Today we looked at how to find the area of a regular polygon given the radius. Remember: if we are given a side length and need to find the area, we will use tangent to find the height. If we are given a radius and need to find the area, we will use sine AND cosine to find the height and base. Here are the notes and the assignment (5 problems are attached). It says assignment #16, but it is really #57.
I also announced my change in office hours on Thursdays. Due to unforeseen scheduling, Thursday’s office hours have been moved from after school to after lunch.
And finally, we have Pi Day information. Monday, March 19 we will celebrate Pi. Bring anything round!
I’m currently working on Moodle benchmark practice for Benchmark 38 on the Area of a Regular Polygon.
Coming up:
- BM 34-38 on Monday, 3/12
- Pi Day on 3/14
Posting for Point Question:
This time you give the question (and solution). Create your own example for finding the area of a regular polygon. Include the number of sides, the angle found, the area of the triangle, and the final area. Use either a side length or radius. And one of these will be used for the benchmark Monday.
March 8, 2007 at 7:29 pm
Find the area of a regular pentagon with side lengths of 5 ft.
(5-2)180/5=108/2=54
2.5tan(54)=3.44
A=1/2(5)(3.44)=8.6
8.6*5=43.01
March 8, 2007 at 8:32 pm
Find the area of a regular decagon with a radius of 6 cm.
Angle:
(11-2)180=1620/11= 147.27/2=73.63
sin73.63=h/6
*6 *6
6sin73.63=h
Base:
cos73.63=a/6
*6 *6
6cos73.63=a
A=1/2(12cos73.63)(6sin73.63)=
6.92 (11)= 76.16 cm2
Mr. F says: This is actually a perfect example of a HENdecagon (11 sides). Good job Sydney!
March 8, 2007 at 9:40 pm
*(12-2)180=1800/12=150/2=75
*Tan75=h/8
*8Tan75=h
*.5(16)(8tan75)=238.85
*Polygon Area=(238.85)(12)=2866.22
March 9, 2007 at 10:45 am
1. (4-2)(180)=360/4=90/2=45
2. Tan45=h/7
3. 7Tan45=h
4. 1/2(14)(7Tan45)=24.5
5. A= (24.5)(4)= 98 sq. units.
March 9, 2007 at 1:29 pm
Make sure you start with a question!
March 9, 2007 at 4:12 pm
find the area of heptagon if the radius is 13cm long
(7-2)180=900
900/7=128.57
128.57/2=64.29—-angle
sine(64.29)=h/13
13sine(64.29)=h
cosine(64.29)=h/13
13cosine(64.29)=h
1/2(26cosine(64.29)(13sine(64.29)=17.35
17.35 x 7 =
March 9, 2007 at 4:13 pm
forgot to leave answer = 121.45
March 9, 2007 at 4:56 pm
Find the area of a regular enneadecagon with a radius of 8ft.
Solution:
ANGLE: (19-2)*180 = 3060/19 = 161.05/2 = 80.53 degrees
HEIGHT: sin80.53 = h/8 –> 8sin80.53 = h
BASE: cos80.53 = a/8 –> 8cos80.53 = a
AREA: 1/2 (16cos80.53) (8sin80.53) = 10.39(19) = 197.34 sq. ft.
March 9, 2007 at 10:13 pm
find area of a octagon with radius of 15in
octagon=8 radius=15in
(8-2)180=1080/8=135/2=67.5
H: sin67.5=h/15 > 15sin67.5=h
B: cos67.5=h/ (15)2> 30cos67.5=B
area=1/2 (15sin67.5)(30cos67.5)8=636.4in squared
March 9, 2007 at 10:19 pm
Find the area of a hexadecagon with a radius of 5 in.
Angle: (16-2)180=2520/16=157.5/2=78.75 degrees
Height: Sin78.75=h/5
5Sin78.75=h
Base: Cos78.75=a/5
5Cos78.75=a *Times base equation by two to get (10Cos78.75). That’s what you plug in for the next equation.
Area: 1/2(10Cos78.75)(5Sin78.75)=4.78
Polygon Area: 4.78(16)=76.54 sq. in.
March 10, 2007 at 6:07 pm
Find the area of the icosihenagon (21-gon) with a radius of 9 inches.
(21-2) 180=3240/21=162.85/2=81.42
Height: sin81.42=h/9
*9 *9
9sin81.42=h
Base: cos81.42=a/9
*9 *9
9cos81.42=a
9(2)=18
18cos81.42=b
A =1/2(18cos81.42)(9sin81.42)= 11.94 (21)
A =250.93 sq. in.
March 10, 2007 at 8:29 pm
Find the area of a regular hexagon with side lengths 10cm.
