Solve Geometry Applications: Triangles, Rectangles, and the Pythagorean Theorem

Solve Applications Using Properties of Triangles

In this section we will use some common geometry formulas. We adapt our problem-solving strategy so that we can solve geometry applications. The geometry formula names the variables and gives us the equation to solve. Since these applications all involve a shape of some kind, it helps to draw the figure and label it with the given information — we add this as the first step of the strategy below.

We will start geometry applications by looking at the properties of triangles. Triangles have three sides and three interior angles. Usually each side is labeled with a lowercase letter to match the uppercase letter of the opposite vertex.

The plural of the word vertex is vertices. All triangles have three vertices. Triangles are named by their vertices: the triangle below is called $\triangle ABC$.

The three angles of a triangle are related in a special way: the sum of their measures is $180\degree$. We read $m\angle A$ as “the measure of angle $A$.” So in $\triangle ABC$ above,

Because the perimeter of a figure is the length of its boundary, the perimeter of $\triangle ABC$ is the sum of the lengths of its three sides:

To find the area of a triangle, we need to know its base and height. The height is a line that connects the base to the opposite vertex and makes a $90\degree$ angle with the base.

The formula for the area of $\triangle ABC$ is $A = \tfrac{1}{2}bh$, where $b$ is the base and $h$ is the height.

Example. The measures of two angles of a triangle are $55$ and $82$ degrees. Find the measure of the third angle.

Let $x =$ the measure of the third angle. Since the sum of the three angle measures is $180\degree$:

We check: $55 + 82 + 43 \overset{?}{=} 180$, and indeed $180 = 180$. ✓ The measure of the third angle is $43$ degrees.

Example. The perimeter of a triangular garden is $24$ feet. The lengths of two sides are four feet and nine feet. How long is the third side?

Let $c =$ the third side. Using $P = a + b + c$ with $P = 24$, $a = 4$, and $b = 9$:

We check: $24 \overset{?}{=} 4 + 9 + 11$, and indeed $24 = 24$. ✓ The third side is $11$ feet long.

Example. 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?

Let $h =$ the height. Using $A = \tfrac{1}{2}bh$ with $A = 90$ and $b = 15$:

We check: $90 \overset{?}{=} \tfrac{1}{2} \cdot 15 \cdot 12$, and indeed $90 = 90$. ✓ The height of the triangle is $12$ meters.

The triangle properties above apply to all triangles. Now we look at one specific type of triangle — a right triangle, which has one $90\degree$ angle, usually marked with a small square in the corner.

Example. One angle of a right triangle measures $28\degree$. What is the measure of the third angle?

The three angles are the right angle, the $28\degree$ angle, and the unknown angle $x$:

We check: $180 \overset{?}{=} 90 + 28 + 62$, and indeed $180 = 180$. ✓ The measure of the third angle is $62\degree$.

Example. The measure of one angle of a right triangle is $20$ degrees more than the measure of the smallest angle. Find the measures of all three angles.

Let $a =$ the first (smallest) angle. Then $a + 20 =$ the second angle, and $90 =$ the third angle (the right angle).

The first angle is $35\degree$. The second angle is $a + 20 = 35 + 20 = 55$ degrees. The third angle is the right angle, $90\degree$. We check: $35 + 55 + 90 \overset{?}{=} 180$, and indeed $180 = 180$. ✓ The three angles measure $35\degree$, $55\degree$, and $90\degree$.

Use the Pythagorean Theorem

We have learned how the measures of the angles of a triangle relate to each other. Now we will learn how the lengths of the sides of a right triangle relate to each other. This relationship is called the Pythagorean Theorem, named for the Greek philosopher and mathematician Pythagoras, who lived around 500 BC.

Before we state the Pythagorean Theorem, we need some vocabulary. Remember that a right triangle has a $90\degree$ angle, marked with a small square in the corner. The side of the triangle opposite the $90\degree$ angle is called the hypotenuse, and each of the other two sides is called a leg.

The Pythagorean Theorem tells how the lengths of the three sides of a right triangle relate to each other. It states that in any right triangle, the sum of the squares of the lengths of the two legs equals the square of the length of the hypotenuse. In symbols: in any right triangle, $a^2 + b^2 = c^2$, where $a$ and $b$ are the lengths of the legs and $c$ is the length of the hypotenuse.

To solve exercises that use the Pythagorean Theorem, we need to find square roots. Recall the definition: if $m = n^2$, then $\sqrt{m} = n$, for $n \ge 0$. For example, $\sqrt{25} = 5$ because $25 = 5^2$. Because the Pythagorean Theorem contains variables that are squared, solving for the length of a side of a right triangle requires square roots.

Example. Use the Pythagorean Theorem to find the length of the hypotenuse of a right triangle whose legs are $3$ and $4$.

Let $c =$ the length of the hypotenuse.

We check: $3^2 + 4^2 \overset{?}{=} 5^2$, that is, $9 + 16 \overset{?}{=} 25$, and indeed $25 = 25$. ✓ The length of the hypotenuse is $5$.

Example. Use the Pythagorean Theorem to find the length of the leg of a right triangle whose other leg is $5$ and whose hypotenuse is $13$.

Let $b =$ the leg of the triangle.

