Triangle Area Problem: Calculate X With Base, Height & Area
Hey guys! Let's dive into a fun math problem involving triangles. We're going to explore how to calculate an unknown value, 'x', when given the base, height, and area of a triangle. It might sound a bit tricky, but don't worry, we'll break it down step by step so it's super easy to understand. This problem is a classic example of how geometry and algebra can come together, and it’s super practical for real-world applications too.
Understanding the Basics of Triangle Area
Before we jump into the specific problem, let's quickly refresh our understanding of how to calculate the area of a triangle. You probably remember the formula: Area = 1/2 * base * height. This formula is fundamental in geometry and is used extensively in various fields, from architecture to engineering. The base of a triangle is any one of its sides, and the height is the perpendicular distance from the base to the opposite vertex (the highest point). Understanding this basic concept is crucial for solving the problem we have at hand. So, with this formula in mind, let's see how we can apply it to solve for 'x' in our triangle problem. Remember, mathematical problem-solving often involves taking a known formula and applying it to a specific scenario, so this is a great skill to develop.
The Triangle Area Formula
The area of a triangle is calculated using the formula: Area = (1/2) * base * height. This straightforward formula is a cornerstone of geometry and is essential for solving various problems related to triangles. To truly master this concept, it's helpful to understand why this formula works. Imagine a rectangle; its area is simply base times height. A triangle can be seen as half of a parallelogram, and a parallelogram can be transformed into a rectangle of the same base and height. Thus, a triangle effectively covers half the area of a corresponding rectangle, leading to the 1/2 factor in the formula. Knowing the "why" behind the formula helps you remember it and apply it confidently in different scenarios. When you encounter triangle problems, always start by identifying the base and height clearly, and then plugging those values into the formula. Remember, the height must be perpendicular to the base. This simple yet powerful formula is your key to unlocking a wide range of geometrical challenges.
Problem Setup: Triangle ABC
Okay, let’s get to the heart of the problem. We have triangle ABC, and we're given some key information. The base AC has a length of (2x + 5), and the height BH (which is perpendicular to the base) measures (x + 7). We also know that the area of the entire triangular region is 55 square units. The main goal here is to find the value of 'x'. This type of problem is a classic example of how algebra and geometry intertwine. We’re using geometric properties—the area of a triangle—and algebraic expressions to represent the dimensions. The beauty of mathematics is in this interplay between different concepts. By setting up the problem correctly, we can use the given information to create an equation that will help us solve for 'x'. So, let's see how we can translate this information into a mathematical equation using the triangle area formula we discussed earlier.
Defining Base and Height
In our triangle ABC, the base AC is given as (2x + 5) and the height BH as (x + 7). It's really important to clearly identify these elements before we proceed. Remember, the base and height are always perpendicular to each other. In this case, BH is the perpendicular height from vertex B to the base AC. Now, let’s think about what these expressions mean. The expressions (2x + 5) and (x + 7) are algebraic representations of the lengths of the base and height, respectively. The 'x' represents an unknown value that we need to find. This is where our algebra skills come into play. By expressing the dimensions in terms of 'x', we can create an equation that relates the base, height, and the area of the triangle. This is a common strategy in math problems: using variables to represent unknowns and then setting up equations to solve for those variables. So, we’ve got our base and height defined; now it’s time to bring in the area and form our equation!
Setting Up the Equation
Now comes the exciting part – translating our geometric information into a mathematical equation. Remember the triangle area formula? Area = 1/2 * base * height. We know the area is 55 square units, the base is (2x + 5), and the height is (x + 7). So, let's plug these values into the formula: 55 = 1/2 * (2x + 5) * (x + 7). This equation is the key to solving for 'x'. It beautifully combines our geometric understanding with our algebraic skills. The left side of the equation represents the given area, and the right side represents the area calculated using the given base and height expressions. Now, our goal is to manipulate this equation using algebraic techniques to isolate 'x' and find its value. This process will involve simplifying the equation, expanding the terms, and rearranging to solve for the unknown. Are you ready to dive into the algebraic manipulation? Let's go!
