Trapazoid geometry

[tompy97]

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

[Walter Whiteley]

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

[kleins...@att.net]

You can turn it into an algebra problem.

Suppose the trapezoid is ABCD with AB parallel to DC. (It is easier to
visualize if you draw it with AB the longer base and the other base DC "above" AB.)

Sketch the altitudes from both C and D - CE and DF. The length of these
altitudes is what you want - call it x.

Using right triangle trig, you can get the length of AF in terms of x
[tan(A) = x/AF so AF = x/tan(A)]
Similarly you can get the length of EB.
Since DCEF is a rectangle you know that the length of EF = DC.

Now you can write an equation in terms of x stating that the length of AB equals the length of AF + length CD + length EB.

Now you have an algebra problem. The numbers are probably not pretty, but it is just a linear equation.

Rich Kleinschmidt

[Peter Chang]

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)]

http://www.idealmath.com