SOLUTION: A classroom is 20 m long, 12 m wide, and 7 m high. Find the distance from the northeast corner of the room to the southwest corner of the room at floor level.

Algebra ->  Pythagorean-theorem -> SOLUTION: A classroom is 20 m long, 12 m wide, and 7 m high. Find the distance from the northeast corner of the room to the southwest corner of the room at floor level.      Log On


   



Question 171038: A classroom is 20 m long, 12 m wide, and 7 m high. Find the distance from the northeast corner of the room to the southwest corner of the room at floor level.
Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
to see a picture of what i'm talking about, go to:
http://www.geocities.com/gonzo89p
and click on "171038 picture"
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your room is 20 meters long.
that would be AD, BC, EH, FG in the picture.
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your room is 12 meters wide.
that would be AE, BF, DH, CG in the picture.
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your room is 9 meters high.
that would be AB, EF, DC, HG in the picture.
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you are saying to find the distance from the northeast corner of the room to the southwest corner of the room at floor level.
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say the top of the picture is north and the bottom of the picture is south.
say the left of the picture is west and the right of the picture is east.
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northeast corner of the room would be at point G
southwest corner of the room would be at point A
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the northeast corner of the room is at ceiling level.
the southwest corner of the room is at floor level.
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this indicates that you want to find the length of the diagonal AG.
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this may not be what you want.
you may have wanted the diagonal on the floor level which would be AH.
doesn't matter.
in finding the length of the diagonal AG, we need to find the length of the diagonal AH anyway.
we'll do both and you pick the one you think you really wanted.
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we need to find the length of the diagonal AH first because that's in the plane of the diagonal AG and can be used to find that.
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the diagonal AH can be found by use of the pythagorean theorem since the triangle ADH is a right triangle (unless the house is skewed the dimensions form a rectangular solid).
the triangle ADH has one leg AD = 20 meters and the other leg DH = 12 meters.
the diagonal AH is the hypotenuse of this right triangle.
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the formula finding the hypotenuse of this right triangle is %28AH%29%5E2+=+%28AD%29%5E2+%2B+%28DH%29%5E2.
this formula becomes:
%28AH%29%5E2+=+%2820%29%5E2+%2B+%2812%29%5E2.
which becomes:
%28AH%29%5E2+=+400+%2B+144.
which becomes:
%28AH%29%5E2+=+544.
take square root of both sides of equation:
sqrt%28%28AH%29%5E2%29+=+sqrt%28544%29.
which becomes:
AH+=+23.32380758 meters.
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if AH is what you were looking for, then you are done.
if you wanted to find AG, then you need to go one step further.
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AH is one of the legs of right triangle AGH.
GH is the other one.
AG is the hypotenuse of this right triangle.
you know GH = 9 meters.
you solved for AH = 23.32380758 meters.
you need to solve for AG.
again, your formula is %28AG%29%5E2+=+%28AH%29%5E2+%2B+%28GH%29%5E2.
substituting known values, your formula becomes:
%28AG%29%5E2+=+%2823.32380758%29%5E2+%2B+%289%29%5E2
which becomes:
%28AG%29%5E2+=+544+%2B+81
which becomes %28AG%29%5E2+=+625
take square root of both sides of the equation:
sqrt%28%28AG%29%5E2%29+=+sqrt%28625%29
which becomes:
AG+=+25
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