Learn on PengiOpenstax Elementary Algebra 2EChapter 3: Math Models

Lesson 3.4: Solve Geometry Applications: Triangles, Rectangles, and the Pythagorean Theorem

Students learn to solve geometry applications involving triangles, rectangles, and the Pythagorean Theorem using properties such as the triangle angle sum (m∠A + m∠B + m∠C = 180°), perimeter and area formulas, and right triangle relationships. This lesson is part of Chapter 3: Math Models in the OpenStax Elementary Algebra 2E textbook and covers setting up and solving algebraic equations from real-world geometric scenarios. Students practice a structured problem-solving strategy to find unknown side lengths, angles, and areas across multiple shape types.

Section 1

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

New Concept

This lesson applies your algebra skills to geometry. You'll use formulas for triangles and rectangles, including the Pythagorean Theorem (a2+b2=c2a^2 + b^2 = c^2), to set up and solve equations for unknown lengths, angles, perimeters, and areas.

What’s next

This is just the beginning. Next, you'll work through interactive examples and tackle practice cards to master solving for sides, angles, and areas.

Section 2

Solving Geometry Applications

Property

HOW TO: Solve Geometry Applications.
Step 1. Read the problem and make sure all the words and ideas are understood. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.
Step 3. Label what we are looking for by choosing a variable to represent it.
Step 4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
Step 5. Solve the equation using good algebra techniques.
Step 6. Check the answer by substituting it back into the equation solved in step 5 and by making sure it makes sense in the context of the problem.
Step 7. Answer the question with a complete sentence.

Examples

  • To find a missing angle in a triangle, draw the triangle, label the known angles, assign a variable xx to the unknown angle, write the equation A+B+x=180A+B+x=180, and solve for xx.
  • To find a rectangle's width given its area and length, you would draw the rectangle, label the length and area, use WW for width, write the formula A=LWA=L \cdot W, substitute the values, and solve for WW.

Section 3

Triangle Properties

Property

For ABC\triangle ABC
Angle measures:
mA+mB+mC=180m\angle A + m\angle B + m\angle C = 180
The sum of the measures of the angles of a triangle is 180180^\circ.

Perimeter:
P=a+b+cP = a+b+c
The perimeter is the sum of the lengths of the sides of the triangle.

Area:
A=12bhA = \frac{1}{2}bh, bb = base, hh = height
The area of a triangle is one-half the base times the height.

Section 4

The Pythagorean Theorem

Property

In any right triangle, a2+b2=c2a^2 + b^2 = c^2.
where a and b are the lengths of the legs, c is the length of the hypotenuse.
A right triangle has one 9090^\circ angle. The side of the triangle opposite the 9090^\circ angle is called the hypotenuse and each of the other sides are called legs.

Examples

  • In a right triangle with legs a=9a=9 and b=12b=12, the hypotenuse cc is found using 92+122=c29^2 + 12^2 = c^2, which simplifies to 81+144=22581 + 144 = 225. Thus, c=225=15c = \sqrt{225} = 15.
  • If a right triangle has a hypotenuse c=17c=17 and one leg a=8a=8, the other leg bb is found using 82+b2=1728^2 + b^2 = 17^2. This gives 64+b2=28964 + b^2 = 289, so b2=225b^2 = 225 and b=15b=15.

Section 5

Properties of Rectangles

Property

  • Rectangles have four sides and four right (9090^\circ) angles.
  • The lengths of opposite sides are equal.
  • The perimeter of a rectangle is the sum of twice the length and twice the width.
P=2L+2WP = 2L+2W
  • The area of a rectangle is the product of the length and the width.
A=LWA = L \cdot W

Examples

  • The perimeter of a rectangle with a length of 15 meters and a width of 8 meters is P=2(15)+2(8)=30+16=46P = 2(15) + 2(8) = 30 + 16 = 46 meters.
  • The area of a rectangle with a length of 12 feet and a width of 7 feet is A=127=84A = 12 \cdot 7 = 84 square feet.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 3: Math Models

  1. Lesson 1

    Lesson 3.1: Use a Problem-Solving Strategy

  2. Lesson 2

    Lesson 3.2: Solve Percent Applications

  3. Lesson 3

    Lesson 3.3: Solve Mixture Applications

  4. Lesson 4Current

    Lesson 3.4: Solve Geometry Applications: Triangles, Rectangles, and the Pythagorean Theorem

  5. Lesson 5

    Lesson 3.5: Solve Uniform Motion Applications

  6. Lesson 6

    Lesson 3.6: Solve Applications with Linear Inequalities

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

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

New Concept

This lesson applies your algebra skills to geometry. You'll use formulas for triangles and rectangles, including the Pythagorean Theorem (a2+b2=c2a^2 + b^2 = c^2), to set up and solve equations for unknown lengths, angles, perimeters, and areas.

