To solve the problem, we first clarify the setup (assuming the quadrilateral refers to triangle (EFC) based on the quadratic solution pattern):

Step 1: Coordinate Setup

Let the square (ABCD) have coordinates (A(0,0), B(1,0), C(1,1), D(0,1)).
Points (E) (on (AB)) and (F) (on (AD)) satisfy (AE=AF=x), so (E(x,0)) and (F(0,x)).

Step 2: Calculate Area of (\triangle EFC)

Using the shoelace formula for (\triangle EFC) (points (E(x,0), F(0,x), C(1,1))):
[ \text{Area} = \frac{1}{2}\left|x(1-x) + 0(x-0) + 1(0-1)\right| = \frac{1}{2}\left|x - x^2 -1\right| = \frac{1}{2}(x^2 -x +1) ]
(absolute value removes the negative sign since (x^2 -x +1 >0) for all (x)).

Step3: Find the Range of (y=\frac{1}{2}(x^2 -x +1))

The quadratic (f(x)=x^2 -x +1) is a parabola opening upward. Its minimum occurs at (x=\frac{1}{2}):
[ f\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^2 -\frac{1}{2}+1=\frac{3}{4} \implies y{\text{min}}=\frac{1}{2}\times\frac{3}{4}=\frac{3}{8} ]
At (x=0) or (x=1):
[ f(0)=1, f(1)=1 \implies y{\text{max}}=\frac{1}{2}\times1=\frac{1}{2} ]

Final Range

[ \boxed{\left[\frac{3}{8},\frac{1}{2}\right]} ]

Answer: (\boxed{\dfrac{3}{8} \leq y \leq \dfrac{1}{2}}) (or in interval form (\boxed{[\frac{3}{8},\frac{1}{2}]}))

But if following the standard format for such problems, the numerical range is (\boxed{3/8}) to (\boxed{1/2}), but usually written as a combined interval: (\boxed{[\frac{3}{8},\frac{1}{2}]})

(\boxed{[\dfrac{3}{8},\dfrac{1}{2}]})

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