shown. The flower garden is a square with side 10 feet. What will the length of the path be?

#figure(figph[The figure is an illustration of a diagonal path of stones through a square garden with 10 foot sides.], alt: "The figure is an illustration of a diagonal path of stones through a square garden with 10 foot sides.", caption: none)

#solutionbox[
14.1 feet
]

Brian borrowed a 20-foot extension ladder to use when he paints his house. If he sets the base of the ladder six feet from the house as shown, how far up will the top of the ladder reach?

#figure(figph[The figure is an illustration of a house that has a ladder against it. The ladder is 20 feet. Its base is positioned 6 feet from the house.], alt: "The figure is an illustration of a house that has a ladder against it. The ladder is 20 feet. Its base is positioned 6 feet from the house.", caption: none)


==== Everyday Math

#strong[Converting temperature] While on a tour in Greece, Tatyana saw that the temperature was 40° Celsius. Solve for #emph[F] in the formula #math.equation(block: false, alt: "C equals the fraction 5 over 9 open parenthesis F minus 32 close parenthesis")[$C = frac(5, 9) ( F − 32 )$] to find the Fahrenheit temperature.

#solutionbox[
#math.equation(block: true, alt: "104 ° F")[$104 "°" #h(0.2em) "F"$]
]

#strong[Converting temperature] Yon was visiting the United States and he saw that the temperature in Seattle one day was 50° Fahrenheit. Solve for #emph[C] in the formula #math.equation(block: false, alt: "F equals the fraction 9 over 5 C plus 32")[$F = frac(9, 5) C + 32$] to find the Celsius temperature.

Christa wants to put a fence around her triangular flowerbed. The sides of the flowerbed are six feet, eight feet and 10 feet. How many feet of fencing will she need to enclose her flowerbed?

#solutionbox[
240 ft
]

Jose just removed the children’s play set from his back yard to make room for a rectangular garden. He wants to put a fence around the garden to keep the dog out. He has a 50-foot roll of fence in his garage that he plans to use. To fit in the backyard, the width of the garden must be 10 feet. How long can he make the other side?


==== Writing Exercises

If you need to put tile on your kitchen floor, do you need to know the perimeter or the area of the kitchen? Explain your reasoning.

#solutionbox[
Answers will vary.
]

If you need to put a fence around your backyard, do you need to know the perimeter or the area of the backyard? Explain your reasoning.

Look at the two figures below.

#figure(figph[A figure of a rectangle with a width that is 2 units and a length that is 8 units and a square with sides that are 4 units.], alt: "A figure of a rectangle with a width that is 2 units and a length that is 8 units and a square with sides that are 4 units.", caption: none)

ⓐ Which figure looks like it has the larger area? Which looks like it has the larger perimeter? #linebreak() ⓑ Now calculate the area and perimeter of each figure. Which has the larger area? Which has the larger perimeter? #linebreak() ⓒ Were the results of part (b) the same as your answers in part (a)? Is that surprising to you?

#solutionbox[
ⓐ Answers will vary. ⓑ The areas are the same. The #math.equation(block: false, alt: "2 times 8")[$2 #h(0.2em) × #h(0.2em) 8$] rectangle has a larger perimeter than the #math.equation(block: false, alt: "4 times 4")[$4 #h(0.2em) × #h(0.2em) 4$] square. #linebreak() ⓒ Answers will vary.
]

Write a geometry word problem that relates to your life experience, then solve it and explain all your steps.


==== Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

#figure(figph[This table has four columns and three rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve a formula for a specific variable. In row 3, the I can was use formulas to solve geometry applications.], alt: "This table has four columns and three rows. The first row is a header and it labels each column, “I can…”, “Confidently,” “With some help,” and “No-I don’t get it!” In row 2, the I can was solve a formula for a specific variable. In row 3, the I can was use formulas to solve geometry applications.", caption: none)

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?