We are tasked with finding the side length of a square inscribed in an isosceles right triangle such that two of the square's vertices lie on the hypotenuse, and the other two vertices lie on the legs of the triangle. Let's proceed step by step:

Step 1: Geometry of the isosceles right triangle

1. Since the triangle is isosceles, the legs are equal, and the angles are $45^\circ$, $45^\circ$, and $90^\circ$.
2. The relationship between the hypotenuse ($h$) and the legs ($l$) of an isosceles right triangle is: $l = \frac{h}{\sqrt{2}}$ Given $h = 3 \, \text{m}$, the length of each leg is: $l = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \, \text{m}$

Step 2: Equation for the side length of the square

Let the side length of the square be $s$. The square is positioned such that:

• Two vertices lie on the legs of the triangle.
• The other two vertices lie on the hypotenuse.

The geometry involves a scaling factor:

1. The square's side $s$ splits each leg into two segments: one segment is $s$, and the other is the remaining portion of the leg after subtracting $s$.
2. The hypotenuse length is also affected by the projection of the square.

Step 3: Solving for $s$

Using the geometry of similar triangles formed by the square and the triangle:

• The smaller triangle above the square is similar to the original isosceles triangle.

From the similarity ratios: $\frac{s}{l - s} = \frac{l - s}{l}$

Expanding and simplifying: $s^2 = l(l - s)$ Substitute $l = \frac{3\sqrt{2}}{2}$: $s^2 = \frac{3\sqrt{2}}{2} \left(\frac{3\sqrt{2}}{2} - s\right)$ $s^2 = \frac{3\sqrt{2}}{2} \cdot \frac{3\sqrt{2}}{2} - \frac{3\sqrt{2}}{2} \cdot s$ $s^2 = \frac{18}{4} - \frac{3\sqrt{2}}{2} s$ $s^2 + \frac{3\sqrt{2}}{2}s - \frac{9}{2} = 0$

Step 4: Solve the quadratic equation

The quadratic equation is: $s^2 + \frac{3\sqrt{2}}{2}s - \frac{9}{2} = 0$

Using the quadratic formula: $s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = \frac{3\sqrt{2}}{2}$, and $c = -\frac{9}{2}$. Substituting: $s = \frac{-\frac{3\sqrt{2}}{2} \pm \sqrt{\left(\frac{3\sqrt{2}}{2}\right)^2 - 4(1)\left(-\frac{9}{2}\right)}}{2(1)}$ $s = \frac{-\frac{3\sqrt{2}}{2} \pm \sqrt{\frac{18}{4} + 18}}{2}$ $s = \frac{-\frac{3\sqrt{2}}{2} \pm \sqrt{\frac{18}{4} + \frac{72}{4}}}{2}$ $s = \frac{-\frac{3\sqrt{2}}{2} \pm \sqrt{\frac{90}{4}}}{2}$ $s = \frac{-\frac{3\sqrt{2}}{2} \pm \frac{3\sqrt{10}}{2}}{2}$ $s = \frac{-3\sqrt{2} \pm 3\sqrt{10}}{4}$

Step 5: Positive root

Since $s > 0$, we take the positive root: $s = \frac{-3\sqrt{2} + 3\sqrt{10}}{4}$

Final Answer:

The side of the square is: $s = \frac{-3\sqrt{2} + 3\sqrt{10}}{4} \, \text{m}$

If you'd like a decimal approximation or a detailed explanation for specific steps, let me know! 😊

Follow-up questions:

1. Why does the similarity of triangles help in this problem?
2. Can the side length of the square be calculated directly using another geometric property?
3. What happens to the square’s side if the hypotenuse length changes?
4. How does the hypotenuse constraint influence the square’s dimensions?
5. Can this problem be generalized for non-isosceles right triangles?

Tip:

Always check if the geometric shapes in a problem involve similar triangles or other proportional relationships; they often simplify complex setups!