Question 894333: A rectangle ABCD, which is 40 inches high by 20 wide, leans so that its top is 4 inches to the right of vertical.
ie corner A (top left) is 4 inches to the right of corner D (bottom left).
By how much must I lift corner C to make the sides AD and BC vertical?
(and how do you work it out accurately?)
The answer will be about 2 inches but that's logic, not maths.
Thank you for your help.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! your logic is correct.
the solution is 2.
i solved it using trigonometry, but it was quite messy and required the use of the calculator. the solution, however, turned out to be exactly 2.
i then looked at it geometrically and, while not obvious, it turns out that the triangles are similar and, since similar triangles have proportional sides, you get a ratio of x/20 = 4/40 and you wind up with x = 2, x being the height you are looking for. in the diagram shown below, x represents the length of DB.
the geometric logic has to do with parallel lines and vertical angles and complementary angles.
it's a bit of a mish mash of geometric relationships.
the following diagram with all the angles that are equal to each other that matter to the solution is shown below.
look below the diagram for further comments.
the bottom line on the geometric analysis is that triangle AFD is similar to triangle DGC because all their angles are equal.
since the triangles are similar, their corresponding sides are proportional.
FA is the side opposite angle FDA
DG is the side opposite angle GCD
AD is the side opposite angle AFD
DC is the side opposite angle DGC
their corresponding sides are proportional.
you get 4/40 = DG / 20
solve for DG and you get DG = 20*4/40 = 80/40 = 2
that's your solution.
point C has to be raised up to the height of point D in order for the rectangle to be vertically aligned.
the height of point D from point G is equal to the length of DG which is equal to 2.
i didn't show you the trigonometric solution because it appears you are in a geometry course rather than a trigonometry course.
the trigonometry solution does confirm the geometric solution however, so no problem there.
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