To solve the problem, we first clarify the setup (assuming the quadrilateral refers to triangle (EFC) based on the quadratic solution pattern):
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)).
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)).
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}
]
[ \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}]})
（免責聲明：本文為本網站出于傳播商業(yè)信息之目的進行轉載發(fā)布，不代表本網站的觀點及立場。本文所涉文、圖、音視頻等資料的一切權利和法律責任歸材料提供方所有和承擔。本網站對此資訊文字、圖片等所有信息的真實性不作任何保證或承諾，亦不構成任何購買、投資等建議，據(jù)此操作者風險自擔。） 本文為轉載內容，授權事宜請聯(lián)系原著作權人，如有侵權，請聯(lián)系本網進行刪除。
