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Magazine Chapter 7: Problem 57
Solve the given problems. All numbers are accurate to at least two significant digits. A computer monitor has a viewing screen that is \(33.8 \mathrm{cm}\) wide and \(27.3 \mathrm{cm}\) high, with a uniform edge around it. If the edge covers \(20.0 \%\) of the monitor front, what is the width of the edge?
Short Answer
Expert verified
The width of the edge is approximately 0.46 cm.
Step by step solution
01
Calculate the Area of the Viewing Screen
First, we need to calculate the area of the part of the monitor that comprises the viewing screen. The viewing screen is a rectangle with a width of \(33.8 \mathrm{cm}\) and a height of \(27.3 \mathrm{cm}\). Use the formula for the area of a rectangle, \( A = ext{width} \times ext{height} \). Thus, the area of the viewing screen is \(33.8 \mathrm{cm} \times 27.3 \mathrm{cm} = 922.74 \mathrm{cm}^2\).
02
Calculate Total Monitor Area with Edge
Next, determine the total area of the monitor, including the edge. If the edge covers \(20.0\%\) of the monitor's front, then \(80.0\%\) of the monitor's area will be the screen. We know the area of the viewing screen is \(922.74 \mathrm{cm}^2\). Let \(x\) be the total area. Set up the equation \(0.8x = 922.74\).
03
Solve for the Total Monitor Area
Solve the equation from Step 2 to find the total area of the monitor: \[x = \frac{922.74}{0.8} \approx 1153.425 \mathrm{cm}^2\].
04
Set Up the Equation for Total Monitor Area
The total monitor area, including the edge, can be represented as \((33.8 + 2w) \times (27.3 + 2w)\), where \(w\) is the width of the edge. This area is equal to \(1153.425 \mathrm{cm}^2\) from Step 3.
05
Solve for Width of the Edge
We solve the equation from Step 4: \((33.8 + 2w)(27.3 + 2w) = 1153.425\). Expand this and simplify to find \(w\). The resulting equation is quadratic: \(4w^2 + 122.2w + 922.74 = 1153.425\). Simplify it to \(4w^2 + 122.2w - 230.685 = 0\). Use the quadratic formula: \(w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4, b = 122.2, c = -230.685\). Calculate \(w\).
06
Calculate Using the Quadratic Formula
Substitute into the quadratic formula: \[w = \frac{-122.2 \pm \sqrt{122.2^2 - 4 \times 4 \times (-230.685)}}{2 \times 4}\]. Calculate inside the square root: \(122.2^2 = 14925.44\) and \(4 \times 4 \times (-230.685) = -922.74\) so the discriminant is: \(14925.44 + 922.74 = 15848.18\). Calculate \(\sqrt{15848.18}\), which is approximately \(125.9\). Thus, \(w = \frac{-122.2 \pm 125.9}{8}\).
07
Determine the Solution for Width
Using the values calculated, solve for \(w\): \[w = \frac{3.7}{8} = 0.4625 \mathrm{cm}\]. Since the width of an edge cannot be negative, we discard the negative result.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equation
A quadratic equation is a type of polynomial equation where the highest exponent of the variable is 2. It follows the general form:
The quadratic formula also helps determine whether the solutions are real and distinct, real and repeated, or imaginary, depending on the value of the discriminant \(b^2 - 4ac\).
In the exercise, solving the quadratic equation led to finding the width \(w\) of the monitor edge.
- \( ax^2 + bx + c = 0 \), where:
- \(a\), \(b\), and \(c\) are constants, with \(a eq 0\)
- \(x\) is the variable
The quadratic formula also helps determine whether the solutions are real and distinct, real and repeated, or imaginary, depending on the value of the discriminant \(b^2 - 4ac\).
In the exercise, solving the quadratic equation led to finding the width \(w\) of the monitor edge.
Area Calculation
Calculating the area of shapes is a fundamental geometric operation.
In the context of the exercise, we calculate the area of two rectangles: the viewing screen and the total monitor area including the edge.
