**9
- 2 Area of a Triangle**
** **
*The formula
to
find the area of a triangle is K = bh/2 where K is the area, b is the
base
and h is the height which is perpendicular to the base. How do we
find the area if we don't have the height?*
* *
*Given *__/__ B,
a,
and c, how can we find the area?
*By dropping down a
perpendicular
from A we note by trig that:*
*sin B = h/c*
*c Sin B = h*
*substituting into
the
area formula we have:*
*K = (ac Sin B)/2*
*We could be given
any
two sides and the included angle and find the area. The other
forms
are:*
*K = (bc Sin A)/2*
*K = (ab Sin C)/2*
*In words, this
means:*
*K = (1/2)*^{ .}
(one side) ^{.} (another side)^{ .} (sine of the
included
angle)

__Sample Problems__
* *

*1) Two sides of a
triangle
have lengths 8 m and 5 m. The included angle measures 53*^{o}.
Find the area.
*Solution:*
*K = (1/2)(8)(5)Sin
53*
*K = 20 Sin 53*
*K = 16.0 m*^{2}
* *

*2) The area of a
triangle
is 20. If a = 6 and b = 12, find all possible measures of *__/__
C.

* *
*Solution:*
*20 = (1/2)(6)(12)
Sin
C*
*20 = 36 Sin C*
*20/36 = Sin C*
*Sin*^{-1}(20/36)
= C
*33.7*^{o} = C
*Since the angle
could
be acute or obtuse, there is another answer.*
*180 - 33.7 = 146.3*^{o}
*C = 33.7*^{o}
or 146.3^{o}
*3) Find the area of
the
following quadrilateral:*
*Solution:*
*To find the area,
we
must split the figure into two triangles. One way is as follows:*
*To find area A:*
*This is a right
triangle,
so one side is the base, the other the height.*
*K = (1/2)(5)(8)*
*K = 20*
*The area of A is 20
square
units.*
* *
*To find the area of
B:*
*We need to find the
length
of the diagonal (in blue)
and the size of the angle included between the blue
length and length of 14.*
*We can find the
length
by using pythagorean's theorem.*
*
_______*
*= \/ 25 + 64*
*= 9.4*
*To find the angle,
we
need to first find the angle included between the sides 8 and 9.4 in
the
right triangle. Using trig we get:*
*tan x = 5/8*
*x = tan*^{-1}(5/8)
*x = 32*^{o}
*Since we know the
big
angle is 75*^{o}, all we have to do is subtract to find the
angle
we need!
*75 - 32 = 43*
*Now we have the
included
angle in triangle B.*
*K = (1/2)(14)(9.4)
Sin
43*
*K = 44.9*
*To find the area of
the
quadrilateral, simply add the two areas:*
*44.9 + 20 = 64.9*
*The area is 64.9
square
units.*

**We
can now head on to the Law of Sines!!**