# Area of a Trapezoid

Let's take a crack at explaining the formula for the area of all trapezoids: 1/2h(b1 + b2).

We will use the most typical definition for "trapezoid": A quadrilateral that has exactly one pair of parallel sides. You can see the parallel sides and the non-parallel sides on the trapezoid below.

There are three important parts of a trapezoid: the bottom base, labeled b1, the top base, labeled b2, and the height, labeled h. You can see all of these in the figure.

The two dotted line segments (which include the non-parallel sides of the trapezoid) are fixed to the bottom horizontal line. You can imagine that both of these line segments can slide left or right and/or rotate on the points shown, stretching or shrinking the top and/or bottom of the trapezoid as they move and changing the angle(s) of one or both of the non-parallel sides. For a little warm-up you can mentally play with this idea in the figure above, sliding and swiveling the non-parallel sides. We don't mess with the parallel sides, though, because "exactly one pair of parallel sides" is essential for "trapezoid-ness." If you like, you can imagine rotating the parallel lines together, so they are not horizontal. As long as they are parallel, this will not affect our explanation.

Let's draw a line inside the trapezoid above that will separate it into two triangles.

You can see that both of the triangles have the same perpendicular height, h. What's interesting is that one of the triangles has b1 as its base, and the other triangle has b2 as its base. And this is key: We want to show that ANY trapezoid can be separated into two triangles (one with a base of b1, and the other with a base of b2) that both have the same height.

Okay, so let us swivel the dotted line on the right outward and leave the dotted line on the left alone.

By doing this, we stretch just the top of the trapezoid (b2). We also stretch that red line that divides the figure into two triangles. Yet, as you can see, we still have two triangles with the same height, and one triangle has a base of b1, while the other has a base of b2.

So far, so good. Now let us swivel and slide both of the dotted lines outward.

Here we have stretched the top and bottom of the trapezoid along with the red dividing line, but we still have two triangles with the same height, one with a base of b1, and the other with a base of b2.

It is possible to swivel and slide our way (inward) into a triangle, a figure that obviously does not meet our definition of "trapezoid." Similarly, when we swivel both dotted lines out far enough, we create a parallelogram (in this case, a rectangle), which also does not meet our definition of "trapezoid" because then the figure would have two pairs of parallel sides. (The formula still works for parallelograms. Since the two bases are congruent, the formula becomes 1/2h(2b) which simplifies to bh.)

And, in fact, we can stop there. We have shown (with the help of your imagination) that every possible trapezoid can be divided into two triangles with the same height, one with a base of b1 and the other with a base of b2.

But, you might ask, what about continuing the outward swiveling of our dotted lines? We can continue to rotate them outward, creating different trapezoids.

The reason we don't need to deal with these examples is because they are all inversions (basically, upside-down cases) of all the possibilities we have already seen. The image below is, for example, an inversion of the second image I showed above.

And, of course, as I mentioned, it does not matter how we rotate our parallel lines. The same explanation applies.

So, every trapezoid can be separated into two triangles with the same height, one with a base of b1 and the other with a base of b2.

To find the area of a trapezoid, then, we add the areas of the triangles. The area of one of the triangles is 1/2(b1)(h), and the area of the other is 1/2(b2)(h). If we add these together, we get 1/2(b1)(h) + 1/2(b2)(h).

And a little fancy distributive property pencilwork reveals 1/2h(b1 + b2).