Rectangle Dimensions Length And Width With Perimeter 26cm And Area 42cm²
Hey guys! Let's dive into a fun geometry problem today. We're going to figure out the length and width of a rectangle when we know its perimeter and area. This is a classic problem that combines a bit of algebra with our understanding of geometric shapes. So, grab your thinking caps, and let's get started!
Understanding the Problem
Okay, first things first, let's break down what we know. We have a rectangle, right? And we're given two crucial pieces of information. The perimeter of this rectangle is 26 cm, and its area is 42 cm². Now, the perimeter is the total distance around the rectangle – imagine walking along all its sides. The area, on the other hand, is the space enclosed within the rectangle. Think of it as the amount of carpet you'd need to cover the floor inside the rectangle.
To solve this, we need to remember the formulas for perimeter and area of a rectangle. Let's say the length of the rectangle is 'l' and the width is 'w'. The formula for the perimeter (P) is: P = 2l + 2w. This makes sense because you have two lengths and two widths that make up the total distance around. The formula for the area (A) is: A = l * w. This is because the area is found by multiplying the length and width.
So, we have two equations:
- 2l + 2w = 26
- l * w = 42
Now, our goal is to find the values of 'l' and 'w' that satisfy both of these equations. This is where our algebraic skills come into play! We'll need to manipulate these equations to isolate one variable and then substitute that value into the other equation. It might sound a bit complicated, but trust me, we'll break it down step-by-step, and you'll see it's not so bad. This is a super useful skill, not just for geometry problems, but for lots of different math and science situations. So, let's get to it and unlock the mystery of this rectangle's dimensions!
Setting Up the Equations
Alright, let's get those equations ready for action! We've already established the two key formulas we need:
- Perimeter: 2l + 2w = 26
- Area: l * w = 42
Now, looking at these, the perimeter equation seems a little bulky with those 2s hanging around. A smart move here is to simplify it. We can divide the entire equation by 2, which will make our numbers smaller and easier to work with. So, let's do that:
(2l + 2w) / 2 = 26 / 2
This simplifies to:
l + w = 13
Awesome! Now our perimeter equation is much cleaner. We have:
- l + w = 13
- l * w = 42
Now, we need a plan of attack. The strategy we're going to use here is called substitution. The idea is to isolate one variable in one equation and then plug that expression into the other equation. This will leave us with a single equation with just one variable, which we can then solve.
Looking at our equations, the first one (l + w = 13) seems easier to manipulate. Let's isolate 'l' in this equation. To do that, we'll subtract 'w' from both sides:
l + w - w = 13 - w
This gives us:
l = 13 - w
Perfect! Now we have an expression for 'l' in terms of 'w'. This is our golden ticket. We're going to take this expression (13 - w) and substitute it in place of 'l' in the area equation (l * w = 42). This is where the magic happens, and we'll see how this substitution helps us crack the problem. So, get ready for the next step – it's where things really start to come together!
Solving for Width (w)
Okay, guys, this is where the real problem-solving fun begins! Remember our plan? We're going to use the substitution method to solve for the width (w) of the rectangle. We've already done the groundwork: we isolated 'l' in the perimeter equation and found that l = 13 - w. Now it's time to put that expression to work.
