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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&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 
[![SquareTri][1]][1]
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**Added**. A tetrahedron formed from the above triangle
and a point over the center of the cube. Just an image&mdash;no computations,
no claims:
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&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 
[![CubeTetra][2]][2]
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<sup>
Tetrahedron with same unit-area isosceles base, and height $3$.
</sup>
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  [1]: https://i.sstatic.net/ZwomP.jpg
  [2]: https://i.sstatic.net/4Mk3c.jpg