We check: $5^2 + 12^2 \overset{?}{=} 13^2$, that is, $25 + 144 \overset{?}{=} 169$, and indeed $169 = 169$. ✓ The length of the leg is $12$.

Example. Kelvin is building a gazebo and wants to brace each corner by placing a $10$-inch piece of wood diagonally as shown, so that the ends of the brace are the same distance from the corner. What is the length of the legs of the right triangle formed? Approximate to the nearest tenth of an inch.

Let $x =$ the distance from the corner. Since both legs are equal,

We check: $(7.1)^2 + (7.1)^2 \approx 10^2$. Yes. ✓ Kelvin should fasten each piece of wood approximately $7.1$ inches from the corner.

Solve Applications Using Rectangle Properties

You may already be familiar with the properties of rectangles. Rectangles have four sides and four right ($90\degree$) angles. The opposite sides of a rectangle are the same length. We refer to one side of the rectangle as the length, $L$, and its adjacent side as the width, $W$.

The distance around this rectangle is $L + W + L + W$, or $2L + 2W$. This is the perimeter, $P$, of the rectangle:

What about the area of a rectangle? Imagine a rectangular rug that is $2$-feet long by $3$-feet wide. Its area is $6$ square feet — there are six unit squares in the rug, arranged in $3$ rows of $2$:

The area is the length times the width. The formula for the area of a rectangle is $A = LW$.

Example. The length of a rectangle is $32$ meters and the width is $20$ meters. What is the perimeter?

Using $P = 2L + 2W$ with $L = 32$ m and $W = 20$ m:

We check: $20 + 32 + 20 + 32 \overset{?}{=} 104$, and indeed $104 = 104$. ✓ The perimeter of the rectangle is $104$ meters.

Example. The area of a rectangular room is $168$ square feet. The length is $14$ feet. What is the width?

Using $A = LW$ with $A = 168$ and $L = 14$:

We check: $168 \overset{?}{=} 14 \cdot 12$, and indeed $168 = 168$. ✓ The width of the room is $12$ feet.

Example. Find the length of a rectangle with perimeter $50$ inches and width $10$ inches.

Using $P = 2L + 2W$ with $P = 50$ and $W = 10$:

We check: $15 + 10 + 15 + 10 \overset{?}{=} 50$, and indeed $50 = 50$. ✓ The length is $15$ inches.

We have solved problems where either the length or the width was given, along with the perimeter or area. Now we will solve problems in which the width is defined in terms of the length. We wait to draw the figure until we have an expression for the width, so that we can label the figure with that expression.

Example. The width of a rectangle is two feet less than the length. The perimeter is $52$ feet. Find the length and width.

Since the width is defined in terms of the length, we let $L =$ length. The width is two feet less than the length, so $L - 2 =$ width.

Using $P = 2L + 2W$ with $P = 52$ and $W = L - 2$:

The length is $14$ feet. The width is $L - 2 = 14 - 2 = 12$ feet. Since $14 + 12 + 14 + 12 = 52$, this checks. ✓ The length is $14$ feet and the width is $12$ feet.

Example. The length of a rectangle is four centimeters more than twice the width. The perimeter is $32$ centimeters. Find the length and width.

Let $W =$ width. The length is four more than twice the width, so $2W + 4 =$ length.

Using $P = 2L + 2W$ with $P = 32$ and $L = 2W + 4$:

The width is $4$ cm. The length is $2W + 4 = 2(4) + 4 = 12$ cm. We check: $P \overset{?}{=} 2(12) + 2(4)$, that is $32 \overset{?}{=} 32$, and indeed $32 = 32$. ✓ The length is $12$ cm and the width is $4$ cm.

Example. The perimeter of a rectangular swimming pool is $150$ feet. The length is $15$ feet more than the width. Find the length and width.

Let $W =$ width. The length is $15$ feet more than the width, so $W + 15 =$ length.

Using $P = 2L + 2W$ with $P = 150$ and $L = W + 15$:

The width of the pool is $30$ feet. The length is $W + 15 = 30 + 15 = 45$ feet. We check: $150 \overset{?}{=} 2(45) + 2(30)$, and indeed $150 = 150$. ✓ The length of the pool is $45$ feet and the width is $30$ feet.

Key terms

vertex/vertices — the corner points of a triangle, where two sides meet; each side is usually labeled with the lowercase letter matching the uppercase letter of the opposite vertex. height (of a triangle) — a line segment connecting the base to the opposite vertex, meeting the base at a $90\degree$ angle. right triangle — a triangle with one $90\degree$ angle, usually marked with a small square. hypotenuse — the side of a right triangle opposite the $90\degree$ angle. leg — either of the two sides of a right triangle that form the right angle. Pythagorean Theorem — in any right triangle, $a^2 + b^2 = c^2$, where $a$ and $b$ are the lengths of the legs and $c$ is the length of the hypotenuse.

This section is adapted from Elementary Algebra 2e, Section 3.4: Solve Geometry Applications: Triangles, Rectangles, and the Pythagorean Theorem 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: recreated the triangle, right-triangle, and rectangle diagrams as accessible inline graphics; omitted the Be Prepared quiz, Self Check checklist, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.