Area of A Triangle
Area of A Triangle
Area of A Triangle
Applications
of
Trigonometry
AREA OF A
TRIANGLE
1
2
=
From elementary geometry, the formula
for the area of the triangle is
length of base
height
Area of a triangle
Given the lengths of two sides and the angle
opposite the third side, we can compute the area
of the triangle using the formula below. Suppose that
a triangle has sides a and b and that the included
angle is .
| |
=
|
\ .
97.2 C =
1
sin
2
A ab C =
( )( )
1
120 170 sin97.2
2
=
10120 =
10120
2.1
4840
=
Now find the area
square yards
Thus the area of the field is 2.1 acres.
The number of acres is found by:
Heron's (or Hero's) formula states that the
area A of a triangle whose sides have lengths a, b,
and c is
where s is the semiperimeter of the triangle:
Proof:
( )( )( )
2 2 2 2
a b c a b c a b c a b c
s s a s b s c
+ + + + + +
=
Lets rewrite the expression under the radical solely in
terms of a, b, and c. We have:
( ) ( ) ( ) ( )
2 2 2 2
b c a b c a a b c a b c ( ( + + + +
=
( (
( ) ( )
2 2
2 2
4 4
b c a a b c
( (
+
= ( (
( (
2 2 2 2 2 2
2 2
4 4
b c bc a a b c bc
( (
+ + +
=
( (
Lets now apply the Law of Cosines to replace
2
a
in each of the expressions in the numerator:
2 2 2
2
4
b c bc a
(
+ +
(
( )
2 2 2 2
2 cos 2
4
b c bc bc c b o + + + +
=
cos
2
bc bc o +
=
( )
1 cos
2
bc o +
=
2 2 2
2
4
a b c bc
(
+
(
( )
2 2 2 2
2 cos 2
4
b c bc b c bc o + +
=
cos
2
bc bc o +
=
( )
1 cos
2
bc o
=
( )( )( )
sin
2
bc
s s a s b s c
o
=
( )( )( )
( ) ( )
1 cos 1 cos
2 2
bc bc
s s a s b s c
o o ( ( +
=
( (
( )
2
2 2
1 cos
4
b c
o
=
2 2 2
sin
4
b c o
=
This gives us
2 2 2
2
a b c
s
+ +
=
23 26 31
40
2
s
+ +
= =
Find the area of a triangle with sides 23 cm, 26 cm,
and 31 cm, to the nearest ten square centimetres.
Solution:
Find s, where
Example 4
Now use
( )( )( )
A s s a s b s c =
( )( )( )
40 40 23 40 26 40 31 =
( )( )( )
40 17 14 9 =
The area is approximately 290 .
2
293cm =
2
cm
Example 5
Suppose that a triangle has sides of length 12
cm, 15 cm, and 11 cm. Use Herons formula to
determine the area of the triangle.
Solution:
In this case we, we have
( )
1
12 15 11
2
s = + +
=19
Therefore, by Herons formula
Area =
2
65.24cm =
( )( )( )
s s a s b s c
( )( )( )
19 19 12 19 15 19 11 =
4256 =
Example 6
The distance as the crow flies from San Jose,
Antique to Bacolod City is 568 mi, from Bacolod City
to Roxas City is 536 mi, and from Roxas City to San
Jose, Antique is 247 mi. What is the area of the
triangular region having these three places as
vertices? (Ignore the curvature of Earth)
Solution
If we let a = 568, b = 536, and c = 247
The semi-perimeter s is
( )
1
568 536 247 675.5
2
= + + =
s
Using Herons formula, the area
( )( )( )
A s s a s b s c =
( )( )( )
675.5 675.5 568 675.5 536 675.5 247 =
2
65900mi ~
CREDITS
1. http://www.ping.be/~ping1339/gonio.htm#
Area-of-a-triangle
2. http://en.wikipedia.org/wiki/File:Triangle.TrigArea.
svg
3. http://en.wikipedia.org/wiki/Heron's_formula