Applying the Area Formula
To set up the equation, we use the formula Area = (1/2) * base * height. We know that the area is 55 square units, the base AC is (2x + 5), and the height BH is (x + 7). Plugging these values into the formula, we get: 55 = (1/2) * (2x + 5) * (x + 7). This equation is the bridge between the geometry of the triangle and the algebra we need to solve for x. Think of it as a mathematical representation of the problem. The left side of the equation, 55, is a fixed value – the known area. The right side is an expression involving x, which represents the area calculated from the given dimensions. Our task now is to work with this equation. We'll first need to simplify the expression on the right side by expanding the product of (2x + 5) and (x + 7). This will involve using the distributive property (often remembered by the acronym FOIL - First, Outer, Inner, Last) and combining like terms. Once we've simplified the equation, we'll have a quadratic equation in terms of x, which we can then solve using various algebraic methods. So, let's roll up our sleeves and start simplifying!
Solving the Equation for x
The next step is to solve the equation we've set up for 'x'. Our equation is 55 = 1/2 * (2x + 5) * (x + 7). First, let's get rid of that fraction by multiplying both sides of the equation by 2. This gives us 110 = (2x + 5) * (x + 7). Now, we need to expand the right side of the equation. Using the distributive property (or the FOIL method), we multiply (2x + 5) by (x + 7). This gives us: 2x * x + 2x * 7 + 5 * x + 5 * 7, which simplifies to 2x² + 14x + 5x + 35. Combining like terms, we get 2x² + 19x + 35. So, our equation now looks like this: 110 = 2x² + 19x + 35. To solve for x, we need to set the equation to zero. We can do this by subtracting 110 from both sides, giving us: 0 = 2x² + 19x - 75. We now have a quadratic equation. To solve this, we can either try to factor it, use the quadratic formula, or complete the square. Let's try factoring first, as it's often the quickest method if it works.
Factoring the Quadratic Equation
Our quadratic equation is 2x² + 19x - 75 = 0. Factoring a quadratic equation involves breaking it down into two binomial factors. We're looking for two expressions (ax + b) and (cx + d) such that when multiplied together, they give us our original quadratic. This can sometimes feel like a puzzle, but with practice, it becomes easier to spot patterns. In this case, we need to find two numbers that multiply to (2 * -75 = -150) and add up to 19. After some trial and error, we find that the numbers 25 and -6 fit the bill. So, we can rewrite the middle term (19x) as 25x - 6x. Our equation now becomes: 2x² + 25x - 6x - 75 = 0. Next, we factor by grouping. We group the first two terms and the last two terms: (2x² + 25x) + (-6x - 75) = 0. Now, we factor out the greatest common factor (GCF) from each group. From the first group, we can factor out x, giving us x(2x + 25). From the second group, we can factor out -3, giving us -3(2x + 25). Notice that both groups now have a common factor of (2x + 25). We can factor this out, leaving us with: (2x + 25)(x - 3) = 0. Now, we have two factors that multiply to zero. This means that either (2x + 25) = 0 or (x - 3) = 0. We can solve these two simpler equations separately to find our possible values for x.
Finding the Solutions for x
We've factored our quadratic equation into (2x + 25)(x - 3) = 0. Now, let's find the solutions for x. If (2x + 25) = 0, we can solve for x as follows: Subtract 25 from both sides to get 2x = -25. Then, divide both sides by 2 to get x = -25/2 or -12.5. This is one possible solution for x. Next, if (x - 3) = 0, we can solve for x by simply adding 3 to both sides, which gives us x = 3. So, we have two possible solutions for x: x = -12.5 and x = 3. But wait! We're dealing with a real-world geometric problem. Can a length be negative? Not really. A length cannot be a negative value in a real-world context. Therefore, we discard the solution x = -12.5 because it doesn't make sense in our triangle problem. This leaves us with x = 3 as our only valid solution. So, after all our algebraic manipulations, factoring, and careful consideration, we've arrived at our answer. The value of x that fits the given conditions is 3.
Checking for Valid Solutions
We found two possible solutions for x: -12.5 and 3. However, it's crucial to check if these solutions are valid in the context of our problem. Remember, we're dealing with the dimensions of a triangle, and lengths cannot be negative. If we substitute x = -12.5 back into the expressions for the base and height, we get: Base AC = 2(-12.5) + 5 = -25 + 5 = -20. Height BH = (-12.5) + 7 = -5.5. Both the base and height would be negative, which is impossible for a real triangle. Therefore, x = -12.5 is not a valid solution. Now, let's check x = 3: Base AC = 2(3) + 5 = 6 + 5 = 11. Height BH = (3) + 7 = 10. Both the base and height are positive values, which makes sense in our geometric context. Additionally, let's plug x = 3 back into the area formula to verify our solution: Area = (1/2) * base * height = (1/2) * 11 * 10 = 55 square units. This matches the given area in the problem, so we can be confident that x = 3 is the correct solution. Always remember to check your solutions in the original context of the problem, especially in geometry, to ensure they make sense.
