College Trigonometry, 2011a

914 Applications of Trigonometry
In Section 11.2, we used the proof of the Law of Sines to develop Theorem 11.4 as an alternate
formula for the area enclosed by a triangle. In this section, we use the Law of Cosines to derive
another such formula - Heron’s Formula.
Theorem 11.6. Heron’s Formula: Suppose a, b and c denote the lengths of the three sides
of a triangle. Let s be the semiperimeter of the triangle, that is, let s = 1 2
(a + b + c). Then the
area A enclosed by the triangle is given by
A = √ s(s − a)(s − b)(s − c)
We prove Theorem 11.6 using Theorem 11.4. Using the convention that the angle γ is opposite the
side c, wehaveA = 1 2ab sin(γ) fromTheorem11.4. In order to simplify computations, we start by
manipulating the expression for A 2 .
A 2 =
( 1
2 ab sin(γ) ) 2
= 1 4 a2 b 2 sin 2 (γ)
= a2 b 2
4
(
1 − cos 2 (γ) ) since sin 2 (γ) =1− cos 2 (γ).
The Law of Cosines tells us cos(γ) = a2 +b 2 −c 2
2ab
, so substituting this into our equation for A 2 gives
A 2 = a2 b 2 (
1 − cos 2 (γ) )
4
[
= a2 b 2 ( a 2 + b 2 − c 2 ) 2
]
1 −
4
2ab
[ (
= a2 b 2 a 2 + b 2 − c 2) ]
2
1 −
4
4a 2 b
[ 2
= a2 b 2 4a 2 b 2 − ( a 2 + b 2 − c 2) ]
2
4
4a 2 b 2
= 4a2 b 2 − ( a 2 + b 2 − c 2) 2
16
= (2ab)2 − ( a 2 + b 2 − c 2) 2
16
( [
2ab − a 2 + b 2 − c 2]) ( 2ab + [ a 2 + b 2 − c 2])
=
16
(
c 2 − a 2 +2ab − b 2)( a 2 +2ab + b 2 − c 2)
=
16
difference of squares.