Squares on a Triangle NAME The Pythagorean theorem states that the sum of the areas of the squares on the legs of a right triangle is equal to the area of the square on the hypotenuse of the right triangle. This is the basis for the formula a 2 + b 2 = c 2, since a 2, b 2, and c 2 are the areas of the actual squares on the sides of the triangle. But what if the triangle is not a right triangle? Open the Squares on a Triangle applet: Go to Illuminations,. Enter the Activities section, and search for squares. Select the Squares on a Triangle applet from the list of search results. 1. Compare the sum of Areas I and II to Area III for different measures of C. Drag the point in the slider to change the measure of C. Points A and B can also be dragged to change the size of the triangle. Record several observations, being sure to test cases for acute, right, and obtuse angle measures of C. MEASURE OF ANGLE C AREA SQUARE I + AREA SQUARE II AREA SQUARE III HOW DO THE AREAS COMPARE? WHAT TYPE OF TRIANGLE IS IT? (ACUTE, OBTUSE, RIGHT) 60º 70º 80º 90º 100º 110º 2. Use your observations to complete the following statements: In an acute triangle, if a, b, and c are the sides of a triangle and c is the longest, then a 2 + b 2 will be. In an obtuse triangle, if a, b, and c are the sides of a triangle and c is the longest, then a 2 + b 2 will be.
Use these two new ideas to answer the following questions. 3. Suppose a triangle has sides of lengths 9, 15, and 17. Is it an acute, obtuse, or a right triangle? 4. Suppose a triangle has sides of lengths 7, 24, and 25. Is it an acute, obtuse, or right triangle? 5. Suppose a triangle has sides of lengths 6, 14, and 16. Is it an acute, obtuse, or right triangle?
Defects in a Triangle You already know the Pythagorean theorem, which is an area relationship among the sides of a right triangle. Wouldn t it be nice if there were an area relationship for any triangle? Actually, there is a nice area relationship among the sides of any triangle. In this next section, you will look for this relationship. Click the Show Defect I button on the applet. You should see something similar to this picture: A similar result will occur when you click the Show Defect II button. Find a way to relate the Areas of Squares I and II and the two Defect Areas to Area III. You may wish to divide your work into cases, looking at acute triangles first and then obtuse triangles. The table below will help to organize your ideas. MEASURE OF ANGLE C AREA SQUARE I + AREA SQUARE II AREA DEFECT I AREA DEFECT II AREA SQUARE III WHAT TYPE OF TRIANGLE IS IT? (ACUTE OR OBTUSE) 60º 70º 80º 90º 100º 110º 120º
6. Briefly summarize what you discovered. Calculating Area Defects The relationships you found for the areas of the squares and defects of the sides of a triangle is called the law of cosines. You may be wondering what trigonometry has do with areas and defects, so let s examine this idea further. 7. Why is the defect area in this diagram equal to a x? 8. Sides b and x are part of a right triangle. What trigonometric function would relate them? (Write the trigonometric equation.) 9. How would you calculate the defect area in this diagram without using x in the calculation? 10. Using ideas similar to those in Questions 7 9, find an expression for the defect area that uses cos C and does not use y.
11. Earlier in your explorations, did you notice that the two defects area always had the same area? Use your expressions for the two defects area to explain why this is true. Putting It All Together the Law of Cosines! 12. Using the ideas you have discovered, write an equation that combines your observations about area into one sentence. (Assume the triangle is acute.) The area of Square III equals. 13. Using a 2, b 2, and c 2 for the areas of the squares and the expressions you found in Questions 9 and 10 for the areas of the defects, rewrite your area relationship into a formula (again, assume the triangle is acute). Check your formula with a classmate.