Henry Ernest Dudeney/Modern Puzzles/131 - The Garden Path/Solution

by : $131$

 * The Garden Path

Solution

 * $66 \tfrac 2 3$ square yards.

Proof
We solve the general case.

Let $L$ yards be the length of the garden.

Let $B$ yards be the breadth of the garden.

Let $C$ yards be the width of the path.

Let $x$ yards be the length of the boundary between the path and the rest of the garden.

Let $A$ square yards be the area of the path.

Let $y$ be the length of one of the sides of the path.

The path is a parallelogram bounded by two sides of length $x$ and two sides of length $y$.

From Area of Parallelogram:
 * $A = C x$

Then we have:

Having done that, we now use the numbers given.

We have:

Inserting these into $(3)$ above:

Finally:
 * $A = C x = 66 \tfrac 2 3$

Hence the area of the path is $66 \tfrac 2 3$ square yards.