To solve this problem, we can apply the Pythagorean theorem, which states:

$a^2 + b^2 = c^2$

where $a$ and $b$ are the lengths of the two shorter sides of a right triangle, and $c$ is the length of the hypotenuse (the longest side).

Given the triangle:

1. The length of one side is $x$.
2. The length of the other side is $x + 1$.
3. The hypotenuse (longest side) is $x + 7$.

So, we can set up the equation:

$x^2 + (x + 1)^2 = (x + 7)^2$

Let's expand and simplify this equation.

Step 1: Expand Each Term

1. $x^2$ remains as $x^2$.
2. $(x + 1)^2 = x^2 + 2x + 1$.
3. $(x + 7)^2 = x^2 + 14x + 49$.

Step 2: Substitute and Simplify

Substitute these into the equation:

$x^2 + (x^2 + 2x + 1) = x^2 + 14x + 49$

Combine like terms:

$2x^2 + 2x + 1 = x^2 + 14x + 49$

Move all terms to one side of the equation to set it to zero:

$2x^2 + 2x + 1 - x^2 - 14x - 49 = 0$

$x^2 - 12x - 48 = 0$

Step 3: Solve the Quadratic Equation

To solve $x^2 - 12x - 48 = 0$, we can use the quadratic formula:

$x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot (-48)}}{2 \cdot 1}$

Simplify under the square root:

$x = \frac{12 \pm \sqrt{144 + 192}}{2}$

$x = \frac{12 \pm \sqrt{336}}{2}$

Calculate $\sqrt{336} \approx 18.3$:

$x = \frac{12 \pm 18.3}{2}$

This gives two possible solutions:

1. $x = \frac{12 + 18.3}{2} = 15.15$ (rounded to 15.2)
2. $x = \frac{12 - 18.3}{2}$ (which would give a negative result, so we discard it as lengths cannot be negative).

So, $x \approx 15.2$ units.

Step 4: Calculate the Other Sides

1. $x = 15.2$
2. $x + 1 = 15.2 + 1 = 16.2$
3. $x + 7 = 15.2 + 7 = 22.2$

Final Answer:

• The length of the short side $x$ is 15.2 units.
• The length of the side $x + 1$ is 16.2 units.
• The length of the hypotenuse $x + 7$ is 22.2 units.

Would you like a further breakdown of each calculation step?

Here are five related questions to extend this topic:

1. What would happen to the side lengths if $x$ was a different value?
2. How does the Pythagorean theorem work for non-integer values?
3. Could there be other types of right triangles with a similar relationship between sides?
4. How can we verify these side lengths satisfy the Pythagorean theorem?
5. How do we use the quadratic formula to solve more complex equations?

Tip: When dealing with right triangles, always verify your results by reapplying the Pythagorean theorem to check if the sides satisfy the equation $a^2 + b^2 = c^2$.