Geometry Math Problems - Perimeters

In these lessons, we will learn to solve geometry math problems that involve perimeter.

Related Pages Geometry math problems involving area Area Formula Geometry math problems involving angles More Algebra Word Problems

Geometry word problems involves geometric figures and angles described in words. You would need to be familiar with the formulas in geometry.

Making a sketch of the geometric figure is often helpful.

Geometry Word Problems Involving Perimeter

Example: A triangle has a perimeter of 50. If 2 of its sides are equal and the third side is 5 more than the equal sides, what is the length of the third side?

Solution: Step 1: Assign variables:

Let x = length of the equal sides Sketch the figure

Step 2: Write out the formula for perimeter of triangle .

P = sum of the three sides

Step 3: Plug in the values from the question and from the sketch.

50 = x + x + x+ 5

Combine like terms 50 = 3x + 5

Isolate variable x 3x = 50 – 5 3x = 45 x = 15

Be careful! The question requires the length of the third side.

The length of third side = 15 + 5 = 20

Answer: The length of third side is 20.

Geometry Math Problem involving the perimeter of a rectangle

The following two videos give the perimeter of a rectangle, a relationship between the length and width of the rectangle, and use that information to find the exact value of the length and width.

Example: A rectangular garden is 2.5 times as long as it is wide. It has a perimeter of 168 ft. How long and wide is the garden?

Example: A rectangular landing strip for an airplane has perimeter 8000 ft. If the length is 10 ft longer than 35 times the width, what is the length and width?

Examples of perimeter geometry word problems This video shows how to write an equation and find the dimensions of a rectangle knowing the perimeter and some information about the about the length and width.

Example: The width of a rectangle is 3 ft less than its length. The perimeter of the rectangle is 110 ft. Find the dimensions.

Perimeter Word Problems

Example: The length of a rectangle is 7 cm more than 4 times its width. Its perimeter is 124 cm. Find its dimensions.

Geometry Math Problem involving the perimeter of a triangle

The following two videos give the perimeter of a triangle, a relationship between the sides of the triangle, and use that information to find the exact value or values of the required side or sides.

Example: Patrick’s bike ride follows a triangular path; two legs are equal, the third is 8 miles longer than the other legs. If Patrick rides 30 miles total, what is the length of the longest leg?

Example: The perimeter of a triangle is 56 cm. The first side is 6 cm shorter than the second side. The third side is 2 cm shorter than twice the length of the first side. What is the length of each side?

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o Identify some critical steps of the process for solving practical geometry problems

o Apply geometry problem-solving techniques to practical situations

Geometry has a variety of real-life applications in everyday situations. In this article, we will learn to apply geometric principles and techniques to solve problems. The key to solving practical geometry problems is translation of the real-life situation into figures, measurements, and other information necessary to represent the situation conceptually. For instance, you already know how to calculate the area of a composite figure; if you were asked to determine how much floor space is available in a certain building with a composite shape, you would simply need to apply the same principles as you would use for calculating the area of a composite figure. Some measurements of the building might, of course, be required, but the same problem-solving techniques apply.

It behooves us to present a basic approach to solving practical geometry problems. This approach is similar to that for solving almost a word problem, but is geared slightly more toward the characteristics of geometry problems in particular.

1. Determine what you need to calculate to solve the problem. In some cases, you may need a length; in others, an area or angle measure. If you are conscious throughout the process of what you need to determine, you can save yourself a significant amount of time.

2. Draw a diagram. Sometimes a straightedge, compass, protractor, or some combination of these tools can be helpful. Even if you only use a rough sketch, however, making a visual representation of the problem can help you organize your thoughts and keep track of important information such as the relationship of line segments and angles as well as the measures thereof.

3. Record all appropriate measurements. If you are calculating an area, for instance, you may need to take measurements of certain lengths (alternatively, these may be provided to you). In either case, record them and mark them in some manner on your diagram.

4. Pay attention to units. Using units of square meters for a length or angle measure can be an embarrassing mistake! Keep careful track of the units you are using throughout the problem. If no units are given, simply use the generic term "units" in place of inches or meters, for example.

5. Divide the figure, if necessary, into manageable portions. If your diagram is a composite figure, it may help to divide the figure into bite-sized portions that you can handle.

6. Identify any appropriate geometric relationships. This step can greatly simplify the problem. Perhaps you can show two triangles to be congruent or similar, or perhaps you can identify congruent segments or angles. Use this step to fill in as much missing information in your diagram as you can.

7. Do the math. At this point, you need to apply what you've learned to analyze the figure and other data to solve the problem. You may, for example, need to apply the Pythagorean theorem, or you may need to calculate the perimeter of a figure. Whatever the details of the problem, you will need to apply your skills in geometry in an appropriate manner.

8. Check your results. Take a look at your answer in the context of your diagram-does your answer make sense? A result of millions of square meters for the area of a figure with dimensions in the range of a few meters should tell you that you've made an error at some point in your analysis.

Not every step of the approach outlined above will be needed in every problem. You must use your best judgment in determining what is necessary to solve the problem in a satisfactory and time-efficient manner. Also, you may not always think to use the exact progression of steps above; the outline is simply a way to describe a systematic approach to problem solving. The remainder of this article provides you the opportunity to test your geometry skills by way of several practice problems. Obviously, these problems do not require you to go out and make any measurements of lengths or angles, but keep in mind that problems you encounter in everyday life may require you to do so!

Practice Problem : The floor plan of a house is shown below. Determine the area covered by the house.

Solution : Let's first divide the diagram of the house into two rectangles and a trapezoid, since we can calculate the area of each of these figures. Using the properties of each figure, we can also fill in some of the unknown information.

Now, the area of the larger rectangle is the product of 40 feet and 20 feet, or 800 square feet. The area of the smaller rectangle is 25 feet times 6 feet, or 150 square feet. The area of the trapezoid is the following:

The height ( h ) is 6 feet, and the two bases ( b 1 and b 2 ) are 8 and 11 feet.

Adding all three areas gives us a total area of the house of 1,007 square feet.

Practice Problem : A hiker is walking up a steep hill. The slope of the hill between two trees is constant, and the base of one tree is 100 meters higher than the other. If the horizontal distance between the trees is 400 meters, how far must the hiker walk to get from one tree to the next?

Solution : Because this problem may be difficult to envision, a diagram is extremely helpful. Notice that the base of the trees differ in height by 100 meters--this is our vertical distance for the walk. The horizontal distance is 400 meters.

Note that we have shown the right angle because horizontal and vertical segments are perpendicular. We can now use the Pythagorean theorem to calculate the distance d the hiker must walk.

Thus, the hiker must walk about 412 meters. Note that although the hiker makes a significant (100 meter) change in elevation over this walk, the difference between the actual distance he walks and the horizontal distance is small--only about 12 meters.

Practice Problem : A homeowner has a rectangular fenced-in yard, and he wants to put mulch on his triangular gardens, as shown below. The inside border of each garden always meets the fence at the same angle. If a bag of mulch covers about 50 square feet, how many bags of mulch should the homeowner buy to cover his gardens?

Solution : We are told in the problem that the inside border of each garden meets the fence at the same angle in every case; thus, we can conclude (as shown below) that the triangles are all isosceles (and that the triangles with the same side lengths are congruent by the ASA condition). We can thus mark each side with an unknown variable x or y .

Recall that the fenced-in area is rectangular; thus the angle in each corner is 90°. We can then solve for x and y using the Pythagorean theorem. Notice first, however, that x and y are the height and base of their respective triangles.

Because the gardens include two of each triangle shape, the total garden area is simply the sum of x 2 and y 2 . (If you do not follow this point, simply use the triangle area formula in each case--you will get the same result.)

Thus, the homeowner needs six bags of mulch (for a total of 300 square feet) to cover his gardens. (Of course, we are assuming here that he must buy a whole number of bags.)

Using Classical Geometric Construction Techniques

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Using Geometric Concepts and Properties to Solve Problems

Introduction.

Often, you will be asked to solve problems involving geometric relationships or other shapes. For real-world problems, those geometric relationships mostly involve measurable attributes, such as length, area, or volume.

Sometimes, those problems will involve the perimeter or circumference, or the area of a 2-dimensional figure.

green elliptical running track

For example, what is the distance around the track that is shown?  Or, what is the area of the portion of the field that is covered with grass?

You may also see problems that involve the volume or surface area of a 3-dimensional figure.  For example, what is the area of the roof of the building that is shown?

building composed of a rectangular prism with a half-cylinder on top

Another common type of geometric problem involves using proportional reasoning.

solving problems using geometry

For example, an artist created a painting that needs to be reduced proportionally for the flyer advertising an art gallery opening. If the dimensions of the painting are reduced by a factor of 40%, what will be the dimensions of the image on the flyer?

In this resource, you will investigate ways to apply a problem-solving model to determine the solutions for geometric problems like these.

A basic problem solving model contains the following four steps:

Solving Problems Using Perimeter and Circumference

You may recall that the perimeter of an object is the distance around the edge of the object. If the object contains circles, then you may need to think about the circumference of a circle, which is the perimeter of the circle.

A tire on a passenger car has a diameter of 18 inches. When the tire has rotated 5 times, how far will the car have traveled?

Image of car and its tire labeled 18 inches for diameter

Step 1 : Read, understand, and interpret the problem.

  • What information is presented?
  • What is the problem asking me to find?
  • What information may be extra information that I do not need?

Step 2 : Make a plan.

  • Draw a picture.
  • Use a formula: Which formula do I need to use? (Hint: Look at your Mathematics Reference Materials)

Step 3 : Implement your plan.

  • What formulas do I need?
  • What information can I interpret from the diagram, table, or other given information?
  • Solve the problem.

Step 4 : Evaluate your answer.

  • Does my answer make sense?
  • Did I answer the question that was asked?
  • Are my units correct?

A cylindrical barrel with a diameter of 20 inches is used to hold fuel for a barbecue cook off. The chef rolls the barrel so that it completes 7 rotations. How many feet did the chef roll the barrel?

Image of barrel with diameter labeled 20 inches

Solving Problems Using Area and Surface Area

You may also encounter real-world geometric problems that ask you to find the area of 2-dimensional figures or the surface area of 3-dimensional figures. The key to solving these problems is to look for ways to break the region into smaller figures of which you know how to find the area.

Mr. Elder wants to cover a wall in his kitchen with wallpaper. The wall is shown in the figure below.

hexagonal wall with dimensions labeled

If wallpaper costs $1.75 per square foot, how much will Mr. Elder spend on wallpaper to completely cover this wall, excluding sales tax?

To solve this problem, let's use the 4-step problem solving model.

Mrs. Nguyen wants to apply fertilizer to her front lawn. A bag of fertilizer that covers 1,000 square feet costs $18. How many bags of fertilizer will Mrs. Nguyen need to purchase?

hexagonal yard with dimensions labeled

Surface Area Problem

After a storm, the Serafina family needs to have their roof replaced. Their house is in the shape of a pentagonal prism with the dimensions shown in the diagram.

pentagonal prism shaped house with roof shaded

To match their new roof, Mrs. Serafina decided to have both pentagonal sides of their house covered in aluminum siding. Their house is in the shape of a pentagonal prism with the dimensions shown in the diagram.

pentagonal prism shaped house with roof shaded

A contractor gave Mrs. Serafina an estimate based on a cost of $3.10 per square foot to complete the aluminum siding. How much will it cost the Serafina family to have the aluminum siding installed?

Solving Problems Using Proportionality

Proportional relationships are another important part of geometric problem solving.

A woodblock painting has dimensions of 60 centimeters by 79.5 centimeters. In order to fit on a flyer advertising the opening of a new art show, the image must be reduced by a scale factor of  1/25.

W hat will be the final dimensions of the image on the flyer?

Image of the sun over a Japanese temple

Measuring Problem

For summer vacation, Jennifer and her family drove from their home in Inlandton to Beachville. Their car can drive 20 miles on one gallon of gasoline. Use the ruler to measure the distance that they drove to the nearest  1/4  inch, and then calculate the number of gallons of gasoline their car will use at this rate to drive from Inlandton to Beachville.

Practice #1

A blueprint for a rectangular tool shed has dimensions shown in the diagram below.

blueprint showing that the length is 4.5 centimeters and the width is 3.5 centimeters, and a scale of 1 centimeter = 2 feet

Todd is using this blueprint to build a tool shed, and he wants to surround the base of the tool shed with landscaping timbers as a border. How many feet of landscaping timbers will Todd need?

Practice #2

A scale model of a locomotive is shown. Use the ruler to measure the dimensions of the model to the nearest 1/4  inch, and then calculate the actual dimensions of the locomotive.

