Lloyd left a comment on a post of mine yesterday, asking:

how do you draw a Irregular quadrilateral trapezoid with fixed dimensions for the two parallel bases and the two legs with no angles given using geometry tools?

top base= 328

bottom base= 223

left leg =220

right leg= 215

How would you answer this question? It’s not trivial. You’ll quickly find that if you do a straight-edge and compass construction, **you’ll need the height of the trapezoid**.

If we let *a, b, c, *and *d* be the side lengths of a trapezoid with *a* and *c* as the bases, can we express the height *h* as

This number *is* constructable, but would take some work to actually construct it on paper. Perhaps we can return to that particular question later. For now, we can let GeoGebra show us the general idea. I’ve made this applet in which you can change the side lengths and the trapezoid will be constructed. I used the height formula above to calculate the height, and the applet shows this value.

*Footnote:*

Just to hit home my usual point one more time, the figure above is ALWAYS a trapezoid, even when sides *b* and *d* just happen to be parallel. Just remember that a parallelogram is a special case of a trapezoid!!

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No, you do not need the height.

Imagine a trapezoid. Draw a line parallel to a side (not a base) from a vertex not on that side. In principle, there are two such lines. One of these is inside the trapezoid. This line, the other side (the one adjacent to the line) and the difference of the bases form a triangle that could be constructed with straightedge and compass by SSS. Next, extend its base and draw through its apex another base. That’s it.

Wow, I just got totally schooled. Thanks, Alexander, for your comment. Your construction is so easy! I feel like an idiot :-). I’ll revise my post.

No reason for self-flagellation. Such blocks happen to the best of us.

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