PROPOSITION XVIII.

In the regular hexagon, whose diameter is three, the area equals three-fourths of the circumference.

In this diagram, the diameter being three, the radius, CD, equals one and a-half. And DH being drawn equal to the breadth of the line of circumference, and therefore equal to one, is consequently equal to two-thirds of CD. Therefore let CD be divided into three equal parts, and DH into two equal parts, and let the small rectangles be completed, and it is manifest that the four rectangles contained in the rectangle ABFG are severally equal to each of the three rectangles contained in DBCE. But DBCE equals one-sixth of the area of the hexagon; therefore the whole area of the hexagon must equal eighteen such rectangles as the three contained in DBCE. And ABFG is one-sixth of the circumference; therefore the whole circumference must be composed of twenty-four such rectangles as the four contained in ABFG. Eighteen is three-fourths of twenty-four, therefore the area of a hexagon, whose diameter is three, equals three-fourths of its circumference, agreeably to the proposition.

PROPOSITION XIX.

In the regular hexagon, whose diameter is four, the area equals the circumference.

In the present diagram, the diameter is divided into four units, therefore CD, the radius, is two, and DH, the breadth of the line of circumference, is made equal to one. CD being double of DH, let CD be divided in the center, and the small rectangles completed. Then it is manifest that the two rectangles contained in ABFG are severally equal to the two contained in DBCE, for they all have equal bases and equal heights by con

struction. But DBCE is equal to one-sixth of the area of the hexagon; therefore the whole area of the hexagon is equal to twelve such rectangles as the two contained in DBCE. And ABFG is one-sixth of the circumference of the hexagon; therefore the whole circumference must be composed of twelve such rectangles as the two contained in ABFG. Area and circumference, each being equal to twelve such rectangles, are equal to each other. Therefore the area of a hexagon, whose diameter is four, is equal to its circumference, agreeably to the proposition.

PROPOSITION XX.

In the regular octagon, whose diameter is one, the area equals one-fourth of the circumference.

D B

Take the diameter of the circle for unit, and then the diameter of the octagon is one, [Def. 17.] And the radius, CD, is half of one. Let CD be produced to H, making DH equal to one, or double CD; and AB being double DB, it is manifest that the rectangle DBCE is equal to each of the four rectangles contained

in the rectangle ABFG, for they have equal bases and equal heights by construction. Lines drawn from the center to each of the eight angles of the octagon, divide the area into eight equal triangles, for the base of each triangle is an equal side of the octagon, and the height or perpendicular of each triangle equals the radius CD. One of the triangles, ABC, is equal to the rectangle DBCE, [Prop. 6.] Therefore the area of the octagon equals eight such rectangles as DBCE. AB being one-eighth of the perimeter of the octagon, and DH being equal to one, the rectangle ABFG constitutes oneeighth of the circumference, [Def. 16.] Therefore the whole circumference must be composed of thirty-two such rectangles as the four contained in ABFG. And the area being equal to eight such rectangles, therefore the area of an octagon, whose diameter is one, is equal to one-fourth of its circumference, agreeably to the proposition.

PROPOSITION XXI.

In the regular octagon, whose diameter is two, the area equals half the circumference.

AB being double DB, it is manifest that each of the two rectangles contained in ABFG is equal to the rectangle DBCE, for they have an equal breadth and equal height by construction. ABFG is oneeighth of the circumference of the octagon; therefore the whole circumference must be

equal to sixteen such rectangles as the two contained in ABFG. The area of the octagon equals eight such rectangles as BDCE, [Prop. 6.] Therefore the area of an octagon, whose diameter is two, equals one-half its circumference, agreeably to the proposition.

PROPOSITION XXII.

In the regular octagon, whose diameter is three, the area equals three-fourths of the circumference.

In the present diagram, diameter being three, the radius CD is one and a half, and DH, the breadth of the line of circumference, is one, and therefore equal to two-thirds of CD. Let CD be divided into three equal parts, and DH into two equal parts, and the small rectangles drawn upon the equal parts will be equal, having equal bases and equal heights. As was shown in the two last propositions, the area of the octagon is equal to eight such rect

angles as DBCE, therefore it must be equal to twenty-four such rectangles as the three contained in DBCE. And ABFG being one-eighth of the circumference of the octagon, the whole circumference must be equal to thirty-two such rectangles as the four contained in ABFG. Twenty-four is three-fourths of thirty-two, therefore the area of an octagon, whose diameter is three, equals three-fourths of its circumference, agreeably to the proposition.

PROPOSITION XXIII.

In the regular octagon, whose diameter is four, the area equals the circumference.

In this diagram, diameter is divided into four units; therefore the radius CD is two, andDH, the breadth of the line of circumference, is one. CD being divided into two equal parts, and AB being double DB, it is manifest that the two rectangles contained in ABFG are equal to the two rectangles in DBCE, for they all have equal bases and equal heights. As already shown in the preceding propositions, the area of the octagon is equal to eight such rectangles as

DBCE; it is therefore equal to sixteen such rectangles as the two contained in DBCE. The rectangle ABFG is one-eighth of the circumference of the octagon, therefore the whole circumference must be equal to sixteen such rectangles as the two contained in ABFG. Therefore the area of an octagon, whose diameter is four, is equal in geometrical quantity to its circumference, agreeably to the proposition.

REMARK.—In all the preceding propositions, diameter has been the given or assumed quantity, and it has been seen to control the relation between area and circumference by one simple and uniform law, whatever may be the form or figure in which the area is presented. We shall now see that when circumference is the given or assumed quantity, it controls the relation between area and diameter by precisely the same simple and uniform law, so rigidly enforced by diameter in the preceding cases.

PROPOSITION XXIV.

In the square, whose circumference is one, the area equals one-fourth of the diameter.

REMARKS. The terms of the proposition may seem to require a few words of explanation, to make them appear consistent with the principles already laid down. Since the unit in geometry always has the form of a cube, and when once assumed, is always indivisible, and invariable in form or size, the question may arise in the mind of the student, how can one constitute a circumference, or inclose area? It is manifest that the one, assumed as the unit, cannot inclose area.

But we may assume smaller units, a cer