Scale : 1 inch = 5 feet

Solving geometric problems, such as those found in art and architecture, is an important skill. As with any mathematical problem, you can use the 4-step problem solving model to help you think through the important parts of the problem and be sure that you don't miss key information.

There are a lot of different applications of geometry to real-world problem solving. Some of the more common applications include the following:

octagonal cup

What is the perimeter of the base of the cup, if the cup is in the shape of an octagonal prism?

aerial photo showing crop circles

The JP Morgan Chase Bank Tower in downtown Houston, Texas, is one of the tallest buildings west of the Mississippi River. It is in the shape of a pentagonal prism. If 40% of each face is covered with glass windows, what is the amount of surface area covered with glass?

image of Van Gogh's Starry Night

The dimensions of Vincent van Gogh's Starry Night are 29 inches by 36 1 4 36\frac{1}{4} inches. If a print reduces these dimensions by a scale factor of 30%, what will be the dimensions of the print?

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9.9: Solve Geometry Applications- Volume and Surface Area (Part 1)

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  • Page ID 5010

Learning Objectives

  • Find volume and surface area of rectangular solids
  • Find volume and surface area of spheres
  • Find volume and surface area of cylinders
  • Find volume of cones

be prepared!

Before you get started, take this readiness quiz.

  • Evaluate x 3 when x = 5. If you missed this problem, review Example 2.3.3 .
  • Evaluate 2 x when x = 5. If you missed this problem, review Example 2.3.4 .
  • Find the area of a circle with radius \(\dfrac{7}{2}\). If you missed this problem, review Example 5.6.12 .

In this section, we will finish our study of geometry applications. We find the volume and surface area of some three-dimensional figures. Since we will be solving applications, we will once again show our Problem-Solving Strategy for Geometry Applications.

Problem Solving Strategy for Geometry Applications

  • Step 1. Read the problem and make sure you understand all the words and ideas. Draw the figure and label it with the given information.
  • Step 2. Identify what you are looking for.
  • Step 3. Name what you are looking for. Choose a variable to represent that quantity.
  • 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 in the problem and make sure it makes sense.
  • Step 7. Answer the question with a complete sentence.

Find Volume and Surface Area of Rectangular Solids

A cheerleading coach is having the squad paint wooden crates with the school colors to stand on at the games. (See Figure \(\PageIndex{1}\)). The amount of paint needed to cover the outside of each box is the surface area , a square measure of the total area of all the sides. The amount of space inside the crate is the volume, a cubic measure.

This is an image of a wooden crate.

Figure \(\PageIndex{1}\) - This wooden crate is in the shape of a rectangular solid.

Each crate is in the shape of a rectangular solid . Its dimensions are the length, width, and height. The rectangular solid shown in Figure \(\PageIndex{2}\) has length 4 units, width 2 units, and height 3 units. Can you tell how many cubic units there are altogether? Let’s look layer by layer.

A rectangular solid is shown. Each layer is composed of 8 cubes, measuring 2 by 4. The top layer is pink. The middle layer is orange. The bottom layer is green. Beside this is an image of the top layer that says “The top layer has 8 cubic units.” The orange layer is shown and says “The middle layer has 8 cubic units.” The green layer is shown and says, “The bottom layer has 8 cubic units.”

Figure \(\PageIndex{2}\) - Breaking a rectangular solid into layers makes it easier to visualize the number of cubic units it contains. This 4 by 2 by 3 rectangular solid has 24 cubic units.

Altogether there are 24 cubic units. Notice that 24 is the length × width × height.

The top line says V equals L times W times H. Beneath the V is 24, beneath the equal sign is another equal sign, beneath the L is a 4, beneath the W is a 2, beneath the H is a 3.

The volume, V, of any rectangular solid is the product of the length, width, and height.

We could also write the formula for volume of a rectangular solid in terms of the area of the base. The area of the base, B, is equal to length × width.

\[B = L \cdot W\]

We can substitute B for L • W in the volume formula to get another form of the volume formula.

\[\begin{split} V &= \textcolor{red}{L \cdot W} \cdot H \\ V &= \textcolor{red}{(L \cdot W)} \cdot H \\ V &= \textcolor{red}{B} h \end{split}\]

We now have another version of the volume formula for rectangular solids. Let’s see how this works with the 4 × 2 × 3 rectangular solid we started with. See Figure \(\PageIndex{3}\).

An image of a rectangular solid is shown. It is made up of cubes. It is labeled as 2 by 4 by 3. Beside the solid is V equals Bh. Below this is V equals Base times height. Below Base is parentheses 4 times 2. The next line says V equals parentheses 4 times 2 times 3. Below that is V equals 8 times 3, then V equals 24 cubic units.

Figure \(\PageIndex{3}\)

To find the surface area of a rectangular solid, think about finding the area of each of its faces. How many faces does the rectangular solid above have? You can see three of them.

\[\begin{split} A_{front} &= L \times W \qquad A_{side} = L \times W \qquad A_{top} = L \times W \\ A_{front} &= 4 \cdot 3 \qquad \quad \; A_{side} = 2 \cdot 3 \qquad \quad \; A_{top} = 4 \cdot 2 \\ A_{front} &= 12 \qquad \qquad A_{side} = 6 \qquad \qquad \; \; A_{top} = 8 \end{split}\]

Notice for each of the three faces you see, there is an identical opposite face that does not show.

\[\begin{split} S &= (front + back)+(left\; side + right\; side) + (top + bottom) \\ S &= (2 \cdot front) + (2 \cdot left\; side) + (2 \cdot top) \\ S &= 2 \cdot 12 + 2 \cdot 6 + 2 \cdot 8 \\ S &= 24 + 12 + 16 \\ S &= 52\; sq.\; units \end{split}\]

The surface area S of the rectangular solid shown in Figure \(\PageIndex{3}\) is 52 square units.

In general, to find the surface area of a rectangular solid, remember that each face is a rectangle, so its area is the product of its length and its width (see Figure \(\PageIndex{4}\)). Find the area of each face that you see and then multiply each area by two to account for the face on the opposite side.

\[S = 2LH + 2LW + 2WH\]

A rectangular solid is shown. The sides are labeled L, W, and H. One face is labeled LW and another is labeled WH.

Figure \(\PageIndex{4}\) - For each face of the rectangular solid facing you, there is another face on the opposite side. There are 6 faces in all.

Definition: Volume and Surface Area of a Rectangular Solid

For a rectangular solid with length L, width W, and height H:

A rectangular solid is shown. The sides are labeled L, W, and H. Beside it is Volume: V equals LWH equals BH. Below that is Surface Area: S equals 2LH plus 2LW plus 2WH.

Example \(\PageIndex{1}\):

For a rectangular solid with length 14 cm, height 17 cm, and width 9 cm, find the (a) volume and (b) surface area.

Step 1 is the same for both (a) and (b), so we will show it just once.

Exercise \(\PageIndex{1}\):

Find the (a) volume and (b) surface area of rectangular solid with the: length 8 feet, width 9 feet, and height 11 feet.

Exercise \(\PageIndex{2}\):

Find the (a) volume and (b) surface area of rectangular solid with the: length 15 feet, width 12 feet, and height 8 feet.

1,440 cu. ft

Example \(\PageIndex{2}\):

A rectangular crate has a length of 30 inches, width of 25 inches, and height of 20 inches. Find its (a) volume and (b) surface area.

Exercise \(\PageIndex{3}\):

A rectangular box has length 9 feet, width 4 feet, and height 6 feet. Find its (a) volume and (b) surface area.

Exercise \(\PageIndex{4}\):

A rectangular suitcase has length 22 inches, width 14 inches, and height 9 inches. Find its (a) volume and (b) surface area.

2,772 cu. in

1,264 sq. in.

Volume and Surface Area of a Cube

A cube is a rectangular solid whose length, width, and height are equal. See Volume and Surface Area of a Cube, below. Substituting, s for the length, width and height into the formulas for volume and surface area of a rectangular solid, we get:

\[\begin{split} V &= LWH \quad \; S = 2LH + 2LW + 2WH \\ V &= s \cdot s \cdot s \quad S = 2s \cdot s + 2s \cdot s + 2s \cdot s \\ V &= s^{3} \qquad \quad S = 2s^{2} + 2s^{2} + 2s^{2} \\ &\qquad \qquad \quad \; S = 6s^{2} \end{split}\]

So for a cube, the formulas for volume and surface area are V = s 3 and S = 6s 2 .

Definition: Volume and Surface Area of a Cube

For any cube with sides of length s,

An image of a cube is shown. Each side is labeled s. Beside this is Volume: V equals s cubed. Below that is Surface Area: S equals 6 times s squared.

Example \(\PageIndex{3}\):

A cube is 2.5 inches on each side. Find its (a) volume and (b) surface area.

Exercise \(\PageIndex{5}\):

For a cube with side 4.5 meters, find the (a) volume and (b) surface area of the cube.

91.125 cu. m

121.5 sq. m

Exercise \(\PageIndex{6}\):

For a cube with side 7.3 yards, find the (a) volume and (b) surface area of the cube.

389.017 cu. yd.

319.74 sq. yd.

Example \(\PageIndex{4}\):

A notepad cube measures 2 inches on each side. Find its (a) volume and (b) surface area.

Exercise \(\PageIndex{7}\):

A packing box is a cube measuring 4 feet on each side. Find its (a) volume and (b) surface area.

Exercise \(\PageIndex{8}\):

A wall is made up of cube-shaped bricks. Each cube is 16 inches on each side. Find the (a) volume and (b) surface area of each cube.

4,096 cu. in.

Contributors and Attributions

Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at http://cnx.org/contents/[email protected] ."

  • 3.4 Solve Geometry Applications: Triangles, Rectangles, and the Pythagorean Theorem
  • Introduction
  • 1.1 Introduction to Whole Numbers
  • 1.2 Use the Language of Algebra
  • 1.3 Add and Subtract Integers
  • 1.4 Multiply and Divide Integers
  • 1.5 Visualize Fractions
  • 1.6 Add and Subtract Fractions
  • 1.7 Decimals
  • 1.8 The Real Numbers
  • 1.9 Properties of Real Numbers
  • 1.10 Systems of Measurement
  • Key Concepts
  • Review Exercises
  • Practice Test
  • 2.1 Solve Equations Using the Subtraction and Addition Properties of Equality
  • 2.2 Solve Equations using the Division and Multiplication Properties of Equality
  • 2.3 Solve Equations with Variables and Constants on Both Sides
  • 2.4 Use a General Strategy to Solve Linear Equations
  • 2.5 Solve Equations with Fractions or Decimals
  • 2.6 Solve a Formula for a Specific Variable
  • 2.7 Solve Linear Inequalities
  • 3.1 Use a Problem-Solving Strategy
  • 3.2 Solve Percent Applications
  • 3.3 Solve Mixture Applications
  • 3.5 Solve Uniform Motion Applications
  • 3.6 Solve Applications with Linear Inequalities
  • 4.1 Use the Rectangular Coordinate System
  • 4.2 Graph Linear Equations in Two Variables
  • 4.3 Graph with Intercepts
  • 4.4 Understand Slope of a Line
  • 4.5 Use the Slope-Intercept Form of an Equation of a Line
  • 4.6 Find the Equation of a Line
  • 4.7 Graphs of Linear Inequalities
  • 5.1 Solve Systems of Equations by Graphing
  • 5.2 Solving Systems of Equations by Substitution
  • 5.3 Solve Systems of Equations by Elimination
  • 5.4 Solve Applications with Systems of Equations
  • 5.5 Solve Mixture Applications with Systems of Equations
  • 5.6 Graphing Systems of Linear Inequalities
  • 6.1 Add and Subtract Polynomials
  • 6.2 Use Multiplication Properties of Exponents
  • 6.3 Multiply Polynomials
  • 6.4 Special Products
  • 6.5 Divide Monomials
  • 6.6 Divide Polynomials
  • 6.7 Integer Exponents and Scientific Notation
  • 7.1 Greatest Common Factor and Factor by Grouping
  • 7.2 Factor Trinomials of the Form x2+bx+c
  • 7.3 Factor Trinomials of the Form ax2+bx+c
  • 7.4 Factor Special Products
  • 7.5 General Strategy for Factoring Polynomials
  • 7.6 Quadratic Equations
  • 8.1 Simplify Rational Expressions
  • 8.2 Multiply and Divide Rational Expressions
  • 8.3 Add and Subtract Rational Expressions with a Common Denominator
  • 8.4 Add and Subtract Rational Expressions with Unlike Denominators
  • 8.5 Simplify Complex Rational Expressions
  • 8.6 Solve Rational Equations
  • 8.7 Solve Proportion and Similar Figure Applications
  • 8.8 Solve Uniform Motion and Work Applications
  • 8.9 Use Direct and Inverse Variation
  • 9.1 Simplify and Use Square Roots
  • 9.2 Simplify Square Roots
  • 9.3 Add and Subtract Square Roots
  • 9.4 Multiply Square Roots
  • 9.5 Divide Square Roots
  • 9.6 Solve Equations with Square Roots
  • 9.7 Higher Roots
  • 9.8 Rational Exponents
  • 10.1 Solve Quadratic Equations Using the Square Root Property
  • 10.2 Solve Quadratic Equations by Completing the Square
  • 10.3 Solve Quadratic Equations Using the Quadratic Formula
  • 10.4 Solve Applications Modeled by Quadratic Equations
  • 10.5 Graphing Quadratic Equations in Two Variables

Learning Objectives

By the end of this section, you will be able to:

  • Solve applications using properties of triangles

Use the Pythagorean Theorem

  • Solve applications using rectangle properties

Be Prepared 3.12

Before you get started, take this readiness quiz.

