The applet below demonstrates the method found in Section 2.6 of the* Śulba-sūtra* of Baudhāyana (*BSS* 2.6) for transforming a rectangle into an isosceles trapezoid of the same area. Click “Go” to advance to the next step and “Reset” when the construction is completed. The dimensions of the given rectangle can be adjusted by sliding points \(A\) and \(D.\) The shorter side of the trapezoid can be adjusted by sliding point \(E.\)

Baudhāyana gave a method for transforming a rectangle into an isosceles trapezoid of area equal to that of the given rectangle. Below is a translation of the original source, which may be found in section 2.6 of *BSS* [Sen and Bag, p. 79].

*If it is desired to reduce one side of a square (that is, to make an isosceles trapezium), the reduced side is to be taken as the breadth (of a rectangular portion to be cut off from the square); the remaining part (of the square) is divided by the diagonal and (one half), after being inverted, is placed on the other side.*

First suppose that rectangle \(ABCD\) has been constructed, perhaps by the previously described method. The area of the given rectangle can be adjusted by sliding points \(A\) and \(D.\) First point \(E\) is marked on side \(AB.\) The length of \(AE\) is to be the smaller base of the isosceles trapezoid. Now point \(F\) can be marked on side \(CD,\) so that \(CF\) has the same length as \(AE.\) Note this forces segment \(AC\) to be parallel to segment \(EF\) and hence triangle \(DEF\) is indeed a right triangle.

As suggested by the original source, the next step is to reflect triangle \(DEF\) about a vertical line, and translate left, so that \(EF\) coincides with \(AC,\) thus constructing triangle \(ACG,\) congruent to \(DEF.\) To do this entirely with ropes, take two ropes of lengths \(DE\) and \(DF.\) Attach one end of the rope of length \(DE\) to point \(A,\) and one end of the rope of length \(DF\) to point \(C.\) Now pulling on the unattached ends of both cords until taut (to the left), they will meet at precisely one point: \(G.\)

The desired trapezoid is now marked by \(AEDG.\) To see it has the same area as rectangle \(ABDC,\) note that by congruence of triangles \(ACG,\) \(DEF,\) and \(BDE,\)

\[{\rm{Area}}(AEDG)={\rm{Area}}(ACG)+{\rm{Area}}(AEFC)+{\rm{Area}}(DEF)\]

\[={\rm{Area}}(AEFC)+{\rm{Area}}(DEF)+{\rm{Area}}(BDE)={\rm{Area}}(ABDC).\]

This can also be seen algebraically. The area of a trapezoid is the average of the base lengths multiplied by the height. That is,

\[{\rm{Area}}(AEDG)=AC\cdot\frac{1}{2}(AE+DG).\]

Now observe that \(DG=CG+CF+FD=CF+2\cdot FD,\) and using that \(AE=CF,\)

\[AE+DG=CF+(CF+2\cdot FD)=2(CF+FD)=2\cdot CD.\] Thus,

\[{\rm{Area}}(AEDG)=AC\cdot\frac{1}{2}(AE+DG)=AC\cdot CD.\]

It is important to note too from this formula that once we have fixed the dimensions of the rectangle, \({\rm{Area}}(AEDG)\) and \(AC\) are both fixed. Once \(AE\) is specified, the trapezoid dimensions are determined completely as \(DG\) can be solved for: \(DG=2\cdot CD-AE.\)

It is also interesting to note the degenerate case, when point \(E\) meets point \(A,\) corresponds to simply cutting up rectangle \(ABDC\) along its diagonal into two triangles and rearranging them.

Observe that this construction can be done entirely with ropes and stakes in the ground: the “heavy lifting” was all contained in the construction of the rectangle. Here we did not need to use the converse of the Pythagorean Theorem to infer that \(DEF\) was a right angle.

**Figure 3.** Construction of a falcon fire altar at an Athirathram ceremony in 2011. Notice the trapezoidal bricks in the lower left corner. (Photo from Wikimedia Commons attributed to Arayilpdas at Malayalam Wikipedia.)