In the diagram below, the length of each of the sides is given in terms of $x$, where $x \in \mathbb{N}.$ Show that there is only one value of $x$ for which this triangle is right-angled. - Junior Cycle Mathematics - Question 14 - 2021

To demonstrate that there is only one value of $x$ for which the triangle is right-angled, we can use the Pythagorean theorem. According to this theorem, if we denote the sides of the triangle as follows:

• Side 1: $x + 4$ (the hypotenuse)
• Side 2: $x - 5$ (one leg)
• Side 3: $x - 4$ (the other leg)

We can set up the equation:

$(x + 4)^2 = (x - 5)^2 + (x - 4)^2$

Now, expanding both sides:

1. Expanding the left side: $(x + 4)^2 = x^2 + 8x + 16$

2. Expanding the right side: $(x - 5)^2 + (x - 4)^2 = (x^2 - 10x + 25) + (x^2 - 8x + 16) = 2x^2 - 18x + 41$

Setting these two expansions equal gives us:

$x^2 + 8x + 16 = 2x^2 - 18x + 41$

Rearranging this leads to:

$0 = x^2 - 26x + 25$

Factoring this quadratic equation, we can find:

$0 = (x - 25)(x - 1)$

Thus, the possible values of $x$ are $x = 25$ and $x = 1$.

Since $x \in \mathbb{N}$, we can now substitute these values back:

1. For $x = 1$:

   • Side 1: $1 + 4 = 5$
   • Side 2: $1 - 5 = -4$ (not valid)
   • Side 3: $1 - 4 = -3$ (not valid)

2. For $x = 25$:

   • Side 1: $25 + 4 = 29$
   • Side 2: $25 - 5 = 20$
   • Side 3: $25 - 4 = 21$

This results in: $29^2 = 20^2 + 21^2$ $841 = 400 + 441$

Therefore, $x = 25$ is the only valid solution, and this shows that there is only one value of $x$ for which the triangle is right-angled.