Answers 293you can double the area, as in D, with 28 hurdles only. If all the hurdles must
be used you might construct it as in Figure E.- THE ROSE GARDEN
Make AD a quarter of the distance
AB, and measure DE and AF each a
quarter of Be. Now, if we make G
the same distance from E that D is
from F, then AG is the correct width
of the path. If the garden is, for
example, 12 ft. by 5 ft., the path will
be 1 ft. wide, yet it cannot always beAG D Egiven in exact figures, though correct
in measurement.- CORRECfING A BLUNDER
The correctness of the diagram can be easily proved, since the squares of 15
and 20 equal the square of 25; the squares of 15 and 36 equal the square of
39; and the squares of 15 and 8 equal the square of 17. Also, 20 + 8 = 28. If
a right-angled triangle had been allowed, the one on the left, 15,25,20, would
itself give the solution, since the height on the base 25 would be 12, and the
median line 12~.
Perhaps our readers would like to try their hand at constructing the general
solution to triangles of this class.[Victor Meally has found a second solution to this problem: an obtuse tri-
angle with base 66, sides of 41 and 85, and altitude of 40. The line bisecting
the base is 58. In this case the altitude is exterior to the triangle, meeting the
extended base line to form a right triangle with a base of 9, and sides of 40 and
41.-M. G.]