Ancient Indian Rope Geometry in the Classroom - Transforming a Rectangle into a Square of Equal Area

A method for constructing a square with area equal to that of a given rectangle is given in both the Śulba-sūtras of Baudhāyana and Āpastamba. This will use the previous technique of finding a square whose area is the difference of the areas of two given squares. Below is a translation of the original text from Āpastamba's Śulba-sūtra, verse 2.7 [Plofker2, p. 22]:

Wishing [to make] an oblong quadrilateral an equi-quadrilateral [square]: Cutting off a [square part of the rectangle] with [its] width, [and] halving the remainder, put [the halves] on two [adjacent] sides [of the square part]. Fill in the missing [piece] with an extra [square]. Its removal [has already been] stated.

Figure 8. This applet demonstrates the technique found in Āpastamba's Śulba-sūtra for transforming a rectangle into a square with equal area. Click "Go" to advance to the next step. The area of the given rectangle may be adjusted by sliding points $B$ and $D.$ The area of square $RTSJ$ may also be changed by sliding point $I.$ The desired area is achieved when point $I$ meets side $CB.$

Consider a rectangle $ADCB.$ Let us assume $AB$ is the smaller side. In the GeoGebra applet in Figure 8, the reader may adjust the dimensions of the rectangle by moving points $B$ and $D.$

First, mark points $H$ and $K$ on sides $AD$ and $CD,$ respectively, so that $AHKB$ forms a square. Then find points $E$ and $M$ so that remaining rectangle $HDCK$ is divided into two congruent rectangles $HEMK$ and $EDCM.$

Next, construct $BKGJ$ to the right of, and adjacent to $AHKB,$ congruent to the rectangles $HEMK$ and $EDCM.$  Now observe that the sum of the areas of $AHKB,$ $HEMK,$ and $BKGJ$ is equal to the area of the original rectangle $ADCB.$

Then construct square $KMFG.$ Now the area of square $AEFJ,$ less the area of $KMFG,$ is equal to the area of the original rectangle $ADCB.$ Thus our goal becomes to find the difference of squares $AEFJ$ and $KMFG.$

We now proceed with the established technique for this. Find where the circular arc with center $J$ and radius $FJ$ intersects the segment $BM.$ In the GeoGebra applet in Figure 8, slide point $I$ along the arc. As point $I$ approaches segment $BM,$ the area of $RTSJ$ approaches the area of $ADCB.$ Indeed, when point $I$ lies on segment $BM,$ we apply the Pythagorean Theorem to triangle $JIB$ and obtain

$${\rm{Area}}(RTSJ)=IB^2=JI^2-JB^2=(JI-JB)(JI+JB).$$

But by congruence, $JI=FJ=EA,$ and $JB=EH$ and therefore,

$$JI-JB=EA-EH=HA=AB$$

and $$JI+JB = EA+EH = EA + DE = AD,$$

from which we infer ${\rm{Area}}(RTSJ)=(AB)(AD),$ as desired.

It is interesting to consider when the constructed square $RTSJ$ will have side length equal to that of $EH.$ For when this occurs, the side $RT$ of $RTSJ$ will lie precisely on side $CB$ of $ADCB.$ Setting $X=AB$ and $Y=AD,$ note that $$EH = \frac{Y-X}{2},$$ and the equation ${\rm{Area}}(RTSJ)=(AB)(AD)$ can be rewritten as $$\left(\frac{Y-X}{2}\right)^2=XY$$ or $$X^2-2XY+Y^2=4XY$$ or $$X^2-6XY+Y^2=0.$$

Viewing the last equation as a quadratic in $X$ and solving for $X/Y$ we obtain

$$X=\frac{6Y-\sqrt{36Y^2-4Y^2}}{2}= (3 – 2\sqrt{2})Y,$$ or

$$\frac{X}{Y}=\frac{AB}{AD}= 3 – 2\sqrt{2}.$$

Consequently, the constructed square $RTSJ$ has side $RT$ lying on side $CB$ of $ADCB$ $-$ equivalently, $RTSJ$ has side length equal to that of $EH = {(AD-AB)}/{2}$ $-$ precisely when ${AB}/{AD} = 3-2\sqrt{2},$ which the reader may verify by changing the dimensions of the original rectangle $ADCB$ in the GeoGebra applet in Figure 8.