(a − x)y = k²,
| x | = | p | , |
| y | q |
k² being the given size of the first, and p and q the base and altitude of the parallelogram which determine the shape of the second of the required parallelograms.
If we substitute the value of y, we get
| (a − x)x = | pk² | , |
| q |
or,
ax − x² = b²,
where a and b are known quantities, taking b² = pk²/q.
The second case (Prop. 29) gives rise, in the same manner, to the quadratic
ax + x² = b².