233. An inverse rule may be formed, sufficiently correct for every purpose, in the following way: If the year consisted of 360 days, or ³/₂ of 240, the subtraction of one-third from any sum per year would give the proportion which belongs to 240 days; and every pound so obtained would be one penny per day. But as the year is not 360, but 365 days, if we divide each day’s share into 365 parts, and take 5 away, the whole of the subtracted sum, or 360 × 5 such parts, will give 360 parts for each of the 5 days which we neglected at first. But 360 such parts are left behind for each of the 360 first days; therefore, this additional process divides the whole annual amount equally among the 365 days. Now, 5 parts out of 365 is one out of 73, or the 73d part of the first result must be subtracted from it to produce the true result. Unless the daily sum be very large, the 72d part will do equally well, which, as 72 farthings are 18 pence, is equivalent to subtracting at the rate of one farthing for 18_d._, or ½_d._ for 3_s._, or 10_d._ for £3. The rule, then, is as follows: To find how much per day will produce a given sum per year, turn the shillings, &c. in the given sum into decimals of a pound (221); subtract one-third; consider the result as pence; and diminish it by one farthing for every eighteen pence, or ten pence for every £3. For example, how much per day will give £224. 14. 0¾ per year? This is 224·703, and its third is 74·901, which subtracted from 224·703, gives 149·802, which, if they be pence, amounts to 12_s._ 5·802_d._, in which 1_s._ 6_d._ is contained 8 times. Subtract 8 farthings, or 2_d._, and we have 12_s._ 3·802_d._, which differs from the truth only about ¹/₂₀ of a farthing. In the same way, £100 per year is 5_s._ 5¾_d._ per day. 234. The following connexion between the measures of length and the measures of surface is the foundation of the application of arithmetic to geometry. [Illustration] Suppose an oblong figure, A, B, C, D, as here drawn (which is called a _rectangle_ in geometry), with the side A B 6 inches, and the side A C 4 inches. Divide A B and C D (which are equal) each into 6 inches by the points _a, b, c, l, m_, &c.; and A C and B D (which are also equal) into 4 inches by the points _f, g, h, x, y_, and _z_. Join _a_ and l, _b_ and _m_, &c., and _f_ and _x_, &c. Then, the figure A B C D is divided into a number of squares; for a square is a rectangle whose sides are equal, and therefore A _a f_ E is square, since A _a_ is of the same length as A _f_, both being 1 inch. There are also four rows of these squares, with six squares in each row; that is, there are 6 × 4, or 24 squares altogether. Each of these squares has its sides 1 inch in length, and is what was called in (215) _a square inch_. By the same reasoning, if one side had contained 6 _yards_, and the other 4 _yards_, the surface would have contained 6 × 4 _square yards_; and so on. [Illustration]