Page:Amusements in mathematics.djvu/196

184 28=65½, and deduct this sum from the area of the large triangle A D B (which we have found to be 76½ acres), what remains must clearly be the area of A B C. That is to say, the area we want must be 76½—65½= 11 acres exactly.

190.—FARMER WURZEL'S ESTATE.

area of the complete estate is exactly one hundred acres. To find this answer I use the following little formula,$$\frac{\sqrt{4ab-(a+b-c)^2}}{4}$$ where $$a$$, $$b$$, $$c$$ represent the three square areas, in any order. The expression gives the area of the triangle A. This will be found to be 9 acres. It can be easily proved that A, B, C, and D are all equal in area; so the answer is 26 + 20 + 18 + 9 + 9 + 9 + 9=100 acres.

Here is the proof. If every little dotted square in the diagram represents an acre, this must be

a correct plan of the estate, for the squares of 5 and 1 together equal 26; the squares of 4 and 2 equal 20; and the squares of 3 and 3 added together equal 18. Now we see at once that the area of the triangle E is 2½, F is 4½, and G is 4. These added together make 11 acres, which we deduct from the area of the rectangle, 20 acres, and we find that the field A contains exactly 9 acres. If you want to prove that B, C, and D are equal in size to A, divide them in two by a line from the middle of the longest side to the opposite angle, and you will find that the two pieces in every case, if cut out, will exactly fit together and form A.

Or we can get our proof in a still easier way. The complete area of the squared diagram is 12x12=144 acres, and the portions 1, 2, 3, 4, not included in the estate, have the respective areas of 12½, 17½, 9½, and 4½. These added together make 44, which, deducted from 144, leaves 100 as the required area of the complete estate.

191.—THE CRESCENT PUZZLE.

to the original diagram, let A C be $$x$$, let C D be $$x-9$$, and let E C be $$x-5$$. Then $$x-5$$ is a mean proportional between $$x-9$$ and $$x$$, from which we find that $$x$$ equals 25. Therefore the diameters are 50 in. and 41 in. respectively.

192.—THE PUZZLE WALL.

answer given in all the old books is that shown in Fig. 1, where the curved wall shuts out the cottages from access to the lake. But in seeking the direction for the "shortest possible" wall most readers to-day, remembering that the shortest distance between two points is a straight line, will adopt the method shown in Fig. 2. This is certainly an improvement, yet the correct answer is really that indicated in Fig. 3. A measurement of the lines will show that there is a considerable saving of length in this wall.

193.—THE SHEEP-FOLD. is the answer that is always given and accepted as correct: Two more hurdles would be necessary, for the pen was twenty-four by one (as in Fig. A on next page), and by moving one of the sides and placing an extra hurdle at each end (as in Fig. B) the area would be doubled. The diagrams are not to scale. Now there is no condition in the puzzle that requires the sheep-fold to be of any particular form. But even if we accept the point that the pen was twenty-four by one, the answer utterly fails, for two extra hurdles are certainly not at all necessary. For example, I arrange the fifty hurdles as in Fig. C, and as the area is increased from twenty-four "square hurdles" to 156, there is now accommodation for 650 sheep. If it be held that the area must be exactly double that of the original pen, then I construct it (as in Fig. D) with twenty-eight hurdles only, and have twenty-two in hand for other purposes on the farm. Even if it were insisted that all the original hurdles must be used, then I should construct it as in Fig. E, where I can get the area as exact as any farmer could possibly require, even if we have to allow for the fact that the sheep might not be able to graze at the extreme ends. Thus we see that, from any