The area \(A\) of a rectangle is determined by multiplying its width by its height:\[ A = \text{width} \times \text{height} \]For the viewing screen, given its dimensions, the area is calculated as:\( 33.8 \, ext{cm} \times 27.3 \, ext{cm} = 922.74 \, ext{cm}^2 \).
This straightforward multiplication gives us the area of the screen, which we later use to find the edge size after determining the total area of the monitor.
In the context of the exercise, we calculate the area of two rectangles: the viewing screen and the total monitor area including the edge.
The area \(A\) of a rectangle is determined by multiplying its width by its height:\[ A = \text{width} \times \text{height} \]For the viewing screen, given its dimensions, the area is calculated as:\( 33.8 \, ext{cm} \times 27.3 \, ext{cm} = 922.74 \, ext{cm}^2 \).
This straightforward multiplication gives us the area of the screen, which we later use to find the edge size after determining the total area of the monitor.
Geometry
Geometry involves the study of shapes, sizes, and dimensions.
In this problem, we deal with rectangular geometry. The monitor and its screen are both rectangles, defined by their width and height. This problem illustrates how additional geometric concepts, like the thickness of an edge, can transform simple geometry into a problem involving quadratic equations. -The screen area and edge alter the dimensions, leading to the equation: \((33.8 + 2w)(27.3 + 2w) = 1153.425\). Here, the added \(2w\) accounts for the uniform borders surrounding the screen, effectively altering the dimensions of the complete rectangle, including the edges.
In this problem, we deal with rectangular geometry. The monitor and its screen are both rectangles, defined by their width and height. This problem illustrates how additional geometric concepts, like the thickness of an edge, can transform simple geometry into a problem involving quadratic equations. -The screen area and edge alter the dimensions, leading to the equation: \((33.8 + 2w)(27.3 + 2w) = 1153.425\). Here, the added \(2w\) accounts for the uniform borders surrounding the screen, effectively altering the dimensions of the complete rectangle, including the edges.
Percentage
Percentages offer a way to express proportions and comparative sizes.
In this exercise, we learn that the edge covers 20% of the monitor's front area.
This information is used to derive the total area of the monitor by knowing that the remaining 80% of the monitor's front is the viewing screen, calculated as 80% of the total area:\[ 0.8 \times \text{total area} = 922.74 \, ext{cm}^2 \]Using the reverse calculation, the complete area we found was:
\[ \text{total area} = \frac{922.74}{0.8} = 1153.425 \, \text{cm}^2 \]This simple percentage concept allows one to connect the area of the viewing screen with the total area of the monitor.
In this exercise, we learn that the edge covers 20% of the monitor's front area.
This information is used to derive the total area of the monitor by knowing that the remaining 80% of the monitor's front is the viewing screen, calculated as 80% of the total area:\[ 0.8 \times \text{total area} = 922.74 \, ext{cm}^2 \]Using the reverse calculation, the complete area we found was:
\[ \text{total area} = \frac{922.74}{0.8} = 1153.425 \, \text{cm}^2 \]This simple percentage concept allows one to connect the area of the viewing screen with the total area of the monitor.
Significant Digits
Significant digits are crucial when it comes to precision in computations. They signify the certainty of measurements.
In computations, the number of significant digits reflects the number of digits believed to be accurate. For this problem's solutions, figures like \(33.8\) and \(27.3\) cm both have three significant digits.
This ensures the resultant calculations for areas and other parts of the problem reflect true precision, like rounding the width of the edge to four significant digits: 0.4625 cm.
In computations, the number of significant digits reflects the number of digits believed to be accurate. For this problem's solutions, figures like \(33.8\) and \(27.3\) cm both have three significant digits.
- It's critical to maintain significant digits throughout calculations to ensure results retain their real-world accuracy.
This ensures the resultant calculations for areas and other parts of the problem reflect true precision, like rounding the width of the edge to four significant digits: 0.4625 cm.
