TCM Geometry 1: Areas of Planar Figures - Part 1

**By: Tao Steven Zheng (鄭濤)**

## **Origin of Chinese Geometry**

Chinese geometry originated from the utilitarian practices of measurement. This included the surveying of land and estimations of the volume of earth used in construction works.

Mojing Kaogongji and Zhuangzi

There was a theoretical side Chinese geometry starting in the Warring States Period (476 - 221 BC). Philosophers of the Mohist, Daoist, and Legalist schools of thought began to ponder upon the nature of reality and its connection to human affairs.

Several paradoxes were recorded by the Daoist philosopher Zhuang Zhou (fl. 4th century BC). These paradoxes were attributed to his longtime friend and dialectician Hui Shi (c. 370 - 310 BC). In the final chapter of Zhang Zhou's work, *Zhuangzi*, what remains of Hui Shi's paradoxical arguments explicitly introduces the concepts of the infinite and the infinitesimal. Hui Shi called what is so vast that it has no exterior, the one of vastness, and what is so small it has no interior, the one of smallness. Hui Shi also created a paradox similar to the paradoxes of the Greek philosopher Zeno of Elea. One states:

一尺之棰，日截其半，萬世不竭

[A rod of one chi halved each day will not cease to be cut after ten thousand generations].

This paradox illustrates the limit of approaching the infinitesimal length. From the above examples, it appears many of these mathematical ideas were littered throughout isolated quotations and anecdotes instead of scientific treatises. Nevertheless, there is one exception: the *Mojing*, or the *Mohist canon*. The *Mojing* can be considered a treasure-trove of scientific thought native to China. What differentiates the Mohists from other schools of thought is their attitude towards the natural sciences. Many schools of thought only appreciate the utility of scientific knowledge; hence, disregarding the need for theory. The Mohist thinkers investigated the mechanics of nature out of the need to understand its fundamental principles. In doing so, the Mohists were driven to define and explain geometric concepts similar to the definitions in Euclid’s elements. With a more theoretical foundation, the Mohists extended their more organized knowledge to experiment and record the principles of simple machines, and ray optics. Nevertheless, with all this achievement, the Mohist school of thought, though considerably influential in its day, was largely dissolved by the end of the Warring States period. During this era of turmoil, kingdoms valued industrial and agricultural knowledge over scientific curiosity. This tragedy in the history of Chinese science was perhaps a reminder that the political environment can chart the destiny of an entire stream of intellectual growth.

## **Area Problems in Ancient China**

Since the study of area originated from the practice of measuring the size of land, it comes to no surprise that all planar figures were referred to as fields (田 *tian*) in ancient China. Chapter 1 of the *Jiuzhang Suanshu*, entitled *fangtian*, includes many problems on computing the areas of planar figures. The formulas for calculating areas were expressed purely in words, with exact formulas for rectangles, triangles, trapezoids, the circle, and the annulus (ring-shape). Approximate formulas for the area of the circle, annulus, and segmented arch were also given. The table below details the name of planar shapes in ancient China.

Within the Chinese theoretical framework of measurement, the area can be called *ji* (積) or *mi* (冪). The first character *ji* (積), which literally translates to "accumulation", refers to the product of two or more numbers. In ancient China, the character *ji* (積) not only refers to the product in an arithmetic sense, but when applied to geometry can refer to an area or a volume. The second character *mi* (冪) describes the geometric notion of area. Hence, it can be said that, pertaining to the computation of areas, the *ji* (積) is the quantitative representation of an area, and the *mi* (冪) is the qualitative visualization of the same area. It is important to note that ancient Chinese mathematicians discuss the units of measurements by the context of the problem. What determines whether the unit of measurement describes a length, an area, or a volume depends on the nature of the geometry problem. Units of area and volume were not distinguished with the terms "square" (平方 *ping fang*) or "cubic" (立方 *li fang*); hence, what would be considered "square *bu*" or "cubic *chi*" are still called “*bu*” and “*chi*”.

## **Area of the Rectangle**

   In ancient China, the rectangle is called a "rectangular field" (方田 *fang tian*), or more simply "field" (田 *tian*). The *Jiuzhang Suanshu* describes how the area of a rectangle is calculated, and refers this as the "rectangular field method" (方田術 *fang tian shu*).

Let

A
,
a
,
b

{\displaystyle A, a, b }

denote the area, width, and length of the rectangle respectively, the formula for computing the area of the rectangle is

# A
a
b

{\displaystyle A = ab }

. The above method introduces two more units of area: the *mu* (畝), which is equal to 240 [square] *bu* (步); and the *qing* (頃), which is equal to 100 *mu*.

### **Problem Study 1: Rectangular Fields**

答曰：一畝。

[1.02] 又有田廣十二步，從十四步。問：為田幾何？

答曰：一百六十八步。

Answer: 1 *mu*.

