Henry Ernest Dudeney/Modern Puzzles/130 - Mr. Grindle's Garden/Solution

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Modern Puzzles by Henry Ernest Dudeney: $130$

Mr. Grindle's Garden
"My neighbour," said Mr. Grindle, "generously offered me, for a garden,
as much land as I could enclose with four straight walls measuring $7$, $8$, $9$ and $10$ rods in length respectively."
"And what was the largest area you were able to enclose?" asked his friend.
Perhaps the reader can discover Mr. Grindle's correct answer.


Solution

Just under $71$ square rods.


Proof 1

Let $\AA$ square rods be the area of the garden.

We are given that $\AA$ is the greatest possible for the given sides.

We also have that the sides of the garden are in arithmetic sequence.


Thus:

\(\ds \AA\) \(=\) \(\ds \sqrt {7 \times 8 \times 9 \times 10}\) Greatest Area of Quadrilateral with Sides in Arithmetic Sequence
\(\ds \) \(=\) \(\ds \sqrt {5040}\)
\(\ds \) \(\approx\) \(\ds 70.99\)

$\blacksquare$


Proof 2

Let $\AA$ square rods be the area of the garden.

We are given that $\AA$ is the greatest possible for the given sides.

From Area of Quadrilateral with Given Sides is Greatest when Quadrilateral is Cyclic, the quadrilateral formed by the sides of the garden is cyclic.

Hence we can apply Brahmagupta's Formula:

$\AA = \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} }$

where $s$ is the semiperimeter:

$s = \dfrac {a + b + c + d} 2$

Thus:

\(\ds s\) \(=\) \(\ds \dfrac {7 + 8 + 9 + 10} 2\)
\(\ds \) \(=\) \(\ds 17\)

and so:

\(\ds \AA\) \(=\) \(\ds \sqrt {\paren {17 - 7} \paren {17 - 8} \paren {17 - 9} \paren {17 - 10} }\)
\(\ds \) \(=\) \(\ds \sqrt {10 \times 9 \times 8 \times 7}\)
\(\ds \) \(=\) \(\ds \sqrt {5040}\)
\(\ds \) \(\approx\) \(\ds 70.99\)

$\blacksquare$


Also see


Sources

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