Regular cause of the similar tri 
Polygons. Cy §¢, we have AB: ab:: 
Fig. 133, 
Fig. 154, 
222 
les OBA, Oba, and OBC, 
O:bo: Lien (18. 4.) 
but AB = BC, therefore ab= dc: For a like reason 
bc=cd, &c. therefore the circumscribmg polygon is 
regular and similar to the inscribed polygon. 
Prop. V. Tueor. 
Regular polygons of the same number of sides in- 
scribed in circles are similar, and are to one another as 
the squares of the radii of the circles. 
Let ABCDEF and adcdef be equilateral hexa- 
gons inscribed in circles ; these will also be equiangu- 
lar, (schol. 2.) and consequently regular ; and because 
all the angles of each polygon are together equal to 
eight right angles (25.1.), the angle A is 4 of eight 
right angles ; and as the same is also true of a, there- 
fore the angles A and a are equal. In like manner it 
appears that B = 6, C= c, &e. and because the sides 
of each figure are all equal, we have FA: AB:: fa: 
ab, &c. therefore the polygons are similar (2 def. = 
Draw BO, CO, bo, co, to the centres: The triangles 
BOC, boc are similar, for the angles at O and o are 
equal, each being + of four right angles, and CO : OB 
:: co:06; therefore COB: cob :: CO® :co*; but 
the triangles COB, co b are manifestly like parts of the 
whole polygons, therefore (1.3.) the polygons are to 
each other as the squares of CO, co, the radii of the 
circles. ! 
Prop. VI. 
A circle being given, two similar polygons may be 
found, the one described about the circle ; and the other 
inscribed in it, which shall differ from one another by 
less than any given space. 
THEoR. 
Let Q be the side of a square, equal to the given 
space ; bisect AC, a fourth part of che circumference 
of the.circle, and again bisect the half of this fourth, and 
roceed in this manner, always bisecting one of the arcs 
pound by the former bisection, until an arc is found of 
which dhe chord AB is less than Q: As this arc will 
be an exact part.of the circumference, if we place chords 
AB, BC, CD, &c. in it each equal to AB, the last will 
terminate at A, and there will be formed a regular po- 
Tygon ABCDE &c. in the circle: 
ext describe about the circle a regular polygon 
abede, &c. of the same number of sides as the inscri- 
bed polygon, and having its sides parallel to those of 
the latter (4.) The difference of these two shall be less 
than the square of the line Q. 
For, draw lines from a and 6 to O the centre, these 
will pass through A and B, as was shewn in the demon- 
stration of Prop. 4, also a line drawn from O to K, the 
poe in which a é touches the circle, will bisect AB in 
, and be perpendicular to it (6. 2.) : complete the dia- 
meter AOE, and join EB. Put P for the circumscri- 
bing polygon, and p for the inscribed polygon ; then, 
‘because the triangles aod, AOB are manifestly like 
parts of P and p, we have P:p::ao06: AOB; (1. 3. 
but these triangles being similar, aob: AOB: : oat: OA%, 
or OK”, (25. 4.); and again, because the triangles O aK, 
EAB are manifestly similar, we have Oat: OK?:: 
EA?; EB*(18. and 23. 4.); therefore P :p:: EA?: EB, 
and by conversion, P: —p::; BA*: EA? —EB* or 
AB’. Now, as a square described about a circle will 
manifestly include within it a polygon of 8, also of 16, 
: 4 
GEOMETRY. 
and of 32 sides, &c. the polygon F will be less than the 
uare of EA ; therefore P — p, the difference of the 
circumscribing and inscribed polygons, will be less than “= 
the square of AB, that is, by construction, less than the 
given space Q. (cor. 3. 3.) ; 
Cor. 1. Because the polygons differ from each other 
more than either differs from the circle, we may infer, 
that a polygon may be described about a circle, and 
also a polygon may be inscribed in a circle, either of 
which shall differ from the circle by less than any given 
space. 
Cor. 2: A space that is less than any pol: what- 
ever, dcersilied aboutia circle, and also re ietna any 
polygon whatever, inscribed in the same circle, must be 
equal to that circle. ; 
Axiom. 
If HBI be any are of a circle, and b H, 61 tangerits Fig. 
at its extremities, the sum of the tangents 5 H, 61 is 
greater than the arc HBI. 
‘Cor. The circumference of a circle is less than the 
perimeter of its circumscribing polygon. 
Prop. VII. Treror. © 
The area of any circle is equal toa rectangle contain 
ed by half the perimeter, and the radius, . 
~- Let ABCD, &c. be a regular polygon inscribed in Fi 
the circle, and a 6c d &c. a similar polygon, descri- 
bed about it, and having the sides ab, bc, &c. parallel 
to AB, BC, &c. eee K perpendicular to AB, ab; 
and OAa, OB4, through the points A, a, and B, & 
Let P be the perimeter of the polygon abed &c. p 
that of the polygon ABCD &c. and Q that of the cir- 
cle. ) ' ; 
The triangle a O b is equal to $a 6x OK, (6. 4.) and 
multiplying these equals by m, the number of sides of © 
the polygon, we have x x trian. a O b=nx 4abx OK; 
but » x trian. a O 6 is manifestly the area of the poly- 
on abcd &c. and nx4ab, or} xn Xa bis equal to — 
fe: therefore the area of the polygon abed &e.is — 
3P x OK; and similarly, the area of the polygon ABCD > 
&e. is 3. px OL. 2: 2 
Because 104 P, therefore 4Q x KO—3 PxKO, 
but } Px KO is the area of polygon abed &c. 
therefore } Q x KO is less than the area of the polygon 
abcd &c. Again, because ,}Q>% p, therefore 70 x _ 
OK>ip x OK; but OK>OI, and 3p x OK> 
pxOI: much more then is }QxXOK>ipxOI, © 
but 4p x O1 is the areaof thepolygon ABCD, &e. there- 
fore } Q x OK is greater than the area of the polygon 
ABCD, &e. Thus it appears, that the rectangle 
‘ 
tained by 4 Q the perimeter of the circle, : 
radius, is greater than any polygon inscribed in the cir- iT 
cle, and less than any polygon described about the’ cir- _ 
cle ; therefore it must be exactly equal to the area of the — 
circle, (Cor. 6.) ‘ 
Prop. VII. Turor.. mt , 
The areas of circles are to each other as the quares es 
°F their radii. ; nee ri 
Let ABCDEF, and abcde f be regular, and similar } 
polygons inscribed in the circles; and as OB? : 0 b%, so. 
et the circle ABCDEF be toa fourth’ ional Q ; 
then because OB? : o 6°: : pol. ABCDEF : pol. abcdef, 
(5.) it follows, that my 
