Problem 25 - A Graphical Approach to Precalculus with Limits

Question

Solve each problem. A square lawn has area 800 square feet. A sprinkler placed at the center of the lawn sprays water in a circular pattern that just covers the lawn. What is the radius of the circle?

Step-by-Step Solution

Answer

The radius of the circle is \( 10\sqrt{2} \) feet.

Step 1: Determine the Length of a Side of the Square

The area of a square is given by the formula \( A = s^2 \), where \( s \) is the side length of the square. Set the area equal to 800 square feet: \( s^2 = 800 \). To find \( s \), take the square root of both sides to get \( s = \sqrt{800} \).

Step 2: Simplify the Square Root

The square root of 800 can be simplified. Notice that 800 can be factored into \( 4 \times 200 = 4 \times 4 \times 50 = 16 \times 50 \). Thus, \( \sqrt{800} = \sqrt{16 \times 50} = 4\sqrt{50} \).

Step 3: Further Simplify the Expression

Simplify \( \sqrt{50} \) which is \( \sqrt{25 \times 2} = 5\sqrt{2} \). Therefore, \( 4\sqrt{50} \) simplifies to \( 4 \times 5\sqrt{2} = 20\sqrt{2} \). So, the side length \( s \) is \( 20\sqrt{2} \) feet.

Step 4: Find the Radius of the Circle

Since the sprinkler covers the entire lawn, the radius of the circle is the same as half of the side length of the square. Thus, the radius \( r \) is \( \frac{20\sqrt{2}}{2} \).

Step 5: Simplify the Expression for the Radius

Divide \( 20\sqrt{2} \) by 2 to get \( 10\sqrt{2} \). Therefore, the radius of the circle is \( 10\sqrt{2} \) feet.

Key Concepts

Understanding the Area of a Square

When we talk about the area of a square, we refer to the amount of space it occupies in two dimensions. A square is a simple geometric figure with all four sides of equal length. The formula to calculate its area is given by \( A = s^2 \), where \( s \) represents the length of one side.

For example, if a square lawn has an area of 800 square feet, we can find the length of each side by rearranging the formula: \( s^2 = 800 \). By taking the square root of both sides, we determine that \( s = \sqrt{800} \). This step introduces us to more complex concepts like simplifying square roots.

Determining the Radius of a Circle Covering a Square

The radius of a circle is a crucial measurement as it defines the distance from the center of the circle to any point on its circumference. In the context of our exercise, we want the radius of the circle the sprinkler creates to just cover the square lawn entirely. This means the radius has to be equal to half the length of the square’s side.

Having found that the side length of the square is \( 20\sqrt{2} \) feet, we use the relationship radius \( r = \frac{s}{2} \). Thus, the radius of the circle will be \( 10\sqrt{2} \) feet. This ensures that the circular spray pattern reaches every edge of the square lawn.

Simplifying Square Roots

Square roots can often be simplified by breaking the number under the root sign into its prime factors, especially when those factors include perfect squares. In our example, it was crucial to simplify \( \sqrt{800} \).

The number 800 can be decomposed into 16 times 50 \( (16 \cdot 50) \), and since 16 is a perfect square (\( 4^2 \)), we can take the square root of 16 to get 4. Therefore, \( \sqrt{800} = 4\sqrt{50} \).

Further simplifying \( \sqrt{50} \) into \( \sqrt{25 \times 2} \), we find \( 25 \) is a perfect square (\( 5^2 \)), simplifying it to \( 5\sqrt{2} \). Hence, \( 4\sqrt{50} = 20\sqrt{2} \). These simplifications make calculations clearer and reveal underlying mathematical relationships.

Applying the Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry, utilized primarily in right triangles. It states that the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides: \( a^2 + b^2 = c^2 \). Though not directly applied in this problem, understanding how geometric principles interplay, like how the sprinkler covers the square, helps reinforce the broader understanding.

Knowing how to manipulate and simplify equations often relies on principles like the Pythagorean Theorem when handling problems involving geometry and algebraic expressions. Even in finding the dimensions of shapes or dealing with square roots, these foundational concepts reinforce each other.