Rectangle Area: Length 4m Longer Than Width, Perimeter 36m
Hey guys! Ever find yourself scratching your head over a rectangle problem? Specifically, one where you know the perimeter, and how much longer the length is compared to the width, but need to find the area? Don't worry, you're not alone! Let's break down this common problem step-by-step, making it super easy to understand.
Understanding the Problem
Before we jump into calculations, let's make sure we grasp the core concepts. This rectangle problem involves a rectangle where the length is 4 meters longer than its width. We also know the perimeter of the rectangle is 36 meters. Our mission? To figure out the rectangle's area. This requires a blend of geometry and basic algebra, but trust me, it's totally manageable!
Defining Perimeter and Area
First, let's refresh our memory on what perimeter and area actually mean for a rectangle:
- Perimeter: The total distance around the outside of the rectangle. Think of it as walking along all four sides; the total distance you walk is the perimeter. The formula for a rectangle's perimeter is P = 2l + 2w, where 'l' is the length and 'w' is the width.
- Area: The amount of space the rectangle covers. Imagine tiling the rectangle with square tiles; the number of tiles you'd need is the area. The formula for a rectangle's area is A = l * w, where 'l' is the length and 'w' is the width.
Key Information from the Problem
Now, let's pinpoint the crucial info given in the problem:
- The length (l) is 4 meters more than the width (w): l = w + 4
- The perimeter (P) is 36 meters: P = 36
With these pieces of information, we're ready to start solving for the area!
Setting Up the Equations
Okay, now for the fun part – turning our word problem into mathematical equations! This is where we use our knowledge of perimeter and the relationship between length and width to create equations we can actually solve.
Expressing Length in Terms of Width
We know from the problem that the length is 4 meters longer than the width. This is a crucial piece of information. We can express this mathematically as:
l = w + 4
This equation is super helpful because it allows us to represent the length using the width, which means we'll have fewer variables to deal with later on. Think of it as translating English into math – we've just translated "the length is 4 meters longer than the width" into a mathematical statement.
Using the Perimeter Formula
We also know that the perimeter of the rectangle is 36 meters. Remember the formula for the perimeter of a rectangle? It's:
P = 2l + 2w
Since we know the perimeter (P = 36), we can substitute that into the formula:
36 = 2l + 2w
This gives us a second equation. Now we have two equations, and that's exactly what we need to solve for two unknowns (the length and the width). This is where the magic of algebra starts to happen!
Solving for Width and Length
Alright, guys, let's get down to business and solve for the width and length. We've got our two equations, and now we're going to use a technique called substitution to crack this problem. It might sound intimidating, but trust me, it's totally doable!
The Substitution Method
The substitution method is a way of solving a system of equations (that's just a fancy term for having more than one equation) by solving one equation for one variable and then substituting that expression into the other equation. Think of it like replacing one thing with something equivalent.
We already have one equation solved for 'l':
l = w + 4
This is perfect! We can substitute this expression for 'l' into our perimeter equation:
36 = 2l + 2w
Replacing 'l' with '(w + 4)' gives us:
36 = 2(w + 4) + 2w
See what we did there? We've now got an equation with only one variable, 'w', which means we can solve for the width!
Solving for the Width (w)
Now, let's simplify and solve for 'w'. First, we distribute the 2:
36 = 2w + 8 + 2w
Combine like terms:
36 = 4w + 8
Subtract 8 from both sides:
28 = 4w
Divide both sides by 4:
w = 7
Woohoo! We've found the width! The width of the rectangle is 7 meters.
Solving for the Length (l)
Now that we know the width, finding the length is a piece of cake! We can use our equation:
l = w + 4
Substitute 'w' with 7:
l = 7 + 4
l = 11
So, the length of the rectangle is 11 meters.
Calculating the Area
We've conquered the hardest part – finding the width and length! Now, calculating the area is the final step, and it's super straightforward. Remember, the area of a rectangle is found by multiplying its length and width.
Applying the Area Formula
The formula for the area of a rectangle is:
A = l * w
We know the length (l) is 11 meters and the width (w) is 7 meters. Let's plug those values into the formula:
A = 11 * 7
Finding the Area
Now, simply multiply 11 and 7:
A = 77
Therefore, the area of the rectangle is 77 square meters. Don't forget the units! Area is always measured in square units.
Final Answer and Summary
Alright, we've made it to the finish line! The area of the rectangle is 77 square meters. Let's quickly recap the steps we took to get there:
- Understood the problem: We identified the given information (perimeter and the relationship between length and width) and what we needed to find (area).
- Set up the equations: We expressed the length in terms of width (l = w + 4) and used the perimeter formula (36 = 2l + 2w).
- Solved for width and length: We used the substitution method to solve for the width (w = 7 meters) and then calculated the length (l = 11 meters).
- Calculated the area: We used the area formula (A = l * w) to find the area (A = 77 square meters).
By breaking down the problem into smaller steps, we were able to solve it easily. Remember, guys, practice makes perfect! The more you work through problems like this, the more confident you'll become in your problem-solving skills.