Problem 95
Question
You measure the side of a square as \(10.4\) inches with a possible error of \(\frac{1}{16}\) inch. Using these measurements, determine the interval containing the possible areas of the square.
Step-by-Step Solution
Verified Answer
The interval containing the possible areas of the square is \([107.1225, 109.2025]\) square inches.
1Step 1: Determine the range of possible lengths
First we determine the possible range of the square's side length by subtracting and adding the measurement error from the actual measurement. Given that the possible error is \(\frac{1}{16}\) inch, the side of the square could be between \(10.4 - \frac{1}{16}\) inches and \(10.4 + \frac{1}{16}\) inches. Simplifying these expressions gives us \(10.35\) inches and \(10.45\) inches, respectively.
2Step 2: Calculate the corresponding areas
Next, we compute the areas corresponding to these measurements. The area of a square with side \(s\) is given by \(s^2\). Substituting \(10.35\) and \(10.45\) for \(s\) provides the possible range for the area of the square. The computations give us \(10.35^2 = 107.1225\) square inches and \(10.45^2 = 109.2025\) square inches.
3Step 3: Provide the Interval
Our final step is to present our answer as an interval. The interval containing the possible areas of the square given that the side was measured to be \(10.4\) inches with a possible error of \(\frac{1}{16}\) inch is \([107.1225, 109.2025]\) square inches.
Key Concepts
Area CalculationSquare MeasurementsMeasurement Error
Area Calculation
The calculation of an area, particularly for basic shapes like squares, is a fundamental concept in geometry and various real-world applications. For a square, the area is determined using the formula:
\[ Area = side^2 \]
where the 'side' refers to the length of one of the square's sides. Since all sides of a square are equal, calculating the area is straightforward once the side length is known. However, if this length is measured using a tool, we must consider the measurement’s accuracy to ensure the calculated area's precision. In the mentioned exercise, we dealt with a side length of 10.4 inches, and by plugging this value into the formula, we would get the area for that specific measurement. Yet, this doesn't account for any margin of error in the measuring process, which is addressed in the next key topic.
\[ Area = side^2 \]
where the 'side' refers to the length of one of the square's sides. Since all sides of a square are equal, calculating the area is straightforward once the side length is known. However, if this length is measured using a tool, we must consider the measurement’s accuracy to ensure the calculated area's precision. In the mentioned exercise, we dealt with a side length of 10.4 inches, and by plugging this value into the formula, we would get the area for that specific measurement. Yet, this doesn't account for any margin of error in the measuring process, which is addressed in the next key topic.
Square Measurements
Working with square measurements introduces a unique consideration: because the area is a function of the side raised to the second power, small errors in measuring the side length can lead to more significant discrepancies once squared.
For instance, a tiny error in measuring a side length when computing a square's area is amplified due to the formula's squaring operation. Hence, precision in measuring the side of a square is crucial for applications involving area calculations, like determining the amount of material needed for construction or the size of a plot of land. This attention to detail ensures that any measurements taken tell an accurate story about the size of the square, informing decisions and predictions based on its area.
For instance, a tiny error in measuring a side length when computing a square's area is amplified due to the formula's squaring operation. Hence, precision in measuring the side of a square is crucial for applications involving area calculations, like determining the amount of material needed for construction or the size of a plot of land. This attention to detail ensures that any measurements taken tell an accurate story about the size of the square, informing decisions and predictions based on its area.
Measurement Error
Measurement error is an unavoidable part of scientific and mathematical calculations, indicating the difference between the measured value and the true value.
In the exercise, the possible measurement error for the square's side is \(\frac{1}{16}\) inch, implying measurement uncertainty. Handling this error requires acknowledging the range within which the true value likely falls. Thus, we calculate the square's area at the minimum and maximum possible side lengths, accounting for this error, to establish an interval for the real area. The interval \([107.1225, 109.2025]\) square inches signifies that the actual area, even with measurement imperfections, is within this range. Understanding measurement error is essential for assessing the reliability of results and making informed conclusions based on those measurements.
In the exercise, the possible measurement error for the square's side is \(\frac{1}{16}\) inch, implying measurement uncertainty. Handling this error requires acknowledging the range within which the true value likely falls. Thus, we calculate the square's area at the minimum and maximum possible side lengths, accounting for this error, to establish an interval for the real area. The interval \([107.1225, 109.2025]\) square inches signifies that the actual area, even with measurement imperfections, is within this range. Understanding measurement error is essential for assessing the reliability of results and making informed conclusions based on those measurements.
Other exercises in this chapter
Problem 93
Solve for the indicated variable. Area of a Trapezoid Solve for \(b\) in \(A=\frac{1}{2}(a+b) h\).
View solution Problem 94
Solve for the indicated variable. Area of a Sector of a Circle Solve for \(\theta\) in \(A=\frac{\pi r^{2} \theta}{360}\).
View solution Problem 96
You measure the side of a square as \(24.2\) centimeters with a possible error of \(0.25\) centimeter. Using these measurements, determine the interval containi
View solution Problem 96
Solve for the indicated variable. Geometric Progression Solve for \(r\) in \(S=\frac{r L-a}{r-1}\).
View solution