Simplify: 1 2 ( 6 h ) . 1 2 ( 6 h ) . If you missed this problem, review Example 1.122 .

Be Prepared 3.13

The length of a rectangle is three less than the width. Let w represent the width. Write an expression for the length of the rectangle. If you missed this problem, review Example 1.26 .

Be Prepared 3.14

Solve: A = 1 2 b h A = 1 2 b h for b when A = 260 A = 260 and h = 52 . h = 52 . If you missed this problem, review Example 2.61 .

Be Prepared 3.15

Simplify: 144 . 144 . If you missed this problem, review Example 1.111 .

Solve Applications Using Properties of Triangles

In this section 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.

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 we 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.

We will start geometry applications by looking at the properties of triangles. Let’s review some basic facts about 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 in Figure 3.4 is called △ A B C . △ A B C .

The three angles of a triangle are related in a special way. The sum of their measures is 180 ° . 180 ° . Note that we read m ∠ A m ∠ A as “the measure of angle A.” So in △ A B C △ A B C in Figure 3.4 ,

Because the perimeter of a figure is the length of its boundary, the perimeter of △ A B C △ A B C 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 ° 90 ° angle with the base. We will draw △ A B C △ A B C again, and now show the height, h . See Figure 3.5 .

Triangle Properties

For △ A B C △ A B C

Angle measures:

  • The sum of the measures of the angles of a triangle is 180 ° . 180 ° .
  • The perimeter is the sum of the lengths of the sides of the triangle.
  • The area of a triangle is one-half the base times the height.

Example 3.34

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

Try It 3.67

The measures of two angles of a triangle are 31 and 128 degrees. Find the measure of the third angle.

Try It 3.68

The measures of two angles of a triangle are 49 and 75 degrees. Find the measure of the third angle.

Example 3.35

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?

Try It 3.69

The perimeter of a triangular garden is 48 feet. The lengths of two sides are 18 feet and 22 feet. How long is the third side?

Try It 3.70

The lengths of two sides of a triangular window are seven feet and five feet. The perimeter is 18 feet. How long is the third side?

Example 3.36

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?

Try It 3.71

The area of a triangular painting is 126 square inches. The base is 18 inches. What is the height?

Try It 3.72

A triangular tent door has an area of 15 square feet. The height is five feet. What is the base?

The triangle properties we used so far apply to all triangles. Now we will look at one specific type of triangle—a right triangle. A right triangle has one 90 ° 90 ° angle, which we usually mark with a small square in the corner.

Right Triangle

A right triangle has one 90 ° 90 ° angle, which is often marked with a square at the vertex.

Example 3.37

One angle of a right triangle measures 28 ° . 28 ° . What is the measure of the third angle?

Try It 3.73

One angle of a right triangle measures 56 ° . 56 ° . What is the measure of the other small angle?

Try It 3.74

One angle of a right triangle measures 45 ° . 45 ° . What is the measure of the other small angle?

In the examples we have seen so far, we could draw a figure and label it directly after reading the problem. In the next example, 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.

Example 3.38

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.

Try It 3.75

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.

Try It 3.76

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.

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 relate to each other. An important property that describes the relationship among the lengths of the three sides of a right triangle is called the Pythagorean Theorem . This theorem has been used around the world since ancient times. It is named after the Greek philosopher and mathematician, Pythagoras, who lived around 500 BC.

Before we state the Pythagorean Theorem, we need to introduce some terms for the sides of a triangle. Remember that a right triangle has a 90 ° 90 ° angle, marked with a small square in the corner. The side of the triangle opposite the 90 ° 90 ° angle is called the hypotenuse and each of the other sides are called legs .

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 we say: in any right triangle, a 2 + b 2 = c 2 , a 2 + b 2 = c 2 , where a and b a and b are the lengths of the legs and c c is the length of the hypotenuse.

Writing the formula in every exercise and saying it aloud as you write it, may help you remember the Pythagorean Theorem.

The Pythagorean Theorem

In any right triangle, a 2 + b 2 = c 2 . a 2 + b 2 = c 2 .

where a and b are the lengths of the legs, c is the length of the hypotenuse.

To solve exercises that use the Pythagorean Theorem, we will need to find square roots. We have used the notation m m and the definition:

If m = n 2 , m = n 2 , then m = n , m = n , for n ≥ 0 . n ≥ 0 .

For example, we found that 25 25 is 5 because 25 = 5 2 . 25 = 5 2 .

Because the Pythagorean Theorem contains variables that are squared, to solve for the length of a side in a right triangle, we will have to use square roots.

Example 3.39

Use the Pythagorean Theorem to find the length of the hypotenuse shown below.

Try It 3.77

Use the Pythagorean Theorem to find the length of the hypotenuse in the triangle shown below.

Try It 3.78

Example 3.40.

Use the Pythagorean Theorem to find the length of the leg shown below.

Try It 3.79

Use the Pythagorean Theorem to find the length of the leg in the triangle shown below.

Try It 3.80

Example 3.41.

Kelvin is building a gazebo and wants to brace each corner by placing a 10 ″ 10 ″ piece of wood diagonally as shown above.

If he fastens the wood 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.

Try It 3.81

John puts the base of a 13-foot ladder five feet from the wall of his house as shown below. How far up the wall does the ladder reach?

Try It 3.82

Randy wants to attach a 17 foot string of lights to the top of the 15 foot mast of his sailboat, as shown below. How far from the base of the mast should he attach the end of the light string?

Solve Applications Using Rectangle Properties

You may already be familiar with the properties of rectangles. Rectangles have four sides and four right ( 90 ° ) ( 90 ° ) 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 , L + W + L + W , or 2 L + 2 W . 2 L + 2 W . 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 squares in the figure.

The area is the length times the width.

The formula for the area of a rectangle is A = L W . A = L W .

Properties of Rectangles

Rectangles have four sides and four right ( 90 ° ) ( 90 ° ) angles.

The lengths of opposite sides are equal.

The perimeter of a rectangle is the sum of twice the length and twice the width.

The area of a rectangle is the product of the length and the width.

Example 3.42

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

Try It 3.83

The length of a rectangle is 120 yards and the width is 50 yards. What is the perimeter?

Try It 3.84

The length of a rectangle is 62 feet and the width is 48 feet. What is the perimeter?

Example 3.43

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

Try It 3.85

The area of a rectangle is 598 square feet. The length is 23 feet. What is the width?

Try It 3.86

The width of a rectangle is 21 meters. The area is 609 square meters. What is the length?

Example 3.44

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

Try It 3.87

Find the length of a rectangle with: perimeter 80 and width 25.

Try It 3.88

Find the length of a rectangle with: perimeter 30 and width 6.

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

Example 3.45

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

Try It 3.89

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

Try It 3.90

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

Example 3.46

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

Try It 3.91

The length of a rectangle is eight more than twice the width. The perimeter is 64. Find the length and width.

Try It 3.92

The width of a rectangle is six less than twice the length. The perimeter is 18. Find the length and width.

Example 3.47

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

Try It 3.93

The perimeter of a rectangular swimming pool is 200 feet. The length is 40 feet more than the width. Find the length and width.

Try It 3.94

The length of a rectangular garden is 30 yards more than the width. The perimeter is 300 yards. Find the length and width.

Practice Makes Perfect

Solving Applications Using Triangle Properties

In the following exercises, solve using triangle properties.

The measures of two angles of a triangle are 26 and 98 degrees. Find the measure of the third angle.

The measures of two angles of a triangle are 61 and 84 degrees. Find the measure of the third angle.

The measures of two angles of a triangle are 105 and 31 degrees. Find the measure of the third angle.

The measures of two angles of a triangle are 47 and 72 degrees. Find the measure of the third angle.

The perimeter of a triangular pool is 36 yards. The lengths of two sides are 10 yards and 15 yards. How long is the third side?

A triangular courtyard has perimeter 120 meters. The lengths of two sides are 30 meters and 50 meters. How long is the third side?

If a triangle has sides 6 feet and 9 feet and the perimeter is 23 feet, how long is the third side?

If a triangle has sides 14 centimeters and 18 centimeters and the perimeter is 49 centimeters, how long is the third side?

A triangular flag has base one foot and height 1.5 foot. What is its area?

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

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

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

One angle of a right triangle measures 33 degrees. What is the measure of the other small angle?

One angle of a right triangle measures 51 degrees. What is the measure of the other small angle?

One angle of a right triangle measures 22.5 degrees. What is the measure of the other small angle?

One angle of a right triangle measures 36.5 degrees. What is the measure of the other small angle?

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.

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 shortest side. The third side is five feet more than the shortest side. The perimeter is 17 feet. Find the lengths of all three sides.

One side of a triangle is three times the shortest 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 two smaller angles of a right triangle have equal measures. Find the measures of all three angles.

The measure of the smallest angle of a right triangle is 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.

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.

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

In the following exercises, solve using the Pythagorean Theorem. Approximate 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 below. How far from the base of the pole should the end of the string of lights be anchored?

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

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

Brian borrowed a 20 foot extension ladder to use when he paints his house. If he sets the base of the ladder 6 feet from the house, as shown below, how far up will the top of the ladder reach?

In the following exercises, solve using rectangle properties.

The length of a rectangle is 85 feet and the width is 45 feet. What is the perimeter?

The length of a rectangle is 26 inches and the width is 58 inches. What is the perimeter?

A rectangular room is 15 feet wide by 14 feet long. What is its perimeter?

A driveway is in the shape of a rectangle 20 feet wide by 35 feet long. What is its perimeter?

The area of a rectangle is 414 square meters. The length is 18 meters. What is the width?

The area of a rectangle is 782 square centimeters. The width is 17 centimeters. What is the length?

The width of a rectangular window is 24 inches. The area is 624 square inches. What is the length?

The length of a rectangular poster is 28 inches. The area is 1316 square inches. What is the width?

Find the length of a rectangle with perimeter 124 and width 38.

Find the width of a rectangle with perimeter 92 and length 19.

Find the width of a rectangle with perimeter 16.2 and length 3.2.

Find the length of a rectangle with perimeter 20.2 and width 7.8.

The length of a rectangle is nine inches more than the width. The perimeter is 46 inches. Find the length and the width.

The width of a rectangle is eight inches more than the length. The perimeter is 52 inches. Find the length and the width.

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

The perimeter of a rectangle is 62 feet. The width is seven feet less than the length. Find the length and the 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.

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

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

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

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

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.

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.

Everyday Math

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?

Jose just removed the children’s playset from his back yard to make room for a rectangular garden. He wants to put a fence around the garden to keep out the dog. 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 length?

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.

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.

  • ⓐ Which figure looks like it has the larger area?
  • ⓑ Which looks like it has the larger perimeter?
  • ⓒ Now calculate the area and perimeter of each figure.
  • ⓓ Which has the larger area?
  • ⓔ Which has the larger perimeter?

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

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

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

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  • Authors: Lynn Marecek, MaryAnne Anthony-Smith, Andrea Honeycutt Mathis
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  • Book title: Elementary Algebra 2e
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Chapter 4: Inequalities

4.5 Geometric Word Problems

It is common to run into geometry-based word problems that look at either the interior angles, perimeter, or area of shapes. When looking at interior angles, the sum of the angles of any polygon can be found by taking the number of sides, subtracting 2, and then multiplying the result by 180°. In other words:

[latex]\text{sum of interior angles} = 180^{\circ} \times (\text{number of sides} - 2)[/latex]

This means the interior angles of a triangle add up to 180° × (3 − 2), or 180°. Any four-sided polygon will have interior angles adding to 180° × (4 − 2), or 360°. A chart can be made of these:

[latex]\begin{array}{rrrrrr} \text{3 sides:}&180^{\circ}&\times&(3-2)&=&180^{\circ} \\ \text{4 sides:}&180^{\circ}&\times&(4-2)&=&360^{\circ} \\ \text{5 sides:}&180^{\circ}&\times&(5-2)&=&540^{\circ} \\ \text{6 sides:}&180^{\circ}&\times&(6-2)&=&720^{\circ} \\ \text{7 sides:}&180^{\circ}&\times&(7-2)&=&900^{\circ} \\ \text{8 sides:}&180^{\circ}&\times&(8-2)&=&1080^{\circ} \\ \end{array}[/latex]

Example 4.5.1

The second angle [latex](A_2)[/latex] of a triangle is double the first [latex](A_1).[/latex] The third angle [latex](A_3)[/latex] is 40° less than the first [latex](A_1).[/latex] Find the three angles.

