Let ABC be an equilateral triangle whose side has length 3. There is a square inscribed in the triangle, with one of its sides lying on side AB. What is the area of the square?

We are given an equilateral triangle \( \triangle ABC \) with a side length of 3 units, and we need to find the area of a square inscribed in the triangle. One of the sides of the square lies on side \( AB \) of the triangle.

### Step 1: Understanding the Problem
The square is inscribed in such a way that one of its sides lies on \( AB \), while the remaining sides of the square touch sides \( AC \) and \( BC \) of the triangle. The goal is to find the area of the square.

### Step 2: Using Geometry and Similar Triangles
Let the side length of the square be \( s \). The height of the equilateral triangle can be calculated using the Pythagorean theorem in one of its 30-60-90 right triangles:
\[
h_{\triangle ABC} = \frac{\sqrt{3}}{2} \times 3 = \frac{3\sqrt{3}}{2}
\]
Thus, the height of the triangle is \( \frac{3\sqrt{3}}{2} \).

Next, observe that the inscribed square divides the triangle into two regions: the square itself and a smaller triangle above it. This smaller triangle is similar to the original equilateral triangle.

### Step 3: Relations Between Heights
The height of the smaller triangle is the total height of the original triangle minus the side length of the square \( s \). Therefore, the height of the smaller triangle is:
\[
h_{\text{small triangle}} = \frac{3\sqrt{3}}{2} - s
\]
Since the smaller triangle is similar to the larger triangle, the ratio of the side lengths of these triangles is the same as the ratio of their heights. Let \( x \) be the side length of the smaller triangle. Then we have the proportion:
\[
\frac{x}{3} = \frac{h_{\text{small triangle}}}{h_{\triangle ABC}} = \frac{\frac{3\sqrt{3}}{2} - s}{\frac{3\sqrt{3}}{2}}
\]
Simplifying the right-hand side:
\[
\frac{x}{3} = 1 - \frac{2s}{3\sqrt{3}}
\]
Thus, the side length of the smaller triangle is:
\[
x = 3\left(1 - \frac{2s}{3\sqrt{3}}\right) = 3 - \frac{2s}{\sqrt{3}}
\]

### Step 4: Solving for \( s \)
At this point, note that the base of the smaller triangle plus the side length of the square must equal the total base of the equilateral triangle:
\[
x + s = 3
\]
Substitute \( x \) from the previous equation:
\[
\left(3 - \frac{2s}{\sqrt{3}}\right) + s = 3
\]
Simplifying:
\[
3 - \frac{2s}{\sqrt{3}} + s = 3
\]
\[
-\frac{2s}{\sqrt{3}} + s = 0
\]
Factor out \( s \):
\[
s\left(1 - \frac{2}{\sqrt{3}}\right) = 0
\]
Solve for \( s \):
\[
s = \frac{3\sqrt{3}}{5}
\]

### Step 5: Calculating the Area of the Square
The area \( A \) of the square is simply the square of its side length:
\[
A = s^2 = \left(\frac{3\sqrt{3}}{5}\right)^2 = \frac{27}{25}
\]

Thus, the area of the square is \( \frac{27}{25} \).

Let me know if you would like more details or further clarification on any step!

### Related Questions:
1. How do you find the height of an equilateral triangle given its side length?
2. Can you explain how similarity of triangles was used in this problem?
3. What are the properties of 30-60-90 triangles used in geometric problems?
4. How do you calculate the area of a square if given its side length?
5. How does the Pythagorean theorem apply to equilateral triangles?

**Tip:** When solving geometry problems involving inscribed shapes, always look for similar triangles or proportional relationships to simplify calculations.

Discover how to calculate the area of a square inscribed in an equilateral triangle with a side length of 3 units. This problem involves geometric concepts such as similarity of triangles, properties of equilateral triangles, and the Pythagorean theorem. Suitable for students in Grades 9-12.

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