. GEOMETRY. 223 
. ABCDEE : pol. ab cdef:: cir. ABCDEF : Q. ns A and B; and if these are expressed by num- 
one the third pa of ‘its ortion is greater than pares will be found by taking the square root of their 
“the first, therefore the fourth is greater than the second; product ; that is, a=4/A x B.* 
Regular 
Polygons. 
—_—— 
ayy des greater than any polygon inscribed in the 
eabcdef. 
. Because the areas of similar polygons described about 
a circle are also to each other as the squares of the radii, 
‘it may be shewn in the same way, that the space Q is 
less any polygon described about the circle abcde’; 
therefore Q must be exactly equal to the circle abcdef 
(2 Cor. 6. } and the circle ABC &c. is to the circle abc 
&c. as OB? to ob. 
« Cor. 1. The perimeters of circles are to each other as 
their radii. Let P be half the circumference of the cir- 
cle ABCDEFP, and R its radius ; also let p and r be half 
‘the circumference and the radius of the other circle ; 
_ then the areas of the circles are equal to the rectangles 
Px R and pxr, (7.); and it has been shewn, that 
Px R:pxr:: R*: r?; therefore, by alternation, P x R: 
v 
° 
t 
R?::pxr:r*; but Px R: R*:: P: R, and pxr:r © 
:ip:r; therefore P: R::p:r; and by alternation, 
‘gp ee 
‘ Con. 2. A circle described with the hypothenuse of 
a right angled triangle as a radius, is equal to two cir- 
cles described with the other two sides as radii, (Fig. 
135.) Let the sides of the triangle be a, 4, and its h: 
thenuse /; and let the circles ibed with these lines 
as radii be A, B, and H; and because 
A:H::a?: hk, 
and B: H:: 67: h?; : 
Therefore A+B: H:: a24-b?; A? (10, 3.) but a*-+-o°= 
_ h*; therefore A4+B=H. - : 
Prop. IX. Pros. 
Having given the surfaces of a regular polygon inscri- 
- bed ina circle, and of a similar polygon described about 
it, to find the surfaces of the inscribed and circumscri- 
|. bed polygons of double the number of sides. 
Let AB be the side of the inscribed polygon, EF pa- 
allel to AB, that of the similar circumscribed polygon, 
and C the centre of the circle. Ifthe chord AM, and 
the tangents AP, BQ be drawn, the chord AM shall be 
the side of an inscribed polygon of double the number 
_ of sides, and PQ=2 PM, side of a similar circum- 
scribed polygon. This being supposed, as the same con- 
struction may be made in all the different angles equal 
to ACM, and as the triangles containedin ACM have to 
_ each other the ratios of the whole polygons, it will be 
sufficient to consider these only... 
_ Let A be the surface of the inscribed polygon, of 
which AB is a side, B the surface of the similar circum- 
scribed polygon, a the surface of the polygon, of which 
AM is a side, and b the surface of the circumscribed po- 
lygon. A and B are supposed known, and it is required 
to determine a and 4. 
L triangles ACD, ACM, are to each other as 
their bases CD, CM, (Cor. 6. 4.) besides they are to 
each other as the polygons A and a, of which they are 
_ like parts; therefore A:a::CD:CM. The triangles 
nCAM,.C are to each other as their bases CA, CE, 
and also as the polygons a and B ; therefore a: B:: 
_ CA:CE. But because of the similar triangles CDA, 
E, we have CD: CM:: CA: CE, (18. 4.) there- 
pas 0? He 1 Bs BO oe Poles 4, one of those . 
we seek, is a mean proportional 
b Was f 
isto be taken. See ALGEBRA, Att, 137. 
een the two known 
2. Because of the common altitude CM, the. triangle 
CPM is to the triangle CPE as PM to PE; but because 
the angle ECM is bisected by CP, PM: PE :; CM : CE 
(17. 4.):: CD: CA:: A: 4, therefore CPM: CPE :: 
A: a, and by inversion and composition, CPM4CPE, 
or CME: CPM: : A+-a: A, and taking the doubles of 
the uents, CME: 2CPM:: A+a: 2A; now 
CME and 2 CPM, or CMPA, are like parts of the poly-~ 
ns B and b, therefore A4+a:2A:: B:b; hence bis 
own, because it is a fourth proportional to the three 
known quantities A-+a,2 A and B, and 6= 2AXB 
Aya" 
Therefore, by means of the polygons A and B, it is easy 
to find the polygons a and 6, which have double the 
number of sides. 
Prop. X. Pros. 
To find nearly the ratio of the diameter of a circle ts 
its circumference. 
The most obvious method, although not the best, is 
to express the diameter by a number, and compute the 
areas of two polygons of the same number of sides, one 
inscribed in the circle, and the other described about it. 
The area of the circle itself will be some quantity be- 
tween these two. If the number of sides be consider- 
able, either, or any quantity between them, will be near- 
ly equal to the area of the circle, And since the area 
is equal to the rectangle contained by the radius and 
half the circumference, if the approximate value of the 
area be divided by the radius, twice the quotient will 
be an approximate value of the circumference. 
Let us suppose the radius to be unity, then the dia- 
meter will be 2, and the side of a square described about 
the circle, will be expressed by the number 4 ; and as the 
inscribed square A, is evidently half the circumscribed 
uare, the area of the inscribed square will be 2, Em~ 
ploying now the formule found in last Reanorttien, viz. 
a=/AXB, b=">% 
on : ~AXa’ 
and making A=2, and B=4, we find a, the area of a 
regular polygon of eight sides inscribed in the circle, 
= 1/2 X 4=1/8=2.8284271, and b the circumscribed 
polygon =3.3137085. 
Putting now A=2.8284271, and B=3.3137085, we 
may hence find a and 4 the imscribed and cireumscri- 
bed Ayers of 16 sides, and so on as in this Table. 
Number of Sides. §_ Inser. Polygon. Circum. Pol. 
4 2.0000000 4:.0000000 
8 2.8284271 3.3137085 
16 3.0614674 3.1825979 
32 3.1214451 3.1517249 
64 3.1365485 . $.1441184 - 
128 $.1403311 3.1422236 
256 3.1412772 3.1417504 
512". 3.1415138 . $.1416321 
1024 $.1415729 3.1416025 
2048 3.1415877 3.1445951 
4096 8.1415914 3.1415938 
8192 3.1415923 3.1415928 
16384 3.1415925 3.1415927 
32768 8.1415926 3.1415926 
* The radical sign 4/ placed ‘over the symbol that expresses any quantity, indicates that the square root of the expression which it affects 
