Rectangle Area And Width Calculation: Solving For T
Hey guys! Ever get those tricky geometry problems that seem like a puzzle? Well, let's break one down together! We're going to dive into a problem involving rectangles, using a little bit of algebra to figure out some dimensions and areas. It might sound intimidating, but trust me, we'll take it one step at a time. So, grab your thinking caps, and let’s get started!
Understanding the Problem: Rectangles ABCD and EFGD
In this geometry problem, we're dealing with two rectangles: ABCD and EFGD. The key here is to carefully dissect the information given and visualize the rectangles. We know that rectangle ABCD has a length (AB) of 2t cm. This means the side AB is expressed in terms of a variable, 't', which we'll need to solve for later. We also have rectangle EFGD, which shares a side (GD) with ABCD. The length of CG is given as 4 cm, and the length of ED is (2t-1) cm. These pieces of information are crucial for setting up our equations and finding the unknowns. Drawing a diagram is super helpful here! Visualizing the rectangles and labeling the sides with the given lengths can make the problem much clearer. It allows us to see the relationships between the sides and how they connect to each other. Think of it like creating a roadmap before a journey; a clear diagram helps us navigate the problem-solving process much more effectively. Remember, in geometry, a good picture is worth a thousand calculations! Before we jump into the math, let’s make sure we understand what the problem is asking. We need to:
- Express the width of rectangle EF (EF) in terms of 't'.
- Express the area of rectangle EFGD in terms of 't'.
- If the area of rectangle EFGD is 88 cm², find the value of 't'.
By understanding these objectives, we can focus our efforts and ensure we're solving for the right things. Okay, with our problem clearly defined, let’s move on to the next step: figuring out how to express the unknowns in terms of 't'.
Expressing the Width of EF in Terms of t
Let's tackle the first part: expressing the width of EF in terms of 't'. This is where our understanding of rectangles and their properties comes in handy. Remember, in a rectangle, opposite sides are equal in length. This means that the length of GD is the same as the length of AB, which we know is 2t cm. Now, look at rectangle EFGD. We know ED is (2t - 1) cm. To find the width EF, we need to recognize that EF is the same as DG. Think of it like this: we're trying to find a missing piece of the puzzle. We know the total length (GD or AB) and a part of it (CG). To find the remaining part (DG or EF), we simply subtract the known part from the total. So, how do we find DG? Well, we know that GD is equal to AB, which is 2t cm. We also know that CG is 4 cm. DG is the remaining part of GD after subtracting CG. Therefore, we can express DG (and thus EF) as:
EF = DG = GD - CG
Substituting the values we know:
EF = 2t - 4 cm
And that’s it! We've successfully expressed the width of EF in terms of 't'. This is a crucial step because it allows us to relate the dimensions of the rectangle to the variable we're trying to solve for. It's like finding the right key to unlock the problem. This expression will be essential when we calculate the area of rectangle EFGD in the next step. Remember, geometry problems often involve breaking down complex shapes into simpler components and using known relationships to find the unknowns. By applying the properties of rectangles and using basic subtraction, we were able to express the width of EF in terms of 't'. Now, let's move on to the next challenge: expressing the area of rectangle EFGD in terms of 't'.
Expressing the Area of EFGD in Terms of t
Now, let's move on to the next part of the problem: expressing the area of rectangle EFGD in terms of 't'. Remember the formula for the area of a rectangle? It's simply Area = length × width. We already know the length ED is (2t - 1) cm, and we just figured out the width EF is (2t - 4) cm. So, to find the area of EFGD, we simply multiply these two expressions: Area of EFGD = ED × EF. Let’s plug in what we know:
Area of EFGD = (2t - 1) × (2t - 4)
Now, we need to expand this expression. Think of it like multiplying two groups of terms. We use the distributive property (sometimes called the FOIL method) to multiply each term in the first group by each term in the second group:
- 2t × 2t = 4t²
- 2t × -4 = -8t
- -1 × 2t = -2t
- -1 × -4 = 4
Now, let's put it all together:
Area of EFGD = 4t² - 8t - 2t + 4
We can simplify this expression by combining the like terms (-8t and -2t):
Area of EFGD = 4t² - 10t + 4 cm²
And there you have it! We've successfully expressed the area of rectangle EFGD in terms of 't'. This quadratic expression represents how the area of the rectangle changes as the value of 't' changes. It's like having a formula that tells us the size of the rectangle for any given value of 't'. This is a major step forward because now we have an equation that relates the area to the variable 't'. In the next part of the problem, we'll use this equation to find the specific value of 't' when we know the area is 88 cm². So, we've conquered expressing the area in terms of 't'. Let's keep the momentum going and move on to solving for 't'!
Finding the Value of t When the Area of EFGD is 88 cm²
Alright, let’s get to the final part of the problem: finding the value of 't' when the area of EFGD is 88 cm². This is where all our hard work pays off! We already have an expression for the area of EFGD in terms of 't': Area of EFGD = 4t² - 10t + 4 cm². Now, we know that the area is 88 cm², so we can set up an equation:
4t² - 10t + 4 = 88
To solve for 't', we need to rearrange this equation into a standard quadratic form (at² + bt + c = 0). Let's subtract 88 from both sides:
4t² - 10t + 4 - 88 = 0
4t² - 10t - 84 = 0
Now, we have a quadratic equation. To make things a bit easier, we can simplify the equation by dividing all the terms by their greatest common divisor, which is 2:
2t² - 5t - 42 = 0
Now, we can solve this quadratic equation using factoring, the quadratic formula, or completing the square. Factoring might be the quickest method if we can find two numbers that multiply to -84 (2 × -42) and add up to -5. After a bit of trial and error, we can find that the numbers are -12 and 7.
So, we can rewrite the middle term (-5t) as -12t + 7t:
2t² - 12t + 7t - 42 = 0
Now, we can factor by grouping:
- Take out the common factor from the first two terms: 2t(t - 6)
- Take out the common factor from the last two terms: 7(t - 6)
Now, we have:
2t(t - 6) + 7(t - 6) = 0
Notice that (t - 6) is a common factor. Let's factor it out:
(t - 6)(2t + 7) = 0
Now, we have two possible solutions for 't':
- t - 6 = 0 => t = 6
- 2t + 7 = 0 => 2t = -7 => t = -3.5
Since 't' represents a length, it cannot be negative. Therefore, we discard the solution t = -3.5. So, the value of t is:
t = 6
We did it! We successfully found the value of 't' when the area of rectangle EFGD is 88 cm². This was a journey through the world of rectangles, algebra, and problem-solving. We started by understanding the problem, then expressed the unknowns in terms of 't', and finally, we solved for 't' using our knowledge of quadratic equations.
Conclusion: Mastering Geometry Problems
So, there you have it! We've successfully navigated this geometry problem step-by-step, and hopefully, you feel a little more confident tackling similar challenges. Remember, the key is to break down the problem into smaller, manageable parts, visualize the shapes and relationships, and use your knowledge of formulas and algebraic techniques. Geometry problems might seem daunting at first, but with a systematic approach and a little bit of practice, you can conquer them all! The most important thing is not to give up. Keep practicing, keep asking questions, and keep exploring the fascinating world of mathematics. And who knows, maybe you'll even start to enjoy these puzzles as much as I do! Keep up the great work, guys, and I'll see you in the next problem-solving adventure!