each of the triangles AGB, BGC, etc., are radii; and since their bases AB, BC, etc., are equal, being chords of equal arcs (Theo. XXIV); it follows that these triangles are mutually equiangular (Theo. XII). But they are also isosceles; hence (Theo. VIII), all the angles at their bases, GBA, GBC, GCB, GCD, etc., are equal; hence, also, the sums, ABC, BCD, etc., that is, all the angles of the polygon, are equal. And its sides have already been shown to be equal. It is, therefore, a regular polygon (Def. 2, Sec. VII).

Hence, if the circumference, etc.

Cor. 1. If the polygon ABCDEF be a hexagon, the angle AGB, being measured by one-third of the semi-circumference, will be one-third of two right angles (Cor., Theo. XXIV). Hence, either of the two equal angles at the base of the triangle, as GBA, will also be one-third of two right angles (Theo. V). Therefore, AB is equal to AG (Cor., Theo. IX); that is, the side of a regular hexagon is equal to the radius of the circumscribed circle.

Cor. 2. A perpendicular, as Gm, let fall from the center of the circumscribed circle on any side of a regular polygon, will be the apothegm of the polygon (Schol., Theo. XXI).

Cor. 3. If from the center G, with the apothegm Gm as radius, a circle be drawn, the side FE, and consequently all the other sides of the regular polygon, will be tangents to that circle (Theo. XXIII), and the circle will be inscribed in the polygon (Def. 7, Sec. VIII).

THEOREM XXIX.

The area of a circle is equal to half the product of its radius by its circumference.

Let ABC be a circle, of which DB or DC is radius.

If we conceive the whole circumference to be divided into equal arcs so small that the points of division may be regarded as infinitely near to each other, the perimeter of the reg

B

E

ular polygon formed by their chords (Theo. XXVIII) will coincide with the circumference, and the area of the polygon will be equal to the area of the circle; also, the apothegm of the polygon (Cor. 2, Theo. XXVIII) will be equal to the radius of the circle. But the area of the polygon will be equal to half the product of its apothegm by its perimeter (Cor., Theo. XXI). Therefore, the area of the circle must be equal to half the product of its radius by its circumference.

Cor. 1. In like manner it may be shown that the area of any sector, as BECD, is equal to half the product of the radius into its arc.

Cor. 2. The area of a segment BEC, less than a semicircle, may be found by subtracting the triangle BDC from the sector BECD. The area of a segment BAC, greater than a semicircle, may be found by adding the triangle BDC to the sector BACD.

THEOREM XXX.

The radius, and one side of a regular inscribed polygon being given, the side of a regular inscribed polygon of twice the number of sides may be found.

Let AB be the side of a regular inscribed polygon in a circle whose radius is CA or CB. Draw CE perpendicular to AB. This will bisect AB and also the arc AEB (Theo. XXV). Hence, the chords AE, EB (Theo. XXVIII), will be sides

E

B

D

of a regular inscribed polygon having twice as many sides as the first.

Now, if the lengths of CB and AB are given, we shall have, in the right-angled triangle CDB, the hypotenuse CB and the side BD; hence, we may find the other side DC (Cor. 2, Theo. XIX). Subtracting this from the radius CE, we shall have DE. Then, in the right-angled triangle BDE, we shall have the two sides BD, DE, from which we can find the hypotenuse EB (Cor. 1, Theo. XIX); which is the side required.

Schol. If the diameter be 1, the side of a regular inscribed hexagon will be (Cor. 1, Theo. XXVIII). From this we can find the side, and consequently the perimeter, of a regular inscribed dodecagon; from that we can find the perimeter of a regular inscribed polygon of 24 sides, etc. By carrying this calculation on, to polygons of an indefinitely large number of sides, it is found that the perimeter, though it increases at every step, never exceeds 3.14159, except by decimal figures beyond the 9; that is, beyond the fifth place of decimals. Hence, since the perimeter ultimately

coincides with the circumference, it follows that the circumference can not differ from 3.14159, except by decimals beyond the fifth place. Disregarding these, as of trifling value, we may conclude that the circumference of a circle whose diameter is 1, is 3.14159. Cor. If the diameter is 1, the area is equal to 3.14159

2

(Theo. XXIX), which by reduction is .7854.

EXERCISES.

1. How many degrees are contained in an arc cut off by the side of an inscribed square? Of a regular inscribed pentagon? Hexagon? Heptagon? etc.

2. What is the area of a sector whose arc is 25 feet, in a circle whose diameter is 16 feet?

3. Prove that in an inscribed quadrilateral the sum of any two opposite angles is 180 degrees (Theo. XXVI).

4. Prove that of two unequal chords the less is the further from the center (Theo. XXV, and Cor. 2, Theo. XIX).

one-half of AB, describe the arc CFD. From B

as a center, with the

E

F

B

D

same radius, describe the arc CED. Join the points

of intersection C and D.

Evans' Geometry.-5

Now, since the opposite sides of the quadrilateral ACBD are equal, it is a parallelogram (Schol., Theo. XIV). But the diagonals of a parallelogram bisect each other (Theo. XV). Therefore, CD bisects AB.

PROBLEM II.

At a given point in a straight line, to erect a perpendicular to that line.

Solution.

Let AB be

an indefinite straight line, and C the given point in it. Take the points D and E equally distant

A D

E B

from C. From D as a center, with a radius greater than DC, describe an arc; and from E as a center, with the same radius, describe another arc intersecting the former at some point F. Now, draw FC, and it will be the perpendicular required.

Join FD and FE. Then, since the triangles DFC, EFC, are by construction mutually equilateral, the angles FCD, FCE, are equal (Theo. XII); they are, therefore, right angles, and CF is perpendicular to AB at the point C (Defs. 2 and 3, Sec. IV.)

PROBLEM III.

To draw a perpendicular to a straight line from a given point without.

Solution. Let C be the given point without the straight line, AB. From C as a center, with Α a radius greater than the

E

B