Concrete Path Area And Lens Equation Calculation

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This comprehensive article will delve into two distinct mathematical problems. The first involves calculating the area of a concrete path surrounding a circular space, a practical application of geometry. The second explores a lens equation, a fundamental concept in physics, requiring algebraic manipulation and problem-solving skills. Let's embark on this mathematical journey together.

H2: Determining the Area of a Concrete Path Around a Circle

H3: Problem Breakdown: Concrete Path Area

Concrete path area problems often appear in practical scenarios, such as landscaping or construction projects. In this specific problem, we are tasked with finding the area of a concrete path that surrounds a circular area with a given diameter. The key to solving this lies in understanding how to calculate the area of a circle and applying the concept to find the difference between two circular areas: the area of the outer circle (including the path) and the area of the inner circle (the circular space). Let's break down the problem step-by-step.

  1. Identify the given information: The diameter of the inner circle is 12 meters, and the width of the concrete path is 2 meters.
  2. Calculate the radius of the inner circle: The radius is half the diameter, so the radius of the inner circle is 12 meters / 2 = 6 meters.
  3. Calculate the radius of the outer circle: The radius of the outer circle is the radius of the inner circle plus the width of the path, which is 6 meters + 2 meters = 8 meters.
  4. Calculate the area of the inner circle: The area of a circle is given by the formula πr², where r is the radius. So, the area of the inner circle is π(6 meters)² = 36π square meters.
  5. Calculate the area of the outer circle: Using the same formula, the area of the outer circle is π(8 meters)² = 64π square meters.
  6. Calculate the area of the concrete path: The area of the concrete path is the difference between the area of the outer circle and the area of the inner circle, which is 64Ï€ square meters - 36Ï€ square meters = 28Ï€ square meters.
  7. Approximate the area using a value for π: Using the approximation π ≈ 3.14159, the area of the concrete path is approximately 28 * 3.14159 square meters ≈ 87.96 square meters.

H3: Detailed Solution and Explanation for Concrete Path

To delve deeper into the solution, let's revisit each step with a more detailed explanation. The calculation of the concrete path area hinges on the fundamental formula for the area of a circle, A = πr². Our initial step involves determining the radii of both the inner circular space and the outer circle encompassing the concrete path. We are given the diameter of the inner circle, which is 12 meters. To find the radius, we simply divide the diameter by 2, yielding a radius of 6 meters for the inner circle. This radius is crucial as it forms the basis for calculating the inner circle's area.

Next, we need to determine the radius of the outer circle. This is where the width of the concrete path comes into play. The path extends 2 meters outward from the inner circle's edge. Therefore, the radius of the outer circle is the sum of the inner circle's radius and the path's width, resulting in 6 meters + 2 meters = 8 meters. This outer radius is essential for calculating the area of the larger circle, which includes both the inner space and the concrete path.

With the radii of both circles established, we can now calculate their respective areas using the formula A = πr². For the inner circle, the area is π(6 meters)² = 36π square meters. For the outer circle, the area is π(8 meters)² = 64π square meters. These calculations provide us with the total areas enclosed by each circle, which are necessary for finding the area of the path itself.

To isolate the area of the concrete path, we subtract the area of the inner circle from the area of the outer circle. This subtraction effectively removes the inner circular space, leaving only the area of the path. Thus, the area of the concrete path is 64π square meters - 36π square meters = 28π square meters. This result represents the exact area in terms of π, which is a common practice in mathematical calculations for precision.

Finally, to obtain a numerical approximation of the area, we substitute the value of π (approximately 3.14159) into our result. This yields 28 * 3.14159 square meters ≈ 87.96 square meters. This approximate value provides a tangible understanding of the path's size, which is useful in practical applications such as material estimation for construction.

H3: Key Takeaways for Calculating Concrete Path Area

This problem highlights the importance of understanding geometric formulas and their application in real-world scenarios. The area calculation for concrete path demonstrates how basic geometric principles can be used to solve practical problems involving areas and dimensions. By breaking down the problem into smaller steps, we can systematically arrive at the solution. This approach of dissecting complex problems into manageable parts is a valuable problem-solving strategy applicable across various mathematical and scientific disciplines. Remember to always carefully identify the given information, apply the correct formulas, and perform the calculations accurately. The use of π and its approximation is also crucial in circle-related calculations, ensuring both precision and practical applicability.

