Rectangle Dimensions: A Tricky Math Problem Solved!

Hey everyone! Let's dive into a fun math problem involving rectangles, their dimensions, and a bit of algebraic trickery. We're given two key pieces of information that relate the width and length of our rectangle. Our goal? To figure out exactly how wide and how long this rectangle is. So, grab your thinking caps, and let's get started!

Understanding the Problem

Before we jump into solving, let's break down what we know. We have a rectangle, and we need to find its dimensions – that’s the length and the width. The problem gives us two clues:

1. Five times the width exceeds four times the length by 3 cm. This means if you take the width and multiply it by 5, the result is 3 cm more than if you multiplied the length by 4.
2. Seven times the length is fourteen more than its perimeter. Remember, the perimeter of a rectangle is the total distance around it. So, twice the length plus twice the width. This clue tells us that if you multiply the length by 7, it's 14 cm more than the perimeter.

Now that we have a good grasp of the information, we can start translating these statements into mathematical equations. This is where the real fun begins!

Setting Up the Equations

Alright, let's turn those wordy clues into neat, manageable equations. This will allow us to use algebra to solve for the unknown length and width.

• Let's use 'w' to represent the width of the rectangle and 'l' to represent the length.

• The first clue, "Five times the width exceeds four times the length by 3 cm," can be written as:

  5w = 4l + 3

  This equation tells us that 5 times the width is equal to 4 times the length, plus an extra 3 cm.

• The second clue, "Seven times the length is fourteen more than its perimeter," needs a little more unpacking. Remember, the perimeter of a rectangle is 2l + 2w. So, we can write this clue as:

  7l = 2l + 2w + 14

  This equation tells us that 7 times the length is equal to the perimeter (2 times length plus 2 times width), plus an extra 14 cm.

Now we have a system of two equations with two unknowns (w and l). We can use different algebraic techniques, like substitution or elimination, to solve for w and l. Let's move on to solving these equations!

Solving the System of Equations

Okay, now for the exciting part: solving for 'w' and 'l'! We have our two equations:

1. 5w = 4l + 3
2. 7l = 2l + 2w + 14

Let's simplify the second equation first:

7l = 2l + 2w + 14

Subtract 2l from both sides:

5l = 2w + 14

Now we have:

1. 5w = 4l + 3
2. 5l = 2w + 14

We can use the substitution or elimination method. Let's use the substitution method. Solve the first equation for w:

5w = 4l + 3

w = (4l + 3) / 5

Now substitute this expression for w into the second equation:

5l = 2 * ((4l + 3) / 5) + 14

Multiply both sides by 5 to get rid of the fraction:

25l = 2 * (4l + 3) + 70

25l = 8l + 6 + 70

25l = 8l + 76

Subtract 8l from both sides:

17l = 76

Divide by 17:

l = 76 / 17

l = 4.47 (approximately)

Now that we have the value for l, we can substitute it back into the equation for w:

w = (4l + 3) / 5

w = (4 * 4.47 + 3) / 5

w = (17.88 + 3) / 5

w = 20.88 / 5

w = 4.18 (approximately)

So, we've found that the length l is approximately 4.47 cm and the width w is approximately 4.18 cm.

Checking Our Solution

It's always a good idea to check our solution to make sure it works with the original problem. Let's plug our values for l and w back into the original equations:

1. 5w = 4l + 3

   5 * 4.18 = 4 * 4.47 + 3

   20.9 = 17.88 + 3

   20.9 ≈ 20.88 (This is close enough, considering we rounded our values)

2. 7l = 2l + 2w + 14

   7 * 4.47 = 2 * 4.47 + 2 * 4.18 + 14

   31.29 = 8.94 + 8.36 + 14

   31.29 ≈ 31.3 (Again, very close!)

Since our values satisfy both equations (with a tiny bit of rounding error), we can be confident in our solution. The length of the rectangle is approximately 4.47 cm, and the width is approximately 4.18 cm.

The Answer

So, after all that math, we've finally arrived at the answer! The dimensions of the rectangle are:

• Length (l): Approximately 4.47 cm
• Width (w): Approximately 4.18 cm

Isn't it satisfying to solve a problem like this? It's a great reminder of how algebra can be used to solve real-world problems. Whether you're calculating dimensions for a construction project or just flexing your brainpower, these skills come in handy. Keep practicing, and you'll become a math whiz in no time!

Key Takeaways

Let's quickly recap the key steps we took to solve this problem:

• Understanding the Problem: Carefully read and understand the information given.
• Setting Up Equations: Translate the word problem into mathematical equations using variables.
• Solving the System: Use algebraic techniques (substitution, elimination) to solve for the unknowns.
• Checking the Solution: Plug the values back into the original equations to verify the answer.

By following these steps, you can tackle similar problems with confidence. Remember, practice makes perfect!

Extra Tips and Tricks

Here are a few extra tips that can help you solve these types of problems more efficiently:

• Draw a Diagram: Visualizing the problem can often make it easier to understand.
• Label Everything: Clearly label your variables and units to avoid confusion.
• Simplify Equations: Before attempting to solve, simplify the equations as much as possible.
• Check Your Work: Always double-check your calculations to avoid errors.

With these tips and a little practice, you'll be solving rectangle dimension problems like a pro in no time!

Conclusion

We have successfully found the dimensions of the rectangle! Remember, breaking down the problem into smaller, manageable steps and translating the word problem into algebraic equations is key. Keep practicing, and you'll be able to solve all sorts of mathematical puzzles! High five! You are awesome! Now you can use this skill in real world applications.