A Rectangle Has An Area Of 217 Square Inches. The Width Is 3 More Than 4 Times The Length. Find The Length And Width, In Inches, Of The Rectangle.
===========================================================
Problem Description
A rectangle has an area of 217 square inches. The width is 3 more than 4 times the length. We need to find the length and width, in inches, of the rectangle.
Step 1: Define the Variables
Let's denote the length of the rectangle as L and the width as W. We are given that the area of the rectangle is 217 square inches, so we can write the equation:
A = L × W
We are also given that the width is 3 more than 4 times the length, so we can write another equation:
W = 4L + 3
Step 2: Substitute the Second Equation into the First Equation
We can substitute the second equation into the first equation to get:
A = L × (4L + 3)
Expanding the right-hand side, we get:
A = 4L² + 3L
Step 3: Substitute the Given Area Value
We are given that the area of the rectangle is 217 square inches, so we can substitute this value into the equation:
217 = 4L² + 3L
Step 4: Rearrange the Equation
We can rearrange the equation to get a quadratic equation in terms of L:
4L² + 3L - 217 = 0
Step 5: Solve the Quadratic Equation
We can solve the quadratic equation using the quadratic formula:
L = (-b ± √(b² - 4ac)) / 2a
In this case, a = 4, b = 3, and c = -217. Plugging these values into the formula, we get:
L = (-(3) ± √((3)² - 4(4)(-217))) / 2(4)
Simplifying the expression, we get:
L = (-3 ± √(9 + 3464)) / 8
L = (-3 ± √3472) / 8
L = (-3 ± 59.12) / 8
We have two possible solutions for L:
L = (-3 + 59.12) / 8
L = 56.12 / 8
L = 7.014
L = (-3 - 59.12) / 8
L = -62.12 / 8
L = -7.765
Since the length cannot be negative, we discard the second solution.
Step 6: Find the Width
Now that we have found the length, we can find the width using the equation:
W = 4L + 3
Substituting the value of L, we get:
W = 4(7.014) + 3
W = 28.056 + 3
W = 31.056
Step 7: Round the Answers
We can round the answers to two decimal places:
L ≈ 7.01
W ≈ 31.06
The final answer is: L ≈ 7.01, W ≈ 31.06
Conclusion
In this problem, we used algebraic techniques to find the length and width of a rectangle with a given area. We defined the variables, substituted the second equation into the first equation, rearranged the equation, solved the quadratic equation, and found the width. The final answer is L ≈ 7.01, W ≈ 31.06.
=====================================
Q: What is the formula for the area of a rectangle?
A: The formula for the area of a rectangle is A = L × W, where A is the area, L is the length, and W is the width.
Q: How do I find the length and width of a rectangle with a given area?
A: To find the length and width of a rectangle with a given area, you need to use the formula A = L × W and the given information about the relationship between the length and width. In this case, we were given that the width is 3 more than 4 times the length, so we could write the equation W = 4L + 3.
Q: What is the quadratic formula?
A: The quadratic formula is a mathematical formula that is used to solve quadratic equations. It is given by:
x = (-b ± √(b² - 4ac)) / 2a
where a, b, and c are the coefficients of the quadratic equation.
Q: How do I use the quadratic formula to solve a quadratic equation?
A: To use the quadratic formula to solve a quadratic equation, you need to identify the values of a, b, and c in the equation. Then, you can plug these values into the quadratic formula and simplify the expression to find the solutions.
Q: What is the difference between the two solutions of a quadratic equation?
A: The two solutions of a quadratic equation are the values of the variable that satisfy the equation. In the case of the quadratic equation 4L² + 3L - 217 = 0, the two solutions are L = 7.014 and L = -7.765. The first solution is the length of the rectangle, while the second solution is not valid because it is negative.
Q: How do I round the answers to a specific number of decimal places?
A: To round the answers to a specific number of decimal places, you can use the rounding rules. For example, to round the answer to two decimal places, you can look at the third decimal place and decide whether to round up or down.
Q: What is the final answer to the problem?
A: The final answer to the problem is L ≈ 7.01, W ≈ 31.06.
Q: Can I use this method to solve other problems with a given area?
A: Yes, you can use this method to solve other problems with a given area. The key is to identify the relationship between the length and width and use the formula A = L × W to find the length and width.
Q: What are some common mistakes to avoid when solving problems with a given area?
A: Some common mistakes to avoid when solving problems with a given area include:
- Not identifying the relationship between the length and width
- Not using the correct formula for the area of a rectangle
- Not solving the quadratic equation correctly
- Not rounding the answers to the correct number of decimal places
Q: How can I practice solving problems with a given area?
A: You can practice solving problems with a given area by working through examples and exercises. You can also try solving problems on your own and checking your answers with a calculator or a friend.
Q: What are some real-world applications of solving problems with a given area?
A: Some real-world applications of solving problems with a given area include:
- Designing buildings and other structures
- Creating art and designs
- Solving problems in physics and engineering
- Working with measurements and dimensions in everyday life
Q: Can I use this method to solve problems with a given perimeter?
A: Yes, you can use this method to solve problems with a given perimeter. The key is to identify the relationship between the length and width and use the formula P = 2L + 2W to find the length and width.
Q: What are some common mistakes to avoid when solving problems with a given perimeter?
A: Some common mistakes to avoid when solving problems with a given perimeter include:
- Not identifying the relationship between the length and width
- Not using the correct formula for the perimeter of a rectangle
- Not solving the quadratic equation correctly
- Not rounding the answers to the correct number of decimal places