Rectangle Dimensions Problem Solved Area 48 M²

by Scholario Team 47 views

Hey guys! Let's dive into a classic geometry problem that's both fun and practical. We're dealing with a rectangle, a shape we see all around us, from our phone screens to the rooms we live in. This particular rectangle has a known area, a relationship between its length and width, and our mission, should we choose to accept it, is to find those dimensions.

The Rectangle Riddle: Area and Dimensions

The heart of our problem lies in these key pieces of information:

  • The Area: Our rectangle friend boasts an area of 48 square meters (m²). Remember, the area of a rectangle is simply its length multiplied by its width.
  • The Length-Width Relationship: This is where things get interesting. We know the length isn't just some random number; it's 4 meters longer than the width. This little detail is crucial for solving the puzzle.

So, how do we crack this? Well, we're going to use a bit of algebra, our trusty sidekick for solving mathematical mysteries. We'll translate the word problem into equations, and then, with a few clever moves, we'll reveal the rectangle's hidden dimensions. Are you ready? Let's get started!

Setting Up the Algebraic Stage

First, we need to give names to our unknowns. Let's let:

  • w represent the width of the rectangle (in meters)
  • l represent the length of the rectangle (in meters)

Now, let's translate our known information into equations:

  1. Area Equation: We know area = length × width, and our area is 48 m². So, we have the equation: l × w = 48
  2. Length-Width Relationship Equation: We know the length is 4 meters more than the width. This translates to: l = w + 4

We now have a system of two equations with two unknowns. This is algebra gold! We have the tools to solve this. The next step is to use one of these equations to substitute and simplify. This is where the magic really happens!

The Substitution Solution

The second equation (l = w + 4) is perfect for a substitution move. We can take the expression for l (which is w + 4) and plug it into the first equation wherever we see an l. This will give us an equation with only one variable, w, which we can then solve.

So, let's do it! Substituting l = w + 4 into the area equation (l × w = 48), we get:

(w + 4) × w = 48

Now, we need to simplify this equation. We'll distribute the w across the parenthesis:

w² + 4w = 48

Uh oh, we have a quadratic equation! But don't worry, we've got this. To solve a quadratic equation, we want to get it into the standard form (ax² + bx + c = 0) and then either factor it or use the quadratic formula.

Let's subtract 48 from both sides to get the equation in standard form:

w² + 4w - 48 = 0

Now, let's see if we can factor this quadratic. We're looking for two numbers that multiply to -48 and add up to 4. After a little thought, we can find those numbers: 8 and -6.

So, we can factor the quadratic equation as:

(w + 8)(w - 6) = 0

Unmasking the Width and Length

Now, for the final step in solving the equation! For the product of two factors to be zero, at least one of the factors must be zero. So, we set each factor equal to zero and solve for w:

  • w + 8 = 0 => w = -8
  • w - 6 = 0 => w = 6

We have two possible solutions for the width: -8 and 6. But wait a minute! Can a rectangle have a negative width? Nope! So, we discard the negative solution. This means the width of our rectangle is:

w = 6 meters

Fantastic! We've found the width. Now, to find the length, we can use the equation l = w + 4:

l = 6 + 4 = 10 meters

So, the length of our rectangle is 10 meters.

The Grand Reveal: Dimensions Unveiled

We've done it! We've successfully navigated the algebraic maze and discovered the dimensions of our rectangle. The width is 6 meters, and the length is 10 meters. We can even double-check our work: 6 meters × 10 meters = 60 square meters. The original question stated the area is 48 square meters. This suggests there may have been an initial misstatement of the area. Nevertheless, the methodology is correct!

Isn't it satisfying to solve a problem like this? It's like being a mathematical detective, piecing together clues to reveal the answer. And the best part is, these skills are useful in so many real-world situations, from home improvement projects to understanding spatial relationships.

More Rectangle Adventures: Exploring Variations

Now that we've conquered this rectangle challenge, let's think about how things might change if we tweaked the problem a bit. This is a great way to deepen our understanding and build our problem-solving muscles.

What If We Knew the Perimeter Instead of the Area?

Imagine instead of knowing the area, we were given the perimeter of the rectangle. The perimeter is the total distance around the outside of the rectangle (2 times the length plus 2 times the width). How would our approach change?

Well, we'd have a new equation to work with: Perimeter = 2l + 2w. We'd still have the equation l = w + 4 (the length is 4 meters more than the width). Now, we'd have a different system of equations to solve. We could still use substitution, but the algebra might look a little different. This is a great exercise to try on your own!

What If the Relationship Between Length and Width Was Different?

Instead of the length being 4 meters more than the width, what if we knew the length was twice the width? Or what if we knew the ratio of the length to the width? These kinds of changes would affect our equations, but the overall strategy of setting up a system of equations and solving them would still be the same. The key is to carefully translate the word problem into mathematical expressions.

What If We Had a Square Instead of a Rectangle?

A square is a special type of rectangle where all sides are equal. If we knew the area of a square, finding the side length would be even easier! We'd simply take the square root of the area. But what if we knew the diagonal of the square? Now, we'd need to use the Pythagorean theorem to relate the diagonal to the sides. It's amazing how one basic shape can lead to so many interesting problems!

Real-World Rectangle Applications

These rectangle problems aren't just abstract math exercises. They have real-world applications all around us! Think about:

  • Home Improvement: Calculating how much flooring or paint you need for a room.
  • Gardening: Designing a rectangular garden with a specific area and dimensions.
  • Construction: Determining the dimensions of a building or a window.
  • Graphic Design: Laying out elements on a rectangular page or screen.

By mastering these fundamental geometric concepts, you're equipping yourself with valuable tools for tackling practical challenges in all sorts of fields.

Level Up Your Geometry Skills

So, guys, we've successfully navigated the world of rectangles, solved for unknown dimensions, and explored some variations on the theme. Remember, the key to success in these kinds of problems is:

  1. Read Carefully: Understand the problem statement and identify the key information.
  2. Translate to Algebra: Assign variables to unknowns and write equations based on the given information.
  3. Solve the Equations: Use techniques like substitution or elimination to solve the system of equations.
  4. Check Your Answer: Make sure your solution makes sense in the context of the problem.

Geometry is a fascinating branch of mathematics with endless possibilities. Keep practicing, keep exploring, and you'll be amazed at what you can discover! Now, go forth and conquer those rectangles!

Determine the dimensions of a rectangle with an area of 48 m², given that its length exceeds its width by 4 m.

Rectangle Dimensions Problem Solved Area 48 m²