Now, AF is contained in Pas many times as HK is contained in P'. If now we have numerical values for p and P, we may obtain the value for P' by formula (1). Again, in the similar triangles ABG, HBL (V. 4), AG: AB :: BL : BH. But AG, AB are equal parts of p and p', and BL, HB equal parts of p' and P'; ... Hence, p:p':: p': P'. p' = VpP'. (2) If, then, we have given a numerical value for p, and have, by formula (1), found one for P', we may find a value for p' by formula (2). Now, we know that the perimeter of a circumscribed square in a circle whose diameter is 1, is 4, and the perimeter of an inscribed square in the same circle is 2/2; hence, substituting these values for Pand pin the formulæ just proved, we have and p' = VpP' = V 8 1/2 =3.0614675. These are perimeters of the regular circumscribed and inscribed octagons; by using them for P and p, we may obtain the values of the perimeters of polygons of 16 sides, and so on. These numbers approach each other as the number of sides increases, until for 8192 sides we have P' = 3.1415928, and p' = 3.1415926. These numbers therefore very nearly represent the length of the circumference of the circle when the diameter is 1; and since, when the diameter is 1, the circumference is ", we have, carrying out the decimal correctly to 8 places, π = 3.14159265+. This decimal has been carried out to several hundred places without coming to an end, and it may be proved by higher mathematics to be incommensurable with 1. The value usually employed in calculations is n = 3.1416. The number 27 is a rough approximation. EXERCISES. 1. In a given circle the inscribed square is equal to one half the circumscribed square. 2. In the figure (VI. 2) the diagonals of the square pass through the centre of the circle, and bisect the angles at E. 3. The regular hexagon inscribed in a circle will contain double the area of an equilateral triangle inscribed in the same circle. 4. Show that in (VI. 5) there are two triangles which possess the required property; also that there is an isosceles triangle of which each of the angles at the base is a third part of the third angle. 5. If from any point of the circumference of a circle straight lines be drawn to the four angles of an inscribed square, the sum of the squares on these lines is double the square on the diameter (II. 9). 6. Describe a circle about a given rectangle. 7. If in (VI. 5) DC be produced to meet the circle in F, then FB is one side of a pentagon. 8. ABCD is a regular pentagon; let AC and BD meet in E; then CB is a mean proportional between CA and CE. 9. The area of a circle is to the area of the regular inscribed hexagon as is to V3. 10. The area of an inscribed polygon C D E B A is a mean proportional between the areas of the inscribed and circumscribed polygons of one half the number of sides (V. 22). 11. The area of a circle whose radius is 1 is π. 12. The area of a sector of a circle is equal to the arc of the sector multiplied by half the radius. 13. The area of a regular inscribed hexagon is of that of a regular circumscribed hexagon. 14. A plane surface may be entirely covered by equal regular polygons of either three, four or six sides, and by no other equal regular polygons. 15. A plane surface may be entirely covered by a combination of squares and regular octagons having equal sides. 16. In a given equilateral triangle, inscribe three circles tangent to each other and to the sides of the triangle. 17. In a given circle, inscribe three equal circles tangent to each other and to the given circle. 18. If the radii of three circles be a, b and c, the radius of a circle containing the same area as the three together is Va2+b2+c2. 19. The chord of 60° is equal to the radius. MENSURATION OF PLANE SURFACES. THE following rules are mostly deduced from preceding propositions: 1. To find the area of a rectangle. Multiply two adjacent sides together. (V. 16.) 2. To find the area of a parallelogram. Multiply one side by the perpendicular distance to the opposite side. (V. 16, Cor. 1.) 3. To find the area of a triangle. I. Multiply the base by one half the altitude. (V. 16, Cor. 2.) II. Multiply together the half sum of the three sides, and the excess of the half sum over the three sides respectively, and take the square root of the product.* Hence, ABC=}BC.AD=4V 4a2c2 - (a2+c2-b2)2. The quantity under the radical sign may be resolved into two factors, [2ac+a2+c2-b2] and [2ас- (a2+c2-b2)], and these again into (a+c+b)(a+c-b) and (b+a-c)(b+c-a). Hence, the last equation becomes ABC=√[a+b+ca+c-b.b+a-c.b+c-a]; and putting Hence the rule. 2 a+b+c 2 2 2 S= , ABC=VS(S-b)(S-c)(S-a.) 2 4. To find the area of a trapezoid. Multiply the sum of the parallel sides by one half the perpen dicular distance between them. (V. 17.) 5. To find the area of a trapezium. Multiply a diagonal by the half sum of the perpendiculars from the opposite angles. 6. To find the area of an irregular polygon. I. Divide it into triangles, and find their areas separately. II. Divide into trapezoids and triangles by a long diagonal and perpendiculars from the opposite angles, and find their areas separately. (V. 17, Sch. 1.) 7. To find the area of a regular polygon. I. Multiply the apothem by one half the perimeter. (VI. 9.) II. Multiply the square of a side by the area of a similar polygon whose side is 1. (V. 20.) The areas of regular polygons whose sides are unity may be calculated once for all, and tabulated. The following table gives some of them to six places : Triangle........... .433013 Heptagon......... 3.633912 Square...... 1.000000 Octagon......... 4.828427 Pentagon........... 1.720477 Decagon ....... 7.694209 Hexagon ......... 2.598076 Dodecagon....... 11.196152 8. To find the circumference of a circle. Multiply the diameter by = 3.1416, or the radius by 2π. (VI. 11, Cor.) 9. To find the diameter from the circumference. Divide the circumference by π. 10. To find the length of an arc of a circle. Say, As 360° is to number of degrees in the arc, so is the whole circumference to the length of the arc. (V. 25.) |