We can solve this problem by setting up an equation based on the Pythagorean Theorem. The Pythagorean Theorem states that for a right triangle with legs $a$ and $b$, and hypotenuse $c$, the following relationship holds:

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

Let:

• $x$ be the length of the shorter leg.
• The longer leg will be $x + 2$ ft (since it's 2 feet more than the shorter leg).
• The hypotenuse will be $2x - 2$ ft (since it's 2 feet less than twice the shorter leg).

Now, using the Pythagorean Theorem:

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

Step 1: Expand both sides

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

Expanding the squares:

$x^2 + (x^2 + 4x + 4) = (4x^2 - 8x + 4)$

$x^2 + x^2 + 4x + 4 = 4x^2 - 8x + 4$

Step 2: Combine like terms

$2x^2 + 4x + 4 = 4x^2 - 8x + 4$

Subtract $2x^2 + 4x + 4$ from both sides:

$0 = 2x^2 - 12x$

Step 3: Factor the quadratic equation

$2x(x - 6) = 0$

This gives us two possible solutions:

$x = 0 \quad \text{or} \quad x = 6$

Since the length of the shorter leg cannot be zero, we conclude that:

$x = 6$

Step 4: Find the lengths of the other sides

• Shorter leg: $x = 6$ ft
• Longer leg: $x + 2 = 6 + 2 = 8$ ft
• Hypotenuse: $2x - 2 = 2(6) - 2 = 10$ ft

Final Answer:

• Shorter leg = 6 ft
• Longer leg = 8 ft
• Hypotenuse = 10 ft

Would you like more details or have any questions?

Here are 5 related questions to explore:

1. How would the solution change if the longer leg were 3 feet more than the shorter leg?
2. Can you find the perimeter and area of the triangular garden?
3. How would you solve the problem if the hypotenuse were exactly twice the shorter leg?
4. What is the ratio between the sides of this right triangle?
5. Can the Pythagorean theorem be applied to non-right triangles?

Tip: When solving geometry problems, always sketch a figure to visualize the relationships among the different sides.