A Painting Measuring 8 Inches By 10 Inches Is Surrounded By A Frame Of Uniform Width. If The Combined Area Of The Painting And The Frame Is 224 Square Inches, Find The Width Of The Frame.

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Introduction

In this article, we will delve into a mathematical problem that involves a painting and its frame. The problem requires us to find the width of the frame given the dimensions of the painting and the combined area of the painting and the frame. We will use algebraic equations to solve this problem and provide a step-by-step solution.

Problem Statement

A painting measuring 8 inches by 10 inches is surrounded by a frame of uniform width. If the combined area of the painting and the frame is 224 square inches, find the width of the frame.

Step 1: Define the Variables

Let's define the width of the frame as x inches. The dimensions of the painting with the frame will be (8 + 2x) inches by (10 + 2x) inches.

Step 2: Calculate the Area of the Painting and the Frame

The area of the painting with the frame is given by the product of its dimensions:

Area = (8 + 2x)(10 + 2x)

We can expand this expression to get:

Area = 80 + 16x + 20x + 4x^2

Combine like terms:

Area = 80 + 36x + 4x^2

Step 3: Set Up the Equation

We are given that the combined area of the painting and the frame is 224 square inches. We can set up an equation using this information:

80 + 36x + 4x^2 = 224

Step 4: Solve the Equation

Subtract 80 from both sides of the equation:

36x + 4x^2 = 144

Subtract 36x from both sides of the equation:

4x^2 = 144 - 36x

Divide both sides of the equation by 4:

x^2 = (144 - 36x)/4

x^2 = 36 - 9x

Rearrange the equation to get a quadratic equation in standard form:

x^2 + 9x - 36 = 0

Step 5: Solve the Quadratic Equation

We can solve this quadratic equation using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 1, b = 9, and c = -36. Plug these values into the formula:

x = (-(9) ± √((9)^2 - 4(1)(-36))) / 2(1)

x = (-9 ± √(81 + 144)) / 2

x = (-9 ± √225) / 2

x = (-9 ± 15) / 2

We have two possible solutions for x:

x = (-9 + 15) / 2 = 3

x = (-9 - 15) / 2 = -12

Since the width of the frame cannot be negative, we discard the solution x = -12.

Conclusion

The width of the frame is 3 inches.

Example Use Case

This problem can be used to teach students about algebraic equations and how to solve them. It can also be used to demonstrate the importance of checking the validity of solutions.

Real-World Application

This problem can be applied to real-world situations where a painting or a picture is surrounded by a frame. For example, a person may want to know the width of the frame to determine how much material to buy for a custom frame.

Mathematical Concepts

This problem involves the following mathematical concepts:

  • Algebraic equations
  • Quadratic equations
  • Solving quadratic equations using the quadratic formula
  • Checking the validity of solutions

Tips and Variations

  • This problem can be modified to involve a different shape, such as a rectangle or a square.
  • The problem can be made more challenging by adding additional constraints or requirements.
  • Students can be asked to create their own problems involving a painting and a frame.
    A Painting and Its Frame: Q&A ================================

Introduction

In our previous article, we solved a problem involving a painting and its frame. We found that the width of the frame is 3 inches. In this article, we will answer some frequently asked questions related to this problem.

Q: What is the formula for the area of a rectangle with a frame?

A: The formula for the area of a rectangle with a frame is:

Area = (length + 2x)(width + 2x)

where x is the width of the frame.

Q: How do I know which solution to choose when solving a quadratic equation?

A: When solving a quadratic equation, you should always check the validity of the solutions. In this case, we discarded the solution x = -12 because it is negative. The width of the frame cannot be negative, so we chose the solution x = 3.

Q: Can I use this problem to teach students about algebraic equations?

A: Yes, this problem can be used to teach students about algebraic equations. It involves solving a quadratic equation and checking the validity of the solutions. This problem can help students understand the importance of algebraic equations in real-world situations.

Q: How can I modify this problem to make it more challenging?

A: You can modify this problem by adding additional constraints or requirements. For example, you can ask students to find the width of the frame given a different combined area or a different shape.

Q: What are some real-world applications of this problem?

A: This problem has several real-world applications. For example, a person may want to know the width of the frame to determine how much material to buy for a custom frame. This problem can also be used to teach students about the importance of algebraic equations in real-world situations.

Q: Can I use this problem to teach students about geometry?

A: Yes, this problem can be used to teach students about geometry. It involves finding the area of a rectangle with a frame, which is a fundamental concept in geometry.

Q: How can I use this problem to teach students about problem-solving?

A: You can use this problem to teach students about problem-solving by asking them to find the width of the frame given different constraints or requirements. This can help students develop their critical thinking skills and learn how to approach problems in a logical and methodical way.

Q: What are some common mistakes students make when solving this problem?

A: Some common mistakes students make when solving this problem include:

  • Not checking the validity of the solutions
  • Not using the correct formula for the area of a rectangle with a frame
  • Not following the correct steps to solve the quadratic equation

Q: How can I assess student understanding of this problem?

A: You can assess student understanding of this problem by asking them to:

  • Solve the quadratic equation and find the width of the frame
  • Check the validity of the solutions
  • Explain the steps they took to solve the problem
  • Apply the problem to real-world situations

Conclusion

In this article, we answered some frequently asked questions related to the problem of finding the width of a frame given the dimensions of a painting and the combined area of the painting and the frame. We hope this article has been helpful in providing additional information and resources for teachers and students.