Note 1.-To find the length of each side, when the area and the perpendicular drawn from the centre to the middle of one of its sides are given.-Divide double the area by the perpendicular, and the quotient is the sum of all the sides. Again, divide the sum of all the sides by the number of sides in the polygon, and the quotient is the length of each side. Note 2.-To find the length of each side when the area only is given.-Divide the area by the number standing opposite its name in the preceding table; then, the square root of this quotient is the length of the side. Note 3.-In some cases it will be possible to find the area of a regular polygon when only one side is given, without using the table above and without recourse to Trigonometry. This can be effected in the case of a regular hexagon. Since by Euclid (I. 15) all the angles at point o are equal to 360°, therefore the angle AOB in the regular hexagon (see 360° figure above)= =60°. 6 And also since OA=0B, therefore the angle OAB angle OBA=60°. Hence, each of the three angles of the triangle OAB is equal to 60°; and therefore the triangle OAB is equiangular, and consequently equilateral. Therefore the area of a regular hexagon is 6 times the area of the equilateral triangle OAB, which can easily be found by Rule 2, Prob. V. Note 4.-In the table which has been given, it will be seen that the multipliers for the area have been carried down to only four figures, but they will be found sufficiently accurate for all practical purposes. Example 1.-Find the area of a regular hexagon, when each side measures 20 ft., and the perpendicular from the centre to the middle of one of the sides is 17.32 ft. Now, the sum of the sides or perimeter of the figure =6×20=120 ft. Then, area of hexagon= (perimeter × perpendicular) × 120 × 17.32=1039.2 sq. ft. Example 2.-Find the area of a regular octagon, each side of which is 30 ft. Now, the multiplier standing opposite 'octagon' in the table is 4.8284. Then, area of octagon square of the side multiplied by 4.8284 I. Formulæ = -302 × 4.8284-900 × 4·8284 Area-perimeter × perpendicular 2 II. Area-side2 x number standing op posite name of polygon. EXAMPLES. (1) Find the area of a regular trigon or equilateral triangle whose side is 2 ft. 6 in. (2) Find the area of a regular hexagon, each side of which is 60 ft. (3) Find the area of a regular heptagon, each side of which measures 30 ft. (4) Each side of a regular octagon is 30 ft., and the perpendicular drawn from the centre to the middle of one of its sides is 36·213 ft.; find its area. (5) The length of each side of a grass-plot, which is in the shape of a regular decagon, is 80 lks.; find the area. (6) Each side of a regular heptagon measures 20 ft., and the perpendicular drawn from the centre to the middle of one of its sides is 20.764 ft.; find the area. (7) Find the expense of walling-in a plot of land, in the shape of a regular hexagon, containing 1039-24 sq. yds., at 7s. 6d. per yd. (8) An entrance hall, which measures. 33 sq. yds. 4 sq. ft. 111.6 sq. in., is to be paved with mosaic-tiles, in the shape of a regular octagon, each side of which is 3 in. ; how many will be required? (9) The perimeter of a regular hexagon is 480 ft., and that of a regular octagon is the same; compare the areas of the two figures. IX. THE IRREGULAR POLYGON. Definition. An irregular polygon is a plane rectilineal figure having three or more sides, but whose sides or angles are not equal. RULE.-To find the area of an irregular polygon. Divide the figure, in the most convenient manner, into triangles, trapeziums, &c. Then the area of the polygon is the sum of the areas of the different figures. |