Problem 42 - Algebra and Trigonometry Real Mathematics, Real People

Question

A landscaper has installed a circular sprinkler that covers an area of 2000 square feet. (a) Find the radius of the region covered by the sprinkler. Round your answer to three decimal places. (b) The landscaper increases the area covered to 2500 square feet by increasing the water pressure. How much longer is the radius?

Step-by-Step Solution

Answer

The radius of the region covered by the sprinkler initially is approximately 25.132 sq. ft., and after increasing the area to 2500 sq. ft., the new radius is about 28.180 sq. ft. Therefore, the increase in radius is approximately 3.048 sq. ft.

Step 1: Find the radius of the initial circular area

Given that the initial area \(A = 2000 \, sq. ft\), we can use the formula for the area of a circle \(A = \pi r^2\) and rearrange to find the radius: \(r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{2000}{\pi}}\).

Step 2: Calculate a numerical value for the initial radius

Using a calculator, find the numerical value for \(r\). Make sure to round to three decimal places as instructed in the problem.

Step 3: Find the new radius when the area is increased

Repeat steps 1-2 with the increased area \(A = 2500 \, sq. ft\). Use the same formula \(r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{2500}{\pi}}\) and calculate a numerical value for the new radius.

Step 4: Find the increase in radius length

Subtract the initial radius from the new radius to find the increase in length. Round to three decimal places, if necessary.

Key Concepts

Understanding the Area of a Circle

When dealing with the area of a circular region, as in our example with the sprinkler system, it's important to understand that this area represents the amount of space inside the circle's boundary. The formula to calculate the area is defined by \( A = \pi r^2 \), where \( A \) stands for the area and \( r \) is the radius of the circle. Here, \( \pi \) is a constant approximately equal to 3.14159, which is the ratio of the circumference of any circle to its diameter.

In simpler terms, you could picture \( \pi r^2 \) as the number of square units that can fit inside the circle. For instance, if a circle has an area of 2000 square feet, that means we could fit 2000 one-foot square tiles entirely within the circle's boundaries.

Radius Calculation from Area

Solving for the radius when we already know the area of a circle requires some manipulation of the area formula. We start with \( A = \pi r^2 \) and work our way backwards to \( r \) by dividing both sides of the equation by \( \pi \) and, subsequently, taking the square root of both sides. This gives us \( r = \sqrt{\frac{A}{\pi}} \). This method allows us to extract the radius from the known area, which is a common strategy when dealing with circular areas in real-world situations, such as planning the reach of a sprinkler system.

Once we apply the radius calculation formula to the initial area of 2000 square feet, we can use a calculator to get the number to the desired precision, in this case, three decimal places.

Using Square Root Transformation

The square root transformation is an essential technique applied when working with areas and radii of circles. By squaring and taking the square root we can move between the area formula and the radius formula. In the initial radius calculation, we employ this transformation to isolate the radius \( r \) and find its value. It's important to handle square root transformation with care, as it deals with reversing a square operation.

For example, when calculating the radius from the increased area of 2500 square feet, we first divide the area by \( \pi \) and then apply the square root. Using square root transformation properly ensures that we get an accurate length for the radius, which can then be rounded to the specified number of decimal places.

Algebraic Manipulation for Comparative Analysis

Algebraic manipulation involves rearranging and simplifying equations to solve for unknowns or compare quantities. In the landscaper's scenario, after computing the radius for both areas (2000 and 2500 square feet), we use algebraic manipulation to find the difference in radius lengths.

This involves subtracting the initial radius from the new radius, which gives us the amount by which the radius increased when the water pressure was enhanced to cover more area. It's a simple yet powerful example of how algebra can be applied to understand real-world changes in a system, such as adjustments to the sprinkler's reach in this case.