[Śulba-Sūtras] Give a method to convert an isosceles trapezium to a rectangle using dissection.

Question

Homi · Accepted Answer

We can convert an isosceles trapezium into a rectangle by cutting a triangle from one end and reattaching it to the other.

Step 1 — Understanding the Isosceles Trapezium

Let us consider an isosceles trapezium ABCD. Its parallel sides are AB and DC. Let the length of the shorter parallel side AB be $a$ . Let the length of the longer parallel side DC be $b$ . Let $h$ be the perpendicular distance between AB and DC. This $h$ is the height of the trapezium. The non-parallel sides AD and BC are equal in length. We can find the area of this trapezium using the formula:

$Area_{trapezium} = \frac{1}{2} \times (AB + DC) \times h$

$Area_{trapezium} = \frac{1}{2} \times (a + b) \times h$

$\boxed{Area_{trapezium} = \frac{(a+b)h}{2} \text{ square units}}$

Diagram 1

Step 2 — Dissecting to Form a Rectangle

We will now cut the trapezium and rearrange its parts. First, we drop perpendiculars from vertices A and B to the base DC. Let these perpendiculars meet DC at points E and F respectively. So, AE and BF are both equal to the height $h$ . This divides the trapezium into three parts:

A right-angled triangle $\triangle$ ADE on the left.
A rectangle AEFB in the middle.
A right-angled triangle $\triangle$ BFC on the right. Since it is an isosceles trapezium, $\triangle$ ADE and $\triangle$ BFC are congruent. The length of the segment EF is equal to the top base $a$ . The lengths of the segments DE and FC are equal.

$DE = FC = \frac{DC - AB}{2}$

$DE = FC = \frac{b - a}{2}$

Now, we cut off the triangle $\triangle$ ADE from the left side. We take this cut triangle $\triangle$ ADE. We move it to the right side of the remaining shape. We place it such that its side AE aligns with the side BF. The vertex A will now be at B, and E will be at F. The side DE of the triangle will extend the base FC. The new shape formed is a rectangle. Let us find the dimensions of this new rectangle. The height of the rectangle is the same as the height of the trapezium.

$Height_{rectangle} = h$

The length of the rectangle is the sum of the length EF and the length FC.

$Length_{rectangle} = EF + FC$

$Length_{rectangle} = a + \frac{b - a}{2}$

$Length_{rectangle} = \frac{2a + b - a}{2}$

$Length_{rectangle} = \frac{a + b}{2}$

The area of this new rectangle is:

$Area_{rectangle} = Length_{rectangle} \times Height_{rectangle}$

$Area_{rectangle} = \frac{a + b}{2} \times h$

$\boxed{Area_{rectangle} = \frac{(a+b)h}{2} \text{ square units}}$

We see that the area of the rectangle is the same as the area of the original trapezium. This method successfully converts an isosceles trapezium into a rectangle by dissection.

Diagram 2

Answer

The method to convert an isosceles trapezium to a rectangle using dissection is as follows: (i) Draw perpendiculars from the endpoints of the shorter parallel side to the longer parallel side. This divides the trapezium into a central rectangle and two congruent right-angled triangles at its ends. (ii) Cut off one of these right-angled triangles. (iii) Move this cut triangle and attach it to the other end of the remaining shape, aligning its height with the side of the central rectangle. This forms a new rectangle whose length is the average of the two parallel sides of the trapezium, and whose width is the height of the trapezium.

Question 32

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