Obliczanie Szerokości Ramy Prostokątnego Zdjęcia
Hey guys! Let's dive into a fun geometry problem. We've got a rectangular photo with sides p dm and q dm, framed by a border of width x dm (check out the picture!). The total area of the photo and the frame is P dm². Our mission, should we choose to accept it, is to figure out the width of the frame (x).
This isn't just about plugging numbers into a formula, it's about understanding how the frame adds to the overall dimensions and, consequently, the area. We'll break down the problem step-by-step, making it super clear and easy to follow. So, grab your pencils (or styluses!), and let's get started. We'll tackle this like pros! The key here is to visualize how the frame expands the original photo, adding width and height on all sides. This expansion directly impacts the overall area, which is what we're going to use to solve for x. Ready to get our math on?
Understanding the Problem: The Photo and Its Frame
Okay, imagine a picture frame – you know, the kind you hang on your wall. Inside, you have your awesome rectangular photo. The frame itself adds a border around the photo, right? This border is the x dm we're trying to find. The original photo has a length of p dm and a width of q dm. When you add the frame, it increases both the length and the width of the entire thing (photo plus frame).
Think about it: the frame adds x dm on each side of the photo. That means the total length of the photo with the frame becomes p + 2x (because you have x on the left and x on the right). Similarly, the total width becomes q + 2x (because you have x on the top and x on the bottom). The total area, P, is simply the area of the entire framed photo, which is the new length multiplied by the new width. Got it? That means P = (p + 2x) * (q + 2x). Now, we're ready to use the data to calculate the width of the frame.
Now, let's look at the two examples provided and how we can use this formula to solve them.
Solving for the Frame Width: Case a) p = 3, q = 5, P = 29.25
Alright, let's get our hands dirty with some numbers! In this case, our photo is 3 dm long and 5 dm wide. The total area of the photo and the frame is 29.25 dm². We need to find x, the width of the frame.
We know that P = (p + 2x) * (q + 2x). Let's plug in the values we have: 29.25 = (3 + 2x) * (5 + 2x*). Now, we need to expand this equation and solve for x. Multiplying the terms on the right side, we get: 29.25 = 15 + 6x + 10x + 4*x².
Simplifying, we have a quadratic equation: 4x² + 16x + 15 = 29.25. Let's move everything to one side: 4x² + 16x - 14.25 = 0. This is a quadratic equation, and we can solve it using the quadratic formula, which is: x = (-b ± √(b² - 4ac)) / 2a, where a = 4, b = 16, and c = -14.25. Let's start solving it. Let's calculate the value under the square root first, so b² - 4ac = 16² - 4 * 4 * -14.25 = 256 + 228 = 484. The square root of 484 is 22. So, we now have x = (-16 ± 22) / 8. We have two possible solutions for x: x = (-16 + 22) / 8 = 6/8 = 0.75 and x = (-16 - 22) / 8 = -38/8 = -4.75. Since the width cannot be negative, we can discard the negative solution. The width of the frame is 0.75 dm. That means the frame is 0.75 dm wide. Congrats, we have solved the first problem! Easy peasy.
We successfully calculated the frame width when the given values are p = 3, q = 5, and P = 29.25. The width of the frame x is 0.75 dm.
Solving for the Frame Width: Case b) p = 4, q = 6, P = 35
Let's keep the good times rolling and solve another one! In this scenario, our photo has a length of 4 dm and a width of 6 dm. The combined area of the photo and the frame is 35 dm². Let's find x.
Again, we start with our trusty formula: P = (p + 2x) * (q + 2x). Plugging in the known values, we have: 35 = (4 + 2x) * (6 + 2x*). Let's expand this equation: 35 = 24 + 8x + 12x + 4x². Simplifying and rearranging, we get another quadratic equation: 4x² + 20x + 24 = 35. Bringing everything to one side: 4x² + 20*x - 11 = 0.
We'll use the quadratic formula again: x = (-b ± √(b² - 4ac)) / 2a, where a = 4, b = 20, and c = -11. First, let's find the value under the square root: b² - 4ac = 20² - 4 * 4 * -11 = 400 + 176 = 576. The square root of 576 is 24. So, we have x = (-20 ± 24) / 8. This time we have two possible solutions, let's find them. x = (-20 + 24) / 8 = 4/8 = 0.5 and x = (-20 - 24) / 8 = -44/8 = -5.5. Again, we can discard the negative solution since the width can't be negative. That leaves us with x = 0.5 dm. The width of the frame is 0.5 dm. Great job! We found the solution for the second problem. Pretty cool, right?
We successfully calculated the frame width when the given values are p = 4, q = 6, and P = 35. The width of the frame x is 0.5 dm.
Final Thoughts: Framing Your Success
Awesome work, everyone! We've successfully navigated the world of framed photos and found the frame widths in both cases. Remember, the key is to understand how the frame affects the overall dimensions and, therefore, the area. By setting up the correct equation and solving for x, we can crack any similar problem that comes our way.
Key Takeaways:
- Always remember to account for the frame adding width on both sides of the length and width of the photo.
- The formula P = (p + 2x) * (q + 2x) is your best friend in these types of problems.
- Don't be afraid of quadratic equations; the quadratic formula is a powerful tool!
Keep practicing, keep exploring, and you'll become a geometry whiz in no time. If you have any more questions, feel free to ask. Keep up the amazing work! You guys rock! See you next time, and happy framing!