Proportions of Circles. until finally the polygon will become equal to the circle, and the perimeter will coincide with the circumference. When this takes place, the line CH, drawn perpendicular to one of the sides, will become equal to the radius of the circle. H The circumferences of circles are to each other as their diameters Let there be two circles whose diameters are AL and FM: then will their circumferences be to each other as AL to FM. F M For, suppose two similar polygons to be inscribed in the circles: their perimeters will be to each other as AL to FM (Th. xxii). Let us now suppose the arcs which subtend the sides of the polygons to be bisected, and new polygons of double the num ber of sides to be formed: their perimeters will still be to each other as AL to FM, and if the number of sides be increased until the perimeters coincide with the circumference, we shall have the circumferences to each other as the diameters AL and FM. THEOREM XXV. The areas of circles are to each other as the squares of their diameters. Area of the Circle. Let there be two circles whose diameters are AL and FM: then will their areas be to each other as the square of AL to the square of FM. M For, suppose two similar polygons to be inscribed in the circles: then will they be to each other as AL to FM2 (Th. xxiii). Let us now suppose the number of sides of the polygons to be increased, by bisecting the arcs, until their perimeters shall coincide with the circumferences of the circles. The polygons will then become equal to the circles, and hence, the areas of the circles will be to each other as the squares of their diameters. Cor. Since the circumferences of circles are to each other as their diameters (Th. xxiv), it follows, that the areas which ire proportional to the squares of the diameters, will also be proportional to the squares of the circumferences THEOREM XXVI. The area of a regular polygon inscribed in a circle, is equal to half the product of the perimeter and the perpendicular let fall from the centre on one of the sides. Let C be the centre of a circle circumscribing the regular polygon, and CD a perpendicular to one of its sides: then will its area be equal to half the product of CD by the perimeter. For, from C draw radii to the vertices of the angles, forming as many Area of Circle. equal triangles as the polygon has sides, in each of which the perpendicular on the base will be equal to CD. Now, the area of one of them, as ACB, will be equal to half the product of CD by the base AB; and the same will be true for each of the other triangles hence, the area of the poly D B gon will be equal to half the product of CD by the perimeter THEOREM XXVII. The area of a circle is equal to half the product of the radius by the circumference. Let C be the centre of a circle: then will its area be equal to half the product of the radius AC by the circumference ABE. For, inscribe within the circle a regular hexagon, and draw CD perpendicular to one of its sides. Then, B E the area of the polygon will be equal to half the product of CD multiplied by the perimeter (Th. xxvi). Let us now suppose the number of sides of the polygon to be increased, until the perimeter shall coincide with the circumference; the polygon will then become equal to the circle, and the perpendicular CD to the radius CA. Hence, the area of the circle will be equal to half the product of the radius by the circumference. Problems. PROBLEMS RELATING TO THE FOURTH BOOK. PROBLEM I. To divide a line into any proposed number of equal parts. Let AB be the line, and let it be equired to divide it into four equal parts. Draw any other line, AC, forming an angle with AB, and take any dis E D A tance, as AD, and lay it off four times on AC. Join C and B, and through the points D, E, and F, draw parallels to CB. These parallels to BC will divide the line AB into parts pro portional to the divisions on AC (Th. xiii): that is, into equal parts. PROBLEM II. To find a third proportional to two given lines. Let A and B be the given lines. Make AB equal to A, and draw AC, making an angle with it. On AC lay off AC equal to B, and join BC: then lay off AD, also equal to A A EC B, and through D draw DE parallel to BC: then will AE be the third proportional sought. For, since DE is parallel to BC, we have (Th. xiii) AB : AC : : AD or AC : therefore, AE is the third proportional sought AE; Problems. PROBLEM III. To find a fourth proportional to the lines A, B, and C. Place two of the lines forming an angle with each other at A; that is, make AB equal to A, and AC equal B; also, lay off AD equal to C. Then join BC, and through D draw A B C E A DE parallel to BC, and AE will be the fourth proportional sought. For, since DE is parallel to BC, we have AB : AC :: AD : AE; therefore, AE is the fourth proportional sought. PROBLEM IV. To find a meun proportional between two given lines, A and B. AC, and it will be the mean proportional sought (Th. xviii. Cor). PROBLEM V. To make a square which shall be equivalent to the sum of two A B BC perpendicular to AB, and make BC equal to B: then draw AC and the square described on AC will be equivalent to the squares on A and B (Th. xii). |