Objective: Find the area of any triangle given at least three pieces of information. Process: Derive...

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 1

2absinC

ExampleFind the area of ∆DEF, if d = 3 cm, e = 8 cm

and F = 35°. Round to the nearest hundredth.

Area 1

2desinF

Area 1

2(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 1

2

a2 sinBsinC

sinA

Formula for

Two Angles and the Included Side

Area 1

2

a2 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 1

2

b2 sinAsinC

sinB

Area 1

2

(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

Example

Find 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 19.6608

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. CBH, h = 3, C = 49, H = 243. DKP, d = 6, k =14, p = 24

4. HYM, h = 4, M = 18, H = 615. DGH, d = 7, g = 9, h = 4

6. CFV, c = 31, F = 27, v = 26

Answer

Answer

Answer

Answer

Answer

Answer

Answer #1:

ABC, a = 3, b = 2, C = 24

Area 1

2absinC

Area 1

2(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 1

2

h2 sinCsinB

sinH

Area 1

2

(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 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 61Y 101

Area 1

2

h2 sinMsinY

sinH

Area 1

2

(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 1

2cv sinF

Area 1

2(31)(26)sin27

Area 183.0units 2

Return to problems.