NOTE: this is a still image, not a movie!

**NP-analysis**

This diagram shows the result of shear *N* then *P* on square ABCD.

Again, the diagram contains several 30˚60˚90˚ triangles so we can find various lengths with relative ease.

We might want to show that B'A', the image of BA under shear *N*, is perpendicular to line *n* (so that line *p* must be chosen to be perpendicular to *n*). This can be done by considering the quadrilateral BB'A'K, which has right-angles at B and K, and showing that BB' = KA'.

We might then want to consider the distance moved by A' under shear *P* (ie A'A'') and the distance of A' from line *p*, to find the shear factor for shear *P*.

Finally, we look at a classic shearing task: PS.