(n-2)180 = (6-2)180 4 x 180 = 720/6 = 120/2 = 60
h = tan 60 = h/5 5 tan 60 = h
a= 1/2 (10) (5 tan 60) = 43.3 x 6 = 259.8cm sq
March 10, 2007 at 9:36 pm
Find the area of a regular octagon with a radius of 20ft.
Angle: (8-2)180= 6×180=1080/8=135/2=67.5°
Height: 20sin67.5°
Base: 20cos67.5°=a (2xa=Base)
40cos67.5°=b
Area: 1/2(40cos67.5°)(20sin67.5°)=141.42×8=1131.37ft.²
March 12, 2007 at 11:56 am
1. (3-2)(180)=180/3=60/2=30
2. Tan30=h/7
3. 7Tan30=h
4. 1/2(14)(7Tan30)=49tan30
5. A= (49tan30)(4)= 196tan30 sq. units.
Mr.F.I do not have a calculator so i had to leave my answere like this…..SORRY
March 12, 2007 at 4:58 pm
Find the area of a regular heptagon if the radius is 20 cm. long.
(7-2)180
5×180=900/7=128.57/2=64.29
sin64.29=h/20
cos64.29=b/20
A=1/2(40cos64.29)(20sin64.29)=156,36×7=
1094.56sq.cm.
March 12, 2007 at 7:11 pm
Find the area of a nonagon with side lengths of 16in.
(16-2)180=2520/16=157.5=78.75
tan78.75=h/16___16tan78.75=h
1/2(9)(16tan78.75)(16)=5791.49in sq.
March 12, 2007 at 7:37 pm
Find the area of a Triacanta with a radius of 12 cm.
(30-2)180=5040/30=168/2=84 degrees
Height: Sin84=h/12
12Sin84=h
Bace: Cos84=a/12
12Cos84=a
24Cos84=b
A=1/2(24Cos84)(12Sin84)(30)=449.09cm. Sq.
March 12, 2007 at 8:02 pm
Find the area of a regular octagon with a radius of 6ft
(8-2)180=1080/8
135/2=67.5
h:sin67.5=h/6
=6sin67.5
cos67.5=a/6
=6cos67.5
A=1/2(12cos67.5)(6sin67.5)=12.73
12.73×8
=101.82sq ft
March 12, 2007 at 8:39 pm
octagon radius 15in
^is 8 and the radius is 15
use n-2 x 180 over n
6 x 180=1080
1080 divided by 8=135 divided by 2= 67.5
Hight : sin67.5=h divide by 15 15sin67.5=h
Base : cos67.5=H divided by (15)2 30cos67.5= B
area :1/2 (15sin67.5)(30cos67.5)8 {636.4in} BAm done
im not sure if i did it right but i still get points for trying….i hope
March 12, 2007 at 8:40 pm
can i do 2 for more points….
March 12, 2007 at 9:23 pm
find the area of hexagon with the radius of 20ft.
1. (6-2)180=720
2. 720/6=120, 120/2=60
3. height: sin60=h/20
=20sin60
4. base: cos60=h/20
=20cos60
5. A=1/2(20sin60)(20cos60)(6)=1500 square ft.
March 13, 2007 at 3:00 pm
What is the area of a regular heptagon with a radius of 20?
(7-2)180=
900/7=
128.57/2=
64.29
sin 64.29=h/20
20sin 64.29=h
cos 64.29=b/20
40cos 64.29=b
A=1/2(40cos64.29)(20sin64.29)=
156,36×7=
1094.56 sq. cm.
March 13, 2007 at 3:35 pm
Find the area of a nonagon with a side length of 12cm.
1- 9-2(180)=1260
2- 1260/9=140
3- 140/2=70
4- h=tan70=6
5- 6tan70=h
6- A=1/2(6tan70)(12)=168.91
7- 168.91(9)=1520.19cmsq.
March 14, 2007 at 9:24 am
find the area of a decagon with a side length of 6.
1.10-2(180)=1440
2.1440/10= 144/2= 72
3.h= (3)tan72
4.A=1/2(6)(3tan72))(10)
5.277.0
March 14, 2007 at 1:27 pm
Find the area of a hexagon with a side length of 10
1. 22-2*180 = 3600
2. 3600/22=163.63/2=81.81
3.h=(5)tan81.81
4.a=2/(10)(5tan81.81)(22)
5. 764.29
Mr. F adds: A hexagon only has 6 sides. This would be the area of a 22-gon. Good example!
March 16, 2007 at 6:30 am
find the area of a hexagon with a side length of 12 cm?
6-2*180=720/6=120/2=60
H=(6)tan60
a=1/2(12)(5tan60)(6)
I didn’t have a calculator on me. That had tan on it