Math Forum / Mathematics / Undergraduate Math / July 2008

Lets assume that the angles are labeled, so you know which go with
which side.

(i) If you have one side (base) and the two adjacent angles to it, then
generally, these will lead to two rays.   Assuming these are not
parallel, then each line parallel to the base will have a unique length
- - - - and the length of the other parallel side will determine a unique
height.

(ii)  Of course, if the two rays you created above are parallel, then
the two parallel sides are equal and there is no unique height.

In case (i) where the lines intersect, then you can:
- - - - find the height of the point of intersection (using trig);
- - - - use the ratio of the two lengths of the parallel sides to scale the
height from the vertex to the second side;
- - - - subtract these two heights to get the perpendicular height between
the two parallel sides.

It is easy to see how this process breaks down completely, in case (ii).

Walter Whiteley

i have a trapazoid of which i know the lengths of the two paralel sides
and all of the angles, is there any way to find the perpendicular
height

Suppose that AB//CD, the other two sides that are not parallel are BC and AD,
Let alpha and beta denote the angle <ADC and <BCD.
Let a , b and h denote AB, CD and the trapezoid altitude.

Draw line BE // AD.

In triangle BCE,
h/tan(alpha) + h/tan(beta) = b - a

So
h = (b - a)/[1/tan(alpha) + 1/tan(beta)]
= (b - a)tan(alpha)tan(beta)/[tan(alpha) + tan(beta)]

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