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You are reading an older version of this FlexBook® textbook: CK-12 Texas Instruments Trigonometry Teacher's Edition Go to the latest version.

6.1: Analyzing Heron’s Formula

Difficulty Level: At Grade Created by: CK-12

This activity is intended to supplement Trigonometry, Chapter 5, Lesson 2.

Time Required: 20 minutes

Activity Overview

In this activity, students will use their graphing calculators to determine the relationship between Heron’s Formula and the basic area formula.

Topics Covered

  • 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.

Associated Materials

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 = \sqrt{s(s - a)(s - b)(s - c)}, and Y1 = \sqrt{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.


Xmin = -1

Xmax = 8

Xscl = 1

Ymin = -1

Ymax = 10

Yscl = 1

Xres = 1

  • 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 (\infty, 00], [3, 4], and [5, \infty) 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 \frac{1}{2}(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.

Problem 2

For this triangle, the area is, A = \frac{1}{2}(15)\left(\frac{4 \sqrt{17}}{3} \right) = 5 \cdot 2 \sqrt{17} = 10 \sqrt{17} \approx 41

Here, students would graph y = \sqrt{x(x - 7)(x - 12)(x - 15)} and y = 10 \sqrt{17} to determine what x, or s, is. The graph of the two functions looks like:

Again, s cannot be negative, so we eliminate the negative point of intersection. Therefore the answer is \left(17, 10 \sqrt{17}\right).

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Date Created:

Feb 23, 2012

Last Modified:

Nov 04, 2014
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