The following applet demonstrates a method, generalized from one in Section 2.7 of the* Śulba-sūtra* of Baudhāyana (*BSS* 2.7), for transforming a rectangle into a triangle of equal 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 \(D.\) Slide point \(X\) to \(X^{\prime}\) to make the area of triangle \(RTY\) equal to the area of \(ABDC.\) Slide point \(Y\) to \(Y^{\prime}\) to make the triangle isosceles.

Baudhāyana considered the problem of transforming a square into an isosceles triangle of equal area in section 2.7 of *BSS*. The original translation is below [p. 79, Sen and Bag].

*If it is desired to transform a square into (an isosceles) triangle, the square whose area is to be transformed is doubled and a pole fixed at the middle of its east side; two cords with their ties fastened to it (the pole) are stretched to the south-western and north-western corners (of the square); portions lying outside the cords are cut off.*

Here we will present a solution to the more general problem of transforming a rectangle into a triangle (not necessarily isosceles) of equal area. Suppose we have constructed rectangle \(ABDC,\) say with the previously described method in this article. Next construct rectangle \(EFHG,\) with twice the height of \(ABDC.\) This is of course easy to do with ropes and stakes once \(A\) is constructed. Note the area of \(EFHG\) is twice that of \(ABDC.\)

Now we find a square with area equal to that of the doubled rectangle. To do this, we appeal to the method found in Section 2.5 of Baudhāyana’s *Śulba-sūtra, *or equivalently, the method found in Section 2.7 of Āpastamba's *Śulba-sūtra. *First mark points \(P_4\) and \(P_5\) so that \(HGP_4P_5\) forms a square. Now add half of the length of segment \(EP_4\) onto \(GH\) to make segment \(GT.\) Now extend square \(HGP_4P_5\) to an “auxiliary square” with side \(GT,\) with other points \(P_i\) as indicated. Now consider a semicircular arc with center \(T\) and radius equal to the length of \(GT\) (the auxiliary square side length). Let \(X\) denote an arbitrary point on this arc. As shown, \(X\) determines a square \(QRTS.\) When \(X\) reaches \(X^{\prime},\) the area of \(QRTS\) is equal to the area of \(EFHG\) (which is twice that of rectangle \(ABDC\) that we started with).

The proof that \({\rm{Area}}(QRTS)={\rm{Area}}(EFHG)\) when \(X^{\prime}\) reaches \(X\) relies on the Pythagorean Theorem. Indeed, assume that \(X^{\prime}\) and \(X\) coincide. We can divide rectangle \(EFHG\) into three smaller sub-rectangles, thus obtaining

\[{\rm{Area}}(EFHG)={\rm{Area}}(EFP_2P_1)+{\rm{Area}}(P_1P_2P_5P_4)+ {\rm{Area}}( P_4P_5HG).\]

We can divide square \(P_1P_3TG\) into two sub-rectangles and two sub-squares with

\[{\rm{Area}}( P_1P_3TG)={\rm{Area}}(P_1P_2P_5P_4)+{\rm{Area}}( P_5P_6TH)+ {\rm{Area}}( P_4P_5HG)+{\rm{Area}}(P_2P_3P_6P_5).\]

Further, by symmetry, \({\rm{Area}}(P_1P_2P_5P_4)={\rm{Area}}(EFP_2P_1),\) which allows us to infer \[{\rm{Area}}( P_1P_3TG)={\rm{Area}}(EFHG)+{\rm{Area}}(P_2P_3P_6P_5).\]

Thus, this becomes a problem of “subtracting squares.” We seek a square with area equal to \[{\rm{Area}}( P_1P_3TG)-{\rm{Area}}(P_2P_3P_6P_5).\] Observe that \[{\rm{Area}}(P_2P_3P_6P_5)=HT^2\] and \[{\rm{Area}}( P_1P_3TG)=P_3T^2=XT^2.\]

The last equalities follow since \(P_3T\) and \(XT\) are radii of the same circular arc. By the Pythagorean Theorem applied to right triangle \(XTH,\) \(XH^2+HT^2=XT^2,\) so that

\[XH^2=XT^2-HT^2={\rm{Area}}( P_1P_3TG)-{\rm{Area}}(P_2P_3P_6P_5)= {\rm{Area}}(EFHG).\]

Finally observe that \(XH\) is the side length of \(QRTS.\) (For more details consult "Ancient Indian Rope Geometry in the Classroom" [Huffman and Thuong].) So we have \[{\rm{Area}}(QRTS)= {\rm{Area}}(EFHG)=2\cdot{\rm{Area}}(ABDC).\]

As a matter of fact, this was the most difficult part of the construction. Now take any point \(Y\) on side \(QS.\) The area of triangle \(RTY\) is half the area of \(QRTS.\) To see this, note that its base and height are both equal to the side length of \(QRTS,\) so that \[{\rm{Area}}(RTY) =\frac{1}{2}{\rm{(Base)(Height)}} =\frac{1}{2}{\rm{Area}}(QRTS)={\rm{Area}}(ABDC).\]

To make triangle \(RTY\) isosceles, simply slide point \(Y\) to \(Y^{\prime},\) the midpoint of segment \(QS.\)

**Figure 4.** A soma cart with an isosceles triangle forepart from an Agnicayana fire sacrifice ritual in 2011 in Panjal, Kerala. (Photo from Wikimedia Commons: https://commons.wikimedia.org/wiki/File:Soma_cart_used_in_Athirathram_2011.JPG)