Pre-Calculus Honors
Book Reference 5.6
Unit 8 Lesson 5: Applications of Area of Oblique Triangles
Objective: ____________________________________________________________
1. Do Now Directions: Mark up the following formulas text with questions,
comments, and reminders. Copy the formulas onto your reference sheet.
2. Group Practice: Complete the following link sheet to find the area of the
hexagon using what you know about special right triangles, right triangle trig, and
area of oblique triangles.
Formulas
Verbal
1
1. Area = bh
2
2. The area of any triangle is one half the product
of the lengths of the two sides times the sine of
their included angle. That is,
A regular octagon inscribed in a circle with a radius of
15 inches.
1
1
1
Area = bcsin A , Area = absinC , Area = acsin B
2
2
2
3. The law of cosines can be used to establish the
following formula for the area of a triangle. This
formula is called Heron’s Area Formula.
Given any triangle with side lengths a, b, and c, the
area of the triangle is given by
Area = s(s - a)(s - b)(s - c) where s =
Diagram
a+b+c
2
Label diagram that represents the verbal.
Algebra
Find the area of the octagon.
Summary: Explain your thinking process as you went through the problem
3. Group Practice:
Directions: Complete the following link sheet to find the area of the
hexagon using what you know about special right triangles, right triangle
trig, and area of oblique triangles.
Formulas
Verbal
1
1. Area = bh
2
2. The area of any triangle is one half the product
of the lengths of the two sides times the sine of
their included angle. That is,
A regular hexagon circumscribed about a circle of a
radius of 12 inches.
1
1
1
Area = bcsin A , Area = absinC , Area = acsin B
2
2
2
3. The law of cosines can be used to establish the
following formula for the area of a triangle. This
formula is called Heron’s Area Formula.
Given any triangle with side lengths a, b, and c, the
area of the triangle is given by
Area = s(s - a)(s - b)(s - c) where s =
Diagram
a+b+c
2
Label diagram that represents the verbal.
Algebra
Find the area of the hexagon.
Summary: Explain your thinking process as you went through the problem
4. Group Practice:
Directions: Complete the following link sheet to answer the questions on
the parallelogram using what you know about special right triangles, right
triangle trig, and area of oblique triangles.
Verbal
A parallelogram has sides of 18 and 26 feet, and an
angle of 39 degrees.
Algebra
Diagram
Draw a diagram that represents the verbal. Label
all dimensions of the parallelogram.
Communication
1. Find the area of the parallelogram.
1. Explain your thinking process as you went
through the problem.
2. Find the length of each diagonal of the
parallelogram.
2. What do you notice about the location of the
longer diagonal and the shorter diagonal?
3. Would you get the same area if you used the
angle not given in the verbal? Explain why
or why not?
Application of Area Problem Set
Do all work on a separate sheet of paper
1. Find the area of regular nonagon circumscribed about a circle with a radius of 10
inches.
2. A parallelogram has sides of 15 and 24 ft., and an angle of 40 degrees. Find the
length of longest diagonal of the parallelogram and the area.
3. Because deer require food, water, cover for protection from weather and predators
and living space for healthy survival, there are natural limits to the number of deer
that given a plot of land can support. Deer populations in parks average 14
animals per square kilometer. If a triangular region with sides of 3 km, 4 km, and
6 km has a population of 50 deer, how close is the population on this land to
average national park population?
4. What equilateral triangle would have the same area as a triangle with sides 6, 8
and 10? Show all work and explain how you arrived at your answer.
5. Use the triangle below to prove the statement A 
1
ac sin B
2
Pre-Calculus Honors Homework: Finish Problem Set
Application of Area Problem Set – Answer Key
é1
ù
1. 9 ê (7.28)(10)ú » 327.6 in 2
ë2
û
2. Longest diagonal
a 2 =152 + 24 2 - 2(15)(24)cos140°
» 36.8 inches
Area
é1
ù
2 ê (15)(24)(sin 40°)ú » 231.4 in 2
ë2
û
3. Area of 3,4,6 triangular area
S = 6.5
A = 6.5(6.5- 3)(6.5- 4)(6.5- 6) » 5.33 km2
Average Population
14 deer
·5.33km 2 » 74.62 deer
2
km
The area is under-populated by approximately 25 deer from the national park
average.
4. Area of 6, 8, 10 right triangle
1
(6)(8) = 24
2
Side length of equilateral triangle with the same area
1
24 = (s)(s)(sin 60°)
2
1
3
24 = s 2
2
2
3 2
24 =
s
4
s » 7.44
1
Step 1: A = bh
2
Step 2: Base = c
h
Step 3: sinB = ® h = asin B
a
1
Step 4: A = acsin B
2
5.