MHB Calculating G's Distance from BH on an ABCD.EFGH Cube

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In an ABCD.EFGH cube with a side length of 8, the distance from point G to line BH is under discussion. The initial calculations yielded $$\frac{8}{3}\sqrt6$$ cm, which does not match any of the provided answer choices. Participants confirm that the cube's structure is understood as having the base ABCD with vertical sides EA, FB, GC, and HD. There is agreement on the interpretation of the cube's geometry. The conversation highlights a potential issue with the answer options provided.
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In an ABCD.EFGH cube whose side length is 8, the distance between the point G and the line BH is ...
A. 4 cm
B. $$4\sqrt2$$ cm
C. $$4\sqrt3$$ cm
D. $$8\sqrt2$$ cm
E. $$8\sqrt3$$ cm

I got $$\frac{8}{3}\sqrt6$$ cm. Do you guys get the same answer?
 
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Monoxdifly said:
In an ABCD.EFGH cube whose side length is 8, the distance between the point G and the line BH is ...
A. 4 cm
B. $$4\sqrt2$$ cm
C. $$4\sqrt3$$ cm
D. $$8\sqrt2$$ cm
E. $$8\sqrt3$$ cm

I got $$\frac{8}{3}\sqrt6$$ cm. Do you guys get the same answer?
I agree with you: $\frac83\sqrt6$. It's odd that in two separate problems the answer does not appear in the list of choices.

I am getting these answers on the assumption that "an ABCD.EFGH cube" means a cube where the base is ABCD, and the upper vertices are above the corresponding lower ones, so that EA, FB, GC and HD are the vertical sides. Presumably that is what is intended?
 
Opalg said:
I am getting these answers on the assumption that "an ABCD.EFGH cube" means a cube where the base is ABCD, and the upper vertices are above the corresponding lower ones, so that EA, FB, GC and HD are the vertical sides. Presumably that is what is intended?

Yes, that is what's intended.
 

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