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

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Modern Puzzles by Henry Ernest Dudeney: $131$

The Garden Path
A man has a rectangular garden, $55$ yards by $40$ yards,
and he makes a diagonal path, one yard wide, exactly in the manner indicated in the diagram.
Dudeney-Modern-Puzzles-131.png
What is the area of the 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:

\(\text {(1)}: \quad\) \(\ds x^2\) \(=\) \(\ds B^2 + \paren {L - y}^2\) Pythagoras's Theorem
\(\text {(2)}: \quad\) \(\ds \dfrac y C\) \(=\) \(\ds \dfrac x B\) Definition of Similar Triangles
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds \dfrac {C x} B\)
\(\ds \leadsto \ \ \) \(\ds x^2\) \(=\) \(\ds B^2 + \paren {L - \dfrac {C x} B}^2\) substituting for $y$ in $(1)$
\(\ds \) \(=\) \(\ds B^2 + L^2 - \dfrac {2 L C x} B + \dfrac {C^2 x^2} {B^2}\) multiplying out
\(\ds \leadsto \ \ \) \(\ds \paren {1 - \dfrac {C^2} {B^2} } x^2 + \dfrac {2 L C} B x - \paren {B^2 + L^2}\) \(=\) \(\ds 0\) gathering terms
\(\ds \leadsto \ \ \) \(\ds \paren {B^2 - C^2} x^2 + 2 B L C x - B^2 \paren {B^2 + L^2}\) \(=\) \(\ds 0\) multiplying through by $B^2$
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \dfrac {-2 B L C \pm \sqrt {\paren {2 B L C}^2 + 4 \paren {B^2 - C^2} B^2 \paren {B^2 + L^2} } } {2 \paren {B^2 - C^2} }\) Quadratic Formula
\(\ds \) \(=\) \(\ds \dfrac {-2 B L C \pm 2 B \sqrt {\paren {L C}^2 + \paren {B^2 - C^2} \paren {B^2 + L^2} } } {2 \paren {B^2 - C^2} }\) extracting $2 B$ from the square root
\(\text {(3)}: \quad\) \(\ds \) \(=\) \(\ds \dfrac {\pm B \sqrt {\paren {B^2 - C^2} \paren {B^2 + L^2} + C^2 L^2} - B L C} {B^2 - C^2}\) simplification


Having done that, we now use the numbers given.

We have:

\(\ds B\) \(=\) \(\ds 40\)
\(\ds L\) \(=\) \(\ds 55\)
\(\ds C\) \(=\) \(\ds 1\)


Inserting these into $(3)$ above:

\(\ds x\) \(=\) \(\ds \dfrac {\pm 40 \sqrt {\paren {40^2 - 1^2} \paren {40^2 + 55^2} + 1^2 55^2} - 40 \times 55 \times 1} {40^2 - 1^2}\)
\(\ds \) \(=\) \(\ds \dfrac {\pm 40 \sqrt {\paren {1599} \paren {4625} + 3025} - 2200} {1599}\)
\(\ds \) \(=\) \(\ds \dfrac {40 \times 2720 - 2200} {1599}\) Positive root needed here
\(\ds \) \(=\) \(\ds 66 \tfrac 2 3\) evaluating

Finally:

$A = C x = 66 \tfrac 2 3$

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

$\blacksquare$


Sources

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