A circle has a circumference of \(62.8\) cm. What is its radius? (Use \(\pi \approx 3.14\))
["How to Find the Radius of a Circle When Circumference is Known", "Understanding the relationship between a circle’s circumference and its radius is key to solving geometry problems efficiently. In this article, we’ll explore how to calculate the radius when the circumference is given—using real-world data such as a circumference of (62.8) cm—with (\pi) approximated as (3.14).", "---", "### What Is Circumference and Why Does Radius Matter?", "The circumference of a circle is the distance around its outer edge. For any circle, the relationship between circumference (C) and radius (r) is defined by the formula:", "[\nC = 2\pi r\n]", "This equation shows that if you know the circumference and the value of (\pi), you can solve for the radius. Knowing the radius is essential in architecture, engineering, design, and everyday measurements.", "---", "### Step-by-Step Solution: Find the Radius from Circumference", "Given:\n- Circumference (C = 62.8) cm\n- Approximate value: (\pi \approx 3.14)", "Step 1: Write down the formula.\n[\nC = 2\pi r\n]", "Step 2: Substitute known values.\n[\n62.8 = 2 \ imes 3.14 \ imes r\n]", "[\n62.8 = 6.28r\n]", "Step 3: Solve for (r).\nTo isolate (r), divide both sides by (6.28):", "[\nr = \frac{62.8}{6.28}\n]", "[\nr = 10 \ ext{ cm}\n]", "---", "### Final Answer", "The radius of a circle with a circumference of (62.8) cm is 10 centimeters.", "---", "### Conclusion", "Finding the radius from circumference is simple once you apply the basic formula (C = 2\pi r). With basic arithmetic and an approximation of (\pi), even complex geometric problems become straightforward. Whether you’re measuring materials, designing objects, or teaching geometry, mastering this calculation empowers accurate problem-solving every day.", "---", "### Key Takeaways:", "- Circumference formula: (C = 2\pi r)\n- Substitute known values and solve for (r).\n- For (C = 62.8) cm and (\pi \approx 3.14), the radius is (10) cm.\n- Understanding this relationship enhances geometry proficiency for real-life applications."]
