<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] Put it on a cartesian plane $A = (0,0); B= (1,0); C=(1,1); D=(0,1)$ and $K = (1,k)$ and $L=(j, 1)$.
Then ... it all comes out in the wash.
$AK = \sqrt{k^2 + 1} = KL = \sqrt{(1-k)^2 + (1-j)^2} = AL = \sqrt{1 + j^2} ; 1>k > 0; 1>j> 0$
So $k^2 + 1 = k^2 + j^2 - 2(k+j) + 2 = 1+ j^2$
So $j = k$ and $k^2 - 4k + 1 = 0$ so $k = 2 \pm \frac{\sqrt{12}}{2}$ and (as $k < 1$) $k = 2 - \sqrt{3}$
So $AK = KL = AL = \sqrt{8 - 4\sqrt{3}}$
So area of triangle is $\frac 12 b*h = \frac 12 * AK * \frac {\sqrt{3}}2*AK = \frac {\sqrt{3}}4*AK^2 = \frac {\sqrt 3}4*(8 - 4\sqrt{3}) = 2\sqrt{3} - 3$.
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