Fig. 176. 
Fig. 176. 
236 , 
Prop. VI. 
Of all ns formed with given sides, the mazi- 
sanintios Sich cae bacilekanedrah 
Let ABCDEFG be the polygon inscribed in a circle, 
and abcdefg that which cannot beso inscribed. Draw 
the diameter EM; join AM, MB, and on a= AB 
make a triangle a fo so that am=AM, and mb=MB, 
and join em. Then by Prop. 4. the polygon EFGAM 
is greater than efgam, unless this last can be in- 
scribed in a circle of which em is the diameter, because 
in this case the two polygons would be equal (Prop. 5.) 
For the same reason the polygon EDCBM is greater 
than e d cb m, excepting the case of the latter admitting 
of being inscribed in a semicircle, so as to make them 
equal. Therefore the whole polygon EFGAMBCDE 
is greater than the whole polygon efg amc de, unless 
they are entirely equal, which cannot happen, because 
the one is supposed to admit of being inscribed in a cir- 
cle, but not the other; therefore the inscribed polygon 
is the greater of the two, and taking away the equal 
triangles AMB, a m 6, there remains the polygon 
ABCDEFG, inscribed in a circle, greater than the other 
polygon abcdefg, which does not admit of such 
inscription, 
Scuotium. It may be demonstrated, as in Prop. V. 
that there can be only one circle, and consequently on- 
ly one maximum polygon that satisfies the question ; 
and this. polygon will have the same surface, in whatever 
order the sides follow-each other. . 
THEoR. 
Prop. VII. Tueor. 
The regular polygon is the greatest of all the isope= 
rimetrical polygons, having the same number of sides, 
For by. Theorem 2, the maximum: polygon has.all its 
sides equal; and by the last Theorem, it may be inscribed 
in a circle, Now no other than a regular polygon has 
these two properties. , ; 
Prop. VIII. Lemma. 
If two circles ADH, ABK, touch each other inter- 
nally at A, andia straight line CD) be drawn from the 
centre of the inner circle, to cut the circumferences in 
B and D, the are AD of the outer circle shall be great- 
er than the arc AB of the inner circle. 
In'the circumferénce of the inner circle, take the arc 
BE=BA, join CE, and through E and D, with a ra- 
dius equal to the radius of ‘the outer cirele, deseribe an’ 
are ED; then -the trilateral figures’ ACD, ECD; will 
manifestly be exactly alike, and the arcs EB, ED ‘will’ 
' touch each other at E. And because the concave line 
Fig. 177. 
ADE, (formed by thearcs AD, ED,) and the are ABE, ' 
have their concavities turned the same way, by an axiom ° 
in geometry, the former is greater than the latter ; there-'' 
fore taking their halves, the arc AD is greater than the 
arc AB, 
Prop. IX. Tueor. 
Of two isoperimetrical regular polygons, that which ! 
has the greater number of sides is the greater. 
gons, C their common centre ; also CA and CAD tho 
GEOMETRY. 
’ and a given area, a regular 
' its perimeter a minimum. 
°°" to a given line } 
Let AB, DE be half the exterior sides of the two poly-’ a piven Tine 2, nc 18 ie Aro 
radii of circles inscribed in the polygons, which are per- Appen 
pendicular to their. sides. Daw én, CE, and Tet CE —— 
meet ABin M. Draw BH paralleltoCE. OnCasa 
centre, describe the arc AF, meeting CE in G; and on 
H as a centre, describe the are AK, meeting CF in I. 
Because the polygons have equal perimeters, the + 
lengths of their sides will be reciprocally as their num- 
ber; and because all the angles at the centre of each 
polygon make four right angles, the angles which the 
sides subtend at the centre, will also be reciprocally as 
= number: hence we have DE : AB:: arc AG : are 
AF. ye tus ‘ . 
The triangles CAM, HAB, are manifestly sintilar, al- . 
so the sectors CAG, HAK, hence Ses 
AB: AM:: are AK: arc AG, 
and since DE ; AB :: arc AG: arc AF, m 
therefore, ex. eg. DE: AM : : are AK: are AF ; 
but DE": AM:: CD: CA; "* - 
therefore are AK : arc AF: : CD: CA. ‘ 
Now the are AK ie greats than, the are AI, which 
again is ter than the arc AF, (preceding Prop. 
therefore CD is greater than CA ; 3 tbe he aditie ‘ 
the circle inscribed in the polygon having the lesser 
angle, or greater number of .sides, is greater than the 
radius of the other polygon; but the pol 
isoperimetrical, and id res of each equal 
radii ; therefore the polygon that has the st - 
ber of sides, has hd peenielh area, Pa, oo 
2 
The circle is 
lygon. roy wy 
' Tt has been proved, that’ if a regia ular, and. i 
polygon have equal perimeters, the a “ 
pa 9 area ;. therefore it only remains_to 
circle with a regular polygon of the same pe 
; Tec AY be half the side ‘polygon, C’ 
tre, In the isoperimetiical ¢ 
ACI, and conséquently the are DE= 
polygon P is-to' the circle C as the triang’ 
the sectot ODE ; now the area of the triangle is JAI x. 
IC, and the area of the sector is } DE x EO: 
P:C::3AIx IC: }DE x EO::IC: EO. Draw 
the tangent EG to meet’ OD in G: e triangles ACI, 
GOE being similar, IC : EO:: Al: GE; therefore, 
P:C!: Al, orare DE:GE::2DEXEO: 1GEx, 
EO;; but DE x EO=sector DOE ; and } GEX EO= 
triangle GOE, therefore P:C: : sector DOE: triangle, _ 
GOE ; now the triangle GOE is greater than the sector: ; 
DOE ; therefore the circlé C is greater than the isope-, 
viene ear Tt ool + Goa 
Cor. A circle contains within a given perimeter the , 
gréatest possible area. . e 
Le faite Sees exinads fom 
Of all polygons, having the. same number of sides, ; 
a polygon is that, ib its has. 
Lop tyia 
t% 
a 
Let A be the given’ area of a polygon ygon, and v its peri- 
* meter ; let a similar polygon have fe peimeter equal 
lar figures: are as the squares a the perimeters, we 9 
‘ have A; Xi: 0? hence AXVo&X xo*,, and v= . 