What’s next

This is just the beginning. Next, you'll work through interactive examples and tackle practice cards to master solving for sides, angles, and areas.

Section 2

Solving Geometry Applications

Property

HOW TO: Solve Geometry Applications.
Step 1. Read the problem and make sure all the words and ideas are understood. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.
Step 3. Label what we are looking for by choosing a variable to represent it.
Step 4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
Step 5. Solve the equation using good algebra techniques.
Step 6. Check the answer by substituting it back into the equation solved in step 5 and by making sure it makes sense in the context of the problem.
Step 7. Answer the question with a complete sentence.

Examples

  • To find a missing angle in a triangle, draw the triangle, label the known angles, assign a variable xx to the unknown angle, write the equation A+B+x=180A+B+x=180, and solve for xx.
  • To find a rectangle's width given its area and length, you would draw the rectangle, label the length and area, use WW for width, write the formula A=LWA=L \cdot W, substitute the values, and solve for WW.

Section 3

Triangle Properties

Property

For ABC\triangle ABC
Angle measures:
mA+mB+mC=180m\angle A + m\angle B + m\angle C = 180
The sum of the measures of the angles of a triangle is 180180^\circ.

Perimeter:
P=a+b+cP = a+b+c
The perimeter is the sum of the lengths of the sides of the triangle.

Area:
A=12bhA = \frac{1}{2}bh, bb = base, hh = height
The area of a triangle is one-half the base times the height.

Section 4

The Pythagorean Theorem

Property

In any right triangle, a2+b2=c2a^2 + b^2 = c^2.
where a and b are the lengths of the legs, c is the length of the hypotenuse.
A right triangle has one 9090^\circ angle. The side of the triangle opposite the 9090^\circ angle is called the hypotenuse and each of the other sides are called legs.

Examples

  • In a right triangle with legs a=9a=9 and b=12b=12, the hypotenuse cc is found using 92+122=c29^2 + 12^2 = c^2, which simplifies to 81+144=22581 + 144 = 225. Thus, c=225=15c = \sqrt{225} = 15.
  • If a right triangle has a hypotenuse c=17c=17 and one leg a=8a=8, the other leg bb is found using 82+b2=1728^2 + b^2 = 17^2. This gives 64+b2=28964 + b^2 = 289, so b2=225b^2 = 225 and b=15b=15.

Section 5

Properties of Rectangles

Property

  • Rectangles have four sides and four right (9090^\circ) angles.
  • The lengths of opposite sides are equal.
  • The perimeter of a rectangle is the sum of twice the length and twice the width.
P=2L+2WP = 2L+2W
  • The area of a rectangle is the product of the length and the width.
A=LWA = L \cdot W

Examples

  • The perimeter of a rectangle with a length of 15 meters and a width of 8 meters is P=2(15)+2(8)=30+16=46P = 2(15) + 2(8) = 30 + 16 = 46 meters.
  • The area of a rectangle with a length of 12 feet and a width of 7 feet is A=127=84A = 12 \cdot 7 = 84 square feet.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 3: Math Models

  1. Lesson 1

    Lesson 3.1: Use a Problem-Solving Strategy

  2. Lesson 2

    Lesson 3.2: Solve Percent Applications

  3. Lesson 3

    Lesson 3.3: Solve Mixture Applications

  4. Lesson 4Current

    Lesson 3.4: Solve Geometry Applications: Triangles, Rectangles, and the Pythagorean Theorem

  5. Lesson 5

    Lesson 3.5: Solve Uniform Motion Applications

  6. Lesson 6

    Lesson 3.6: Solve Applications with Linear Inequalities