Proposition 5. Theorem.

419. The area of a regular polygon is equal to half the product of its perimeter and apothem.

Hyp. Let P denote the perimeter and R the apothem OH, of the regular polygon ABCDEF.

To prove area ABCDEF = 1⁄2 P × R.

Proof. Join OA, OB, OC, etc.

The polygon is divided into equals whose bases are the equal sides of the polygon and whose common altitude is the apothem.

Then,

area OAB

=

AB × OH,

the area of a ▲ = the product of its base and altitude (366).

Similarly,

area OBC = 1⁄2 BC × OH,

and so on for all the As of the polygon.

... area OAB + area OBC + etc. = (AB+ BC + etc.) OH.

... area of the polygon ABCDEF = { P × R.

Q. E. D.

420. COR. The area of any polygon that circumscribes a circle is equal to half the product of its perimeter and the radius of the circle.

THE MEASURE OF THE CIRCLE.

421. DEF. A variable quantity, or simply a variable, is a quantity which may have an indefinite number of different successive values.

A constant is a quantity whose value does not change in the same discussion.

422. Limit. When the successive values of a variable, under the conditions imposed upon it, approach more and more nearly to the value of some constant quantity, which it can never equal, yet from which it may be made to differ by as small as we please, the constant is called the limit of the variable; the variable is said to approach indefinitely to its limit.

DE

B

423. For example, suppose a point to start at A and move along AB towards B under the condition that, during successive seconds of time, the point moves first half the distance from A to B, that is to C; then half the remaining distance, or to D, then half the remaining distance, or to E, and so on indefinitely. Then the distance from A to the moving point is a variable whose limit is the distance AB. For, however long the point may move, under these conditions, there will always remain some distance between it and the point B, so that the distance from A to the moving point can never equal AB; but as the moving point can be brought as near as we please to B, its distance from A can be made to differ from the distance AB by as little as we please.

If we call the distance AB 2, then the distance the point moves the first second will be 1, the distance moved the next second will be, the distance moved the third second will be, and so on.

Therefore the whole distance from A

to the moving point at the end of n seconds is the sum of n terms of the series

1+1+1+ś+ etc.

Now it is evident that, however many terms of this series are taken, the sum can never equal 2; but by taking a great number of terms the sum can be made to differ from 2 by as little as we please. Hence we say that the limit of the sum of the series as the number of terms is indefinitly increased is 2.

The limit of the fraction, as is indefinitely increased, is zero; for, by increasing x at pleasure, 1 may be made to approach as near as we please to the value zero, but can never be made exactly equal to zero.

424. Principle of Limits. THEOREM. If two variables are always equal and each approaches a limit, the two limits must be equal.

For, two variables that are always equal have always the same common value, that is, they are really but a single variable; and it is clear that a single variable cannot at the same time approach indefinitely to two unequal limits.

425. COR. If two variables, while approaching their respective limits, are always in the same ratio, their limits are in the same ratio.

For, if x and y are two variables in the constant ratio m, so that x= my, then the two variables x and my are always equal to each other, and their limits are equal (424); therefore, if a and b are the limits of x and y respectively, we have amb; that is, a and b are in the same constant ratio m.

EXERCISE.

The apothem of a regular pentagon is 3 and a side is 2: find the perimeter and area of a regular pentagon whose apothem is 6.

Proposition 6. Theorem.

426. Every convex curve is less than any enveloping line whatever that has the same extremities.

Hyp. Let ABC be a convex curve, and ADHC any line enveloping it and terminating at A and C.

To prove

that ABC ADHC. D

Proof. Of all the lines enveloping the area ABC, there must be at least

one line shorter than any other.

Now ADHC cannot be this line.

H

B

A

For, draw the tangent AH to the curve ABC.

Then,

the line AHC <the line ADHC,

since the st. line AH is < the curve ADH (Ax. 10).

... ADHC is not the shortest line.

In the same way it may be shown that no other enveloping line can be the shortest.

... the curve ABC is less than any enveloping line. Q E.D.

427. COR. 1. The circumference of a circle is less than the perimeter of any circumscribed polygon, and greater than the perimeter of any inscribed polygon.

428. COR. 2. The perimeter of a regular inscribed polygon is less than the perimeter of a regular inscribed polygon of double the number of sides.

429. COR. 3. The perimeter of a regular circumscribed polygon is greater than the perimeter of a regular circumscribed polygon of double the number of sides.

Proposition 7. Theorem.

430. If a regular polygon be inscribed in a given circle, and another regular polygon be circumscribed about the same circle, and if the number of sides of the polygons be increased indefinitely, then the perimeter of each of the polygons approaches the circumference of the given circle as its limit, and the area of each approaches the area of the circle as its limit.

Hyp. Let O be the centre of the given O, and AB, CD homologous sides of the regular inscribed and circumscribed polygons of the same number of sides; let p and P denote the perimeters, s and S the areas of the inscribed and circumscribed polygons respectively.

F

D

Το prove that the limit of p and P is the Oce of the O, the limit of s and S is the area of the O.

and

Now let the number of sides be increased indefinitely, always remaining the same in each polygon; then the angle FOC will diminish indefinitely, and the pt. C will approach the pt. F, as near as we please.

and

... OC will approach OF as its limit,

OC' will approach OF2 as its limit.