Problem 84 - Elementary Algebra

Question

Set up an algebraic equation and use it to solve the following. If a circle has an area of \(32 \pi\) square centimeters, then find the length of the radius.

Step-by-Step Solution

Answer

The radius is \(4\sqrt{2}\) cm.

Step 1: Recall the Formula for the Area of a Circle

To find the radius of a circle, we start by recalling the formula used to calculate the area of a circle. The formula is given by: \[ A = \pi r^2 \]where \( A \) is the area and \( r \) is the radius of the circle.

Step 2: Substitute the Given Area into the Formula

We know that the area of the circle is \( 32 \pi \) square centimeters. Substitute \( 32 \pi \) for \( A \) in the area formula:\[ 32 \pi = \pi r^2 \]

Step 3: Simplify the Equation

Divide both sides of the equation by \( \pi \) to simplify:\[ 32 \pi \ \div \pi = r^2 \]This simplifies to:\[ 32 = r^2 \]

Step 4: Solve for the Radius

Take the square root of both sides of the equation to solve for \( r \):\[ r = \sqrt{32} \]Simplify \( \sqrt{32} \) by expressing 32 as a product of perfect squares:\[ \sqrt{32} = \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4 \sqrt{2} \]

Step 5: Finalize the Solution

Thus, the radius of the circle is \( 4 \sqrt{2} \) centimeters.

Key Concepts

Area of a Circle

The area of a circle is a measure of the space enclosed within its boundaries. To calculate this area, we use a special formula: \[ A = \pi r^2 \]Here, \( A \) represents the area, and \( r \) is the radius of the circle.
Understanding this formula is important because it connects the circle’s area to its radius through the mathematical constant \( \pi \), which is approximately 3.14159. \( \pi \) is a unique number that relates to the circle’s geometry, especially with its circumference and area.
This formula highlights the importance of the radius, as the area increases with the square of the radius. Meaning, if you double the radius, you quadruple the area!

Solving for Radius

When given a circle's area, one common task is to find the radius. This process involves algebraic manipulation. We start with:\[ A = \pi r^2 \]and solve for \( r \).
If the area \( A \) is provided, say \( 32\pi \), you substitute this into the formula:\[ 32\pi = \pi r^2 \]To isolate \( r^2 \), divide both sides by \( \pi \):\[ 32 = r^2 \]
Next, to solve for \( r \), take the square root:\[ r = \sqrt{32} \]By simplifying \( \sqrt{32} \), you recognize it can be broken down into\[ \sqrt{16} \times \sqrt{2} = 4\sqrt{2} \]Thus, the radius \( r \) of this circle with the area of \( 32\pi \) is \( 4\sqrt{2} \) centimeters.
This shows how powers and roots are used in algebra to solve problems involving circles.

Algebra Problems

Algebra is fundamental for solving equations that include variables, like when dealing with geometric figures. When working with algebra problems, the key steps are:

• Identifying what you know, such as the area of a circle.
• Substituting the known values into the relevant formulas.
• Using algebraic operations to isolate and solve for the unknown variable, in this case, the radius.
• Simplifying expressions, like finding the square root or simplifying radicals.

Using these steps methodically allows you to solve for variables even when they seem hidden in a formula.
In the context of geometry, where precise calculations are crucial, mastering these algebraic techniques ensures you can tackle questions about areas, volumes, or other properties effectively.