Rectangle Area Problem: Comparing Widths

Hey guys! Let's dive into a super interesting math problem involving rectangles and their areas. We're going to break it down step by step so it's crystal clear. This problem involves comparing the widths of two rectangles that have the same area but different lengths. So, grab your thinking caps, and let's get started!

Understanding the Problem

Okay, so here's the deal. We have two rectangles that are special because they both cover the same amount of space – we call this the area. The problem tells us this area is ‘a’ square meters. Now, the first rectangle has a length of ‘b’ meters. The second rectangle is a bit different; its length is longer, specifically 14 meters longer than the first one. Our mission? To figure out how many times smaller the width of the second rectangle is compared to the first rectangle. This involves understanding the relationship between area, length, and width, and then using some simple algebra to find our answer. It's like a puzzle, and we're going to solve it together!

Defining the Variables

Before we jump into calculations, let's define our variables clearly. This will help us keep track of everything and make the process smoother. So, we've got:

• Area of both rectangles: a m²
• Length of the first rectangle: b m
• Length of the second rectangle: b + 14 m

Now, we need to figure out the widths. Let's call the width of the first rectangle w₁ and the width of the second rectangle w₂. Our ultimate goal is to find the ratio of w₁ to w₂, which will tell us how many times smaller the second width is compared to the first. Remember, the area of a rectangle is calculated by multiplying its length and width. This is the key to solving our problem, so let's keep this in mind as we move forward. With our variables defined, we're ready to start setting up some equations and cracking this problem!

Setting Up the Equations

Alright, now that we've got our variables sorted out, it's time to build some equations. Remember that the area of a rectangle is just its length multiplied by its width. Since both rectangles have the same area, ‘a’, we can set up two equations:

1. For the first rectangle: a = b * w₁
2. For the second rectangle: a = (b + 14) * w₂

These equations are the backbone of our solution. They tell us how the lengths and widths of our rectangles relate to their shared area. The next step is to use these equations to find expressions for the widths, w₁ and w₂. Once we have those, we can compare them and figure out how many times smaller w₂ is compared to w₁. Think of it like this: we're translating the word problem into the language of math, and these equations are our first sentences. With these in place, we're well on our way to finding the answer!

Solving for the Widths

Now comes the fun part – solving for the widths! We've got our two equations:

1. a = b * w₁
2. a = (b + 14) * w₂

Let's start with the first equation. To find w₁, we need to isolate it on one side of the equation. We can do this by dividing both sides by b:

• w₁ = a / b

Easy peasy, right? Now, let's tackle the second equation. We want to find w₂, so we'll divide both sides of the equation by (b + 14):

• w₂ = a / (b + 14)

Great! We've now found expressions for both w₁ and w₂ in terms of a and b. This is a major step forward. We know how to calculate each width if we have the values of a and b. But remember, our goal isn't just to find the widths themselves; we want to compare them. Specifically, we want to know how many times smaller w₂ is than w₁. To do that, we'll need to find the ratio of w₁ to w₂. So, let's move on to the next step and figure out that ratio.

Finding the Ratio of Widths

Okay, we're on the home stretch now! We've got expressions for w₁ and w₂:

• w₁ = a / b
• w₂ = a / (b + 14)

To find out how many times smaller w₂ is than w₁, we need to calculate the ratio w₁ / w₂. This is like asking, "If w₁ is this big, how many of w₂ would fit into it?"

So, let's set up the division:

(w₁ / w₂) = (a / b) / (a / (b + 14))

Dividing by a fraction is the same as multiplying by its reciprocal, so we can rewrite this as:

(w₁ / w₂) = (a / b) * ((b + 14) / a)

Now, we can simplify this expression. Notice that ‘a’ appears in both the numerator and the denominator, so we can cancel them out:

(w₁ / w₂) = (b + 14) / b

This is our final ratio! It tells us exactly how many times smaller the width of the second rectangle is compared to the width of the first rectangle. The ratio is (b + 14) / b. This means that w₁ is (b + 14) / b times larger than w₂, or conversely, w₂ is (b / (b + 14)) times smaller than w₁. We've cracked the code! This problem shows us how math can help us compare different shapes and sizes, even when they have something in common, like their area.

Conclusion

So, there you have it! We've successfully navigated through this rectangle area problem. We started by understanding the problem, defining our variables, and setting up equations. Then, we solved for the widths of the rectangles and, most importantly, found the ratio that compares them. The final answer is that the width of the second rectangle is (b + 14) / b times smaller than the width of the first rectangle.

This kind of problem is a fantastic way to sharpen our math skills and understand how different geometric concepts relate to each other. It's not just about formulas; it's about thinking logically and breaking down a problem into manageable steps. I hope you found this explanation helpful and that it boosted your confidence in tackling similar challenges. Keep practicing, keep exploring, and most importantly, keep having fun with math! You guys nailed it! 🚀