The relationships described in equation form are as follows:

[latex]A_2 = 2A_1 \text{ and } A_3 = A_1 - 40^{\circ}[/latex]

Because the shape in question is a triangle, the interior angles add up to 180°. Therefore:

[latex]A_1 + A_2 + A_3 = 180^{\circ}[/latex]

Which can be simplified using substitutions:

[latex]A_1 + (2A_1) + (A_1 - 40^{\circ}) = 180^{\circ}[/latex]

Which leaves:

[latex]\begin{array}{rrrrrrrrrrr} 2A_1&+&A_1&+&A_1&-&40^{\circ}&=&180^{\circ}&&&\\ &&&&4A_1&-&40^{\circ}&=&180^{\circ}&&\\ \\ &&&&&&4A_1&=&180^{\circ}&+&40^{\circ}\\ \\ &&&&&&A_1&=&\dfrac{220^{\circ}}{4}&\text{or}&55^{\circ} \end{array}[/latex]

This means [latex]A_2 = 2 (55^{\circ})[/latex] or 110° and [latex]A_3 = 55^{\circ}-40^{\circ}[/latex] or 15°.

Another common geometry word problem involves perimeter, or the distance around an object. For example, consider a rectangle, for which [latex]\text{perimeter} = 2l + 2w.[/latex]

Example 4.5.2

If the length of a rectangle is 5 m less than twice the width, and the perimeter is 44 m long, find its length and width.

[latex]L = 2W - 5 \text{ and } P = 44[/latex]

For a rectangle, the perimeter is defined by:

[latex]P = 2 W + 2 L[/latex]

Substituting for [latex]L[/latex] and the value for the perimeter yields:

[latex]44 = 2W + 2 (2W - 5)[/latex]

Which simplifies to:

[latex]44 = 2W + 4W - 10[/latex]

Further simplify to find the length and width:

[latex]\begin{array}{rrrrlrrrr} 44&+&10&=&6W&&&& \\ \\ &&54&=&6W&&&& \\ \\ &&W&=&\dfrac{54}{6}&\text{or}&9&& \\ \\ &\text{So}&L&=&2(9)&-&5&\text{or}&13 \\ \end{array}[/latex]

The width is 9 m and the length is 13 m.

Other common geometric problems are:

Example 4.5.3

A 15 m cable is cut into two pieces such that the first piece is four times larger than the second. Find the length of each piece.

[latex]P_1 + P_2 = 15 \text{ and } P_1 = 4P_2[/latex]

Combining these yields:

[latex]\begin{array}{rrrrrrr} 4P_2&+&P_2&=&15&& \\ \\ &&5P_2&=&15&& \\ \\ &&P_2&=&\dfrac{15}{5}&\text{or}&3 \end{array}[/latex]

This means that [latex]P_2 =[/latex] 3 m and [latex]P_1 = 4 (3),[/latex] or 12 m.

For questions 1 to 8, write the formula defining each relation. Do not solve.

  • The length of a rectangle is 3 cm less than double the width, and the perimeter is 54 cm.
  • The length of a rectangle is 8 cm less than double its width, and the perimeter is 64 cm.
  • The length of a rectangle is 4 cm more than double its width, and the perimeter is 32 cm.
  • The first angle of a triangle is twice as large as the second and 10° larger than the third.
  • The first angle of a triangle is half as large as the second and 20° larger than the third.
  • The sum of the first and second angles of a triangle is half the amount of the third angle.
  • A 140 cm cable is cut into two pieces. The first piece is five times as long as the second.
  • A 48 m piece of hose is to be cut into two pieces such that the second piece is 5 m longer than the first.

For questions 9 to 18, write and solve the equation describing each relationship.

  • The second angle of a triangle is the same size as the first angle. The third angle is 12° larger than the first angle. How large are the angles?
  • Two angles of a triangle are the same size. The third angle is 12° smaller than the first angle. Find the measure of the angles.
  • Two angles of a triangle are the same size. The third angle is three times as large as the first. How large are the angles?
  • The second angle of a triangle is twice as large as the first. The measure of the third angle is 20° greater than the first. How large are the angles?
  • Find the dimensions of a rectangle if the perimeter is 150 cm and the length is 15 cm greater than the width.
  • If the perimeter of a rectangle is 304 cm and the length is 40 cm longer than the width, find the length and width.
  • Find the length and width of a rectangular garden if the perimeter is 152 m and the width is 22 m less than the length.
  • If the perimeter of a rectangle is 280 m and the width is 26 m less than the length, find its length and width.
  • A lab technician cuts a 12 cm piece of tubing into two pieces such that one piece is two times longer than the other. How long are the pieces?
  • An electrician cuts a 30 m piece of cable into two pieces. One piece is 2 m longer than the other. How long are the pieces?

Answer Key 4.5

Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Teaching and Learning

Elevating math education through problem-based learning, by lisa matthews     feb 14, 2024.

Elevating Math Education Through Problem-Based Learning

Image Credit: rudall30 / Shutterstock

Imagine you are a mountaineer. Nothing excites you more than testing your skill, strength and resilience against some of the most extreme environments on the planet, and now you've decided to take on the greatest challenge of all: Everest, the tallest mountain in the world. You’ll be training for at least a year, slowly building up your endurance. Climbing Everest involves hiking for many hours per day, every day, for several weeks. How do you prepare for that?

The answer, as in many situations, lies in math. Climbers maximize their training by measuring their heart rate. When they train, they aim for a heart rate between 60 and 80 percent of their maximum. More than that, and they risk burning out. A heart rate below 60 percent means the training is too easy — they’ve got to push themselves harder. By combining this strategy with other types of training, overall fitness will increase over time, and eventually, climbers will be ready, in theory, for Everest.

solving problems using geometry

Knowledge Through Experience

The influence of constructivist theories has been instrumental in shaping PBL, from Jean Piaget's theory of cognitive development, which argues that knowledge is constructed through experiences and interactions , to Leslie P. Steffe’s work on the importance of students constructing their own mathematical understanding rather than passively receiving information .

You don't become a skilled mountain climber by just reading or watching others climb. You become proficient by hitting the mountains, climbing, facing challenges and getting right back up when you stumble. And that's how people learn math.

solving problems using geometry

So what makes PBL different? The key to making it work is introducing the right level of problem. Remember Vygotsky’s Zone of Proximal Development? It is essentially the space where learning and development occur most effectively – where the task is not so easy that it is boring but not so hard that it is discouraging. As with a mountaineer in training, that zone where the level of challenge is just right is where engagement really happens.

I’ve seen PBL build the confidence of students who thought they weren’t math people. It makes them feel capable and that their insights are valuable. They develop the most creative strategies; kids have said things that just blow my mind. All of a sudden, they are math people.

solving problems using geometry

Skills and Understanding

Despite the challenges, the trend toward PBL in math education has been growing , driven by evidence of its benefits in developing critical thinking, problem-solving skills and a deeper understanding of mathematical concepts, as well as building more positive math identities. The incorporation of PBL aligns well with the contemporary broader shift toward more student-centered, interactive and meaningful learning experiences. It has become an increasingly important component of effective math education, equipping students with the skills and understanding necessary for success in the 21st century.

At the heart of Imagine IM lies a commitment to providing students with opportunities for deep, active mathematics practice through problem-based learning. Imagine IM builds upon the problem-based pedagogy and instructional design of the renowned Illustrative Mathematics curriculum, adding a number of exclusive videos, digital interactives, design-enhanced print and hands-on tools.

The value of imagine im's enhancements is evident in the beautifully produced inspire math videos, from which the mountaineer scenario stems. inspire math videos showcase the math for each imagine im unit in a relevant and often unexpected real-world context to help spark curiosity. the videos use contexts from all around the world to make cross-curricular connections and increase engagement..

This article was sponsored by Imagine Learning and produced by the Solutions Studio team.

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Solving equations

Here you will learn about solving equations, including linear and quadratic algebraic equations, and how to solve them.

Students will first learn about solving equations in grade 8 as a part of expressions and equations, and again in high school as a part of reasoning with equations and inequalities.

What is solving an equation?

Solving equations is a step-by-step process to find the value of the variable. A variable is the unknown part of an equation, either on the left or right side of the equals sign. Sometimes, you need to solve multi-step equations which contain algebraic expressions.

To do this, you must use the order of operations, which is a systematic approach to equation solving. When you use the order of operations, you first solve any part of an equation located within parentheses. An equation is a mathematical expression that contains an equals sign.

For example,

\begin{aligned}y+6&=11\\\\ 3(x-3)&=12\\\\ \cfrac{2x+2}{4}&=\cfrac{x-3}{3}\\\\ 2x^{2}+3&x-2=0\end{aligned}

There are two sides to an equation, with the left side being equal to the right side. Equations will often involve algebra and contain unknowns, or variables, which you often represent with letters such as x or y.

You can solve simple equations and more complicated equations to work out the value of these unknowns. They could involve fractions, decimals or integers.

What is solving an equation?

Common Core State Standards

How does this relate to 8 th grade and high school math?

  • Grade 8 – Expressions and Equations (8.EE.C.7) Solve linear equations in one variable.
  • High school – Reasoning with Equations and Inequalities (HSA.REI.B.3) Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

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How to solve equations

In order to solve equations, you need to work out the value of the unknown variable by adding, subtracting, multiplying or dividing both sides of the equation by the same value.

  • Combine like terms .
  • Simplify the equation by using the opposite operation to both sides.
  • Isolate the variable on one side of the equation.

Solving equations examples

Example 1: solve equations involving like terms.

Solve for x.

Combine like terms.

Combine the q terms on the left side of the equation. To do this, subtract 4q from both sides.

The goal is to simplify the equation by combining like terms. Subtracting 4q from both sides helps achieve this.

After you combine like terms, you are left with q=9-4q.

2 Simplify the equation by using the opposite operation on both sides.

Add 4q to both sides to isolate q to one side of the equation.

The objective is to have all the q terms on one side. Adding 4q to both sides accomplishes this.

After you move the variable to one side of the equation, you are left with 5q=9.

3 Isolate the variable on one side of the equation.

Divide both sides of the equation by 5 to solve for q.

Dividing by 5 allows you to isolate q to one side of the equation in order to find the solution. After dividing both sides of the equation by 5, you are left with q=1 \cfrac{4}{5} \, .

Example 2: solve equations with variables on both sides

Combine the v terms on the same side of the equation. To do this, add 8v to both sides.

7v+8v=8-8v+8v

After combining like terms, you are left with the equation 15v=8.

Simplify the equation by using the opposite operation on both sides and isolate the variable to one side.

Divide both sides of the equation by 15 to solve for v. This step will isolate v to one side of the equation and allow you to solve.

15v \div 15=8 \div 15

The final solution to the equation 7v=8-8v is \cfrac{8}{15} \, .

Example 3: solve equations with the distributive property

Combine like terms by using the distributive property.

The 3 outside the parentheses needs to be multiplied by both terms inside the parentheses. This is called the distributive property.

\begin{aligned}& 3 \times c=3 c \\\\ & 3 \times(-5)=-15 \\\\ &3 c-15-4=2\end{aligned}

Once the 3 is distributed on the left side, rewrite the equation and combine like terms. In this case, the like terms are the constants on the left, –15 and –4. Subtract –4 from –15 to get –19.

Simplify the equation by using the opposite operation on both sides.

The goal is to isolate the variable, c, on one side of the equation. By adding 19 to both sides, you move the constant term to the other side.

\begin{aligned}& 3 c-19+19=2+19 \\\\ & 3 c=21\end{aligned}

Isolate the variable to one side of the equation.

To solve for c, you want to get c by itself.

Dividing both sides by 3 accomplishes this.

On the left side, \cfrac{3c}{3} simplifies to c, and on the right, \cfrac{21}{3} simplifies to 7.

The final solution is c=7.

As an additional step, you can plug 7 back into the original equation to check your work.

Example 4: solve linear equations

Combine like terms by simplifying.

