History of Hindu Mathematics: A Source Book v2.c17.s19.21

v2.c17.s19.21

Let *a*, *b*, *c*, *h*, denote respectively the face, base, equal sides and altitude of the known isosceles trapezium and let *a'*, *b'*, *c'*, *h'*, denote the corresponding quantities of the required isosceles trapezium. Then, since the two trapeziums are equal in area and altitude, we must have $h'=h,$ $b'+a'=b+a, 1$ and $c'²-((b'-a')/2)²=h²,$ or $c'+(b'-a')/2c'-(b'-a')/2=h²,$ whence $c'-(b'-a')/2=r,$ and $c'+(b'-a')/2=h²/r,$ *r* being any rational number. Then $c'=½(h²/r+r), 2$ $b'-a'=(h²/r-r). 3$