Page 288 – Area of Triangles
Surveyors calculate measures of
distances and angles so that they
can represent boundary lines of
parcels of land.
B
202 ft
82.5º
C
124.5º
180.25 ft
201.5 ft
75º
D
97º
A
The diagram at the right is a plot
of John and Renee Walter’s land.
What is the area of the region to the
nearest square foot?
161º
125 ft
E
158 ft
Area Formula
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K will represent the area.
K = ½bh
Notice in the triangle that sin A = h/c
So, h = c sin A
Substitute into the area formula and
you get K = ½bc sin A
You can also use:
K = ½ab Sin C & K = ½ac Sin B
Use these when you know 2 sides
and the included angle
C
b
a
h
A
c
B
Find the area of triangle ABC if a = 7.5, b = 9 and
C = 100º. Round to the nearest tenth.
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K = ½ab Sin C
= ½(7.5)(9)sin 100º
= (33.75)(0.9848)
= 33.237
The area is about 33.2 sq units.
B
7.5
100º
C
9
A
If you know the measure of one side and
two angles.
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Remember, K = ½bc sinA
The Law of Sines states:
So, b = c sin B
sin C
K = ½a² sinB sinC
sin A
b = c
sin B
sin C
.
.s
K = ½c² sinA sin B
sinC
K = ½b² sinA sinC
sinB
Find the area of triangle ABC if a = 18.6, A =
19º20’, and B = 63º50’. Round to the nearest
tenth.
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Find C: 180 – (19º20’ + 63º50’) = 96º50’
The find the area of the triangle:
B
63º50’
K = ½a² sinB sinC
sinA
A
= ½(18.6)² sin63º50’ sin96º50’
sin19º20’
= ½(345.96)
18.6
19º20’
(0.8975)(0.9929)
(0.3311)
= 465.6
So the are a of the triangle is about 465.6 sq units
C
If you know 3 sides of the triangle you can use the Law of
cosines and the formula. Find the area of ABC if a = √2, b
= 2, and c = 3.
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First, solve for A using the Law of Cosines
a² = b² + c² - 2bc Cos A
(√2)² = 2² +3² - 2(2)(3) Cos A
2 = 4 + 9 – 12 Cos A
2 = 13 – 12 Cos A
0.9167 = Cos A
23º33’ = A
C
√2
2
A
3
Now find the area:
K = ½(2)(3) sin 23º33’
= 3(0.3995)
= 1.199
So the area is about 1.2 sq units
B
Hero’s Formula
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If you know the measures of all three sides of
the triangle you can also use Hero’s Formula
K = √ s(s - a)(s - b)(s - c)
s = ½(a + b + c)
Use Hero’s Formula to find the area of ABC
if a = 20, b = 30 and c = 40.
First find s:
s = ½(20 + 30 + 40)
= ½(90) = 45
Then use Hero’s Formula:
K = √45(45 – 20)(45 – 30)(45 – 40)
= √45(25)(15)(5)
= √84,375
= 290.5
So the area of the triangle is
about 290.5 sq units.
To solve some applications, you may have
to use more than formula.
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Separate the region into
triangles.
a
B
202 ft
82.5º
C
124.5º
180.25 ft
Now the area of the region
is the sum of the areas of
the triangles.
e
c
201.5 ft
75º
D
97º
Let a be the side with measure
202, b is 158, c is 201.5, and
e is 180.25.
A
161º
125 ft
E
158 ft
b
Area of each triangle
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= ½ac sin B
= ½(202)(201.5)sin 82.5º
= (20,351.5)(0.9914)
= 20,176.5
a
B
202 ft
82.5º
C
124.5º
180.25 ft
e
c
201.5 ft
= ½eb sin D
= ½(180.25)(158) sin 75º
= 14,239.75)(0.9659)
= 13,754.2
75º
D
97º
A
This gives you the areas of ABC and CDE
161º
125 ft
E
158 ft
b
Area of ACE
Use the Law of Cosines to find AC
and CE
AC = √202² + 201.5² - 2(202)(201.5)cos82.5º
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= 266.0
a
B
202 ft
82.5º
C
124.5º
180.25 ft
20.176.5
e
c
201.5 ft
13,754.2
CE = √180.25² + 158² - 2(180.25)(158) cos 75º
75º
D
= 206.7
The use Hero’s Formula:
s = ½(125 + 266 + 206.7) = 298.85
97º
A
161º
125 ft
E
158 ft
b
K = √298.85(298.85 – 125)(298.85 – 266)(298.85 – 206.7)
= √298.85(173.85)(32.85)(92.15)
= 12,540.9
So the area of the land is 20,176.5 + 13,754.2 + 12,540.9 or 46,472 sq ft
Assignment
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Page 293 – 294
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#12 –20, 27 - 30