Henry Ernest Dudeney/Puzzles and Curious Problems/202 - A Garden Puzzle/Solution

Puzzles and Curious Problems by Henry Ernest Dudeney: $202$

A Garden Puzzle

The four sides of a garden are known to be $20$, $16$, $12$ and $10$ rods,

and it has the greatest possible area for these sides.

What is the area?

Solution

Approximately $194.4$ square rods.

Proof

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:

\(\ds s\)

\(=\)

\(\ds \dfrac {20 + 16 + 12 + 10} 2\)

\(\ds \)

\(=\)

\(\ds 29\)

and so:

\(\ds \AA\)

\(=\)

\(\ds \sqrt {\paren {29 - 20} \paren {29 - 16} \paren {29 - 12} \paren {29 - 10} }\)

\(\ds \)

\(=\)

\(\ds \sqrt {9 \times 13 \times 17 \times 19}\)

\(\ds \)

\(=\)

\(\ds \sqrt {37 \, 791}\)

\(\ds \)

\(\approx\)

\(\ds 194.4\)

$\blacksquare$

Also see

• $130$ - Mr. Grindle's Garden in Dudeney's Modern Puzzles collection of $1926$, which is the same as this but for the specific numbers.

Sources

• 1932: Henry Ernest Dudeney: Puzzles and Curious Problems ... (previous) ... (next): Solutions: $202$. -- A Garden Puzzle
• 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $268$. A Garden Puzzle