[1.02] Suppose there is a field (田 *tian*) of width 12 *bu* wide and length 14 *bu*. Question: what is the area of the field?

Answer: Answer: 168 [square] *bu*.

**Solution for *Jiuzhang Suanshu* [1.01]**

Substitute the given width and length into the area formula:

A

=
15

b
u
×
16

b
u

=
240

b

u
2

.

{\displaystyle \begin{align}
A &= 15 \; bu \times 16 \; bu \\
&= 240 \; bu^2.
\end{align}}

Since

1

m
# u
240

b

u
2

{\displaystyle 1 \; mu = 240 \; bu^2 }

, the area of the rectangular field is

1

m
u

{\displaystyle 1 \; mu }

.

**Solution for *Jiuzhang Suanshu* [1.02]**

Substitute the given width and length into the area formula:

A

=
12

b
u
×
14

b
u

=
168

b

u
2

.

{\displaystyle
\begin{align} A &= 12 \; bu \times 14 \; bu \\
&= 168 \; bu^2.
\end{align} }

## **Out-In Complementary Method**

One versatile method that the Chinese mathematicians used to derive the areas of shapes is called the “out-in complementary principle” (出入相補 *chu ru xiang bu*). This method can also be generalized to prove the equivalence of areas of dissimilar shapes, by showing that the “excess” (盈 *ying*) may be dissected and rearranged to fill-in the deficiency (虚 *xu*). To demonstrate the out-in complementary principle, consider a rectangle divided into uneven quadrants as shown in Figure 4-2. Colour one quadrant red and its diagonally opposite quadrant blue. Then draw a diagonal from one pair of opposite corners that divides the larger rectangle into two equivalent right triangles. This produces two additional pairs of smaller triangles with equal areas. This fact shows, rather visually and intuitively, that the red and blue rectangles have equal areas despite having different dimensions. The out-in complementary method can be used to derive other geometric formulas related to right triangles.

## **Area of the Triangle**

In ancient China, the triangle was called *gui tian* (圭田); however, the character gui (圭) does not mean triangle. A *gui* (圭) is a pentagonal jade tablet carried by monarchs and officials in court or during rituals. These tablets were long handheld objects with a triangular pointed tip (often times equilateral or isosceles triangles).

Chinese mathematicians derived the area of a triangle by applying the "out-in complementary principle" (出入相補 *chu ru xiang bu*). Figure 3 illustrates two triangles of base

a

{\displaystyle a}

 and height

b

{\displaystyle b}

. Let the area of the triangles be

A

{\displaystyle A}

. For the triangle on the left, the half-triangle excess (盈 *ying*) can be cut, detached, and then attached to fill-in the deficiency (虚 *xu*), forming a rectangle of base

1
2

a

{\displaystyle \frac{1}{2} a }

 and height

b

{\displaystyle b}

. Thus, the area of the original triangle is equal to the area of the newly formed rectangle:

# A

1
2

a
×
b

{\displaystyle A = \frac{1}{2} a \times b }

For the triangle on the right, when the two excess areas (盈 *ying*) are cut, detached, and then filled-into each deficiency areas (虚 *xu*), it forms a rectangle of base

a

{\displaystyle a}

 and height

1
2

b

{\displaystyle \frac{1}{2} b }

. Once again, the area of the original triangle is equal to the area of the rectangle:

# A
a
×

1
2

b

{\displaystyle A = a \times \frac{1}{2}b }

Both cases gives the same formula for the area of a triangle:

# A

1
2

a
b

{\displaystyle A= \frac{1}{2} ab }

### **Problem Study 2: Triangular Fields**

答曰：一百二十六步。

[1.26] 又有圭田廣五步、二分步之一，從八步、三分步之二。問：為田幾何？
答曰：二十三步、六分步之五。

Answer: 126 [square] *bu*.

[1.26] Again, suppose there is a triangular field of base

5

1
2

{\displaystyle 5\frac{1}{2} }

*bu* and height

8

2
3

{\displaystyle 8\frac{2}{3} }

*bu*. Question: what is the area of the field?

Answer:

23

5
6

{\displaystyle 23\frac{5}{6} }

[square] *bu*.

**Solution for *Jiuzhang Suanshu* [1.25]**

Substitute the given width and length into the area formula:

A

=

1
2

×
12

b
u
×
21

b
u

=
126

b

u
2

.

{\displaystyle \begin{align}
A &= \frac{1}{2}\times 12 \; bu \times 21 \; bu \\
&= 126 \; bu^2.
\end{align}}

**Solution for *Jiuzhang Suanshu* [1.26]**

Substitute the given width and length into the area formula:

A

=

1
2

×
5

1
2

b
u
×
8

2
3

b
u

=
23

5
6

b

u
2

.

