As shown in the preceding proposition, OE OF can be made as small as we please, though not equal to zero;

P

P

and

since does not increase, (OEOF), or its equal PP,

OE

can be decreased at pleasure.

OE

Since P is always greater than the circumference, and p is always less than the circumference, the difference between the circumference and either perimeter is less than the difference P —p, and can consequently be made as small as we please, but not equal to zero.

The circumference is therefore the common limit of the two perimeters as the number of sides is indefinitely increased.

Q.E.D.

II. To Prove that the areas of the polygons approach the area of the circle as a limit, when the number of sides is indefinitely increased.

Proof. Let S and s stand for the areas of the circumscribed and inscribed polygons respectively.

As the number of sides is indefinitely increased, CD approaches zero as a limit, as does also CF, and consequently

CF

[The remainder of the proof is similar to that of Case I. of this proposition.]

Q.E.D.

767. EXERCISE. If, as is shown in § 766, the difference between P and p can be made as small as we please, why is not p the limit of P? (See definition of limit.)

PROPOSITION XII. PROBLEM

768. Given the perimeters of a regular inscribed polygon and of a similar circumscribed polygon, to find the perimeters of regular inscribed and circumscribed polygons of double the number of sides.

Let AB be a side of a regular inscribed polygon of n sides, CD (parallel to AB) a side of a regular circumscribed polygon of n sides,

AE a side of a regular inscribed polygon of 2 n sides,

FG a side of a regular circumscribed polygon of 2 n sides. Required to find the perimeters of the inscribed and circumscribed polygons of 2 n sides.

Call the perimeter of the inscribed

the perimeter of the circumscribed

polygon of n sides p, polygon of n sides P,

the perimeter of the inscribed polygon of 2 n sides p',

Since p and P are given, Formula I. gives the value of P'; then from Formula II. the value of p' can be derived.

Q.E.F.

769. EXERCISE. The side of an inscribed square is 3√1⁄2 and the side of a circumscribed square is 6. Find the sides of regular octagons inscribed in and circumscribed about the same circle.

770. EXERCISE. Find the perimeters of regular dodecagons (12-sided polygons) inscribed in and circumscribed about a circle having a diameter 12 in. long.

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

Let ABCDEF be a regular polygon.

To Prove that its area is equivalent to one half the product of its perimeter and apothem.

Suggestion. The altitude of each ▲ is the apothem, and the polygon is equivalent to the sum of the triangles.

772. COROLLARY. The area of any circumscribed polygon is equal to one half the product of its perimeter and the radius of its inscribed circle.

773. EXERCISE. The perimeter of a polygon circumscribed about a circle having a 5 ft. radius, is 32 ft. What is its area?

774. EXERCISE.

The side of a regular hexagon is 6 in. Find its

area. [Suggestion. First find its apothem.]

PROPOSITION XIV. THEOREM

775. The area of a circle is equal to one half the product of its circumference and radius.

Proof. Circumscribe a regular polygon ABC about the circle Area ABC perimeter x apothem. (?)

xyz.

=

If the number of sides of the polygon is increased, the area changes as does also the perimeter, and yet the area is always equal to perimeter x apothem. So the two members of the above equation may be regarded as two variables that are always equal. Since each is approaching a limit, their limits must be equal. [§ 341.]

The limit of area ABC = area of circle.

The limit of the perimeter = circumference. (?)
The apothem is constant and equals the radius.
.. area xyz = 1⁄2 circumference × radius.

Q.E.D.

777. EXERCISE.

is 628.32 ft.

The radius of a circle is 100 ft. and its circumference Find its area.

778. EXERCISE. The area of a sector is 68 sq. in., and its radius is 8 in. How long is its arc ?

779. EXERCISE. The area of a circle is 100 sq. ft. The area of a sector of this circle is 12 sq. ft. How many degrees in the arc of the

sector?

780. EXERCISE. The area of a circle is 10 sq. ft. Find the area of a segment whose arc contains 60°.

Suggestion. Find the area of the sector having arc = 60°. Subtract the area of the triangle formed by the chord and the radii from the area of the sector.

781. EXERCISE. The circumference of a circle is 94.248 ft. The side of an inscribed equilateral triangle is 15√3 ft. Find the area of the circle.

782. EXERCISE. The area of a circle is 314.16 sq. in. The perimeter of a regular inscribed hexagon is 60 in. Find the circumference of the circle.

783. EXERCISE.

Find the area of the part of the circle of § 782 lying between its circumference and the perimeter of a regular hexagon inscribed in the circle.