The hypotenuse of a right-angled triangle is 1 metre less than twice the shortest side. If the third side is 1 metre more more than the shortest side, find the sides of the triangle.

Explanation

To solve the problem, we will use the information given about the sides of the right-angled triangle and apply the Pythagorean theorem. Let's denote the shortest side as \( s \).

### Step-by-Step Solution:

1. **Define the sides of the triangle:**
- Let the shortest side be \( s \) (in meters).
- The hypotenuse is given as "1 metre less than twice the shortest side," which can be expressed as:
\[ h = 2s - 1 \]
- The third side is described as "1 metre more than the shortest side," which can be expressed as:
\[ t = s + 1 \]

2. **Apply the Pythagorean theorem:**
According to the Pythagorean theorem, for a right-angled triangle:
\[ h^2 = s^2 + t^2 \]
Substituting the expressions for \( h \) and \( t \):
\[ (2s - 1)^2 = s^2 + (s + 1)^2 \]

3. **Expand both sides:**
- Left side:
\[ (2s - 1)^2 = 4s^2 - 4s + 1 \]
- Right side:
\[ s^2 + (s + 1)^2 = s^2 + (s^2 + 2s + 1) = 2s^2 + 2s + 1 \]

4. **Set the equation:**
Now we have:
\[ 4s^2 - 4s + 1 = 2s^2 + 2s + 1 \]

5. **Rearrange the equation:**
Move all terms to one side:
\[ 4s^2 - 4s + 1 - 2s^2 - 2s - 1 = 0 \]
Simplifying gives:
\[ 2s^2 - 6s = 0 \]

6. **Factor the equation:**
Factor out \( 2s \):
\[ 2s(s - 3) = 0 \]

7. **Solve for \( s \):**
This gives us two solutions:
\[ 2s = 0 \quad \text{or} \quad s - 3 = 0 \]
Thus, \( s = 0 \) or \( s = 3 \).

8. **Determine the valid solution:**
Since the shortest side cannot be zero (as it would not form a triangle), we take:
\[ s = 3 \]

9. **Find the lengths of the other sides:**
- The hypotenuse \( h \):
\[ h = 2s - 1 = 2(3) - 1 = 6 - 1 = 5 \]
- The third side \( t \):
\[ t = s + 1 = 3 + 1 = 4 \]

10. **Conclusion:**
The sides of the triangle are:
\[ s = 3 \text{ m}, \quad t = 4 \text{ m}, \quad h = 5 \text{ m} \]

### Final Answer:
The sides of the triangle are 3 m, 4 m, and 5 m.