308. Given the radius and the side of a regular inscribed polygon, to compute the side of the regular inscribed polygon of double the number of sides.

Given AB, a side of the regular inscribed polygon, and OC= R, the radius of the circle.

To compute AC, a side of an inscribed polygon of double the number of sides.

Draw OC; then since it bisects the arc ACB, it will bisect AB at right angles (by 147).

C

A

B

E

D

Produce CO to D and draw AD; then since ≤ CAD is inscribed in a semicircle (by 177), it is a right angle.

Then (by 228) AC is a mean proportional between CD and CE, or

AC2= CD x CE = CD (CO- EO) = 2 R(REO)

Substituting this in the value for AC, we have

AC2 = R(2 R – √ 4 R2 — AB”),

or

AC =√R (2 R −√4 R2 – AB3).

PROPOSITION XII. PROBLEM.

309. To compute the ratio of the circumference of a circle to its diameter.

It is known (from 303) that

C = 2πR.

If, therefore, we take a circle whose radius is unity, we have

C = 2π, or

= 14

that is, a semicircle of unit radius.

C;

Hence the semiperimeter of each inscribed polygon is an approximate value of #, and the semiperimeter of each circumscribed polygon is also an approximate value of . Therefore, if by constantly increasing the number of sides of these polygons the approximate values of become practically identical, we know that as the circle lies between the inscribed and circumscribed polygons, this coincident value for can be taken as the semicircumference of the circle of unit radius.

If we begin with the square we know that the side is the hypotenuse of an isosceles right triangle, the two equal sides being radii; hence

or

AB (in 308)=√2 = 1.4142136,

semiperimeter = 2.824272.

Then each side of the circumscribed square is the diameter or twice the radius = 2, and the semiperimeter will be 4.

From the final equation in Prob. XI. it is easy to compute the semiperimeter of a polygon of 8 sides; then from the final equation in Prob. X. can be computed the semiperimeter of a circumscribed polygon of 8 sides; and so on.

In the following table are given the semiperimeters of inscribed and circumscribed polygons:

The figures in face type show the approximation.

310. SCHOLIUM. By the aid of simpler methods the value of has been computed to more than eight hundred places of decimals.

The first twenty figures of the result are

For all practical purposes it is sufficient to take = 3.1416.

EXERCISES.

1. If the radius of a circle is 4, find its circumference and area.

2. If the circumference of a circle is 30, find its radius and area.

3. If the diameter of a circle is 26, find the length of an arc of 72°.

4. If the radius of a circle is 12, find the area of a sector whose central angle is 80°.

5. If the apothem of a regular hexagon is 4, find the area of the circumscribing circle.

MAXIMA AND MINIMA.

311. Of quantities of the same kind, the one which is the greatest is called the Maximum, and the least is called the Minimum.

312. Isoperimetric figures are those which have equal perimeters.

PROPOSITION XIII. THEOREM.

313. Of all triangles formed with two given sides, that in which these sides include a right angle is the maximum.

Let ABC and A'BC be two triangles having the sides AB and BC equal to the sides A'B and BC respectively, and let the angle ABC be a right angle.

To prove that

area ABC area A'BC.

Draw A'D perpendicular to BC.

Then since (by 52) the oblique line A'B is greater than the perpendicular A'D, we have

AB> A'D.

But AB and A'D are the altitudes of the triangles ABC and A'BC, and as they have the same base, that triangle is the greater which has the greater altitude, or

area ABC area A'BC.

Q. E. D.

PROPOSITION XIV. THEOREM.

314. Of isoperimetric triangles having the same base, that which is isosceles is the maximum.

Let ABC and A'BC be two isoperimetric triangles having the same base BC, and let the triangle ABC be isosceles.

To prove that

area ABC area A'BC.

Produce AB to D, making AD = AB, and draw CD.

Since B, C, and D are equally distant from A, a circle with A as a centre could be drawn through B, C, and D, of which BD would be the diameter.

The angle BCD would therefore (by 177) be a right angle. Draw AF and A'G parallel to BC, take A'E equal to A'C, and draw BE.

Since the triangles ABC and A'BC are isoperimetric,

AB+ AC = A'B+ A'C=A'B+ A'E.