Problem 32 - Algebra 1

Question

You are designing a game for a school carnival. Players will drop a coin into a basin of water, trying to hit a target on the bottom. The water is kept moving randomly, so the coin is equally likely to land anywhere. You use a rectangular basin twice as long as it is wide. You place the blue rectangular target an equal distance from each end. Express the two dimensions of the target in terms of the variables \(x\) and \(y\) THE GRAPH CANNOT COPY

Step-by-Step Solution

Answer

The dimensions of the target in terms of the given variables \(x\) and \(y\) are \(x = 2y/3\) for the length and \(y\) for the width.

Step 1: Understanding the Problem

Let's define the dimensions of the basin first. It's stated that the basin is rectangular and twice as long as it is wide. So if we let \(y\) represent the width, the length would then be \(2y\). The target is also rectangular and placed an equal distance from each end, meaning that it spans the width, but not the complete length of the basin.

Step 2: Assigning Values to Variables

Since the target is as wide as the basin, its width is \(y\). As for its length, we need to realize that the remaining length of the basin (after placing the target) is equally divided into two parts. If we let the length of the target be \(x\), the total length of the basin, \(2y\), can be split into three equivalent parts. Therefore, \( x = 2y/3 \). This makes sense because the target only takes up a third of the total length of the basin.

Step 3: Expressing the Target's Dimensions

Now that we have the dimensions in terms of the variables \(x\) and \(y\), the target's dimensions are \(x\) in length and \(y\) in width.

Key Concepts

Geometry

Geometry plays a crucial role in solving problems related to shapes, sizes, and configurations. In this exercise, we're dealing with a rectangular basin and target inside it. A rectangle is a four-sided shape where opposite sides are parallel and equal in length. The basin here is a rectangle where the length is twice the width, so understanding how its geometry behaves helps us to figure out the placement and size of the target.
In practical terms, geometry aids us in visualizing the problem. Imagine the basin as a large rectangle and the target as another, smaller rectangle placed inside. By visualizing this setup, we can determine the necessary dimensions to ensure the target fits snugly and symmetrically inside the basin. This visualization is an invaluable tool in algebraic expressions because it connects abstract mathematical ideas to tangible, real-world structures.

Variables

Variables are symbols, often letters, used in algebra to represent unknown quantities or values that can change. In this exercise, we use the variables \(x\) and \(y\). They serve as placeholders to express the dimensions of the target and basin.
Using variables allows us to formulate flexible solutions applicable to various scenarios. For instance:

• \(y\) represents the width of the basin and also the target, indicating a fixed dimension for both.
• \(x\) represents the target's length, which we derived in relation to \(y\).

By defining these variables, we can easily rewrite dimensions and solve for them depending on what we know or change in the problem.
Variables are powerfully useful because they simplify complex problems, like breaking down the rectangular dimensions of the basin and target in this game setting, making calculations manageable and understandable.

Rectangular Dimensions

Rectangular dimensions refer to the length and width of a rectangle. These dimensions define the shape and size of a rectangle, key factors when calculating area or perimeter. In our case, the basin and the target inside it are both rectangles.
The problem states the basin's length is twice its width. Hence, if \(y\) is the width, then the overall length of the basin is \(2y\).
As for the target’s dimensions, we determined:

• The target's width is the same as the basin's, \(y\).
• The length of the target, \(x\), fits into the basin's length as \(2y/3\). This makes the length one-third of the basin’s length because it's placed equally from each end.

Understanding how to calculate these dimensions using algebraic expressions is important. It helps in figuring out precisely how much space the target will occupy, ensuring accurate and balanced placement inside the basin. Such computations are critical in applications where geometric accuracy matters greatly, such as in design or architecture.