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Could You Pass This Test?

Thump;821891; said:
I interpret "intersect" to mean they cross each other like an "X."

Then that is not correct.

i.e. in right Triangle ABC, there is no crossing, but there are three intersections. (one of which is a perpendicular one)

I think the key to this is in the question, what angle is created by the perpendicular segments in question.
 
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Buckeye Buh Nim;821897; said:
Then that is not correct.

i.e. in right Triangle ABC, there is no crossing, but there are three intersections. (one of which is a perpendicular one)

I think the key to this is in the question, what angle is created by the perpendicular segments in question.

From my math textbook:

Intersecting lines- Lines that cross at exactly one point.
 
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Jeffcat;821570; said:
i have to laugh at myself there because somewhere along the lines i added 9+16 to get 27 when it is 25 so the answer is 5. i honestly believe i am becoming dyslexic in some way because i added it originally as 19+6 and the other day i put the milk on the cereal rack and the cereal in the fridge.:crazy:
The question is even more of a gimme than that. It's not necessary to add anything if you recognize the 3:4:5 right triangle.
 
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the difference here is that lines in the problem exist in a 3-dimensional space. consider a line, A, that is perpendicular to a plane. now imagine another line, B, on that plane that does not intersect A. A and B are still perpendicular, even though the lines do not intersect.
 
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OSU_Buckguy;821958; said:
the difference here is that lines in the problem exist in a 3-dimensional space. consider a line, A, that is perpendicular to a plane. now imagine another line, B, on that plane that does not intersect A. A and B are still perpendicular, even though the lines do not intersect.
True. Another way to think of it is that the angle between skew lines (lines that don't intersect) is the angle between one of the lines and a line parallel to the other line that intersects the first line.
 
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OSU_Buckguy;821958; said:
the difference here is that lines in the problem exist in a 3-dimensional space. consider a line, A, that is perpendicular to a plane. now imagine another line, B, on that plane that does not intersect A. A and B are still perpendicular, even though the lines do not intersect.

I understand what they are asking for, I think they are incorrect in the phrasing of the question. The segments do not intersect, and are not perpendicular. A projection of those lines may be, but those particular segments as specified are not perpendicular. A better question would be to define the planes and to prove the planes intersect, but not the segments.

FWIW, it is just an easy copout from trying to prove the statement.
 
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Buckeye Buh Nim;821995; said:
I understand what they are asking for, I think they are incorrect in the phrasing of the question. The segments ... are not perpendicular. A projection of those lines may be, but those particular segments as specified are not perpendicular.

You are being far too strict with your reading of a particular dictionary definition here. By another definition, Merriam's says perpendicular means "being at right angles to a given line or plane" and says nothing about intersecting: those segments/lines are at right angles even without actually touching one another.
 
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BayBuck;822086; said:
You are being far too strict with your reading of a particular dictionary definition here. By another definition, Merriam's says perpendicular means "being at right angles to a given line or plane" and says nothing about intersecting: those segments/lines are at right angles even without actually touching one another.

At what plane does this angle exist? Can you name the angle?
 
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Buckeye Buh Nim;822138; said:
At what plane does this angle exist? Can you name the angle?

As has been said, for the purposes of determining a 90-degree right angle without direct intersection of a line segment, you can extrapolate the lines into planes using line B-B1 or D-D1 as a reference.
 
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