Chapter 8: Problem 59
Use the formula for the area of a rectangle and the Pythagorean Theorem to solve. A small television has a picture with a diagonal measure of 10 inches and a viewing area of 48 square inches. Find the length and width of the screen.
Short Answer
Expert verified
The possible length and width dimensions of the television are 6 inches by 8 inches or 8 inches by 6 inches.
Step by step solution
01
Calculation of possible dimensions
Since the area of a rectangle is acquired by multiplying its length and width, let's denote the length as \(L\) and width as \(W\). From the area formula \(L \times W = 48\), and the equation \(a^2 + b^2 = c^2\) (Pythagorean theorem) assuming \(L = a\), \(W = b\), and \(10 = c\) , we get \(L^2 + W^2 = 100\). We now have two equations to solve for \(L\) and \(W\).
02
Isolate one of the variables
From \(L \times W = 48\), we isolate \(L\) by dividing both sides of the equation by \(W\). So, we get \(L = 48/W\).
03
Substitution
Now, we can substitute \(L = 48/W\) into the equation \(L^2 + W^2 = 100\), we get \((48/W)^2 + W^2 = 100\) or \(2304/W^2 + W^2 = 100\). Re-arranging terms so that we have all like terms on the same side gives \(W^4 - 100W^2 + 2304 = 0\).
04
Solving Quadratic Equation
In step 3, a quadratic equation is obtained in terms of \(W^2\). With \(x = W^2\), it turns into \(x^2 - 100x + 2304 = 0\). One can solve this using the quadratic formula or factoring, yielding potential solutions of \(x = 36\) and \(x = 64\). Therefore, \(W^2\) could be 36 or 64.
05
Determining Final Values
Taking the square root of 36 or 64, we get \(W = 6\) or \(W = 8\). Substituting these two values back into \(L = 48/W\), one finds corresponding values for \(L = 8\) or \(L = 6\). Thus the dimensions are probably 6 inches by 8 inches, or vice-versa, so let's generalize:If \(W = 6\) (width = 6 inches), then \(L = 8\) (length = 8 inches).If \(W = 8\) (width = 8 inches), then \(L = 6\) (length = 6 inches).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Area of a Rectangle
The area of a rectangle is a fundamental concept in geometry, closely tied with basic arithmetic. It helps us figure out how much space is contained within a rectangle's boundaries. To find a rectangle's area, you simply multiply its length by its width. This is expressed in the formula:
Understanding this formula is crucial when solving real-world problems, such as determining how much material is needed to cover a surface. In our case, knowing the area and connecting it with other properties like the diagonal helps us piece together more about the rectangle's shape.
- Area = Length × Width
Understanding this formula is crucial when solving real-world problems, such as determining how much material is needed to cover a surface. In our case, knowing the area and connecting it with other properties like the diagonal helps us piece together more about the rectangle's shape.
Solving Quadratic Equations
Quadratic equations are a key part of algebra and appear frequently in various mathematical and practical scenarios. In this problem, we encounter an equation involving the square of a variable. The equation derived in the solution, namely
- \(W^4 - 100W^2 + 2304 = 0\)
- \(x^2 - 100x + 2304 = 0\)
- \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Diagonal Measurement
The diagonal of a rectangle is a fascinating measurement that involves a blend of geometry and algebra. It serves as a crucial link between a rectangle's sides and its hypotenuse. By applying the Pythagorean Theorem, which indicates that
- \(a^2 + b^2 = c^2\)
- \(L^2 + W^2 = 100\)