[Śulba-Sūtras] Give a method to convert a rectangle into an isosceles triangle by dissection.

Question

Homi · Accepted Answer

We will cut the rectangle into two pieces and rearrange them to form an isosceles triangle that has the same area.

Step 1 — Setting up the Problem

Let us start with a rectangle. Let its length be $L$ and its width be $W$ . The area of this rectangle is $L \times W$ . We want to change this rectangle into an isosceles triangle. An isosceles triangle is a triangle with two sides of equal length. Its area is calculated as $(1/2) \times \text{base} \times \text{height}$ . We need the area of the triangle to be equal to the area of the rectangle. So, we must have $L \times W = (1/2) \times \text{base} \times \text{height}$ . Let us choose the height of our new triangle to be the width of the rectangle, $W$ . Then, we can write the equation as: $L \times W = \frac{1}{2} \times \text{base} \times W$ We can cancel $W$ from both sides of the equation. $L = \frac{1}{2} \times \text{base}$ So, the base of our isosceles triangle must be $2L$ . We will construct an isosceles triangle with base $\mathbf{2L}$ and height $\mathbf{W}$ .

Step 2 — The Dissection Process

Let us draw our rectangle and label its corners A, B, C, and D. Let side AB be the length $L$ , and side BC be the width $W$ . First, we cut the rectangle along its diagonal AC. This divides the rectangle into two identical right-angled triangles. These are $\triangle ABC$ and $\triangle ADC$ .

Diagram 1

Now, we extend the side AB in a straight line to a new point E. We make sure that the length of BE is equal to the length of AB. So, BE = $L$ .

Diagram 2

Next, we take the triangle $\triangle ADC$ . We move this triangle and place it next to $\triangle ABC$ . We place it so that side AD of $\triangle ADC$ sits exactly on top of side BC of $\triangle ABC$ . Also, side DC of $\triangle ADC$ will now align with the extended line segment BE. So, point D moves to C, point C moves to E, and point A moves to B. The triangle $\triangle ADC$ becomes $\triangle BCE$ . The new shape formed by combining $\triangle ABC$ and $\triangle BCE$ is a large triangle, $\triangle ACE$ . The base of this new triangle is AE. $AE = AB + BE$ $AE = L + L$

$\boxed{AE = 2L}$ The height of this triangle is the perpendicular distance from C to AE, which is BC. $Height = BC$ $\boxed{Height = W}$ Since point C is directly above point B, and B is the midpoint of AE, the triangle $\triangle ACE$ is an isosceles triangle. Its two equal sides are AC and CE. Let us check the area of this new isosceles triangle. Area of $\triangle ACE$ = Area of $\triangle ABC$ + Area of $\triangle BCE$ . Area of $\triangle ABC$ : $= \frac{1}{2} \times \text{base} \times \text{height}$ $= \frac{1}{2} \times AB \times BC$ $= \frac{1}{2} \times L \times W$ Area of $\triangle BCE$ : $= \frac{1}{2} \times \text{base} \times \text{height}$ $= \frac{1}{2} \times BE \times BC$ $= \frac{1}{2} \times L \times W$ Total area of $\triangle ACE$ : $= \left(\frac{1}{2} \times L \times W\right) + \left(\frac{1}{2} \times L \times W\right)$ $= L \times W$ $\boxed{\text{Area of } \triangle ACE = L \times W}$ This is the same as the area of the original rectangle. So, we have successfully converted the rectangle into an isosceles triangle by dissection.

Diagram 3

Answer

(i) The method involves cutting the rectangle along its diagonal and rearranging one of the resulting triangles. (ii) The rectangle ABCD (length L, width W) is cut along diagonal AC, forming $\triangle ABC$ and $\triangle ADC$ . (iii) Side AB is extended to E such that BE = L. $\triangle ADC$ is moved to form $\triangle BCE$ . The resulting shape $\triangle ACE$ is an isosceles triangle with base $2L$ and height $W$ , having the same area as the original rectangle.

Question 27

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