Using steps to solve, you know that the goal is to isolate x to one side of the equation. In order to do this, you must begin by subtracting from both sides of the equation.

\begin{aligned} & 2x+5=15 \\\\ & 2x+5-5=15-5 \\\\ & 2x=10 \end{aligned}

Continue to simplify the equation by using the opposite operation on both sides.

Continuing with steps to solve, you must divide both sides of the equation by 2 to isolate x to one side.

\begin{aligned} & 2x \div 2=10 \div 2 \\\\ & x= 5 \end{aligned}

Isolate the variable to one side of the equation and check your work.

Plugging in 5 for x in the original equation and making sure both sides are equal is an easy way to check your work. If the equation is not equal, you must check your steps.

\begin{aligned}& 2(5)+5=15 \\\\ & 10+5=15 \\\\ & 15=15\end{aligned}

Example 5: solve equations by factoring

Solve the following equation by factoring.

Combine like terms by factoring the equation by grouping.

Multiply the coefficient of the quadratic term by the constant term.

2 x (-20) = -40

Look for two numbers that multiply to give you –40 and add up to the coefficient of 3. In this case, the numbers are 8 and –5 because 8 x -5=–40, and 8+–5=3.

Split the middle term using those two numbers, 8 and –5. Rewrite the middle term using the numbers 8 and –5.

2x^2+8x-5x-20=0

Group the terms in pairs and factor out the common factors.

2x^2+8x-5x-20=2x(x + 4)-5(x+4)=0

Now, you’ve factored the equation and are left with the following simpler equations 2x-5 and x+4.

This step relies on understanding the zero product property, which states that if two numbers multiply to give zero, then at least one of those numbers must equal zero.

Let’s relate this back to the factored equation (2x-5)(x+4)=0

Because of this property, either (2x-5)=0 or (x+4)=0

Isolate the variable for each equation and solve.

When solving these simpler equations, remember that you must apply each step to both sides of the equation to maintain balance.

\begin{aligned}& 2 x-5=0 \\\\ & 2 x-5+5=0+5 \\\\ & 2 x=5 \\\\ & 2 x \div 2=5 \div 2 \\\\ & x=\cfrac{5}{2} \end{aligned}

\begin{aligned}& x+4=0 \\\\ & x+4-4=0-4 \\\\ & x=-4\end{aligned}

The solution to this equation is x=\cfrac{5}{2} and x=-4.

Example 6: solve quadratic equations

Solve the following quadratic equation.

Combine like terms by factoring the quadratic equation when terms are isolated to one side.

To factorize a quadratic expression like this, you need to find two numbers that multiply to give -5 (the constant term) and add to give +2 (the coefficient of the x term).

The two numbers that satisfy this are -1 and +5.

So you can split the middle term 2x into -1x+5x: x^2-1x+5x-5-1x+5x

Now you can take out common factors x(x-1)+5(x-1).

And since you have a common factor of (x-1), you can simplify to (x+5)(x-1).

The numbers -1 and 5 allow you to split the middle term into two terms that give you common factors, allowing you to simplify into the form (x+5)(x-1).

Let’s relate this back to the factored equation (x+5)(x-1)=0.

Because of this property, either (x+5)=0 or (x-1)=0.

Now, you can solve the simple equations resulting from the zero product property.

\begin{aligned}& x+5=0 \\\\ & x+5-5=0-5 \\\\ & x=-5 \\\\\\ & x-1=0 \\\\ & x-1+1=0+1 \\\\ & x=1\end{aligned}

The solutions to this quadratic equation are x=1 and x=-5.

Teaching tips for solving equations

  • Use physical manipulatives like balance scales as a visual aid. Show how you need to keep both sides of the equation balanced, like a scale. Add or subtract the same thing from both sides to keep it balanced when solving. Use this method to practice various types of equations.
  • Emphasize the importance of undoing steps to isolate the variable. If you are solving for x and 3 is added to x, subtracting 3 undoes that step and isolates the variable x.
  • Relate equations to real-world, relevant examples for students. For example, word problems about tickets for sports games, cell phone plans, pizza parties, etc. can make the concepts click better.
  • Allow time for peer teaching and collaborative problem solving. Having students explain concepts to each other, work through examples on whiteboards, etc. reinforces the process and allows peers to ask clarifying questions. This type of scaffolding would be beneficial for all students, especially English-Language Learners. Provide supervision and feedback during the peer interactions.

Easy mistakes to make

  • Forgetting to distribute or combine like terms One common mistake is neglecting to distribute a number across parentheses or combine like terms before isolating the variable. This error can lead to an incorrect simplified form of the equation.
  • Misapplying the distributive property Incorrectly distributing a number across terms inside parentheses can result in errors. Students may forget to multiply each term within the parentheses by the distributing number, leading to an inaccurate equation.
  • Failing to perform the same operation on both sides It’s crucial to perform the same operation on both sides of the equation to maintain balance. Forgetting this can result in an imbalanced equation and incorrect solutions.
  • Making calculation errors Simple arithmetic mistakes, such as addition, subtraction, multiplication, or division errors, can occur during the solution process. Checking calculations is essential to avoid errors that may propagate through the steps.
  • Ignoring fractions or misapplying operations When fractions are involved, students may forget to multiply or divide by the common denominator to eliminate them. Misapplying operations on fractions can lead to incorrect solutions or complications in the final answer.

Related math equations lessons

  • Math equations
  • Rearranging equations
  • How to find the equation of a line
  • Solve equations with fractions
  • Linear equations
  • Writing linear equations
  • Substitution
  • Identity math
  • One step equation

Practice solving equations questions

1. Solve 4x-2=14.

GCSE Quiz False

Add 2 to both sides.

Divide both sides by 4.

2. Solve 3x-8=x+6.

Add 8 to both sides.

Subtract x from both sides.

Divide both sides by 2.

3. Solve 3(x+3)=2(x-2).

Expanding the parentheses.

Subtract 9 from both sides.

Subtract 2x from both sides.

4. Solve \cfrac{2 x+2}{3}=\cfrac{x-3}{2}.

Multiply by 6 (the lowest common denominator) and simplify.

Expand the parentheses.

Subtract 4 from both sides.

Subtract 3x from both sides.

5. Solve \cfrac{3 x^{2}}{2}=24.

Multiply both sides by 2.

Divide both sides by 3.

Square root both sides.

6. Solve by factoring:

Use factoring to find simpler equations.

Set each set of parentheses equal to zero and solve.

x=3 or x=10

Solving equations FAQs

The first step in solving a simple linear equation is to simplify both sides by combining like terms. This involves adding or subtracting terms to isolate the variable on one side of the equation.

Performing the same operation on both sides of the equation maintains the equality. This ensures that any change made to one side is also made to the other, keeping the equation balanced and preserving the solutions.

To handle variables on both sides of the equation, start by combining like terms on each side. Then, move all terms involving the variable to one side by adding or subtracting, and simplify to isolate the variable. Finally, perform any necessary operations to solve for the variable.

To deal with fractions in an equation, aim to eliminate them by multiplying both sides of the equation by the least common denominator. This helps simplify the equation and make it easier to isolate the variable. Afterward, proceed with the regular steps of solving the equation.

The next lessons are

  • Inequalities
  • Types of graph
  • Coordinate plane

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Table of Content

  • Introduction
  • Related Work
  • State-Transformer Paradigm and Generalized APGD-Solving Approach
  • The Proposed Algorithm for APGD Solving
  • Experiments

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Solving Algebraic Problems with Geometry Diagrams Using Syntax-Semantics Diagram Understanding

Litian Huang , Xinguo Yu , Lei Niu * , Zihan Feng

Faculty of Artificial Intelligence in Education, Central China Normal University, Wuhan, 430079, China

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(This article belongs to this Special Issue: Cognitive Computing and Systems in Education and Research )

Computers, Materials & Continua 2023 , 77 (1), 517-539. https://doi.org/10.32604/cmc.2023.041206

Received 14 April 2023; Accepted 12 August 2023; Issue published 31 October 2023

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1  Introduction

Solving Algebraic Problems with Geometry Diagrams (APGDs) is a challenging task in artificial intelligence due to the complexity and diversity of geometric relations that exist between geometric objects. APGDs are typically narrated by both textual descriptions and geometry diagrams, making it a multimodal reasoning task that requires a joint understanding of both modalities [ 1 ]. The additional information provided by the diagram, such as the relative location of lines and points, makes it essential for solvers to be able to parse the diagram. Furthermore, APGDs often require extra theorem knowledge in the problem-solving process. Although significant progress has been made in developing algorithms for solving math word problems [ 2 , 3 ], the research on solving APGDs is still limited. This presents a significant research challenge and opportunity to develop algorithms that can automatically solve APGDs, potentially providing valuable applications in education, such as intelligent tutoring systems.

In recent years, various methods have been developed to address APGD solving. These methods can be primarily categorized into two main types: Sequence-to-sequence (Seq2seq) methods and relation-centric methods. Seq2seq methods, such as Neural Geometric Solver (NGS) [ 4 ] and Geoformer [ 5 ], show the feasibility of using Seq2seq solution generation for solving APGDs. However, these methods currently suffer from limited readability and interpretability of the generated solution steps, which are often represented as sequential structures that do not resemble natural language. Additionally, the accuracy of these methods in solving APGDs remains a challenge, with existing methods often failing to achieve high accuracy rates. These limitations hinder the potential of using Seq2seq methods for effectively tutoring students in APGD solving.

Except for Seq2seq methods, the majority of current algorithms for solving APGDs, which belong to relation-centric methods, can be divided into two primary steps: problem understanding and symbolic solving. Similar to the algorithms used in solving arithmetic word problems [ 3 ], the symbolic solver in solving APGDs heavily relies on the relations of the output of problem understanding. The ability to obtain the necessary geometric knowledge for problem-solving from geometry diagrams is a critical issue. As a result, significant research focus on developing algorithms for understanding APGDs. As demonstrated by previous research [ 6 ], understanding APGDs involves two primary tasks: text understanding and diagram understanding. Both tasks are critical, with diagram understanding playing a key role in acquiring the knowledge from the diagram for solving the APGD. The challenge lies in effectively extracting and utilizing advanced knowledge embedded within the problem text and diagram. For text understanding, previous studies [ 1 , 7 ] showed that methods based on syntax semantic models can be successfully applied to extracting geometric knowledge in the problem text. In contrast, diagrams offer supplemental geometric information that complements the data not explicitly stated in the problem text. To achieve diagram understanding, the previous methods can be divided into two primary categories: rule-based method [ 8 – 10 ] and machine learning-based method [ 11 ]. Both rule-based and machine learning-based methods primarily focus on identifying basic elements and labels in geometry diagrams, which serves as a fundamental step toward generating simple geometric relations. However, a deep understanding of geometric relations is necessary for accurate problem-solving. Unfortunately, there are currently few studies that focus on achieving a deep understanding of geometric relations in geometry diagrams. Thus, there remains a need for a more comprehensive and effective approach to understanding diagrams in APGDs.

In this study, a three-phase scheme for solving APGDs is proposed, consisting of three phases: applying the state-transformer paradigm, employing the generalized APGD-solving approach, and developing a specific APGD-solving algorithm. This scheme underscores a progressive algorithm design process, transitioning from abstract concepts to concrete implementations, enabling a systematic and modular blueprint for constructing APGD-solving approaches. With the structured guidance of the three-phase scheme, a specialized APGD-solving algorithm is designed, encompassing the following key components: 1) the text understanding part employs the Syntax-Semantics ( S 2 ) model [ 1 , 7 ] for extracting geometric relations from problem text; 2) the diagram understanding part takes parsed diagrams as inputs and uses a vectorized S 2 model to extract basic geometric relations; 3) derived geometric relations are generated based on the diagramet theory proposed by Xia et al. [ 6 ]. After integrating all extracted relations, the comprehensive set of relations is fed into existing solvers. Fig. 1 illustrates the process of solving a given APGD by the proposed algorithm, which is designed to ensure a more in-depth understanding of the problem and holds the potential to provide more accurate and comprehensive solutions to APGDs. The experiments are conducted on datasets of APGDs from both primary and secondary school levels, including geometry calculation problems and shaded area problems, demonstrating that the proposed APGD-solving method significantly improves problem-solving accuracy.

images

Figure 1: The process that the proposed algorithm solves the given APGD

In summary, this study contributes to solving APGDs by:

1.    Proposing a novel three-phase scheme, specifically designed to systematize and modularize APGD-solving approaches. This scheme bridges the gap between abstract problem-solving concepts and their practical implementation, allowing for more efficient and effective exploration of the APGD-solving process.

2.    Development of a unique algorithm, the first of its kind to lay special emphasis on diagram understanding in the context of APGDs. The algorithm employs the vectorized S 2 model for extracting basic geometric relations and leverages diagramet theory-based method to generate derived geometric relations. This combination fosters a more comprehensive understanding of the geometric diagrams, thereby enhancing the algorithm’s overall problem-solving capability.

