PROBLEM 15. To find the difference between the perimeters of two regular polygons of the same number of sides, one inscribed in, and the . other circumscribed about a given circle. Let ABC... be a regular polygon inscribed in the circle ABC, and HIJ...a regular polygon of the same number of sides circumscribed about it, formed by drawing tangents to the o l to the sides of ABC.... Draw OAH, OBI, also BF I to AH. Then FI = difference between AB and HI. Cut off Op, Oq, Or, &c. equal to AH or BI : join pq, qr, &c. then pq = FI. (1. 1) Now complete the polygon par. Then the perimeter of this polygon is equal to the difference between the perimeters of the polygons ABC, CoR. 1. If the polygons are formed by continual bisections of the circumference, then the perimeter of the polygon par is less than that of a square whose side is XY. Now XY is double of Ox, and .. double of NR; but NR is < AN, for 2 NAR is <L OAR and ..<LARN; i. XY is < AB; ... the difference between the perimeters of the given polygons, and .. between either of them and the given circle is less than the perimeter of the square on a side of the inscribed polygon. (111. C) COR. 2. Hence a polygon can be inscribed in a circle whose perimeter shall differ from that of the circumference of the circle by less than any assignable magnitude. So also can a polygon be circumscribed about a circle, &c. PROBLEM 16. Let BFGHK be a polygon circumscribed about a circle. Join the centre Q with the angular points of the polygon, and also with the points of contact. Then the area of 6 BFQ is = } rectangle (radius QM, and BF), (1. 32) so area of a FG Q is = } rectangle (radius, and FG), &c. &c. .. area of polygon = } sum of the rectangles contained by (radius, and BF), (radius, and FG) &c. = } rectangle (radius, and a straight line = the perimeter). (11. 1) COR. Now let the points of contact of the sides of the polygon be determined by continual bisections of the circumference, then ultimately the area of the polygon will differ from that of the circle by less than any assignable magnitude, as also will the perimeter of the polygon from the circumference of the o. Hence the area of the circle is = } rectangle (radius, and a straight line = the circumference). BOOK V. RATIO AND PROPORTION. INTRODUCTORY. Two magnitudes of the same kind may be compared by considering how many times one contains the other. If one does not contain the other a certain number of times exactly, it may happen that each contains some other magnitude a certain number of times, and then the two given magnitudes may be compared. If each of two magnitudes does not contain some other magnitude a certain number of times, they are called incommensurable. In order to institute a comparison, one of them may be divided into a number of equal parts as small as we please, and parts equal to these may be taken from the other till there remains less than one of these parts, and therefore all the parts together thus taken away differ from the whole magnitude by less than one of these parts. Hence the former magnitude may be compared with one which differs from the latter by less than any assignable quantity. DEFINITION. The first of four magnitudes is said to have the same ratio to the second which the third has to the fourth when, if the first be divided into any number whatever of equal parts and the third be divided into the same number of equal parts, the second contains the same integral number of the former parts as the fourth does of the latter. PROBLEM 15. Let ABC...be a regular polygon inscribed in the circle ABC, and HIJ...a regular polygon of the same number of sides circumscribed about it, formed by drawing tangents to the o ll to the sides of ABC.... Draw OAH, OBI, also BF || to AH. Then FI = difference between AB and HI. Cut off Op, Oq, Or, &c. equal to AH or BI: join pa, qr, &c. then pq = FI. . (1.1) Now complete the polygon par. Then the perimeter of this polygon is equal to the difference between the perimeters of the polygons ABC, |