if x be the width, x (x + ½) = 3,630. Solving this Quadratic, we find x = 60. Hence the dimensions are 60, 60½.
Twelve answers have been received—seven right and five wrong.
C. G. L., Nabob, Old Crow, and Tympanum assume that the number of yards in the length of the path is equal to the number of square-yards in the garden. This is true, but should have been proved. But each is guilty of darker deeds. C. G. L.'s "working" consists of dividing 3,630 by 60. Whence came this divisor, oh Segiel? Divination? Or was it a dream? I fear this solution is worth nothing. Old Crow's is shorter, and so (if possible) worth rather less. He says the answer "is at once seen to be 60 × 60½"! Nabob's calculation is short, but "as rich as a Nabob" in error. He says that the square root of 3,630, multiplied by 2, equals the length plus the breadth. That is 60.25 × 2 = 120½. His first assertion is only true of a square garden. His second is irrelevant, since 60.25 is not the square-root of 3,630! Nay, Bob, this will not do! Tympanum says that, by extracting the square-root of 3,630, we get 60 yards with a remainder of 30/60, or half-a-yard, which we add so as to make the oblong 60 × 60½. This is very terrible: but worse remains behind. Tympanum proceeds thus:—"But why should there be the half-yard at all? Because without it there would be no space at all for flowers. By means of it, we find reserved in the very centre a small plot of ground, two yards long by half-a-yard wide, the only