BY JAMES JOHONNOT (ADAPTED) In the year 1781 the war was chiefly carried on in the South, but the North was constantly troubled by bands of Tories and Indians, who would swoop down on small settlements and make off with whatever they c... Read more of A Brave Girl at Children Stories.caInformational Site Network Informational
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THE YORKSHIRE ESTATES.

(Patchwork Puzzles)
I was on a visit to one of the large towns of Yorkshire. While walking
to the railway station on the day of my departure a man thrust a
hand-bill upon me, and I took this into the railway carriage and read it
at my leisure. It informed me that three Yorkshire neighbouring estates
were to be offered for sale. Each estate was square in shape, and they
joined one another at their corners, just as shown in the diagram.
Estate A contains exactly 370 acres, B contains 116 acres, and C 74
acres.
Now, the little triangular bit of land enclosed by the three square
estates was not offered for sale, and, for no reason in particular, I
became curious as to the area of that piece. How many acres did it
contain?


Answer:

The triangular piece of land that was not for sale contains exactly
eleven acres. 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 of (4 x 370 x 116)
-(370 + 116 - 74) squared--that is a quarter of the square root of 1936, which
is one-quarter of 44, or 11 acres. 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.
[Illustration:
A
|
| .
| .
|5 .
| 7 .
E +--------- +C .
| | ` . .
| | `. .
|4 |4 ` . .
| 7 | ` ..
D----------+----------------- B
F
]
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 C. 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. Now, although 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 precision.
The area of our triangle ADB is clearly half of 9 x 17, or 761/2 acres.
The area of AEC is half of 5 x 7, or 171/2 acres; the area of CFB is half
of 4 x 10, or 20 acres; and the area of the oblong EDFC is obviously 4 x
7, or 28 acres. Now, if we add together 171/2, 20, and 28 = 651/2, and
deduct this sum from the area of the large triangle ADB (which we have
found to be 761/2 acres), what remains must clearly be the area of ABC.
That is to say, the area we want must be 761/2 - 651/2 = 11 acres exactly.










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