14 SOUTHERN CALIFORNIA ACADEMY OF SCTENGES 
7. Produce BD to S making DS=O m. 
8. Draw ST parallel to DE and equal to Bm. 
9. Draw BT, intersecting BT at Q. 
10. Produce FG to I making GI equal to QT. 
11. Biseet FI at V and from the center O locate the points 
abe, and: ds makanlo©)a——0))—O <= @/d—_ hae 
12. Through a, b, ec, and d draw the:sides of the square W X 
NOZ: 
DEMONSTRATION. 
mince BC is the are of 60°) Hm, the Sine of 60° equals 
V3 
9) 
Since Gm equals Fm and Gm F being a right angle, FG 
a cy 
radius < 
equals radius < 
9 
Since DS is 1-5 of BS and the triangles S B T and DB Q are 
similar, Q T equals 1-5 of BT. 
Since DE equals radius B E=radius & 5 and ST equals 
1% of BE. 
Hence BT equals % V/30 and Q T=1-10V/ 30, 
5/5 \/30 
Therefore FI equals radius , = XO 
2 NE 
WO 
—— = 1.22474487 
2 
30 
== —_ 94772255 
10 
Sum == 1.77246742 
V/ 3.1415926 =—— LEAD 38o 
Difference — 00001357 
Since the area of a circle equals the product of the square of 
its radius and 3.1415926-+ or P, it follows that when the radius 
is 1 the area—P, and the side of the required squareVP. If 
the radius be any other number greater or less than 1, denoted 
by R, the side of the square is Rv P. 
The negligible difference, .00001357 for the length of the side 
of the square for the length for a 122 feet circle is less than 
1-100 of an inch. 
