Section 9.4 – The Area of a Triangle

166
Section 9.4 – The Area of a Triangle
Recall the formula for finding the area of a triangle;
Area of a Triangle
Let b be the base of the triangle and let h be the height or altitude of the
triangle that intersects the base at a right angle. Then, the area K of the
triangle is:
K=
1
bh
2
Note that the base b is not necessarily the length of the side b in the
triangles we have worked with in this chapter.
Suppose we are not given the height or the altitude of the triangle. How do
we find the area of the triangle? The answer will depend on what
information we are given. If we have SAS triangle, we can use trigonometry
to find the height of the triangle. If we have SSS triangle, then we will need
to derive a formula for finding the area of a triangle using just the sides.
Objective 1;
Find the Area of SAS Triangle.
In finding the area of a triangle where two of the sides and the included
angle are known, we have two cases to consider.
Case 1:
Acute Triangle
B
c
Case 2:
Obtuse Triangle
B
a
a
c
A
C
b
A
C
b
Construct h as the height of the triangle such that it intersects side b at a
right angle. Thus, side b will be the base.
167
Case 1:
Acute Triangle
Case 2:
B
Obtuse Triangle
B
c
a
a
h
h
A
c
C
A
b
C
b
In both cases, looking at the right triangle formed by ∠C, sides a and h:
sin(C) =
opp
hyp
=
h
a
which implies h = asin(C)
Since the area of a triangle is K =
1
bh
2
€
=
1
b(asin(C))
2
€
=
1
bh,
2
we can replace h by asin(C):
1
absin(C)
2
We can repeat this process for the other two sides.
Area of a Triangle
Let ∆ABC be any triangle. Then, the area of the triangle is given by:
1
1)
K = absin(C)
2)
K=
3)
K=
2
1
bcsin(A)
2
1
acsin(B)
2
In other words, "the area of the triangle is one-half the product of the two
sides times the sine of the included angle."
Find the area of the following triangles:
Ex. 1
∠A = 32.5˚, b = 4.5 m, c = 5.6 m
Solution:
First, we draw a figure. Notice that
B
we do have two sides and the
c = 5.6 m
a
included angle so we can plug
our values into the formula:
A = 32.5˚
C
1
K = bcsin(A)
b = 4.5 m
=
2
1
(4.5)(5.6)sin(32.5˚)
2
= 12.6sin(32.5˚) = 6.769975… ≈ 6.77
Thus, the area is 6.77 m2.
168
Ex. 2
∠C = 108.7˚, a = 14.5 ft, b = 17.3 ft
Solution:
First, we draw a figure. Notice that
B
we do have two sides and the
c
a = 14.5 ft
included angle so we can plug
our values into the formula:
A
108.7˚ = C
1
K = absin(C)
b = 17.3 ft
=
2
1
(14.5)(17.3)sin(108.7˚)
2
= 125.425sin(108.7˚) = 118.803… ≈ 118.8
Thus, the area is 118.8 ft2.
Objective 2:
Find the Area of SSS Triangle.
We need to formula for the area of the triangle given only the lengths of the
side. We will start with the formula we derived previously for the area when
given two sides and the included angle:
1
C
K = absin(C)
(rewrite C as 2( ))
=
=
=
2
1
C
absin(2( ))
2
2
1
C
C
ab•2sin( )cos(
2
2
2
C
C
absin( )cos( )
2
2
€
2
(use the double angle formula for sine)
)
(simplify)
€
Recall example 6 from the previous section that sin(
€
€
C
2
cos( € ) =
absin(
€
s(s−c)
€ab
where s =
1
(a
2
C
2
(s−a)(s−b)
ab
)=
+ b + c). Thus,
C
C
(s−a)(s−b)
s(s−c)
)cos( ) = ab
€ab
ab
2
2
s(s−a)(s−b)(s−c)
= s(s−a)(s−b)(s−c)
ab
€ (simplify)
€ = ab
This formula is know as Heron's formula
€
€
€
€
Heron's Formula €
€ a, b, and c be the lengths of the triangle and s be
Let
the semi-perimeter of the triangle. Then the area of the
triangle is
1
K = s(s−a)(s −b)(s−c) where s = (a + b + c)
2
a
c
b
and
169
Find the area of the following triangles:
Ex. 3a
Ex. 3b
9.2 cm
4.8 cm
5.00 m
5.00 m
Ex. 3c
5.00 m
3.95 in
2.1 in
7.3 cm
Solution:
a)
First, calculate the semi-perimeter:
s = (4.8 + 7.3 + 9.2)/2 = 21.3/2 = 10.65 cm
Now, plug into Heron's Formula:
A = 10.65(10.65 −4.8)(10.65 −7.3)(10.65 −9.2)
= 10.65(5.85)(3.35)(1.45)
= 302.6334...
= 17.3963… ≈ 17.4 cm2
b)
First, calculate the semi-perimeter:
s = (5 + 5 + 5)/2 = 15/2 = 7.5 cm
Now, plug into Heron's Formula:
A = 7.5(7.5−5)(7.5−5)(7.5−5)
= 7.5(2.5)(2.5)(2.5)
(simplify)
= 117.1875
= 10.825… ≈ 10.83 m2
c)
First, calculate the semi-perimeter:
s = (3.95 + 3.95 + 2.1)/2 = 10/2 = 5 in
Now, plug into Heron's Formula:
A = 5(5−3.95)(5−3.95)(5−2.1)
= 5(1.05)(1.05)(2.9)
= 15.98625
= 3.9982… ≈ 4 in2
3.95 in
(simplify)
(simplify)