This activity is intended to supplement Trigonometry, Chapter 5, Lesson 2.
Time Required: 20 minutes
In this activity, students will use their graphing calculators to determine the relationship between Heron’s Formula and the basic area formula.
- Finding the area of a triangle
- Points of intersection
- Interpreting a graph
Teacher Preparation and Notes
- Make sure students have cleared Y= menu before starting.
- You may need to remind students how to TRACE and find points of intersection.
Problem 1: The 3, 4, 5 right triangle
- Students should know this is a right triangle, with hypotenuse 5. Make sure that they have that the legs are 3 and 4.
- The area of this triangle is 6.
- With Heron’s Formula, A=s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−√, and Y1=x(x−3)(x−4)(x−5)−−−−−−−−−−−−−−−−−−√ students might get confused with the parenthesis. Make sure all students change their window to the dimensions to the right before graphing.
- The graph is to the right. Have students analyze the domain and range and why there are blank spaces in the graph. The domain is (∞,00], [3, 4], and [5,∞) and the range is all real numbers greater than zero. If you have students zoom in further, they will see that there are no x−intercepts and one y intercept at (0, 0).
Y2=6 represents the area of this triangle. The horizontal line crosses the graph at (-0.435, 6) and (6, 6). The first point, however is invalid because x cannot be negative here. Ask students why. Explain that x is actually s and that s cannot be negative, because by definition it is 12(a+b+c).
- Specifically, the point (6, 6) represents (s, area). So, from this system of equations, we have determined what s is such that we have the correct area, in this case, also 6. If we can find the area in more that one manner, this will always work as a way to solve for s.