Problem 24 - Intermediate Algebra

Question

Use an area formula to find the unknown measurement. See Example 2. First Aid. \(\quad\) A rectangular band-aid covers \(2 \frac{1}{2}\) in. \(^{2}\) of skin. If the width of the band-aid is \(\frac{5}{8}\) in., find its length.

Step-by-Step Solution

Answer

The length of the band-aid is 4 inches.

Step 1: Understand the Problem

We have a band-aid in the shape of a rectangle and we need to find its unknown length. The area covered by the band-aid is given as \(2 \frac{1}{2}\) square inches, and its width is \(\frac{5}{8}\) inches.

Step 2: Convert Mixed Numbers to Improper Fractions

Convert the area from \(2 \frac{1}{2}\) to an improper fraction: \(2 \frac{1}{2} = \frac{5}{2}\).

Step 3: Use the Area of a Rectangle Formula

Recall that the area of a rectangle (\(A\)) is given by \(A = \text{length} \times \text{width}\). Substitute the known values into the formula: \(\frac{5}{2} = \text{length} \times \frac{5}{8}\).

Step 4: Solve for the Length

To isolate the length, divide both sides of the equation by \(\frac{5}{8}\): \[ \text{length} = \frac{\frac{5}{2}}{\frac{5}{8}} \].

Step 5: Simplify the Expression

To simplify \(\frac{\frac{5}{2}}{\frac{5}{8}}\), multiply by the reciprocal: \(\frac{5}{2} \times \frac{8}{5}\). Cancel out the 5s: \(\frac{5}{2} \times \frac{8}{5} = \frac{8}{2} = 4\).

Step 6: Verify the Calculation

Double-check the calculation to ensure there are no errors. Start with the given area \(2 \frac{1}{2} = \frac{5}{2}\) and width \(\frac{5}{8}\) to verify that the length \(4\) satisfies the area equation.

Key Concepts

Mixed Numbers to Improper Fractions

Mixed numbers, like 2 \( \frac{1}{2} \), are common in everyday measurements. To work with mixed numbers in algebraic calculations, sometimes it's useful to convert them into improper fractions. This conversion makes it much easier to apply mathematical operations, such as multiplication or division.

To convert a mixed number to an improper fraction:

• Multiply the whole number by the denominator of the fractional part. In our example, multiply 2 by 2, which equals 4.
• Add the numerator of the fractional part to this product. Using our numbers, 4 plus 1 equals 5.
• Write this sum over the original denominator. So, 2 \( \frac{1}{2} \) becomes \( \frac{5}{2} \).

This improper fraction now expresses the same value as the mixed number but is more convenient for algebraic work.

Solving Equations

Solving equations is a vital part of algebra that allows us to find unknown values. Once you have an equation, like \( \frac{5}{2} = \text{length} \times \frac{5}{8} \), you need to isolate the variable (here, 'length') to find its value.

For this, you can use these steps:

• Identify the operation that is affecting the variable. In this case, 'length' is being multiplied by \( \frac{5}{8} \).
• To isolate 'length', perform the opposite operation. Divide both sides by \( \frac{5}{8} \) to cancel the multiplication. This gives: \( \text{length} = \frac{\frac{5}{2}}{\frac{5}{8}} \).
• Simplify this expression. You do this by multiplying by the reciprocal, transforming it into \( \frac{5}{2} \times \frac{8}{5} \).
• Carry out the multiplication. Cancel out common terms and simplify to find \( \frac{8}{2} = 4 \).

Solving equations often involves reversing operations to isolate the variable, turning complex expressions into simple solutions.

Geometry

Geometry often involves understanding shapes and their properties. In this problem, we dealt with a rectangle, which is a four-sided shape with opposite sides equal and right angles.

The area of a rectangle gives us insight into how much surface it covers, and it's calculated by multiplying its length by its width: \( A = \text{length} \times \text{width} \).

Understanding rectangles and their properties helps us connect algebra with geometry:

• For any rectangle, once you know two dimensions (length and width), the area tells you how much space the rectangle covers.
• When you have the area and one dimension, you can reorder the formula to find the unknown dimension. For example, with a known area \( 2 \frac{1}{2} \text{ in}^2 \) and width \( \frac{5}{8} \text{ in} \), the length can be found by rearranging the area formula.

Geometry formulas like these allow us to solve practical problems, such as determining materials needed to cover a certain space.