stated as ordered pairs in the first quadrant of the coordinate plane; from this information, the lengths of a base and a height must be found in order to use the familiar area formula. So first we must identify one side as the base and determine its length. Then the height of the triangle can be found, and the familiar formula for determining the area of a triangle can be applied. This solution process depends on successful execution of several algebraic procedures that typically are learned only in high school mathematics (e.g. applying the distance formula, finding the equation of a line, finding the slope and equation of a perpendicular line to a given line, finding the intersection point of two lines). Were these the strategies you employed? The problem teaches us an important lesson: a formula is only useful if one has the “ingredients” of the formula. It also raises an interesting question: are there other formulas for the area of a triangle that are
3. EXPLORING AREAS OF TRIANGLES 579 better suited to some problems (meaning, they define the area in terms of some data other than the base and the height)? We now see that this is the case. 3.1. Side Lengths and Angles. One method for finding the area of a triangle depends on knowing the lengths of two sides of a triangle and the measure of the angle between them. Proposition 1 . If a triangle has side lengths of A , B , and C , and opposite angles with measures α , β , and γ as in Figure 11, then the area of the triangle is given by area = 1 2 BC sin α = 1 2 AC sin β = 1 2 AB sin γ. Figure 11. Angles and sides Proof. From Figure 11 we see that if we take B to be the base of the triangle, then the altitude is A sin γ . Since the area is half of the base times the height, we have that the area is 1 2 AB sin γ . The other equations are provided similarly. Your Turn 13 . Use the formula above to determine the area of the triangle given in the Let’s Go 5. Compare this answer with your previous answer. (Hint: Use the Law of Cosines to find the measure of an angle.)
580 13. MEASUREMENT 3.2. Heron’s Formula. Heron’s Formula , due to Heron of Alexandria, gives the area of an arbitrary triangle in terms of the lengths of the three sides 3 . The proof is elementary, requiring only the Law of Cosines and repeated use of “difference-of-two-squares” factoring. 4 Theorem 2 . Heron’s Formula. Given a triangle with side lengths A , B , and C , let S be half of the perimeter ( S = A + B + C 2 ). The area of the triangle is given by area = p S ( S - A )( S - B )( S - C ) . Proof. First, by the Law of Cosines, we have C 2 = A 2 + B 2 - 2 AB cos γ , so cos γ = A 2 + B 2 - C 2 2 AB . 3 Heron of Alexandria lived sometime around A.D. 75. He was interested largely in applied mathematics problems, and in his Dioptra describes how one may use elementary geometry to build tunnels. However, he is best known for his work Metrica , which contains the formula bearing his name.