- Membership
- MAA Press
- Meetings
- Competitions
- Community
- Programs
- Students
- High School Teachers
- Faculty and Departments
- Underrepresented Groups
- MAA Awards
- MAA Grants

- News
- About MAA

Required the Lengths of the Sides of an Equilateral Triangle inscribed in a Circle whose Diameter is 200 perches, with a general Theorem for all such Questions

Solution of the above problem

10.00 ….. 3.142 …… 200

200

1000)628400

Lenght of the periphery 628.400

1/3 of the length of the periphery 209.466

1/3 of 1/3 of the periphery 69.822

2

2/3 of 1/3 of the periphery 139.644

Length of the Sides required 349.110

Note: In Banneker's journal there is a big X crossing out the last five numbers in his calculations and adjacent to each of the 175's in his figure is 349.110

This example deals with the problem of finding the lengths of the sides of an inscribed equilateral triangle in a circle of diameter 200 perches. A perch, a synonym of a rod, is a unit of measurement equal to 16½ feet. There are 320 perches in a mile. A surveyor could find the number of acres in a rectangular piece of land by multiplying the length in perches by the width in perches and dividing the product by 160.

What did Banneker do to solve this problem? First of all he calculated the circumference, which he calls periphery, by multiplying an approximation of \(\pi\) (3.142) by 200. He did this by multiplying 3.142 by 200 to get 628.400 and then dividing that by 1000 to get 628.400 (How often have do we, as teachers, pull our hair out when we see our students multiplying or dividing by powers of 10?) He then finds 1/3 the length (misspelled in his journal) of the circumference and later 2/9 of the circumference. He adds those numbers together to get 349.110, but clearly he knew that the side of the equilateral triangle had to be less than the diameter. He crossed out his work and then labeled the sides of triangle 175 which is approximately half of 349.110.

Using the properties of a 30º-60º-90º triangle and a circle of radius 100, one can see that the length of the side of the equilateral triangle is 173.205 which shows that Banneker's solution is within 1% of the actual one. The side of the equilateral triangle is the square root of 3 times the radius of the circumscribed circle.

How did Banneker figure this out without using the properties of a 30º-60º-90º triangle or trigonometry? Banneker calculated 5/9 of the circumference to get 349.110 and by taking half of that he essentially computed 5/18 of the circumference. Since (5/18)(2\(\pi\))100 is approximately equal to 174.533, Banneker's method is quite good. It would have been even better if he had taken exactly half of 349.110 to get 174.555. Banneker's method essentially uses (5/9)\(\pi\) to approximate the square root of 3. Perhaps this approximation was a rule of thumb that surveyors used in the 18th century.

John F. Mahoney (Benjamin Banneker Academic High School), "Benjamin Banneker's Inscribed Equilateral Triangle - Transcription of Banneker's Problem," *Convergence* (July 2010)