Quadrature of the Triangle

Problem

Figure 1. Diagram of Qin Jiushao's triangle problem

This classic problem from the Shushu Jiuzhang (數書九章) is called ”Quadrature of the Triangle“ (三斜求積術 sān xié qiú jī shù).

《數書九章·田域類2》 問沙田一段，有三斜，其小斜一十三里，中斜一十四里，大斜一十五里。 里法三百步，欲知為田幾何？ 答曰：田積三百一十五頃。 術曰：以少廣求之。以小斜冪並大斜冪，減中斜冪，餘，半之，自乘於上。以小斜冪乘大斜冪，減上，餘，四約之為實。一為從隅，開平方得積。

Shushu Jiuzhang (Problems on Fields 2) There is a field of sand with three sides. The short side is 13 li, the medium side is 14 li, and the long side is 15 li. The li rule is 300 bu. How large is the field? Answer: The area of the field is 315 qing. Method: Take the square of the short side squared and add it to the square of the long side, then subtract the square of the medium side. Halve the remainder, then square the result and put this above. Take the square of the short side and multiply it to the square of the long side. Subtract what is above, and let one-quarter of this remainder be the radicand. The coefficient of the quadratic term is 1. Extract the square root to obtain the area.

Solution

In the Shushu Jiuzhang, Qin Jiushao gave the formula:

$$A = \sqrt{\frac{1}{4} \left[a^2 c^2 - {\left(\frac{a^2+c^2-b^2}{2} \right)}^{2} \right]}$$

where $A$ is the area of a triangle, $a$ is the short side, $b$ is the middle side, and $c$ is the long side. 

$$A^2 = \frac{1}{4} \left[a^2 c^2 - {\left(\frac{a^2+c^2-b^2}{2} \right)}^{2} \right]$$

$$A^2 - 7056 = 0$$

$$A = 84$$

The area Qin Jiushao gave is 315 qing. This is calculated by knowing that 1 li is 300 bu, 1 mu is 240 sq.bu, and 1 qing is 100 mu.

$$84 \;sq.li \left(\frac{300 \; bu}{1 \;li} \right) \left(\frac{300 \; bu}{1 \;li} \right) \left(\frac{1 \; mu}{240 \; sq.bu} \right) \left(\frac{1 \; qing}{100 \; mu} \right) = 315 \; qing$$

Derivation

Figure 2. Dissection of a scalene triangle

Let $c$ be the base, $h$ be the height, and $A$ be the area of the triangle, then

$$A = \frac{1}{2} ch.$$

From the diagram (Figure 2), determine two equations for $h^2$:

Equation 1 $h^2 = a^2 - d^2$

Equation 2 $h^2 = b^2 - {(c-d)}^{2}$

Equate both equations and solve for $d$:

$$a^2 - d^2 = b^2 - (c^2 - 2dc + d^2)$$

$$c^2 + a^2 - b^2 = 2dc$$

$$d = \frac{c^2+a^2-b^2}{2c}.$$

Substitute this result into Equation 1 and take the positive square root:

$$h^2 = a^2 - {\left(\frac{c^2+a^2-b^2}{2c}\right)}^{2}$$

$$h = \sqrt{a^2 - {\left(\frac{c^2+a^2-b^2}{2c}\right)}^{2}}.$$

The area of the triangle is therefore

$$A = \frac{1}{2}c \sqrt{a^2 - {\left(\frac{a^2+c^2-b^2}{2} \right)}^{2}}$$

or

$$A = \sqrt{\frac{1}{4} \left[a^2 c^2 - {\left(\frac{a^2+c^2-b^2}{2} \right)}^{2} \right]}.$$

Additional Note

One can obtain Heron's formula from Qin Jiushao's formula. Note that Heron of Alexandria (c. 10 AD - c. 70 AD) derived his formula with a different approach, using an inscribed circle.

$$A^2 = \frac{1}{4} \left[ac + \left(\frac{a^2+c^2-b^2}{2} \right) \right] \left[ac - \left(\frac{a^2+c^2-b^2}{2} \right) \right]$$

$$A^2 = \frac{1}{4} \left[\frac{2ac+a^2+c^2-b^2}{2}\right] \left[\frac{2ac - (a^2+c^2-b^2)}{2} \right]$$

$$A^2 = \frac{1}{4} \left[\frac{(a^2+2ac+c^2)-b^2}{2}\right] \left[\frac{b^2 - (a^2 - 2ac +c^2)}{2} \right]$$

$$A^2 = \frac{1}{4} \left[\frac{(a+c+b)(a+c-b)}{2}\right] \left[\frac{(b+a-c)(b-(a-c))}{2} \right]$$

$$A^2 = \frac{a+b+c}{2} \cdot\frac{a-b+c}{2} \cdot \frac{a+b-c}{2}\cdot \frac{-a+b+c}{2}$$

Define the semi-perimeter as $s =\frac{a+b+c}{2}$, then

$$\begin{cases} s-a = \frac{-a+b+c}{2} \\ s-b = \frac{a-b+c}{2} \\ s-c = \frac{a+b-c}{2} \end{cases}$$

Therefore,

$$A^2 = s (s-b) (s-c) (s-a)$$

and

$$A = \sqrt{s (s-a) (s-b) (s-c)}.$$