The triangular piece of land that was not for sale
contains exactly eleven
Of course it is not difficult to find
the answer if we follow the eccentric and tricky tracks of intricate
trigonometry; or I might say that the application of a well-known
formula reduces the problem to finding one-quarter of the square root
(4 × 370 × 116)
(370 + 116 - 74)²—that
is a quarter of the square root of 1936, which is one-quarter of 44, or
But all that the reader really requires to know is the
Pythagorean law on which many puzzles have been built, that in any
right-angled triangle the square of the hypotenuse is equal to the sum
of the squares of the other two sides.
I shall dispense with all
"surds" and similar absurdities, notwithstanding the fact that the
sides of our triangle are clearly incommensurate, since we cannot
exactly extract the square roots of the three square areas.
In the above diagram ABC represents our triangle.
ADB is a right-angled triangle, AD measuring 9 and BD measuring 17,
because the square of 9 added to the square of 17 equals 370, the known
area of the square on AB.
Also AEC is a right-angled triangle, and the
square of 5 added to the square of 7 equals 74, the square estate on A
Similarly, CFB is a right-angled triangle, for the square of 4 added
to the square of 10 equals 116, the square estate on BC.
the sides of our triangular estate are incommensurate, we have in this
diagram all the exact figures that we need to discover the area with
The area of our triangle ADB is clearly half of
9 × 17, or 76½ acres.
of AEC is half of 5 × 7, or 17½
acres; the area of CFB is half of 4 × 10,
or 20 acres; and the area of the oblong EDFC is obviously
4 × 7, or 28 acres.
Now, if we add together
17½, 20, and 28
= 65½, and deduct this sum from the area of the large
triangle ADB (which we have found to be 76½ acres), what
remains must clearly be the area of ABC.
That is to say, the area we
want must be 76½ - 65½ = 11