We're going to substitute (13 - w) in place of 'l' in the area equation, which is l * w = 42. So, let's do it:
(13 - w) * w = 42
See how we've replaced 'l' with our expression? Now we have an equation that only involves 'w'. This is fantastic! But before we can solve for 'w', we need to simplify this equation a bit. We'll start by distributing the 'w' across the parentheses:
13w - w² = 42
Now, this looks like a quadratic equation! Don't let that scare you. We can handle this. To make it look more familiar, let's rearrange the terms to get it into the standard quadratic form (ax² + bx + c = 0). We'll subtract 42 from both sides and rearrange the terms:
-w² + 13w - 42 = 0
To make things even easier, let's multiply the entire equation by -1 to get rid of the negative sign in front of the w² term. This will make it look even more like the quadratics we're used to seeing:
w² - 13w + 42 = 0
Great! Now we have a standard quadratic equation. We can solve this using a few different methods, like factoring, completing the square, or using the quadratic formula. In this case, factoring looks like the easiest approach. So, let's see if we can find two numbers that multiply to 42 and add up to -13. Think about the factors of 42... they are 1 and 42, 2 and 21, 3 and 14, and 6 and 7. Hmm... 6 and 7 look promising! Since we need them to add up to a negative number, and multiply to a positive number, we'll use -6 and -7. Let's see if it works:
(-6) * (-7) = 42 (-6) + (-7) = -13
Perfect! So, we can factor our quadratic equation like this:
(w - 6)(w - 7) = 0
Now, to find the solutions for 'w', we set each factor equal to zero:
w - 6 = 0 or w - 7 = 0
Solving for 'w' in each case, we get:
w = 6 or w = 7
So, we have two possible values for the width! This means we're on the right track. But remember, we also need to find the length. Don't worry; we're not done yet. We'll use these values of 'w' to find the corresponding values of 'l' in the next step. We're getting closer to cracking this rectangle's code!
Finding the Length (l)
Awesome work, team! We've successfully found two possible values for the width (w): 6 cm and 7 cm. Now, the next logical step is to figure out the corresponding lengths (l) for each of these widths. Remember that we have a handy equation relating l and w: l = 13 - w. This is the equation we derived from the perimeter formula, and it's going to be our key to unlocking the length.
Let's take each value of 'w' and plug it into this equation:
Case 1: w = 6 cm
l = 13 - w l = 13 - 6 l = 7 cm
So, when the width is 6 cm, the length is 7 cm.
Case 2: w = 7 cm
l = 13 - w l = 13 - 7 l = 6 cm
And when the width is 7 cm, the length is 6 cm.
Wait a minute... do you notice something interesting here? We got the same two numbers, just swapped! This is actually a good sign. It tells us that we've likely done our calculations correctly. Think about it: a rectangle with a length of 7 cm and a width of 6 cm is essentially the same rectangle as one with a length of 6 cm and a width of 7 cm – we've just rotated it. The dimensions are the same.
So, what does this mean for our answer? Well, it means that we have found the dimensions of the rectangle! The length is 7 cm, and the width is 6 cm (or vice versa, it doesn't really matter which we call length and which we call width). We've solved the problem! But before we celebrate too much, let's do a quick check to make sure our answer makes sense in the context of the original problem.
Verification and Final Answer
Alright, guys, we've done the heavy lifting, but it's always a good idea to double-check our work to make sure everything adds up. We found that the length (l) of the rectangle is 7 cm and the width (w) is 6 cm. Now, let's plug these values back into our original equations for perimeter and area to see if they hold true.
Perimeter Check:
Remember, the perimeter formula is P = 2l + 2w. Let's substitute our values:
P = 2(7 cm) + 2(6 cm) P = 14 cm + 12 cm P = 26 cm
Yes! Our calculated perimeter matches the given perimeter of 26 cm. That's a great sign.
Area Check:
The area formula is A = l * w. Let's plug in our values again:
A = 7 cm * 6 cm A = 42 cm²
Awesome! Our calculated area also matches the given area of 42 cm². This confirms that our solution is correct! We've successfully found the dimensions of the rectangle that satisfy both the perimeter and area conditions.
So, after all that algebraic maneuvering and careful calculations, we've arrived at our final answer. The length of the rectangle is 7 cm, and the width is 6 cm. We did it! This problem is a great example of how we can use math to solve real-world geometric puzzles. It combines our understanding of geometric formulas with our algebraic skills to find a solution. Hopefully, this walkthrough has helped you understand the process and given you some confidence in tackling similar problems in the future.
Remember, practice makes perfect! The more you work through problems like this, the more comfortable you'll become with the concepts and the techniques involved. Keep up the great work, and happy problem-solving!
