area of oblique triangles

8/12/2019 Area of Oblique Triangles

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CALCULUS


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Oblique Triangles

Oblique triangles do not have any 90oangles

45o

72o

63o

120o

24o

46o

A

B

C

A

B

C

a

b

c

a

bc

Acute angled

Obtuse angled


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METHOD 1A

B D C

If we want to find the

area of a but the height

(altitude) is unknown

Stage 1

(small )

In ACD a

bh

h = b sinc

hin

Stage 2

(big )

In ABC

A = (base )(height)

But we just determined height

from using a smaller part of

the triangle

Substitute height

A = ( a )( b sinc )

A = a b sinc


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This is used to find the area of oblique triangles if

you know two sides and the included angle

A

CB

48o

25 cm

c

b

a

A = a c sinB

A= (30cm)(25 cm) sin 48o

A= (30cm)(25 cm) (0.743)

A= (30cm)(25 cm) (0.743)

A= 278.625 cm2

EXAMPLE:


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METHOD 2

When the two angles and a side is known. The area of the triangle

is equal to the square of one side and twice the sine of the oppositeangle multiplied by the sine of the other angles.

FORMULA:

Given side A and the other three angles

Area =

Given side B and the other three angles

Area =

Given side C and the other three angles

Area =

2 sin

sin

2sin

2 sin sin2sin

2 sin sin2sin


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EXAMPLES

1. B SOLUTION:Area = 2 sin sin

2sin 522

=(6.9 )2 sin 8048

2sin 52 = 13.73

A C

2. E SOLUTION:

Area = 2 sin sin 2 sin

Y =(7.5 )2 sin3075

2 sin 75 = 13.12

S

3. O SOLUTION:

Area =2 sin sin

sin

=(7.5)2 sin3075

2 sin 75 = 9.32

D G


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METHOD 3 If all three sides are known, the area can be determined using

HERONS Formula.

FORMULA:

Lets represent half of the perimeter of a triangle and a, b and c

as its sides then,

Where: s =

The area of herons formula is stated as:

Area =

++2 : ++2

( )


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Case 2 Find the area of a triangle given 3 sides

24 m

20 m10 m

Herons Formula

Step 1: Find the semi-perimeter

2

cbas

2

cbas

2102420 mmms

ms 27

Step 2: area formula

))()(( csbsassA ))()(( csbsassA

)1027)(2027)(2427(27 A

)17)(7)(3(27A

9639A 22.98 mA

B

C A


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2

cbas

))()(( csbsassA

Herons Formula

52 m

65 m

28 m

B

C

A

2

655228 mmms

ms 5.72

)5.7)(5.20)(5.44(5.72A

9375.496035A

2704.298mA


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METHOD 4

When the vertices are given, this method is called the

Matrix = adcd

Using the determinants, we have the two formulas:

METHOD A

1 1 12 2 13 3 1= 1(23)- 1(2 3)+ 1(23 (32)

The area of this would be : Area =1

2


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METHOD B:

1 1 12 2 1

3

3 1

1 1 12 2 1

3

3 1

= 12+13+23 32+13+12

Area =

1

2


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1. (3,4) SOLUTION:Area =

1

2

Q = 12 21A = 10 5 u2

(6, -2) (8, 1)

3 4 16 2 18 1 1

D = x1(y2

y3)

y1(x2

x3) + 1(x2y3

y3x3)= 3(-2 - 1) 4(6-8) + 1(6 (-2)(8))

= -9 + 8 +22

D = 21

EXAMPLES

1. (4,6) SOLUTION:Area =

1

2 =

1

2

A = 2u2(9,3)

(7,5)

4 6 19 3 17 5 1

D = 4(3 - 5) 6(9 - 7) + 1(9)(5) (3)(7)

= -8 12 + 24

D = 4


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END

area of oblique triangles

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