The easiest way is to arrange the eighteen matches
as in Diagrams 1 and 2, making the length of the perpendicular AB equal
to a match and a half.
Then, if the matches are an inch in length,
Fig. 1 contains two square inches and Fig. 2 contains six square
inches—4 × 1½.
The second case (2) is a little more difficult to
The solution is given in Figs. 3 and 4. For the purpose of
construction, place matches temporarily on the dotted lines.
Then it will be seen that as 3 contains five equal equilateral
triangles and 4 contains fifteen similar triangles, one figure is three
times as large as the other, and exactly eighteen matches are used.