This document outlines Leland Bartlett's proof that the area of a circle can be calculated by summing the areas of many triangles circumscribing the circle. It begins with area formulas and trigonometry lessons. Then, it shows the calculations for 4, 8, 16, 32, 1024, and 32768 triangles approximating a circle with diameter 6 and radius 3. The sums converge on the expected area and circumference values with very small differences. In the end, it is shown that the height of the triangles approximates the radius, proving the area formula. Leland Bartlett provides training in various topics like Microsoft Office and databases as a professional consultant and trainer.

1. RIP-ROAR TRAINING & CONSULTING
By
Leland Bartlett

2. TODAY’S PROOF IS HOW:
= 𝜋𝑟21
2
𝑏ℎ

3. HOW WE ARE GOING TO PROCEED
• Reminder of Area Formulas
• How the idea came to me
• It’s all in the triangles
• Quick Trigonometry Lesson
• Parameters to work towards
• Area of
• 4 triangles
• 8 triangles
• 16 triangles
• 32 triangles
• 1024
• Summary
• Agreement
Proof
• Ta-dah
• Additional Info
• My Info

4. REMEMBER THESE FORMULAS?
Square RectangleS1
S2
S1
S2
Triangle
h
b
Circle
A=S1 * S2A=S1 * S2
A =
𝟏
𝟐
𝒃𝒉
A = * r2

5. octagon
IT’S ALL IN THE ANGLES, THE TRI-
ANGLES
dodecagonsquare

6. IT’S ALL IN THE ANGLES, THE TRI-ANGLES
1
2
3
4
5
6
7
8
1 2
3
4
5
6
7
8
9
10
11
12
1
2
3
4

7. TRIANGLES AND SOME TRIGONOMETRY
In order to calculate the triangles, a couple trigonometry functions. Sin( ) and Cos( )
A
Sin(A) = the side opposite of the angle over the hypotenuses. The hypotenuses is the longest side of a
triangle. So we have to divide the triangles in ½. This is also needed to be done in order to calculate the
height of the triangle. Additionally, when doing the calculation for the area of the triangle, the base needs to
be multiplied by 2 in order to get the area of the triangle correct. The cos(A) would use the adjacent side, the
height over the hypotenuses. Also, note, the angle A needs to be divided in half as well. So, an angle of 90
deg, the angle used with sin and cos would be 45 deg.
(h)eight
Base
a
In the triangle There is angle A the whole angle then the angle a being used to
calculate the height and base. In the example, A = 45 deg and then angle a =
22.5 deg. The sides of the angle are equal to the radius from our circle or 3.
Sin(a) = Opp/Hyp  Sin(22.5) = b/3  b = 3(sin(22.5))
Cos(a) = Adj/Hyp  Cos(22.5) = h/3  h = 3(cos(22.5))
b
3
45o
22.o
90o

8. WHAT WE ARE SHOOTING FOR. . .
D = 6
r = 3
Circumference = 𝜋* 6 => 18.8495559215 sq in
Area = 𝜋 * r2 => 𝜋 * 9 => 28.274333882308 sq in
Diameter = 6 inches
Radius = 3 inches

9. 4 -- TRIANGLES
Base:
Sin(45) = b/3  Base = 4.242640687
Height:
COS(45) = h/3  h = 2.121320344
Area:
4*[½ (4.242640687) * (2.121320344)] = 18
Difference: 10.27
Circumference:
4* 4.242640687 = 16.97056275
Difference = 1.878993173
900
450
b
3
h

10. 8 -- TRIANGLES
Base:
Sin(22.5) = b/3  Base = 2.296100594
Height:
Cos(22.5) = h/3  h = 2.771638598
Area:
8*[½ (2.296100594 ) * (2.771638598 )] = 25.45584
Difference: 2.81848976
Circumference:
8* 2.296100594 = 18.36880475
Difference: 0.480751168
b
3 h

11. 16 -- TRIANGLES
Base:
Sin(11.25) = b/3  Base = 1.170541932
Height:
Cos(11.25) = h/3  h = 2.942355841
Area:
16*[½ (1.170541932 ) * (2.942355841 )] = 27.55321
Difference: 0.721126752
Circumference:
16* 1.170541932 = 18.72867091
Difference: 0.120885008
2
3
4
5
67
8
1
9
10
11
12
13
14
15
16
22.5
11.25
Base
b
3
h

12. 32-- TRIANGLES
Base:
Sin(5.625) = b/3  Base = 0.588102842
Height:
Cos(5.625) = h/3  h = 2.98555418
Area:
32*[½ (0.588102842) * (2.98555418)] = 28.09301
Difference: 0.181327512
Circumference:
32* 0.588102842 = 18.81929094
Difference: 0.030264978
2 3
4
5
6
7
8
1
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32 11.25
5.625
Base
b
3
h