3.    Demonstrating the effectiveness of the proposed method through experiments on real-world datasets. The method led to significant enhancements in the accuracy of problem-solving, ranging from around 4% to 10% across different datasets and problem types. It exhibited a remarkable performance, particularly in complex problem goals like shaded area calculation, underscoring its potential for robust and effective problem-solving in the domain of APGDs.

The remainder of this paper is organized as follows: Section 2 provides an overview of related work; Section 3 illustrates the proposed three-phase scheme, with a focus on the paradigm phase and approach phase; Section 4 describes the proposed algorithm for solving APGDs; Section 5 presents the experimental results; and Section 6 concludes the paper and discusses future work.

2  Related Work

In this section, two main aspects of the literature related to the study are discussed. The first aspect provides an overview of existing methods for solving APGDs, including geometry calculation problems and shaded area problems. Research on both Seq2seq-based methods and relation-centric methods will be covered, highlighting their similarities and differences in addressing APGDs. The second aspect focuses on diagram understanding, which is particularly relevant to this study, as it aims to improve the overall performance of APGD-solving algorithms by comprehensively understanding the geometry diagram. In this part, various techniques and methods for parsing and understanding geometry diagrams are reviewed, which form the basis for generating geometric relations required in relation-centric methods. By examining the state-of-the-art methods in these areas, the foundation for the proposed method is laid, and its novelty in comparison to the existing research is demonstrated.

2.1 Methods of Solving APGDs

In recent years, two primary types of methods have emerged for solving APGDs: Seq2seq methods and relation-centric methods. Seq2seq methods have shown promise in solving APGDs. Chen et al. [ 4 ] introduced the GeoQA dataset and proposed NGS, which utilizes a co-attention mechanism to fuse text and diagram representations, predicting explainable programs based on the cross-modal representation. Chen et al. [ 5 ] advanced this research by constructing the UniGeo benchmark. They also proposed a unified geometric transformer framework called Geoformer, which is capable of handling geometry calculation and proof reasoning simultaneously. Despite these advancements, the main limitations of these Seq2seq methods include the limited interpretability and generalization capabilities and insufficient accuracy rates in their solutions.

Unlike Seq2seq methods, relation-centric methods aim to identify and utilize the underlying relationships and structures present in APGDs. The following studies showcase some notable advancements in relation-centric methods for APGD-solving tasks. G-ALIGNER by Seo et al. [ 8 ] and its subsequent improvement, GEOS [ 9 ], are pioneering attempts that align visual elements with their textual descriptions in APGDs. However, their approach essentially reduces the task to an optimization problem, aiming to find which choice satisfies all constraints. This approach lacks the reasoning process involved in actual problem-solving. Lu et al.’s Inter-GPS [ 10 ] utilized formal language and symbolic reasoning to address the complexities of APGD-solving tasks. However, their method focuses too heavily on the semantic aspects of the problems, leading to less accurate relation extraction from the given text. Yu et al. [ 12 ] proposed a two-phase algorithm for understanding and solving text-diagram function problems with impressive accuracy. Despite its strengths, the method falls short in terms of the interpretability of solutions, which is a crucial aspect of APGD-solving tasks. Alvin et al.’s approach called GeoShader [ 13 ] and Feng et al.’s approach [ 14 ] are both tailored to tackle shaded area problems, formalizing and solving such tasks efficiently. Despite their effectiveness for this specific problem type, these methods exhibit limitations in their applicability, as they are primarily designed to solve shaded area problems. While these studies each provide significant advancements in the field, a common limitation across all methods is their reliance on inputting simple geometric relations into the solver. This reliance hinders their ability to deeply understand geometry diagrams and extract more complex geometric relations, which in turn, affects the overall accuracy of problem-solving.

The proposed method is developed after an in-depth examination of existing methods and addressing identified strengths and weaknesses. Emphasizing the importance of diagram understanding, it extracts and interprets geometric relations from diagrams through dedicated procedures. By integrating relations from both text and diagrams, the method facilitates a profound understanding of APGDs. Symbolic reasoning is employed in the final problem-solving phase, utilizing the consolidated representation from previous phases for robust solution generation. The method thereby not only enhances APGD-solving efficiency but also ensures the interpretability of solutions, applicable across various problem types.

2.2 Methods of Understanding Geometry Diagrams

Understanding geometry diagrams is a crucial and necessary step in relation-centric methods for solving APGDs. The ability to accurately comprehend and interpret diagrams is essential for the subsequent identification of geometric elements and relations, which ultimately aids in problem-solving tasks. Seo’s foundational work [ 8 , 9 ] primarily relies on computer vision techniques, identifying simple relationships within diagrams but lacking comprehensive interpretability. Some studies [ 1 , 15 , 16 ] utilized numerical verification-based methods to extract relations from diagrams. While effective for specific scenarios, these methods lack a universal strategy for relation extraction, which can limit their applicability to more diverse problem sets. Xia et al. [ 6 ] introduced the diagramet theory for K−12 education, a promising concept still in its nascent stage and requiring further development. Zhang et al. [ 11 ] offered PGDPNet, an end-to-end deep learning model, but it lacks interpretability, limiting the transparency in relation identification. These studies collectively demonstrate the increasing importance and feasibility of automating the process of understanding geometry diagrams, which holds great potential for applications in education and intelligent tutoring systems. However, existing methods struggle to deeply comprehend advanced geometric relations within diagrams and rely on traditional methods that do not effectively handle the diversity of geometric styles and the complex relationships between primitives. These shortcomings restrict the accuracy and overall effectiveness of automatic APGD-solving methods.

As a precursor to the current research, Huang et al. [ 17 ] introduced a uniform vectorized S 2 model for automatic APGD understanding. This foundational method provided a simultaneous approach to both text and diagram understanding, which distinguished it from traditional methods. Building upon this foundational work, the present research introduces significant advancements, notably in efficiency, accuracy, and scalability. It employs an innovative three-phase scheme for APGD-solving that fosters a more systematic problem-solving approach, better equipped to handle complex problems, and enhances the interpretability of the solution process. These improvements represent a clear progression from the prototype-like algorithm of the previous work. Furthermore, this study strives to advance relation-centric methods in APGD-solving tasks, addressing the limitations noted in previous studies. A more robust problem-understanding method based on the vectorized S 2 model is proposed, which transforms problem text and diagrams into geometric relations automatically. Emphasizing interpretability, the proposed method utilizes symbolic reasoning and incorporates theorem knowledge as conditional rules, facilitating step-by-step reasoning, and enhancing the performance of APGD-solving algorithms. This advancement marks a significant improvement over previous research, making a considerable contribution to the field of automatic APGD-solving.

3  State-Transformer Paradigm and Generalized APGD-Solving Approach

In this study, a three-phase scheme (paradigm, approach, and algorithm) is adopted for solving APGDs. This scheme establishes a bridge between abstract concepts and specific implementations, offering a hierarchical and systematic framework for the research. Firstly, the state-transformer paradigm is established, serving as the foundation for the problem-solving approach design. Next, a generalized APGD-solving approach under the state-transformer paradigm is illustrated, outlining the solving process that encompasses multiple methods. Lastly, details of the algorithmic implementation are delved into, with the development and optimization of techniques to efficiently navigate through the states and transformers, ultimately generating a reliable and accurate solution to the given APGD. Through the adoption of this three-phase scheme, a comprehensive and coherent exploration of the research topic is facilitated. Each phase builds upon the previous one, ultimately resulting in a well-rounded and effective problem-solving method. This section presents the details of the state-transformer paradigm and APGD-solving approach.

3.1 State-Transformer Paradigm

The first phase, the state-transformer paradigm, is inspired by the state-action paradigm proposed by Yu et al. [ 3 ]. The state-transformer paradigm, as shown in Fig. 2 , is a general framework for solving APGDs. It consists of various states representing different phases of the APGD-solving process and transformers representing different operations that enable transitions between these states. The core idea behind the state-transformer paradigm is to systematically explore the state space by applying different transformers, enabling the algorithm to effectively navigate from an initial problem state to a desired solution state.

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Figure 2: The state-transformer paradigm for solving APGDs

Definition 1 ( State ) A state signifies the diverse phases or conditions that the APGD problem-solving process traverses. Each state is characterized by its distinctive input and output formats, capturing a particular facet of the problem-solving process.

Definition 2 ( Transformer ) A transformer corresponds to the operations or procedures affecting transitions between different states. It encapsulates the distinct algorithms that convert the output from one state into the input of another, thus allowing for the modularity and reusability of the components of the APGD-solving approach.

The shift from the original state-action paradigm to the state-transformer paradigm has been made in this study to emphasize the crucial role of transformers in enabling transitions between different states during the APGD-solving process. For a comprehensive understanding of the original state-action paradigm and its definitions, refer to the study [ 3 ].

3.2 Generalized APGD-Solving Approach

In the second phase, the state-transformer paradigm is elaborated on by describing a generalized approach for solving APGDs. This approach encapsulates the common characteristics and processing steps found in various methods. Fig. 3 below illustrates the transitions between different states using the transformers in the proposed approach. An elliptical node represents a state, and an arrow represents a transformer.

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Figure 3: APGD-solving approach under state-transformer paradigm

The key components of the proposed approach include:

•   Input Problem: The original APGD, including the text and diagrams.

•   Parsed Problem: The problem after parsing, which includes structured data of the APGD. These structured data encapsulate the essential information extracted from both the problem text and diagrams, organized in a systematic and structured manner conducive to further processing and problem-solving.

•   Basic Relations: A group of geometric relations that present fundamental connections between elements like points, lines, and shapes, including adjacency, collinearity, parallelism, etc.

•   Integrated Relations: A group of geometric relations that integrate basic and derived relations, offering a more comprehensive understanding of geometric problems.

•   Equation Set: A collection of mathematical equations generated from geometric relations.

•   Computable Sequence: A sequence that can be directly computed to obtain the final answer.

•   Output Answer: The final answer to the APGD.

Transformers:

•   Parsing: Parse the text and diagrams into structured data called parsed problem.

•   Diagram Understanding: Analyze a geometric diagram to identify and interpret geometric primitives and symbols, ultimately generating basic geometric relations that capture the visual information in the diagram.

•   Text Understanding: The process of extracting and interpreting geometric keywords and entities from the textual description of the APGD. The outcome is basic geometric relations extracted from the problem text.

•   Derived Relations Generation: The process of combining basic relations extracted from text and diagram understanding to generate derived relations that are more complex than basic relations.

•   Relation Processing: Convert geometric relations into equations.

•   Symbolic Solving: Solve equations and get the solution to the APGD.

•   Seq2seq Solution Generation: Use the Seq2seq method to directly generate the computable sequence.

•   Computation: Computes the final answer from the computable sequence.

Table 1 presents concrete examples illustrating how this generalized approach can be applied to various methods: Seq2seq methods directly embed the text and diagrams through transformer a , then decode the embeddings using transformer h to generate the computable sequence, and finally utilize transformer i to obtain the final answer. Traditional relation-centric methods parse geometry diagrams with transformer a , acquire basic relations through transformers b and c , convert basic relations into equations using transformer f , and input equations into a symbolic solver to obtain the final answer through transformer g .

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In this study, a state called Integrated Relations is introduced. By considering both basic and derived relations, a comprehensive representation of the diagram is obtained. In contrast to traditional relation-centric methods, the proposed method employs transformer d to generate derived relations and fuses them with basic relations to obtain integrated relations. Then, these integrated relations are converted into equations using transformer e , enabling more accurate and efficient solutions when utilized.

Through this section, the state-transformer paradigm and its application in the approach for APGD solving are introduced. In the next section, a detailed description of the algorithm design that stems from the APGD-solving approach will be provided.

4  The Proposed Algorithm for APGD Solving

This section introduces the proposed algorithm for solving APGDs, which demonstrates a concrete implementation of the generalized APGD-solving approach.

4.1 Algorithm Outline

In this section, a three-step algorithm based on the APGD-solving approach to address APGDs is presented. The first step involves acquiring a set of basic geometric relations through text understanding and diagram understanding. The second step generates derived geometric relations based on basic geometric relations, while the third step transforms all geometric relations into equations that are then input into a symbolic solver to determine the unknown value, which serves as the solution to the APGD. These three steps form the proposed algorithm, as illustrated in Algorithm 1. To implement the tasks of the algorithm, three procedures are employed: Procedure 1 for text understanding, Procedure 2 for diagram understanding, and Procedure 3 for derived relations generation.

