101. By Exs. 5 and 6, § 94, we see that the area of any regular polygon equals the rectangle on the halfperimeter and the apothem, or frequently expressed as half the product of perimeter and apothem. The same holds true no matter how many sides the polygon may have.

If successive polygons, doubling in number of sides, be inscribed within a circumference, the perimeter will continually increase, approaching the circumference of the circle as its limit, and the apothem will increase, approaching the radius of the circle as its limit.

If successive polygons, doubling in number of sides, be circumscribed about a circumference, the perimeter will continually decrease, but the apothem will remain constant.

In each case the area equals the half-product of perimeter and apothem.

When we have reached 32768 sides for each, the areas of the circumscribed and inscribed regular polygons (in terms of the square on the radius) agree to seven decimal places. The area of the circle lies between the two areas, and the circumference of the circle is the common limit upon which the perimeters of the inscribed and circumscribed polygons converge; we therefore say that the area of a circle equals half the product of its circumference and radius. If A represent the area of a circle, C its circumference, and R its radius,

A = C. R.

Expressed in general language:

The area of a circle equals the half product of the circumference and the radius.

Hence may also be described as the ratio of the circumference to the diameter of a circle, as well as the ratio of the area of a circle to the square on its radius.

103. THEOREM. The perimeter of ANY polygon inscribed within a circle is LESS than the circumference, and the perimeter of ANY polygon circumscribed about a circle is GREATER than the circumference of the circle.

(a) Each side of the inscribed polygon is a straight line, joining two points, hence will

be

shorter than the arc which it subtends. The sum of these chords - which is the perimeter of the polygon-will therefore be less than the sum of the subtended arcs which is the circumference of the circle.

Q. E. D.

FIG. 169.

(b) By Ex. 6, § 94, the area of a circumscribed polygon equals the half-product of the perimeter and the radius of the circle.

A' = 1⁄2 PR.

By § 101, the area of the circle equals the half-product of its circumference and radius.

A = CR.

But the area of the circumscribed polygon is greater than the area of the circle.

. . 1 PR > ¦ CR, or P> C.

Q. E. D.

Exercises. 1. Show that the area of any sector equals half the product of the included arc and the radius.

2. Find the areas of sectors of 30°, of 45°, of 120°, of 300°, of 390°, of 480°, and of 540°.

3. The areas of circles are to each other as the squares of their radii, or the squares of their diameters, or the squares of their circumferences, or the squares of any corresponding lines.

4. If at the extremities of an arc AB, which determines a sector of a circle, tangents be drawn intersecting at I, AI + IB will be greater than the arc AB.

5. Show that if two circumferences intersect, the subtended arcs on each side of the common chord will be unequal, and the enveloping one will be the longer.

Show how to determine by tangents at A or B which curve will be the enveloping one. Discuss the problem thoroughly.

A

FIG. 170.

A

M

FIG. 171.

B

104. Problems.-1. To inscribe a regular decagon in a circle. Analysis. If AB were the side of a regular inscribed decagon, and if we should join the vertices of the

polygon to the centre, we should have ten

triangles formed, each equal to ACB.

The ▲ ACB and ABD being mutually equiangular, are similar.

FIG. 172.

CA BD

AB DA

Which means that the radius will be separated into mean and extreme ratio.

Construction. By Prob. 13, § 91, separate the radius of the circle in which the decagon is to be inscribed, into parts such that the ratio of the whole radius to the larger part equals the ratio of the larger part to the smaller.

From K lay off chords equal to CQ;

they will form the sides of the required decagon.

C

K

FIG. 173.

2. Show that if the alternate vertices of a decagon be joined, a regular pentagon will be formed.

3. Making use of § 88, show how to find the diagonals of a regular pentagon, having given the length of one side.

4. Show how a regular hexagon and a regular decagon may be used to inscribe a regular polygon of 15 sides, and hence one of 30 sides.

5. Show that the square on the side of a regular inscribed pentagon equals the sum of the squares on a

side of the regular inscribed decagon and on the radius of the circle.

6. Having given the side of a regular decagon, find the corresponding radius.

7. Find the area of a regular inscribed dodecagon in terms of the square on the radius.

FIG. 174.

8. Show how to make a five-pointed star and determine the sum of the angles at the points.

1. The three altitudes of a triangle are concurrent lines.

B

FIG. 175.

Suggestion. Through the vertices of the given triangle draw lines parallel to the opposite sides.

2. A quadrangle, the opposite sides of which are equal, is a parallelogram.

3. Find the area of a triangle the sides of which are 12, 16, and 20. Find the length of the median from the largest angle. Determine the radii of the circumscribed and inscribed circles.

4. If from any point in the base of an isosceles triangle lines be drawn parallel to the equal sides, a parallelogram will be formed, the perimeter of which will equal the sum of the sides parallel, to which lines have been drawn.