13. 1024- TRIANGLES Base:
Sin(0.17578125) = b/3  b = 0.018407741
Height:
Cos(0.17578125) = h/3  h = 2.999985881
Area:
1024*[½ (0.018407741) * (2.999985881)] = 28.27416
Difference: 0.0001774190043129
Circumference:
1024* 0.018407741 = 18.84952635
Difference: 0.00002956987579949550
2 3
4
5
6
7
8
1
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32

14. TABLE OF VALUES
Area 28.27433388
Circumference 18.84955592
A-Diff Base Height Area Circumferance C - Diff # Triangles Agle 1/2 Angle
10.2743338823081000 4.242640687 2.121320344 18 16.97056275 1.878993173 4 90 45
2.8184897595924300 2.296100594 2.771638598 25.45584 18.36880475 0.480751168 8 45 22.5
0.7211267520216750 1.170541932 2.942355841 27.55321 18.72867091 0.120885008 16 22.5 11.25
0.1813275119856680 0.588102842 2.98555418 28.09301 18.81929094 0.030264978 32 11.25 5.625
0.0001774190043129 0.018407741 2.999985881 28.27416 18.84952635 0.000029570 1024 0.3515625 0.17578125

15. IF WE COULD SUM UP ALL THE TRIANGLES. . .
D = 6
r = 3
Circumference = 𝜋* 6 => 18.8495559215
Calculated 1024 = 18.84955589
Area = 𝜋 * r2 => 𝜋 * 9 => 28.274333882308
Calculated: 28.27415646
Radius = 3
Height = 2.999985881
Diff of = .000014119

16. WOULD YOU AGREE. . .
D = 6
r = 3
Circumference = 𝜋* 6 => 18.8495559215
Sum of Bases<1024> = 18.84955589
Area = 𝜋 * r2 => 𝜋 * 9 => 28.274333882308
Sum of the area of Triangles: 28.27415646
Radius = 3
Height = 2.999985881
• The sum of the base of the triangles is a good approximation
of the circumference of the circle?
• The height of the triangle is a pretty good approximation of
the radius of the circle?

17. SO, IF YOU AGREE – THIS IS HOW IT WORKS?
½ 2r h = r2
Circumference = 𝜋* 6 => 18.8495559215
Sum of B <1024> = 18.84955589
C = D *
D = 2r
½ b h = r2
½ C h = r2
½ D h = r2

18. HOW DOES IT WORK – FINAL STEPS?
r2 = r2
h = 2.999985881
h ≈ r
D = 6
r = 3
2𝑟 h
2
= r2
r h = r2
r r = r2

19. TA-DAH!!!
2 =

20. ADDITIONAL INFO . . .
Area 28.27433388
Circumference 18.84955592
A-Diff Base Height Area Circumferance C - Diff # Triangles Agle 1/2 Angle
10.2743338823081000 4.242640687 2.121320344 18 16.97056275 1.878993173 4 90 45
2.8184897595924300 2.296100594 2.771638598 25.45584412 18.36880475 0.480751168 8 45 22.5
0.7211267520216750 1.170541932 2.942355841 27.55320713 18.72867091 0.120885008 16 22.5 11.25
0.1813275119856680 0.588102842 2.98555418 28.09300637 18.81929094 0.030264978 32 11.25 5.625
0.0001774190043129 0.018407741 2.999985881 28.27415646 18.84952635 0.000029570 1024 0.3515625 0.17578125
0.0000001732610713 0.000575243 2.999999986 28.27433371 18.84955589 2.88768E-08 32768 0.010986328 0.00549316
28.2743338823081000 ########### 2.999999997 0 0 18.84955592 65536 0.005493164 0.00274658

21. LELAND BARTLETT SUMMARY
• I am a professional consultant, consultant trainer and trainer. I've been doing both
soft skills and hard skills training for over 30 years.
• I provide the following training:
• instructor-led-training
• individual training
• eLearning
• On-the-job training
• Blended learning
• Self-paced learning
Microsoft Office
• Excel Advanced
• Access Advanced
• Word Advanced
• PowerPoint Advanced
• OneNote Advanced
• Outlook Advanced
• Vision
• Project
Others
• MSSQL
• Adobe Captivate
• Camtasia . . .more

22. QUESTIONS?
• Contact Leland Bartlett by:
• Phone: 682-667-1243
• email: leland.bartlett@outlook.com
• Leave me a comment in the comment section of this video.