Problem 52 - Prealgebra

Question

GEOMETRY Use the formula for the area of a circle \(A=\pi r^{2}\) where \(A\) represents the area, \(r\) represents the radius, and \(\pi\) is approximately equal to 3.14 , to find the radius of the circle with an area of 28.26 square inches.

Step-by-Step Solution

Answer

The radius of the circle is 3 inches.

Step 1: Understand the Given Formula

We are given the formula for the area of a circle: \( A = \pi r^2 \). Here, \( A \) is the area, \( r \) is the radius, and \( \pi \) is approximately 3.14.

Step 2: Substitute the Known Values

We know the area \( A \) is 28.26 square inches, so substitute 28.26 for \( A \) in the formula: \[ 28.26 = 3.14 \times r^2 \]

Step 3: Solve for \( r^2 \)

To find \( r^2 \), divide both sides of the equation by 3.14: \[ r^2 = \frac{28.26}{3.14} \]Calculate the division: \[ r^2 \approx 9 \]

Step 4: Solve for \( r \)

Now, take the square root of both sides to find \( r \): \[ r = \sqrt{9} \]\[ r = 3 \]

Step 5: Verify the Solution

Substitute \( r = 3 \) back into the area formula to check: \[ A = \pi r^2 = 3.14 \times 3^2 = 3.14 \times 9 = 28.26 \]The calculated area matches the given area, confirming that our solution is correct.

Key Concepts

Geometry

Geometry is a fascinating branch of mathematics that explores shapes, sizes, and the properties of space. Understanding the fundamental shapes, like circles, is crucial, as these form the building blocks for more complex geometrical concepts. In the context of finding a circle's area, we consider the circle as a two-dimensional shape defined by its center point and distance called the radius. The importance of geometry lies not only in solving problems but also in developing spatial reasoning and analytical skills.

When examining a circle, the main focus is usually on parameters such as the radius and the diameter, which relates directly to the radius as twice its length. These measurements help in computing the circle's circumference and area. Learning geometry helps us appreciate how these concepts interrelate and how they apply to real-world problems, from designing wheels to architectural planning. By connecting with these core concepts, students can better understand how to navigate geometric challenges with confidence.

Radius Calculation

Calculating the radius of a circle is essential in various mathematical contexts and practical applications. When given the area of a circle, you can easily derive its radius using the formula for the area of a circle:

• Formula: \( A = \pi r^2 \), where \( A \) stands for the area and \( r \) denotes the radius.
• If the area is known, rearrange the formula to solve for \( r^2 \): \( r^2 = \frac{A}{\pi} \).
• Finally, find the radius by taking the square root of \( r^2 \).

For instance, if the area of a circle is 28.26 square inches, dividing this by \( \pi \) approximately 3.14 gives \( r^2 \approx 9 \). By taking the square root, you find the radius to be 3 inches.

Understanding the calculation process not only confirms mathematical accuracy but also fosters deeper insight into how measurements are fundamental to other mathematical operations.

Mathematical Formulas

Mathematical formulas are invaluable tools that assist in solving a wide array of problems. They provide a reliable method for calculating outcomes in an organized and systematic way. In geometry, formulas help quantify properties like area, volume, and surface area, which would be challenging to deduce from visual estimation alone.

The formula for a circle's area, \( A = \pi r^2 \), is a prime example of how knowledge of formulas helps solve geometric problems efficiently. By plugging in known values, formulas turn abstract concepts into tangible results. It's important to understand each component of a formula:

• \( A \) indicates the area or the space enclosed within the circle.
• \( r \) represents the radius, a fundamental measurement needed for the calculation.
• \( \pi \) is a constant (approximately 3.14) that relates the diameter of a circle to its circumference.

By mastering these formulas, consistently practicing them, and applying their principles, students can enhance their problem-solving capabilities across various mathematical disciplines.