1. ## [ASK] G to BH

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. $\displaystyle 4\sqrt2$ cm
C. $\displaystyle 4\sqrt3$ cm
D. $\displaystyle 8\sqrt2$ cm
E. $\displaystyle 8\sqrt3$ cm

I got $\displaystyle \frac{8}{3}\sqrt6$ cm. Do you guys get the same answer?

2. ## Re: [ASK] G to BH

Since there is no image, I will describe what I think you are looking for. If we have the cube sitting on a flat surface, there are four vertices on the top and four vertices on the bottom (against the flat surface). If we start at the top and start labeling clockwise, then move to the bottom (right below the label A) and again start labeling clockwise (looking from above), we wind up with A is above E, B is above F, C is above G, and D is above H. So, BH is the diagonal from the top back left corner to the bottom front right corner. The distance from G to BH is the distance from G to the middle of the cube, which is half the length of BH. Let's look at the triangle BFH. This is a right triangle. We know the length of BF = 8. Let's calculate the length of FH. That is $\displaystyle 8\sqrt{2}$. So, BH has length $\displaystyle \sqrt{3\cdot 64} = 8\sqrt{3}$. Half of that is choice C.

3. ## Re: [ASK] G to BH

Originally Posted by SlipEternal
The distance from G to BH is the distance from G to the middle of the cube, which is half the length of BH.
In the middle? Shouldn't the distance from a point to a line is if we make a line from the point to the line so that both lines are perpendicular?

4. ## Re: [ASK] G to BH

Originally Posted by Monoxdifly
In the middle? Shouldn't the distance from a point to a line is if we make a line from the point to the line so that both lines are perpendicular?
Yes, that occurs in the exact middle of the cube.

5. ## Re: [ASK] G to BH

But how? BCH isn't an isosceles triangle.

6. ## Re: [ASK] G to BH

Originally Posted by Monoxdifly
But how? BCH isn't an isosceles triangle.
What is your point? Not all right triangles are isosceles.

7. ## Re: [ASK] G to BH

Sorry, I mean BGH. My point is, if a line perpendicular to BH goes to the vertex G, the intersect point of that line to BH must not be exactly in the middle of BH since GH doesn't equal BG.

8. ## Re: [ASK] G to BH

Originally Posted by Monoxdifly
Sorry, I mean BGH. My point is, if a line perpendicular to BH goes to the vertex G, the intersect point of that line to BH must not be exactly in the middle of BH since GH doesn't equal BG.
We are clearly discussing different cubes. I told you how I was visualizing it. If you are visualizing a different cube, you either need to give the specific labeling or if it is the same cube then it works exactly as I described.

9. ## Re: [ASK] G to BH

Well, never mind. It's already solved.