Why Checking Solutions is Important
Checking our solutions is a critical step in mathematical problem-solving, and it's especially important in problems that involve real-world applications, like our triangle problem. In mathematics, we often manipulate equations and perform algebraic operations to find potential solutions. However, these operations can sometimes introduce extraneous solutions, which are solutions that satisfy the equation but don't make sense in the original context. In our case, we found two solutions for x, but one of them, x = -12.5, resulted in negative lengths for the triangle's base and height. This is a clear indication that it's not a valid solution because lengths can't be negative in the real world. By checking our solutions, we ensure that we're providing an answer that's not only mathematically correct but also logically sound. This practice helps us avoid errors and develop a deeper understanding of the problem. So, always make it a habit to check your solutions and ask yourself if they make sense in the given situation.
The Final Answer
So, after all the calculations and checks, we've arrived at our final answer. The value of x that satisfies the conditions of the problem is x = 3. This means that the base AC of the triangle is 2(3) + 5 = 11 units, and the height BH is 3 + 7 = 10 units. We successfully used the area formula, set up an equation, solved a quadratic equation, and, most importantly, checked our solutions to ensure they made sense in the real world. This problem is a great example of how math can be used to solve practical problems and how different mathematical concepts, like geometry and algebra, can work together. I hope you guys found this explanation helpful and that you feel more confident tackling similar problems in the future. Keep practicing, and you'll become a math whiz in no time!
Recap of the Solution
Let's quickly recap the steps we took to solve this problem: 1. We understood the problem and identified the given information: the base AC as (2x + 5), the height BH as (x + 7), and the area as 55 square units. 2. We recalled the formula for the area of a triangle: Area = (1/2) * base * height. 3. We set up the equation using the given information: 55 = (1/2) * (2x + 5) * (x + 7). 4. We simplified the equation by multiplying both sides by 2 and expanding the product of the binomials. 5. We rearranged the equation into a quadratic equation: 2x² + 19x - 75 = 0. 6. We factored the quadratic equation into (2x + 25)(x - 3) = 0. 7. We found the possible solutions for x: x = -12.5 and x = 3. 8. We checked the validity of the solutions in the context of the problem and discarded x = -12.5 because lengths cannot be negative. 9. We confirmed that x = 3 is the correct solution, which gives a base of 11 units and a height of 10 units. This step-by-step approach is a powerful way to tackle mathematical problems. By breaking the problem down into smaller, manageable steps, we can approach it methodically and increase our chances of finding the correct solution. Remember, math is like building with blocks – each step builds on the previous one!
Final Thoughts
So there you have it, guys! We've successfully navigated through this triangle area problem, and hopefully, you've picked up some valuable problem-solving skills along the way. Remember, math isn't just about memorizing formulas; it's about understanding the concepts and applying them in different situations. The key to mastering math is practice, practice, practice! The more problems you solve, the more comfortable you'll become with different techniques and approaches. Don't be afraid to make mistakes, either. Mistakes are a natural part of the learning process. When you encounter a challenge, try to break it down into smaller steps, just like we did in this problem. And most importantly, have fun with it! Math can be a fascinating subject, especially when you see how it connects to the real world. Keep exploring, keep learning, and keep challenging yourselves. You've got this!
Practice Makes Perfect
To truly master the concepts we've covered, it's essential to practice solving similar problems. The more you practice, the more comfortable and confident you'll become in your problem-solving abilities. Try finding similar triangle area problems in textbooks, online resources, or even create your own variations. For example, you could change the given area or the expressions for the base and height and see if you can still solve for x. You could also explore problems where you need to find the base or height given the area and one of the dimensions. Varying the types of problems you tackle will help you develop a deeper understanding of the concepts and techniques involved. Remember, each problem you solve is a step forward in your mathematical journey. So, grab a pencil, some paper, and start practicing! You'll be amazed at how much you can improve with consistent effort. Happy problem-solving!