Transcript No Slide Title

Objective: Find the area of any triangle given at
least three pieces of information.
Process: Derive several formulas to allow use
of given information (to avoid rounding errors).
By: Anthony E. Davis
Summer 2003
Where to start?
If you have two sides
and the included angle
then click here.
If you have two angles
and the included side
then click here.
If you have all three
sides then click here.
b
C
a
B
C
a
c
a
Not Quite Sure? - Try Some Practice Problems.
b
Two sides and the Included angle
See derivation.
b
Look at formula.
Try an example.
Return to choices.
a
C
Two angles and the included side
See derivation.
Look at formula.
Try an example.
Return to choices.
B
a
C
All three sides
Look at formula.
c
b
Try an example.
Return to choices.
a
Not Quite Sure?
•Draw a triangle.
•Label the information given.
•Match this triangle with one of the
three shown. Remember all triangles
and all variables are arbitrarily drawn
so rotation may be necessary.
•Return to choices
Derivation
A
c
b
h
know Area=1/
•We
2(base)(height).
B
C
•Let a represent the base.
a
•Using right triangle trigonometry, sin C = h/b
•Solve for h, h = b sin C.
•Replace values in area formula: Area = 1/2 a (bsin C)
•Hence the Area given two sides and the included angle
is any of the following:
Area = 1/2 ab sin C
Area = 1/2 bc sin A
Area = 1/2 ac sin B
Formula
for
Two Sides and the Included
Angle
1
Area  absin C
2
Example
Find the area of ∆DEF, if d = 3 cm, e = 8 cm and
F = 35°. Round to the nearest hundredth.
1
Area  desin F
2
1
Area  (3)(8)sin 35
2
2
Area  6.88cm
Derivation
A
c
b
h
•We know Area = 1/2 (base)(height) B
a
•Let a represent the base
•Find the third angle by subtracting the two known
from 180.
•From right triangle trigonometry, sin C = h/b
•Solve for h, h = b sin C
•Replacing values, Area = 1/2 a (bsin C).
C
Derivation (cont.)
A
c
b
h
B
C
a
•However we are only given one side, so we need to
substitute the ‘a’ or ‘b’ out. (Let say the ‘b’).
•From the Law of Sines, sin A/a = sin B/b. We know
A, B and ‘a’ so we will solve for ‘b’.
•Solve for ‘b’, b = (a sin B)/(sin A)
•Replace, Area = 1/2 a ((a sin B)/(sin A)) sin C
•Thus we have the following
1 a2 sin Bsin C
Area 
2
sin A
Formula
for
Two Angles and the Included
Side
2
1 a sin Bsin C
Area 
2
sin A
Example
Find the area of ∆CAB if b = 7 ft., C = 42º, and
B = 28º. Round your answer to the nearest tenth.

Angle A = 180 - B - C



A  180  28  42

A  110
2
1 b sin Asin C
Area 
2
sin B
1 (7) 2 sin 110sin 42
Area 
2
sin 28
Area  32.8 ft2
Formula
for
All Three Sides
(Heron’s Formula)
Area  S ( S  a )( S  b)( S  c)
a+b+c
where S (semiperimeter) =
2
Example
Find the area of an equilateral triangle having legs of
length 3.2 mm. Round your answer to two decimal
places.
abc
S
, in this case a = b = c = 3.2
2
3.2 + 3.2 + 3.2
S=
2
S  4.8mm Area  (4.8)(4.8  3.2)(4.8  3.2)(4.8  3.2)
Area  19.6608
Area  4.43mm2
Practice Problems
Directions: Find the area of each triangle using the given
information. Round only your final answer to the nearest
tenth. You may click on the question to see the solution.
1. ABC, a = 3, b = 2, C = 24
2.
3.
4.
5.
6.
CBH, h = 3, C = 49, H = 24
DKP, d = 6, k =14, p = 24
HYM , h = 4, M = 18, H = 61
DGH, d = 7, g = 9, h = 4
CFV, c = 31, F = 27, v = 26
Answer
Answer
Answer
Answer
Answer
Answer
Answer #1:
ABC, a = 3, b = 2, C = 24
1
Area  absin C
2
1
Area  (3)(2)sin 24
2
2
Area  1.2units
Return to problems.
Answer #2
CBH, h = 3, C = 49, H = 24
B  180  49  24
B  107
1 h 2 sin Csin B
Area 
2
sin H
1 (3)2 sin 49sin 107
Area 
2
sin 24
Area  8.0units 2
Return to problems.
Answer #3
DKP, d = 6, k = 14, p = 24
d k p
S
2
Area  S(S  d)(S  k)(S  p)
6  14  24
S
Area  22(22  6)(22  14)(22  24)
2
S  22units
Area  5632
Cannot take square root of negative number.
These three sides (6, 14, 24) do not form a triangle.
Remember the two smaller sides must add to more
than the third side.
Return to problems.
Answer #4
HYM, h = 4, M =18, H = 61
Y  180  18  61
Y  101
1 h 2 sin Msin Y
Area 
2
sin H
1 (4)2 sin18sin101
Area 
2
sin 61
Area  2.8units 2
Return to problems.
Answer #5
DGH, d = 7, g = 9, h = 4
d g h
2
7 94
S
2
S  10units Area  S(S  d)(S  g)(S  h)
S
Area  10(10  7)(10  9)(10  4)
Area  180
Area  13.4units2
Return to problems.
Answer #6
CFV, c = 31, F = 27, v = 26
1
Area  cv sin F
2
1
Area  (31)(26)sin 27
2
Area  183.0units2
Return to problems.