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4.2 Geometric Relations Extraction

This subsection outlines the methods for extracting geometric relations in detail. It encompasses three components: (1) text understanding, (2) diagram understanding, and (3) derived relations generation. Before delving into text understanding and diagram understanding, it is crucial to preprocess the raw text and diagram separately to obtain structured representations suitable for further analysis. For the original text, parsing and annotation methods mentioned in [ 1 ] are employed to transform it into a sequence of tokens with associated Part-Of-Speech (POS) labels. This structured text representation serves as the input for text understanding. For the diagram, a combination of techniques, including the Hough transform [ 18 ] and object detector such as RetinaNet [ 19 ], is used to extract geometric primitives (points, lines, angles, arcs, circles), labels (textual description of geometric primitives) and symbols (vertical and parallel symbols, etc.). After obtaining the structured diagram representation, it becomes the input for diagram understanding, which helps build a comprehensive understanding of the problem and extract essential geometric relations for solution generation.

4.2.1 Text Understanding

In this part, an S 2 model-based method [ 1 , 7 ] is introduced to implement text understanding for APGDs. The syntax elements of the S 2 models consist of POS patterns, while the semantic elements are formed by keyword structures. The S 2 models are employed in Procedure 1 for extracting geometric relations from the text. Procedure 1 operates effectively once a suitable pool of S 2 models has been prepared.

Definition 1 ( S 2 Model ) An S 2 model, or syntax-semantics model, can be represented by a triplet m = ( K ,   P ,   R ) , where K stands for semantic keyword structures, P denotes POS labels, and R denotes the output geometric relations. The collection of all prepared S 2 models is symbolized by M = { m i = ( K i ,   P i , R i ) | i = 1 ,   2 ,   … ,   n } , and is referred to as the pool of S 2 models for APGDs.

In Fig. 4 , an example of an S 2 model and its components are provided. The S 2 model identifies the keyword structure K and the corresponding POS labels P within the parsed text T . It then replaces the elements in the matched sections of T with the geometric relation template R , generating the output.

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Figure 4: Example of an S 2 model and its components

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As shown in Procedure 1, the text understanding process starts with initializing an empty group of basic geometric relations Σ T . Then, the pool of prepared S 2 models M = { m i | i = 1 , 2 , … , n } is loaded. For each model in the pool, the algorithm attempts to match the model with the portions of the parsed problem text. For every matched portion, the geometric entities in the text are used to instantiate the elements in the corresponding S 2 model. Finally, an instance of the relation of the matched model is added to the set Σ T . This process iterates through all the models in the pool, ensuring that all geometric relations are extracted from the given input text. The output of this procedure is a set of geometric relations that are used for further analysis.

4.2.2 Diagram Understanding

This part introduces a method for understanding diagrams using a vectorized model called the Syntax-Semantics Diagram S 2 D model. The S 2 D model is particularly useful for analyzing the geometric diagram because it converts the diagram into a form that is suitable for efficient matching. Consequently, vectorization reduces the dimensionality of the data and speeds up the matching process, resulting in a more accurate and robust understanding of the diagram.

The S 2 D model extends the concepts of semantic keyword structures and syntactic POS labels from the S 2 model to geometric diagrams. In the S 2 D model, the structure of geometric primitives defines the type of geometric primitives that are placed in each position of the vector, thus providing the underlying semantic structure of the diagram. The geometric primitives reveal information about the type of geometric primitive in each location, thus providing the syntactic structure of the diagram. In addition, some geometric relations require numerical validation of the geometric primitives within them (e.g., perpendicularity, bisection). Therefore, the matching functions are added to determine whether the numerical relationships between geometric primitives match the model.

Definition 2 ( S 2 D Model ) An S 2 D model is represented by a quadruplet m ¯ = ( V , G P , F , R ) . V denotes the structure of geometric primitives, G P denotes the geometric primitives, F represents the matching functions, R represents the output geometric relations. The collection of all prepared S 2 D models is symbolized by M ¯ = { m ¯ i = ( V i ,   G P i , F i , R i ) | i = 1 ,   2 ,   … ,   n } , and is referred to as the pool of S 2 D models for APGDs.

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In Fig. 5 , a visual representation of an S 2 D model and its components, based on an example diagram D , is provided. The geometric primitives G P and its structure V are combined to form a single vector, which will be utilized for matching with the portions of the diagrams. The matching function set F contains primitives that require numerical verification and corresponding functions. The geometric relations set R which contains the relation templates are provided as the output.

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Figure 5: Example of an S 2 D model and its components

As shown in Procedure 2, the process begins by initializing an empty set Σ D . Next, the pool of prepared S 2 D models M ¯ is loaded. The models are clustered to create a list of tensors Γ M ¯ . The geometry diagram is encoded into the vector form E D . Based on Γ M ¯ and E D , a list Γ D which consists of candidate tensors to be matched is generated, and an empty list W is initialized. For each tensor in Γ D , the method attempts to match the candidates with the corresponding models in Γ M ¯ . When a candidate matches a model, the pair of candidate and model is added to the list W . After all candidates have been processed, the algorithm iterates through each matched candidate in W , decodes the candidate, and adds it to the set Σ D . This diagram understanding method efficiently extracts geometric relations from the given diagram input. The overview of the diagram understanding method is shown in Fig. 6 .

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Figure 6: The method of diagram understanding based on S 2 D models

In the following discussion, the details of Steps 3 to 7 in Procedure 2 are further explored.

Step 3: S 2 D Model Clustering

Given a pool of prepared S 2 D models M ¯ , each model comprises the structure of geometric primitives, multiple primitives, matching functions, and the geometric relations provided as output. As there are five distinct primitives, the primitive-count vector of model m ¯ i can be defined as N m ¯ i = [ n p , n l , n a , n s , n c ] , with n p , n l , n a , n s , n c representing the count of points, lines, angles, arc segments, and circles included in m i D , respectively. The models in M ¯ are clustered based on their primitive-count vector, resulting in a cluster set C = { c j | j = 1 , 2 , … , k } . The primitive-count vectors of C can be expressed as Φ = { N c j | j = 1 , 2 , … , k } , where N c j represents the primitive-count vector of all S 2 D models in c j .

As a result, all vectorized S 2 D models in the cluster c j can be created as tensor γ j M ¯ . Then, a list of tensors Γ M ¯ = { γ j M ¯ | j = 1 , 2 , … , k } can be generated, making it feasible for model matching.

This clustering process enables the prepared S 2 D models to be represented as tensors, which are essential for matching with the unique features of diagrams and for enhancing the matching speed and accuracy.

Step 4: Diagram Encoding

The geometric diagram complements the textual description by providing additional geometric information for problem-solving. To enable efficient matching with the S 2 D models, the geometric diagram should be represented in vector form as well. Inspired by the vector graph representations in [ 20 ], a bilayer undirected graph is designed to model the point-line-angle relationships. In the first layer, each segment represents a point in the diagram, and each element within a segment corresponds to a line that passes through that point. Since each line is associated with two points, it appears in two segments. Pointers to the other end of the line are kept in the elements of the segmented vector, which facilitates graph traversal. Additionally, a separate vector containing the lengths of each line is used to encode the line length information. In the second layer, each segment corresponds to a line in the diagram, and each element within a segment represents an angle formed by that line. Similarly, an additional vector that contains the degrees of the angles is used to include the angle information of the diagram. By using this bilayer undirected graph, the geometry diagram is encoded into the vector form E D consisting of vector tables.

Fig. 7 shows an example of encoding the geometry diagram into vector tables. The left side of Fig. 7 shows an undirected graph that represents the topological structure between geometric primitives, while the right side shows the corresponding vector table. The Fig. 7 demonstrates how points are connected to form lines within the diagram, while the lower table reveals how lines interact with each other to form angles in the same diagram. The “segment-descriptor” in the table header represents the number of edges connected to the corresponding vertex (e.g., the segment-descriptor of A is 2, which means that vertex A is connected to two edges, A C and A M ). The “cross-pointer” represents the indices of the vertices connected to the corresponding vertex (e.g., the cross-pointer of A is 4, indicating that A is connected with vertex C , which has an index of 4, to form the edge A C ).

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Figure 7: Example of encoding the geometry diagram into vector tables

The next step will use this vector table to generate candidate vectors. The advantage of this method is that it ensures that the primitives in the generated candidate vectors exist in the geometry diagram, which can avoid the occurrence of invalid candidate vectors and speed up the matching process.

Step 5: Candidate Vectors Generation

A combination function is introduced to generate a diverse range of candidate vectors for a specified geometric diagram. Given the encoded representation of a geometric diagram, E D , the combination module stochastically selects primitives from E D to construct candidate vectors by the set of primitive-count vectors Φ of Γ M ¯ . For each N c j ∈ Φ , the combination module can identify numerous vector sequences sharing identical primitive counts:

γ j D = C o m b ( E D , N c j ) (1)

where C o m b represents a combination function that locates all candidate vectors with the same primitive counts and groups them into a tensor.

Ultimately, a collection of candidate tensors can be produced as Γ D = { γ j D | j = 1 , 2 , … , k } .

Step 6: S 2 D Model Matching

In the model-matching process, the objective is to identify matched models in γ j M ¯ corresponding to a given candidate in γ j D . To ascertain whether the model matches the candidate, a two-step process is employed: anchoring and numerical verification.

Definition 3 ( Anchoring ) Anchoring is the initial step in the S 2 D model matching process, which aims to find matched models whose primitive structures are identical to the primitive structure of a given candidate vector. Successful anchoring between the two vectors implies that they share the same topology in the geometry diagram.

To implement the anchoring process, the simplest method would be to directly compare each primitive in the candidate vector with the corresponding primitive in the model vector. However, this method requires considering the order of primitives during matching, resulting in the generation of numerous candidate vectors with varying primitive orders, which in turn reduces matching efficiency. Therefore, this study adopts a method that transforms vector matching into graph matching. During the matching process, only the topological structure of the primitives within the vectors needs to be considered, without considering the order of the primitives. This significantly reduces the number of candidate vectors generated and improves matching efficiency.

To perform the anchoring process between candidate vectors and model vectors using graph matching, first, both the candidate vector and the model vector are converted into graph structures, with these graphs hierarchically divided based on points, lines, and angles. Next, matching between the two graphs is carried out by focusing on the outdegree of nodes. For each node in both the candidate graph and the model graph, a list is created, consisting of the node’s outdegree and the outdegrees of its child nodes. By comparing these lists of outdegrees for each node in the candidate vector graph with the corresponding lists in the model vector graph, it becomes possible to identify one-to-one correspondences between nodes. When two nodes from the candidate graph and the model graph have identical outdegree lists, it indicates that they share a one-to-one correspondence. Finally, if every node in the two graphs has a one-to-one correspondence, a mapping between the points in the candidate vector and the points in the model vector is obtained as below:

μ : G P h ′ → G P i (2)

where μ is a mapping between the primitives in the two vectors, G P h ′ and G P i are the primitives in the h -th candidate vector to be matched and the vector of S 2 D model m ¯ i , respectively. Due to the symmetry characteristics of geometric diagrams, the obtained mapping may not be unique at times. In such cases, it is essential to eliminate unreasonable combinations and select the appropriate mapping for decoding matched candidates.

An illustrative example of the anchoring process is provided in Fig. 8 , which visually demonstrates the aforementioned steps. The anchoring process is performed between the S 2 D model vector and the candidate vector composed of points ( A , M , B ), lines ( M A , M B ), and angle ( ∠ B M A ). After the successful anchoring process, a mapping between the points ( A , M , B ) in the diagram and those in the S 2 D model is obtained.

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Figure 8: Example of the anchoring process based on graph matching

If the anchoring process fails, it indicates that the model vector and the candidate vector do not match. However, if the anchoring process is successful, the next step is to perform numerical verification for the primitives. This is necessary because some geometric relations cannot be confirmed solely based on topological relationships. Using the primitive mapping obtained from the anchoring process, the variables in the matching function of the S 2 D model are replaced with the corresponding primitives from the candidate vector. Let f ( x 1 , x 2 , … , x n ) be one of the matching functions of m ¯ i , where x 1 , x 2 , … , x n are the variables representing a portion of primitives in G P i . After substituting the corresponding primitives from the candidate vector using the mapping function μ , the verification is considered successful if the following condition is satisfied:

f ( y 1 , y 2 , … , y n ) = 0 (3)

where y 1 , y 2 , … , y n represents a portion of primitives in G P h ′ .

In the case of Fig. 8 , by using the mapping, the lengths of lines and the measure of angle are substituted into the matching functions F in S 2 D model. Then, it is verified whether the lengths of segments A M and M B are equal and if ∠ B M A measures 180 degrees.

To sum up, if both the anchoring process and numerical verification are completed, it indicates that the candidate vector matches the S 2 D model. Following the above process, all the successfully matched candidate vectors and their corresponding models are added to a list W for decoding.

