Solve for X: Triangular Lengths in an Isosceles Right Triangle with Area 12.5 cm²

The problem involves finding the side lengths of an isosceles right triangle given its area. Let's proceed with the solution.

Since the triangle is an isosceles right triangle, the two legs are equal, and the area $A$ is provided by the formula: $A = \frac{1}{2} \cdot \text{base} \cdot \text{height}$ For this triangle, both the base and the height are $x + 5$.

The area is given as 12.5, so we set up the equation: $12.5 = \frac{1}{2} \cdot (x + 5) \cdot (x + 5)$ $12.5 = \frac{1}{2} \cdot (x + 5)^2$

Multiply both sides by 2 to solve for $(x + 5)^2$: $25 = (x + 5)^2$

Take the square root of both sides: $x + 5 = \sqrt{25}$ $x + 5 = 5$

Solve for $x$: $x = 5 - 5$ $x = 0$

Therefore, the length of each leg of the triangle is $x + 5 = 5$ cm.

For the hypotenuse $c$, use the Pythagorean theorem $c^2 = a^2 + a^2$, where $a = x + 5 = 5$: $c^2 = 5^2 + 5^2 = 25 + 25 = 50$ $c = \sqrt{50} = 5\sqrt{2}$

Thus, the lengths of the sides of the triangle are $5$, $5$, and $5\sqrt{2}$.

Therefore, the correct solution is $5, 5, 5\sqrt{2}$.