College Trigonometry, 2011a

11.3 The Law of Cosines 915
A 2 =
=
=
=
=
(
c 2 − [ a 2 − 2ab + b 2]) ([ a 2 +2ab + b 2] − c 2)
16
(
c 2 − (a − b) 2)( (a + b) 2 − c 2)
16
(c − (a − b))(c +(a − b))((a + b) − c)((a + b)+c)
16
(b + c − a)(a + c − b)(a + b − c)(a + b + c)
16
(b + c − a) (a + c − b) (a + b − c) (a + b + c)
· · ·
2
2
2
2
perfect square trinomials.
difference of squares.
At this stage, we recognize the last factor as the semiperimeter, s = 1 a+b+c
2
(a + b + c) =
2
. To
complete the proof, we note that
(s − a) = a + b + c
2
− a = a + b + c − 2a
2
= b + c − a
2
Similarly, we find (s − b) = a+c−b
2
and (s − c) = a+b−c
2
. Hence, we get
A 2 =
(b + c − a)
2
·
(a + c − b)
2
= (s − a)(s − b)(s − c)s
·
(a + b − c)
2
·
(a + b + c)
2
so that A = √ s(s − a)(s − b)(s − c) as required.
We close with an example of Heron’s Formula.
Example 11.3.3. Find the area enclosed of the triangle in Example 11.3.1 number 2.
Solution. We are given a =4,b = 7 and c = 5. Using these values, we find s = 1 2
(4 + 7 + 5) = 8,
(s − a) =8− 4 = 4, (s − b) =8− 7=1and(s − c) =8− 5 = 3. Using Heron’s Formula, we get
A = √ s(s − a)(s − b)(s − c) = √ (8)(4)(1)(3) = √ 96 = 4 √ 6 ≈ 9.80 square units.