Question 1209520
Here's how to find the area of triangle QAU:

**1. Visualize the problem:**

It's helpful to draw a diagram.  You have a square SQUR with side length *x*.  Triangle SQE is equilateral, meaning all its sides are also of length *x*.  Point A is likely somewhere related to the geometry of the square and triangle.  We need more information about the location of point A to solve the problem.  Let's assume A is the intersection of lines RE and QU.

**2. Find the coordinates of the points (assuming a coordinate system):**

Let's place the square in a coordinate system.  Put S at the origin (0,0), Q at (x,0), U at (x,x), and R at (0,x).

Since SQE is equilateral, point E will be at (x/2, (x√3)/2).

**3. Find the equations of lines RE and QU:**

* **Line RE:**  We have points R(0,x) and E(x/2, (x√3)/2).  The slope of RE is:
   m_RE = [(x√3)/2 - x] / [x/2 - 0] = (√3 - 2)x / x = √3 - 2

   The equation of line RE (using point-slope form) is:
   y - x = (√3 - 2)(x - 0)
   y = (√3 - 2)x + x
   y = (√3 - 1)x

* **Line QU:** We have points Q(x,0) and U(x,x).  The slope of QU is undefined since it is a vertical line. The equation of the line is x=x.

**4. Find the coordinates of point A (the intersection of RE and QU):**

Since QU is the vertical line x = x, the x-coordinate of point A is *x*.  Substitute *x* for x in the equation of line RE to find the y-coordinate:
y = (√3 - 1)x

So, point A is at (x, (√3 - 1)x).

**5. Find the area of triangle QAU:**

The base of triangle QAU is QU, which has length *x*.  The height of the triangle is the perpendicular distance from A to the line QU. Since QU is a vertical line, the height is the horizontal distance from x to x, which is 0. The height of the triangle is the difference in the y-coordinates of A and Q. Thus the height is (√3 - 1)x - 0 = (√3 - 1)x.

Area of triangle QAU = (1/2) * base * height
Area = (1/2) * x * (√3 - 1)x
Area = (x²(√3 - 1))/2

Therefore, the area of triangle QAU is (x²(√3 - 1))/2.