286 Answers- THE GARDEN PATH
The area of the path is exactly 66¥! square yards, which is clearly seen if
you imagine the little triangular piece cut off at the bottom and removed to
the top right-hand comer. Here is the
proof. The area of the garden path is 5S"
55 X 40 = 2,200. And (53YJ X 40) +
66¥.J also equals 2,200. Finally, the
sum of the squares of 53YJ and 40
must equal the square of 66¥!, as it 4P
does.
The general solution is as follows:
Call breadth of rectangle B, length of
rectangle L, width of path C, and
length of path x.Then x= ± By(B2 - C2)(B2 B2_C2 + LZ) + C2LZ - BCL
In the case given above x = 66¥!, from which we find the length, 53 YJ.
- THE GARDEN BED
Bisect the three sides in A, B, and
E. If you join AB and drop the perpen-
diculars AD and BC, then ABCD will
be the largest possible rectangle and
exactly half the area of the triangle.
The two other solutions, FEAG and
KEBH, would also serve (all these
rectangles being of the same area)
except for the fact that they would G
enclose the tree. This applies to any
triangle with acute angles, but in the
case of a right-angled triangle there are only two equal ways of proceeding.
- A PROBLEM FOR SURVEYORS
A rectilinear figure of any number of sides can be reduced to a triangle of