I don't know if this is the optimal, but an isosceles triangle
with base and height $\sqrt{2}$ overlaps 
$2 \left(\sqrt{2}-1\right) \approx 0.828427$
when placed as below,
and so improves over $\frac{3}{4}$:
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[![SquareTri][1]][1]
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  [1]: https://i.sstatic.net/ZwomP.jpg