The Science of the Shulba: A Study in Early Hindu Geometry c12.8

c12.8

From the longer side *AC* cut off a portion *CE* equal to the breadth *CD* of the rectangle. Complete the square *CDFE*. Divide the remaining part *ABFE* of the given rectangle into two halves by the line *GH*. Take the remoter half *ABHG* and place it after inversion on the other side of the square *CDFE* in the position *DG'H'F*. Complete the square *CGPG'*, by adding the portion *HPH'F*. The given rectangle is easily found to be equal to the difference of the two squares *G'PG'C* and *HPH'F*. This difference can be found by the method taught before. That is, draw a circle with centre *G'* and radius *G'P* cutting *DH* at *M*. Draw *MN* perpendicular to *G'P*. Then $G'N² = G'M² - MN² = G'P² - HP²$ so that *G'N* is the side of the square which is equivalent to the given rectangle *ABDC*.