Here we shear triangle ABC with a horizontal invariant line through C (ie through C and the point D). The resulting triangle has a vertical 'base' of 3 units and a horizontal 'height' of 4 units.

Note that we can also approach the task in a more static way, by simply splitting our original triangle along CD (see right). This produces two triangles, both with CD as base (4 units) and with heights of 1 unit and 2 units. Thus the total area is half x 4 x 1 + half x 4 x 2 = half x 4 x 3 unit squares. This could be thought of as an algebraic approach which uses the distributive law.

We could also split ABC into two triangles using a vertical cut through B
- what expression would this give us for the area?
Some related tasks can be found here.

In the movie on the next page we move point A only, in much the same way as we had earlier moved point C only, but this time along a line parallel to BC.