Henry Ernest Dudeney/Modern Puzzles/130 - Mr. Grindle's Garden/Solution/Proof 2

by : $130$

 * Mr. Grindle's Garden

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:

and so: