Problem 59 - Algebra and Trigonometry

Question

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.

Step-by-Step Solution

Answer

The possible length and width dimensions of the television are 6 inches by 8 inches or 8 inches by 6 inches.

Step 1: 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 \(W\).

Step 2: 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 \(L = 48/W\).

Step 3: 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 \(W^4 - 100W^2 + 2304 = 0\).

Step 4: 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.

Step 5: 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).

Key Concepts

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:

• Area = Length × Width

For example, if you have a rectangle with a length of 6 inches and a width of 8 inches, its area would be calculated as 6 multiplied by 8, which equals 48 square inches.
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

• \(W^4 - 100W^2 + 2304 = 0\)

can be transformed by letting \(x = W^2\), turning it into a simpler quadratic equation:

• \(x^2 - 100x + 2304 = 0\)

This type of equation can be solved using various methods such as factoring or applying the quadratic formula:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

In our provided solution, it hints that this equation can either be solved by using the quadratic formula or factoring, which identifies the potential solutions of \(x = 36\) and \(x = 64\), leading to further insights about the width and length of the rectangle when taking square roots. Solving such equations is a crucial skill in understanding the relationships in mathematical problems, like our rectangle's dimensions.

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

• \(a^2 + b^2 = c^2\)

where \(c\) represents the hypotenuse, also known as the diagonal in the case of rectangles, we can find unknown side lengths given the diagonal length. In our exercise, the diagonal is given as 10 inches, so the equation is structured to find the rectangle’s side measurements as:

• \(L^2 + W^2 = 100\)

Understanding and being able to calculate diagonal measurements by combining the properties of right triangles is critical, especially when both the area and perimeter can lead to different insights about a rectangle's dimensions. The Pythagorean Theorem enables us to solve for these components effectively, generating a comprehensive view of the rectangles involved in various problems.