Rectangle Dimensions: Solving Perimeter & Area Problem
Hey there, math enthusiasts! Today, we're diving into a classic geometry problem that combines perimeter and area to unlock the dimensions of a rectangle. We've got a rectangle with a perimeter of 24 meters and an area of 35 square meters. Our mission? To find its length and width. Buckle up, because we're about to embark on a mathematical adventure!
Understanding the Basics: Perimeter and Area
Before we jump into solving the problem, let's refresh our understanding of perimeter and area. These two concepts are fundamental to understanding the size and shape of two-dimensional figures, and they play a crucial role in various real-world applications, from calculating fencing needs to determining the amount of paint required for a wall.
Perimeter, in simple terms, is the total distance around the outside of a shape. Imagine walking along the edges of a rectangle; the total distance you walk is the perimeter. For a rectangle, the perimeter is calculated by adding up the lengths of all four sides. Since a rectangle has two pairs of equal sides (length and width), the formula for the perimeter (P) is:
P = 2 * (length + width)
Think of it as adding the length and width together, and then doubling the result to account for the two lengths and two widths.
Now, let's talk about area. Area is the amount of space enclosed within a shape. It's like measuring the amount of carpet needed to cover a floor, or the amount of land within a fenced yard. For a rectangle, the area (A) is calculated by multiplying its length and width:
A = length * width
It's a straightforward calculation that gives us the two-dimensional space occupied by the rectangle.
In our problem, we're given the perimeter (24 meters) and the area (35 square meters) of the rectangle. We need to use these two pieces of information to figure out the individual values of the length and width. This is where the fun begins!
Setting Up the Equations: Translating Words into Math
The key to solving this problem lies in translating the given information into mathematical equations. We know the perimeter and the area, and we have formulas for both. This means we can create two equations with two unknowns (length and width), which is a classic setup for solving a system of equations. Let's break it down step by step.
First, let's represent the length of the rectangle with the variable 'l' and the width with the variable 'w'. This is a standard practice in algebra, as it allows us to manipulate these unknown quantities using mathematical operations. It's like giving the length and width nicknames so we can talk about them more easily in our equations.
Now, let's use the perimeter information. We know the perimeter is 24 meters, and we have the formula P = 2 * (l + w). So, we can write our first equation as:
2 * (l + w) = 24
This equation states that twice the sum of the length and width is equal to 24. It's a direct translation of the given information into a mathematical statement. We can simplify this equation by dividing both sides by 2, which gives us:
l + w = 12
This simplified equation tells us that the sum of the length and width is 12 meters. It's a more manageable form of the perimeter information that we can use in our calculations.
Next, let's use the area information. We know the area is 35 square meters, and we have the formula A = l * w. So, our second equation is:
l * w = 35
This equation states that the product of the length and width is equal to 35. It captures the area information in a mathematical form. Now we have two equations:
- l + w = 12
- l * w = 35
This is a system of two equations with two unknowns. We have two pieces of information, and we have two variables to solve for. This is a classic mathematical puzzle, and we're well on our way to cracking it!
Solving the System: Unraveling the Dimensions
Now that we have our system of equations, the next step is to solve for the length (l) and width (w). There are several methods we can use to solve a system of equations, such as substitution, elimination, or graphing. For this problem, the substitution method is a particularly elegant approach.
The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This eliminates one variable, leaving us with a single equation with a single unknown, which is much easier to solve. Let's see how it works in our case.
Looking at our equations:
- l + w = 12
- l * w = 35
It seems easier to solve the first equation for one of the variables. Let's solve for 'l'. To isolate 'l', we can subtract 'w' from both sides of the equation:
l = 12 - w
Now we have an expression for 'l' in terms of 'w'. This is the key to the substitution method. We can now substitute this expression for 'l' into the second equation. This means replacing 'l' in the second equation with '12 - w'. Our second equation becomes:
(12 - w) * w = 35
This equation now has only one variable, 'w'. We've successfully eliminated 'l' using substitution. Now, let's simplify and solve for 'w'. Distribute the 'w' on the left side:
12w - w² = 35
This is a quadratic equation, which is an equation of the form ax² + bx + c = 0. To solve it, we need to rearrange it into standard form by moving all terms to one side:
0 = w² - 12w + 35
Now we have a quadratic equation in standard form. There are several ways to solve quadratic equations, such as factoring, using the quadratic formula, or completing the square. In this case, factoring is the easiest approach.
We need to find two numbers that multiply to 35 and add up to -12. These numbers are -5 and -7. So, we can factor the quadratic equation as:
(w - 5) * (w - 7) = 0
For this equation to be true, one or both of the factors must be equal to zero. This gives us two possible solutions for 'w':
- w - 5 = 0 => w = 5
- w - 7 = 0 => w = 7
So, the width 'w' can be either 5 meters or 7 meters. Now we need to find the corresponding values for the length 'l'. We can use the equation we derived earlier: l = 12 - w.
If w = 5, then l = 12 - 5 = 7
If w = 7, then l = 12 - 7 = 5
Notice that we get the same two values for length and width, just swapped. This makes sense because the length and width are interchangeable in a rectangle. The longer side is usually considered the length, but it doesn't really matter which one we call which. So, we have found our dimensions!
The Solution: Unveiling the Rectangle's True Form
After all our mathematical maneuvering, we've arrived at the solution! The dimensions of the rectangle are 7 meters and 5 meters. This means the rectangle has a length of 7 meters and a width of 5 meters, or vice versa. We've successfully solved the rectangle riddle!
Let's double-check our answer to make sure it satisfies the given conditions. The perimeter should be 24 meters, and the area should be 35 square meters.
- Perimeter: 2 * (length + width) = 2 * (7 + 5) = 2 * 12 = 24 meters (Correct!)
- Area: length * width = 7 * 5 = 35 square meters (Correct!)
Our solution checks out! We've found the dimensions of the rectangle that satisfy both the perimeter and area conditions. It's always a good idea to verify your answer, especially in math problems. It gives you confidence that you've arrived at the correct solution and haven't made any calculation errors along the way.
So, there you have it! We've successfully determined the dimensions of a rectangle given its perimeter and area. This problem highlights the power of using algebraic equations to represent geometric relationships. By translating the word problem into mathematical equations, we were able to solve for the unknowns and unlock the rectangle's true form. Math is awesome, isn't it? Keep practicing, and you'll become a master problem-solver in no time!