{\displaystyle \begin{align}
A &= \frac{1}{2}\times 5\frac{1}{2} \; bu \times 8\frac{2}{3}\; bu \\
&= 23\frac{5}{6} \; bu^2.
\end{align}}

### **Problem Study 3: Quadrature of the Triangle**

The following problem comes from the *Shushu Jiuzhang* (數書九章). Expanded form of Heron's formula.

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：田積三百一十五公頃。
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。
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.

**【注释】**

[1] 幂：。 The *mi* usually refers to an area, here it is the square.

## **Area of the Trapezoid**

The *Jiuzhang Suanshu* describes two types of trapezoids: the right trapezoidal field (邪田 *xie tian*), and the isosceles trapezoidal field (箕田 *ji tian*). The area all trapezoids can be calculated using the same formula, so the distinction is purely visual. The method for calculating the area of trapezoidal states:

In a similar fashion to the triangular field, Chinese mathematicians relied on the "out-in complementary method" (出入相補 *chu ru xiang bu*) to determine the trapezoid area formula.

    Let

a

{\displaystyle a}

 be the upper base,

b

{\displaystyle b}

 be the lower base, and

h

{\displaystyle h}

 be the height of the trapezoid. Make a cut through half the height of the right trapezoid, parallel to the upper and lower bases, and then draw a perpendicular line that intersects the mid-line.  Invert the excess portion (盈) and connect it to empty space (虚) such that it forms a rectangle. This rectangle is equal in area to the right trapezoid. The rectangle has the same height of the trapezoid and a width that is equal in length to the average of the upper and lower base. Thus, the area of the trapezoid is

# A

(

a
b

2

)

×
h

{\displaystyle A = \left(\frac{a+b}{2}\right) \times h }

Alternatively, make a cut through half the height of the right trapezoid, parallel to the upper and lower bases. Invert the upper portion and connect it to the lower portion such that it forms a rectangle. This rectangle is equal in area to the right trapezoid. The rectangle has half the height of the trapezoid and a width that is equal in length to the sum of the upper and lower base. Thus, the area of the trapezoid is

# A

h
2

×
(
a
b
)

{\displaystyle A = \frac{h}{2} \times (a+b) }

    Both methods give the same result

# A

1
2

(
a
b
)
h

{\displaystyle A = \frac{1}{2}(a+b) h }

The isosceles trapezoid (箕田 *ji tian*) can be derived the same area formula using the "out-in complementary principle".

### **Problem Study 3: Trapezoidal Fields**

Problem 27 and Problem 28

答曰：九畝一百四十四步。
 [1.28] 又有邪田，正廣六十五步，一畔從一百步，一畔從七十二步。問：為田幾何？
答曰：二十三畝七十步。

術曰：并兩邪而半之，以乘正從若廣。又可半正從若廣，以乘并，畝法而一。

Answer: 23 *mu* 70 [square] *bu*.
Right trapezoidal field method (邪田 *xie tian*): Add the two bases and halve, then multiply by the height (or width). Another method is to multiply half the height (or width) by the sum of the bases. Divide by the number of [square] *bu* per *mu*.

**Solution for *Jiuzhang Suanshu* [1.27]**

Substitute the upper base, lower base, and height of the right trapezoid (*xie tian*) into the formula:

A

=

1
2

(
30

b
u
42

b
u
)
×
64

b
u

=
2304

b

u
2

.

{\displaystyle \begin{align}
A &= \frac{1}{2}(30 \; bu + 42 \; bu) \times 64 \; bu \\
&= 2304 \; bu^2.
\end{align}}

Convert the area in terms of *mu* and *bu*:

2304

b

u
2

240

b

u
2

/

m
u

=
9

144
240

m
u

{\displaystyle \frac{2304 \; bu^2}{240 \; bu^2/mu} = 9\frac{144}{240} \; mu

}

The area of right trapezoid is 9 *mu* 144 [square] *bu*.

**Solution for *Jiuzhang Suanshu* [1.28]**

Substitute the upper base (one vertical base), lower base (another vertical base), and width of the right trapezoid (*xie tian*) into the formula:

A

=

1
2

(
100

b
u
72

b
u
)
×
65

b
u

=
5590

b

u
2

.

{\displaystyle \begin{align}
A &= \frac{1}{2}(100 \; bu + 72 \; bu) \times 65 \; bu \\
&= 5590 \; bu^2.
\end{align}}

Convert the area in terms of *mu* and *bu*:

5590

b

u
2

240

b

u
2

/

m
u

=
23

70
240

m
u

{\displaystyle \frac{5590 \; bu^2}{240 \; bu^2/mu} = 23\frac{70}{240} \; mu

}

The area of right trapezoid is 23 *mu* 70 [square] *bu*.