<s>[INST] Let $ABCD$ be a square with sidelength 1 and $AKL$ is an equilateral triangle where $K$ lies on BC and L lies on CD. What is the area for AKL? I tried to use the Pythagorean Theorem to find one side of $AKL$, but there isn't enough information. What am I missing?
 [/INST] Let $a$ be the sidelength of the triangle $AKL$. Then $K$ and $L$ are the intersection points between the circle $\omega$ centered in $A$ with radius $a$ and $BC$ and $CD$ respectively. Since both the circle $\omega$ and the square $ABCD$ are symmetric with respect to the line $AC$ then $|BK| = |DL|$.
Let $b = |BK|$. Then by applying the Pythagorean Theorem on $ABK$ we get
$$a^2 = 1 + b^2$$
and by applying it on $KCL$ we get
$$a^2 = 2(1-b)^2 = 2 - 4b + 2b^2$$
Subtracting the two equations and solving for $b$ we get
$$b = 2 \pm \sqrt{3}.$$
Note that $0 < b < 1$ so we have to discard the option $2+\sqrt{3}$. Then
$$b = 2 - \sqrt{3}.$$
Now
$$|AKL| = |ABCD| - |ABK| - |KCL| - |ALD| = 1 - \dfrac{b}{2} - \dfrac{(1-b)^2}{2} - \dfrac{b}{2} = \dfrac{1-b^2}{2}$$
$$ = \boxed{2\sqrt{3} - 3}$$
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