The volume of a right cylinder is v = πr2h. if we have an oblique cylinder, like in the figure, what is the volume of a cross-section? a) πr2h b) 4πr2h c) 1 10 πr2h d) 1 2 πr2h

Since you did not attach any picture we cannot say for sure what is the correct answer, but we can discuss the options in order to find the most probable correct answer.

First of all, according to the Cavalieri's principle, an oblique cylinder has the same volume as a right cylinder with the same base surface area and same height.
A cross-section of an oblique cylinder will be a small right cylinder with the same base surface area and a height as small as possible.

I guess the oblique cylinder has height h and it is divided into many (probably 10) cross-sections.

Option A: πr2h
This is exactly the volume of the right cylinder, therefore, unless you are given a cross-section of height h (which would be too easy), this won't be the correct answer.

Option B: 4πr2h
This is 4 times the right cylinder. Again, here the height of the cross-section should be 4h, but it doesn't sound like a possible data (too easy again).

Option C: 1 10 πr2h
Here comes a n issue with the notation: I think the right number you meant to write is (1/10)·πr2h and not 110·πr2h.
If I am right, this means that your oblique cylinder of height h is divided into 10 cross-sections, and therefore the volume of each of these cross-sections will be a tenth of the volume of the oblique cylinder, which means 1/10·πr2h.

Option D: 1 2 πr2h
Here, we have the same notation issue as before. I think you meant (1/2)·πr2h.
Here, your oblique cylinder height h should be divided into only 2 cross-sections. Now, we said the cross-section's height should be the smallest as possible, so an oblique cylinder divided only into two pieces doesn't sound good.

Therefore, the most probable correct answer will be C) (1/10)·πr2h

Skew lines(segments) are lines(segments) in 3 dimensions that do not intersect the given segment and are not parallel to it.  ae, bf, eg, and fh are your skew segments.  i hope that : )