H2: Solving the Lens Equation

H3: Problem Breakdown: Lens Equation

The second part of this article focuses on the lens equation, a fundamental concept in optics. The equation given is $ rac{1}{u}+ rac{1}{v}= rac{1}{f}$, where u represents the object distance, v represents the image distance, and f represents the focal length of the lens. This equation describes the relationship between these three variables for a thin lens. We are tasked with two sub-problems:

(i) Find u in terms of f and v. This requires algebraic manipulation to isolate u on one side of the equation. (ii) Find the value of u when v = 6. This involves substituting the given value of v into the rearranged equation from part (i) and solving for u.

H3: Detailed Solution and Explanation for Lens Equation

Let's delve into the solution for each part of the lens equation problem. Part (i) requires us to express u in terms of f and v. This involves algebraic manipulation of the given equation, $ rac{1}{u}+ rac{1}{v}= rac{1}{f}$. Our goal is to isolate u on one side of the equation. The first step is to subtract $ rac{1}{v}$ from both sides of the equation, which gives us $ rac{1}{u} = rac{1}{f} - rac{1}{v}$. This step isolates the term containing u on one side.

Next, we need to combine the fractions on the right-hand side of the equation. To do this, we find a common denominator, which is fv. We rewrite the fractions with this common denominator: $ rac{1}{u} = rac{v}{fv} - rac{f}{fv}$. Combining the fractions, we get $ rac{1}{u} = rac{v-f}{fv}$. This step simplifies the equation and prepares us for the final isolation of u.

To solve for u, we take the reciprocal of both sides of the equation. This means flipping the fractions, which gives us $u = rac{fv}{v-f}$. This is the solution for part (i), expressing u in terms of f and v. This formula is crucial for calculating the object distance given the focal length and image distance of a lens.

Part (ii) of the problem asks us to find the value of u when v = 6. This involves substituting v = 6 into the equation we derived in part (i). So, we have $u = rac{f(6)}{6-f}$, which simplifies to $u = rac{6f}{6-f}$. Notice that to find a specific value for u, we would also need a value for the focal length, f. Without a specific value for f, we can only express u in terms of f. This highlights the importance of having sufficient information to solve a problem completely. If a value for f were given, we would simply substitute it into this equation to find the numerical value of u.

H3: Key Takeaways for Solving Lens Equations

This problem demonstrates the importance of algebraic manipulation in solving equations. The lens equation solving process involves multiple steps, including isolating variables, combining fractions, and taking reciprocals. Each step requires careful attention to detail to avoid errors. This problem also highlights the importance of understanding the relationships between variables in a physical equation. The lens equation describes a fundamental relationship in optics, and being able to manipulate and solve it is crucial for understanding lens behavior. Furthermore, this problem illustrates the concept of solving for a variable in terms of other variables, which is a common technique in mathematics and science. The ability to express one variable as a function of others is essential for analyzing and predicting the behavior of systems.

This article has explored two distinct mathematical problems: calculating the area of a concrete path and solving the lens equation. Both problems require a strong understanding of fundamental concepts and the ability to apply them in different contexts. The combined mathematical problem-solving approach highlighted in this article demonstrates the importance of breaking down complex problems into smaller, manageable steps. By carefully identifying the given information, applying the correct formulas or equations, and performing the calculations accurately, we can arrive at the solution. These problem-solving skills are not only valuable in mathematics but also in various other fields, including science, engineering, and everyday life. The ability to analyze problems, develop strategies, and execute them effectively is a hallmark of a strong problem-solver. This article serves as a testament to the power of mathematical thinking and its ability to unlock solutions to a wide range of challenges. The interplay of geometric and algebraic concepts further emphasizes the interconnectedness of mathematical disciplines, fostering a holistic understanding of the subject.