Problem 19 - Algebra 2 and Trigonometry

Question

The sum of the lengths of the legs of a right triangle is 20 feet. a. If \(x\) is the measure of one of the legs, represent the measure of the other leg in terms of \(x .\) b. If \(y\) is the area of the triangle, express the area in terms of \(x .\) c. Draw the graph of the function written in \(\mathbf{b}\) . d. What are the dimensions of the triangle with the largest area?

Step-by-Step Solution

Answer

The triangle has maximum area when both legs are 10 feet.

Step 1: Understand the Problem

We want to find expressions related to a right triangle with two legs. The sum of these legs is 20 feet, and we have to express in terms of one leg's measure.

Step 2: Formulate Equation for Leg Measurement

Let the legs of the triangle be represented as \(x\) and \(20 - x\). If \(x\) is the measure of one leg, the other leg must be \(20 - x\) because their sum is given as 20 feet.

Step 3: Expression for Area in Terms of \(x\)

The area \(y\) of a right triangle with legs \(x\) and \(20-x\) is given by \[y = \frac{1}{2} \times x \times (20-x) = \frac{1}{2} (20x - x^2) = 10x - \frac{x^2}{2}.\]

Step 4: Draw the Graph of the Area Function

The function is \(y = 10x - \frac{x^2}{2}\). This is a quadratic equation in standard form, representing a downward-opening parabola. To draw the graph, note that it is symmetric about its vertex, which can be found using \(x = -\frac{b}{2a}\) where \(a = -\frac{1}{2}\) and \(b = 10\).

Step 5: Find the Maximum Area

The vertex of the quadratic equation occurs at \[x = -\frac{10}{2 \times -\frac{1}{2}} = 10.\] At \(x = 10\), the other leg is \(20 - 10 = 10\). Thus, the triangle has maximum area when both legs are 10 feet.

Key Concepts

Triangle Area

The area of a right triangle is an important concept in geometry. A right triangle has one 90-degree angle, and the two sides that form this angle are known as the 'legs' of the triangle. The third side is known as the hypotenuse. The formula to find the area of any triangle is \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\), where the base and the height are the lengths of the legs in a right triangle. In our exercise, with legs being \(x\) and \(20 - x\), the area becomes \(\text{Area} = \frac{1}{2} \times x \times (20 - x)\). It's simplified to \(10x - \frac{x^2}{2}\). This specific formula helps us to express and calculate the triangle's area using one single variable.

Quadratic Equation

A quadratic equation is a mathematical expression that equates a polynomial of degree two to zero. It follows the form \ax^2 + bx + c = 0\, where \(a\), \(b\), and \(c\) are constants, and \(x\) represents an unknown. The equation of the area of our triangle, \(y = 10x - \frac{x^2}{2}\), can be rearranged into a standard quadratic form by rewriting it as \(-y = \frac{-x^2}{2} + 10x\), which is similar in structure to the standard form. This equation represents how the area changes with respect to the value of \(x\), one of the right triangle's legs.

Parabola

When graphed, a quadratic equation forms a curve known as a parabola. Parabolas can open upwards or downwards. In this exercise, our equation \( y = 10x - \frac{x^2}{2}\) forms a downward opening parabola, since the leading coefficient (\(-\frac{1}{2}\)) is negative.
Parabolas are symmetric around a vertical line known as the axis of symmetry, which passes through its vertex, the point of the maximum or minimum value. Understanding this symmetry helps in sketching the graph accurately.

Maximizing Area

The process of maximizing or minimizing the value of a quadratic function involves locating its vertex. Since in our context, we are dealing with a downward-opening parabola, the vertex will provide us with the maximum area of the triangle.
To find this vertex, we use the vertex formula: \[ x = -\frac{b}{2a} \] where \(a = -\frac{1}{2}\) and \(b = 10\). Calculating gives us \(x = 10\). This means both triangle legs are 10 feet long when the area is maximized.
So, maximizing involves identifying such critical points where the area achieves its largest possible value.