Areas
In this section, we will use three new formulas for the area of a triangle. Here, S  the
area of triangle ABC.
The first formula is used when you know two of the sides of the triangle and the angle
between these sides (SAS).
Area of a Triangle
1
ab sin C
2
1
S  ac sin B
2
1
S  bc sin A
2
S
EXAMPLE 1: In triangle ABC, a  12.5 cm, b  10.7 cm, and C  58.2 . Find the area.
Solution: S 
1
1
ab sin C   12.5  10.7  sin 58.2  56.8 cm 2 .
2
2
The next formula is used when you know two angles and one side of the triangle. (ASA)
It is derived by using the Law of Sines along with the first formula.
Area of a Triangle
a 2 sin B sin C
2 sin A
2
b sin A sin C
S
2 sin B
2
c sin A sin B
S
2 sin C
S
EXAMPLE 2: If A  45, B  65 , and c  32 inches, find the area of triangle ABC.
Solution: The remaining angle C  180  45  65  70 . The area is
S
32 2 sin 45 sin 65
 350 square inches.
2 sin 70
The last area formula is used when you know the lengths of all three sides of the
triangle (SSS). This formula is known as Heron’s formula, after a Greek mathematician.
Area of a Triangle
S  s(s  a)(s  b)(s  c) ,
1
Where s  (a  b  c)
2
is the semiperimeter of the
triangle
EXAMPLE 3: Suppose that a  12, b  27, and c  34 . Find the area of triangle ABC.
12  27  34 73

 36.5 . The area is then
2
2
S  36.5(36.5  12)(36.5  27)(36.5  34)  146 square units.
Solution: The semiperimeter is s 