\fdO'^ - Old Forge Coal Mines

148 GEOMETRY AND TRIGONOMETRY.
3
(293) The arc intercepted equals - of 4, or 3 quadrants.
As the inscribed angle is measured by one-half the intercepted
arc, we have — = 1— quadrants as the size of the angle.
2 7
(294) Four right angles -^ y = 4 X ^, or 14 equal
sectors.
(295) Since 24 inches equals the perimeter, we have — ,
o
or
^
3 inches, as the length of each side or chord.
Then, 2 X y ( ^ ) + 3.62* = 7.84 inches diameter. .
(296) Given, or 5.25 inches. A O
^(7=^=1^,
zndiAP= 13 inches. (See Fig. 5.)
The required distance between the arcs D D' is equal
to OA-\-AP-0 P. In the
^^ iLux ^„ ,
24
right-angled triangle yi (7 (9, we
have
Fig- 5. inches.
or (9 (7 = -/ 169 - 27. 5625 = 11. 9
Likewise, CP^ Va P' - A C = 11.9. OP=OC+CP
= 11.9 + 11.9 = 23.8 inches. O A -\- A P= 13 + 13 = 26
inches. 26 — 23.8 = 2.2 inches. Ans.
(297) Given ^ /*= 13 inches, OA=S inches, and
A C= 5.25 inches. Fig. 6.
0C= VTW' - AT' = - --""^^
i/S' 5.25"=
6.03 inches.
CP= VA~P'-A~r' = 11.9 inches.
0P= O C + CP= 6.03 + 11.9 =
17.93 inches.
fig. c.
DD'= OA -f AP- 0P= 8 + 13 - 17.93 = 3.07 inches. Ans.
(298) AB=U inches, and A£= 3^ inches, Fig. 7.
CE =. E D is a mean proportional between the segments