A plot consists of a rectangular garden measuring 8 m by 10 m, surrounded by a path of constant width, as shown in the diagram - Junior Cycle Mathematics - Question 9 - 2015

Setting the total area equal to 143 m², we have:

$4x^2 + 36x + 80 = 143$

This simplifies to:

$4x^2 + 36x - 63 = 0$

In Elaine's diagram, the total width of the plot, including the path, is given by:

• Width = $8 + 2x$ m
• Length = $10 + 2x$ m

The area given by Elaine is the product of the length and the width equal to the total area:

$(8 + 2x)(10 + 2x) = 143$

Expanding this gives:

$80 + 16x + 20x + 4x^2 = 143$

This simplifies to:

$4x^2 + 36x - 63 = 0$

We can factor the equation as follows:

$4x^2 + 36x - 63 = 0$

Factoring yields:

$(2x - 3)(2x + 21) = 0$

Thus:

Either $x = \frac{3}{2} = 1.5$ or $x = -21$ (not valid as width must be positive). Therefore, the width of the path is $1.5$ m.

Tony's approach involves testing values for x:

Let’s choose $x = 1, 2, 3$:

• For $x = 1$: $Area = (8+2(1))(10+2(1)) = (8+2)(10+2) = 10 imes 12 = 120 ext{ m}^2$
• For $x = 2$: $Area = (8+2(2))(10+2(2)) = (8+4)(10+4) = 12 imes 14 = 168 ext{ m}^2$
• For $x = 3$: $Area = (8+2(3))(10+2(3)) = (8+6)(10+6) = 14 imes 16 = 224 ext{ m}^2$

Since the total area is $143$ m², the closest estimates are between $x = 1$ and $x = 2$. Tony might conclude by estimating $x = 1.5$.

Answer: Elaine's method is best.

Reason: Although all three methods yield the correct result, Elaine's method is the quickest and simplest. Tony's method relies on trial and error rather than systematic solving, while Kevin's method involves more complex calculations.