objective: find the area of any triangle given at least three pieces of information
DESCRIPTION
Area of Triangles. 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. b C - PowerPoint PPT PresentationTRANSCRIPT
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?
c b
a
B C
a
b
C
a
Not Quite Sure? - Try Some Practice Problems.
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
Two sides and the Included angle
See derivation.
Look at formula.
Try an example.
Return to choices.
B C
a
Two angles and the included side
See derivation.
Look at formula.
Try an example.
Return to choices.
c b
a
All three sides
Look at formula.
Try an example.
Return to choices.
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•We know Area=1/2(base)(height).•Let a represent the base.•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
A
c b
B C
a
h
Formula for
Two Sides and the Included Angle
Area 12absinC
ExampleFind the area of ∆DEF, if d = 3 cm, e = 8 cm
and F = 35°. Round to the nearest hundredth.
Area 12desinF
Area 12
(3)(8)sin35
Area 6.88cm2
Derivation
•We know Area = 1/2 (base)(height)•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).
A
c b
B C
a
h
Derivation (cont.)
•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
A
c b
B C
a
h
Area 12a2 sinBsinC
sinA
Formula for
Two Angles and the Included Side
Area 12a2 sinBsinC
sinA
ExampleFind 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
A180 28 42
A110 Area 12b2 sinAsinC
sinB
Area 12
(7)2 sin110 sin 42sin 28
Area 32.8 ft2
Formula for
All Three Sides(Heron’s Formula)
( )( )( )a+b+c where S (semiperimeter) =
2
Area S S a S b S c
ExampleFind the area of an equilateral triangle having legs of length 3.2 mm. Round your answer to two decimal
places.
S a b c2
, in this case a = b = c = 3.2
S =3.2 + 3.2 + 3.2
2S 4.8mm Area (4.8)(4.8 3.2)(4.8 3.2)(4.8 3.2)
Area 19.6608Area 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 = 242. CBH, h = 3, C = 49 , H = 243. DKP, d = 6, k =14, p = 244. HYM, h = 4, M = 18, H = 615. DGH, d = 7, g = 9, h = 46. CFV, c = 31, F = 27 , v = 26
Answer
Answer
Answer
Answer
Answer
Answer
Answer #1:
ABC, a = 3, b = 2, C = 24
Area 12absinC
Area 12
(3)(2)sin 24
Area 1.2units 2
Return to problems.
Answer #2
CBH, h = 3, C = 49 , H = 24
B 180 49 24B 107
Area 12h2 sinCsinB
sinH
Area 12
(3)2 sin 49 sin107sin 24
Area 8.0units 2
Return to problems.
Answer #3
DKP, d = 6, k = 14, p = 24
S d k p2
S 6 14 24
2S 22units
Area S(S d)(S k)(S p)
Area 22(22 6)(22 14)(22 24)
Area 5632Cannot 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 61Y 101
Area 12h 2 sinMsin Y
sinH
Area 12
(4)2 sin18 sin101sin61
Area 2.8units 2
Return to problems.
Answer #5
DGH, d = 7, g = 9, h = 4
S d g h2
S 7 9 4
2S 10units Area S(S d)(S g)(S h)
Area 10(10 7)(10 9)(10 4)
Area 180
Area 13.4units 2
Return to problems.
Answer #6
CFV, c = 31, F = 27 , v = 26
Area 12cv sinF
Area 12
(31)(26)sin27
Area 183.0units 2
Return to problems.