Download Unit 6 Lesson 5 Area With Trig

Honors Geometry
Unit 6 Lesson 5
Finding Areas with Trigonometry
Objectives
• I can use trigonometry to find the surface
area of a regular polygon
• I can use trigonometry to find the area of a
triangle.
Application
• We can use right triangles to find the surface
areas of two figures
– Oblique triangles – triangles with no right angle
– Regular polygons – n-gons with all sides and
angles congruent
Vocabulary
PQRST is a regular pentagon
• Center of a regular polygon – the
center of a circle circumscribed
around the polygon (X)
• Radius of a regular polygon –
radius of the circumscribed circle
(
) XQ
• Apothem – segment from the
polygon center to a side,
perpendicular to that side(XN )
• Central angle of a regular
polygon – angle with vertex @
polygon center, joining two
polygon vertices (QXR )
Find the measure of
a central angle of
PQRST
A pentagon is a regular polygon with 5 sides.
Thus, the measure of each central angle of
360
pentagon PQRST is
or 72.
5
In the figure, regular hexagon ABCDEF is inscribed in
Find the measure of a central angle.
A. mDGH = 45°
B. mDGC = 60°
C. mCGD = 72°
D. mGHD = 90°
A.
B.
C.
D.
A
B
C
D
Formula
Use the Formula for the Area of a Regular
Polygon
Find the area of the
regular hexagon. Round
to the nearest tenth.
Step 1 Find the measure of a central
angle.
A regular hexagon has 6 congruent
central angles, so
Step 2 Find the apothem.
Apothem PS is the height of isosceles
ΔQPR. It bisects QPR, so mSPR = 30.
It also bisects QR, so SR = 2.5 meters.
ΔPSR is a 30°-60°-90° triangle with a
shorter leg that measures 2.5 meters, so
30°
2.5
≈ 65.0 m2
Practice
What is the area of a regular hexagon with
side length of 8 centimeters? Round to the
nearest tenth if necessary.
A. 48 cm2
B. 144 cm2
C. 166.3 cm2
D. 182.4 cm2
A.
B.
C.
D.
A
B
C
D
Practice
What is the area of a square with an
apothem length of 14 inches? Round to
the nearest tenth if necessary.
A. 784 in2
B. 676 in2
C. 400 in2
D. 196 in2
A.
B.
C.
D.
A
B
C
D
Practice
Find the area of a regular triangle with
a side length of 18.6 meters.
A. 346 m2
B. 299.6 m2
C. 173 m2
D. 149.8 m2
A.
B.
C.
D.
A
B
C
D
Next Application…
• Area of an oblique triangle
– Given two sides of any triangle and the measure
of an angle between them
– Use trigonometry to find its surface area
• Recall previous formula for the area of a
triangle: A = ½ bh
We will use an obtuse triangle
B
c
A
h
a
D
b
• Label sides a, b, and c, opposite their
corresponding angles
• Draw a height, h, inside
C
B
Next…
c
h
A
a
C
D
b
• In order to use A = ½ bh, we need b and h, but all
we know are a, b, and the measure of angle C (for
example) we need “h”!
B
• Look at triangle BDC inside:
– How can we write a trig ratio
using sides h and a?
– We can use this to
sin C 
solve for “h”!
h
h
a
D
a
C
So Far we have…
h
sin C 
a
• Solve this for “h”: h = a sin C
• Now we have the info we need to use A = 1/2bh!
• A = ½ bh
• A = ½ a b sin C
substitute “a sin C” for “h”
IN CONCLUSION
• The area of an oblique triangle is one-half
the product of the lengths of two sides,
times the sine of their included angle!
• For any triangle, ABC
Area = ½ bc sinA = ½ ab sinC = ½ ac sinB
Practice
• Find the area of a triangular lot having two
sides of lengths 90m and 52m and an
included angle of 102°.
• Draw it:
52
102
90
• Area = ½ (90)(52) sin 102
≈ 2288.87 m2
Practice
• Find the area of a triangle with sides 6 and
10 and an included angle of 110° Round to
the nearest hundredth.
• Area = 28.19
Practice
• Find the area of a triangle with side lengths
92 and 30 with an included angle 130°.
• Area = 1057.14