Step 7: Matched Candidates Decoding

In the final decoding phase, for each candidate vector in the list W , the corresponding S 2 D model’s geometric relation template is instantiated using the previously obtained mapping from the anchoring process. By substituting the primitives from the candidate vector into the model’s geometric relation template, the template is effectively instantiated with the specific primitives, resulting in a concrete geometric relation that reflects the original input diagram. These instantiated geometric relations are then collected into a set Σ D . The set Σ D represents the basic geometric relations obtained after performing diagram understanding on the input diagram D .

4.2.3 Derived Relations Generation

In this part of the process, the set of basic geometric relations Σ 0 is obtained by integrating geometric relations Σ T and Σ D , which are extracted from text understanding and diagram understanding:

Σ 0 = Σ T ∪ Σ D (4)

Most basic features of the geometry diagram, including the quantity and position of geometric primitives, can be described by the basic geometric relations in Σ 0 . However, these basic geometric relations alone are insufficient to solve the geometric problem. Therefore, it becomes necessary to consider derived relations, which are generated from the basic geometric relations, to describe more advanced geometric features.

Following the diagramet theory proposed in [ 6 ], a derived relations generation model is introduced that can define a type of diagramet and generate the corresponding derived geometric relations. To constrain the model’s scope, a diagram corpus U containing all diagrams from plane geometry theorems is defined, along with a pool of diagramets for U .

The generation process of derived relations is outlined in Procedure 3. First, a pool of diagramet models for the geometry problem is loaded. The model searches the pool based on the relations in Σ 0 and identifies instances of diagramets. Next, a mapping between primitives in the basic relations and primitives in the relation representation of the diagramet is established, leading to the generation of the derived relation. Finally, all obtained derived relations are collected into Σ 1 . This comprehensive set of derived relations, combined with the basic geometric relations, provides a more complete and accurate representation of the geometric problem, allowing for a deeper understanding of the underlying geometric features and relations.

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5  Experiments

5.1 Experimental Settings

5.1.1 Datasets and Evaluation Metrics

Experiments are conducted on four datasets: Geometry Calculation Problems for Primary School (GCP-PS), Geometry Calculation Problems for Secondary School (GCP-SS), Shaded Area Problems for Primary School (SAP-PS), and Shaded Area Problems for Secondary School (SAP-SS). In the GCP-PS dataset, 217 problems are collected from multiple versions of primary school math textbooks (including Beijing Normal University Press, People’s Education Press, and Jiangsu Education Press). For GCP-SS, the Geometry3K dataset [ 10 ] is used, containing 3,002 problems. Regarding the SAP-PS dataset, 120 problems are collected from various versions of primary school math textbooks, while for SAP-SS, Feng’s dataset [ 14 ] containing 192 problems is utilized. All problem texts have been translated into English.

To evaluate the performance of the proposed method, accuracy on different datasets is considered. To facilitate the evaluation of the solutions, all problems are transformed into a single-choice question format with four numerical choices. For the proposed method, if the obtained numerical result has the smallest absolute difference with a choice corresponding to the ground truth, the answer is considered correct. If the numerical result has the same absolute difference with multiple choices, a random selection is made from these choices. If the method fails to produce a result, a random choice is selected from the four choices.

5.1.2 Baselines

In the following section, the performance of the proposed method, referred to as PROPOSED, will be compared with several existing baseline methods on the four datasets. The aim of this comparison is to evaluate the effectiveness and robustness of PROPOSED against established approaches in the field.

GEOS [ 9 ] is the first automated system that solves SAT geometry problems by combining text understanding and diagram interpretation. The approach identifies a formal problem description compatible with both problem text and diagram and then feeds it into a geometric solver to determine the correct answer.

Inter-GPS [ 10 ] is a geometry problem-solving approach that leverages formal language and symbolic reasoning. It parses problem text and diagrams into formal language using rule-based text parsing and neural object detection. Inter-GPS incorporates theorem knowledge as conditional rules and performs symbolic reasoning step-by-step. As a key component of the approach, it features a theorem predictor, which is responsible for inferring the theorem application sequence, ultimately leading to a more efficient and reasonable search path.

FengAlg [ 14 ] is a method specifically designed for solving shaded area problems. It focuses on constructing equations to generate readable solutions, addressing the challenges associated with the diverse expressions of such problems and the complex relationships between shaded areas and other areas. By acquiring a system of equations and using a variety of techniques to construct them from inputs, FengAlg offers a concise and understandable solving process.

Due to GEOS and Inter-GPS being specialized in solving geometry calculation problems, and FengAlg being focused on shaded area problems, PROPOSED will be compared with GEOS and Inter-GPS on GCP-PS and GCP-SS, and with FengAlg on SAP-PS and SAP-SS.

5.1.3 Implementation Details

In the experimental process, both the proposed method and the baselines follow a similar process that starts by taking the problem, including the textual description and associated geometry diagram, as input. Each method employs a phase for understanding the problem text and the diagram separately, extracting relevant geometric relations. These relations are subsequently integrated into a unified representation that is used as input to the symbolic solver.

During the experiments, different symbolic solvers are employed for two types of problems.

For geometry calculation problems, the symbolic solver proposed in [ 10 ] is adopted, which takes the geometric relations extracted from PROPOSED as input. This solver has proven effective in handling various geometric calculations and provides a robust solution for problems in this category.

For shaded area problems, the symbolic solver from Feng’s method [ 14 ] is selected, which also takes the geometric relations obtained by PROPOSED as input. This solver is specifically designed to address the challenges associated with computing shaded areas in geometric diagrams. It considers the unique aspects of such problems and delivers accurate results accordingly.

5.2 Experimental Results

5.2.1 Comparisons with Baselines

In the evaluation, a comparison of PROPOSED with several baseline methods on four datasets is presented: GCP-PS and GCP-SS for Table 2 , and SAP-PS and SAP-SS for Table 3 . Considering that geometry calculation problems encompass multiple problem goals (length, angle, area, and ratio), the performance of several methods is measured in terms of solving accuracy across different problem goals on GCP-PS and GCP-SS datasets. On the other hand, shaded area problems focus solely on area calculation; hence, only the overall solving accuracy is considered on SAP-PS and SAP-SS datasets.

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Table 2 showcases the solving accuracy of PROPOSED and the baselines on both GCP-PS and GCP-SS datasets. PROPOSED outperforms all other methods, achieving an accuracy of 81.5% and 67.6% on the GCP-PS and GCP-SS datasets, respectively. It can be observed that the improvement of PROPOSED over the other baselines on the GCP-PS dataset is not as significant as on the GCP-SS dataset. The reason for this difference is that the GCP-PS dataset contains problems related to primary school geometry, which involve relatively simple geometric relations. In contrast, the GCP-SS dataset includes problems of secondary school geometry, which require more complex geometric relations to solve. This demonstrates that PROPOSED exhibits a greater advantage when tackling problems involving complex geometric relations. Notably, PROPOSED shows a significant improvement in the problem goal Area compared to the baselines. The substantial improvement in the problem goal Area can be attributed to the fact that area calculation typically necessitates a more extensive set of geometric relations. PROPOSED is particularly effective in extracting these more geometric relations, which contributes to its enhanced performance in the problem goal Area.

Table 3 compares PROPOSED with FengAlg on the SAP-PS and SAP-SS datasets. PROPOSED again surpasses the baseline method, achieving an accuracy of 68.1% on the SAP-PS dataset and 62.9% on the SAP-SS dataset. FengAlg, as a baseline, shows a reasonable performance but is still outperformed by PROPOSED.

In summary, PROPOSED demonstrates superior performance across all datasets and problem goals, indicating its effectiveness in solving APGDs. The results also highlight the improvements achieved by PROPOSED, particularly in the more challenging problem goals such as Area.

5.2.2 Ablation Study

In the ablation study, the examination of the impact of various components of PROPOSED on the overall accuracy is divided into two cases: (1) comparing the performance of using basic relations and integrated relations, and (2) comparing the performance with and without text understanding and diagram understanding. The results are presented in Table 4 .

images

In Case 1, the performance of the method when using only basic geometric relations (Basic) vs. the proposed approach of utilizing integrated geometric relations (Integrated) is compared. The method using integrated relations significantly improves the accuracy across all datasets, with a 6.9% increase in overall accuracy. Notably, the improvements in GCP-SS and SAP-SS datasets are more pronounced, indicating that integrated relations play a more substantial role in solving problems that require more complex relations.

In Case 2, the importance of text understanding and diagram understanding in PROPOSED is assessed. In the case of Text & Diagram w/o, the results show that relying solely on the text without the diagram is insufficient for solving APGDs accurately. On the other hand, when only diagram understanding is included (Text w/o & Diagram), the accuracy improves substantially, indicating that relying on the information from the diagrams alone can still solve a portion of APGDs. The highest accuracy is achieved when both text understanding and diagram understanding are combined (Text & Diagram), emphasizing the importance of utilizing both components.

5.2.3 Threats to Validity

Threats to validity are potential weaknesses in the design or execution of a study that could impact the credibility of the results. In the case of this research, certain limitations are present that could pose threats to both the internal and external validity of the findings.

In terms of internal validity, certain limitations are identified that directly affect the accuracy of the results. Two specific instances where PROPOSED encounters these limitations are illustrated in Fig. 9 . The first example shows a situation where the problem requires the construction of auxiliary lines within the diagram for its solution. Currently, PROPOSED lacks the capability to draw auxiliary lines in the geometry diagrams, a vital step in solving specific types of problems. This constraint contributes to an incomplete solution in this case. The second example highlights a failure case where PROPOSED is unable to interpret the implicit information present within the text. As a result, it fails to establish a connection between calculating the geometric area and the actual problem-solving goal. This points out a limitation in the text understanding component, necessitating improvements to better handle implicit information and effectively link textual data with the geometric diagram.

images

Figure 9: Failure examples of PROPOSED

Regarding the external validity, the method is designed to handle a broad range of APGDs. Nevertheless, its effectiveness may vary based on the complexity and specificity of the problem. The current research primarily targets standard geometric problems typically encountered in primary school textbooks. The applicability of PROPOSED to other types of problems is yet to be thoroughly tested, posing a potential threat to the generalizability of the findings.

Future efforts will aim to mitigate these threats to validity, with a focus on improving the performance and broad applicability of PROPOSED.

6  Conclusion

In this paper, the problem of solving APGDs is addressed by employing an algorithm designed based on a systematic and modular three-phase scheme. The state-transformer paradigm is first applied to model the problem-solving process, which effectively represents the intermediate states and transformations that occur during the process. This paradigm paves the way for a structured approach to problem-solving and facilitates the integration of various algorithms and techniques. Next, the generalized APGD-solving approach is employed, which provides a high-level strategy for extracting and utilizing geometric knowledge from both textual descriptions and geometry diagrams. This approach ensures the effective extraction of relevant information and lays the foundation for developing specific APGD-solving algorithms. Lastly, a specific APGD-solving algorithm is developed that incorporates the S 2 model for extracting geometric relations from the problem text and the S 2 D model for relation extraction from diagrams. In addition, a derived geometric relations generation method is proposed to extract derived relations from the basic geometric relations. The proposed method enables a more in-depth understanding of the problem and leads to a more accurate and comprehensive solution to APGDs. The experimental results on real-world datasets of primary and secondary school level problems demonstrate that the proposed method significantly improves APGD problem-solving accuracy across various problem types, including geometry calculation problems and shaded area problems.

However, there are limitations to the current approach, as demonstrated by the failure cases. These cases indicate areas for future improvement, such as developing the ability to construct auxiliary lines in diagrams and enhancing the understanding of implicit textual information. Future work will focus on addressing these limitations and further enhancing the proposed method’s capabilities to provide more accurate and comprehensive solutions to APGDs. This will not only contribute to the advancement of research in artificial intelligence but also pave the way for valuable applications in education, such as intelligent tutoring systems.

Acknowledgement: Sincere gratitude is extended to the fellow researchers for their invaluable expertise and insightful guidance throughout this study.

Funding Statement: This work is supported by the National Natural Science Foundation of China (No. 61977029) and the Fundamental Research Funds for the Central Universities, CCNU (No. 3110120001).

Author Contributions: Study conception and design: Litian Huang and Xinguo Yu; data collection: Litian Huang and Zihan Feng; analysis and interpretation of results: Litian Huang and Lei Niu; draft manuscript preparation: Litian Huang, Xinguo Yu and Lei Niu. All authors have reviewed the results and given their approval for the final version of the manuscript.

Availability of Data and Materials: The data used in this study is available from the corresponding author upon reasonable request.

Conflicts of Interest: The authors declare that they have no conflicts of interest to report regarding the present study.

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