TCM Geometry 2: Areas of Planar Figures - Part 2

**By: Tao Steven Zheng 鄭濤**

## **Area of the Circle**

The land surveyors of ancient China quickly realized that not all plots of land were in the shape of triangles or rectilinear figures! They had to figure the areas of land that are shaped like circles, circular segments, and rings. To determine the areas of these rounded figures, Chinese mathematicians gave formulas that approximated the area. The formulas were expressed in terms of measurements that were easily attainable in practice. Instead of using the radius of a circle, Chinese mathematicians opted for using the diameter (徑 *jing*) or circumference (周 *zhou*) in their calculations. For the area of a circle, the *Jiuzhang Suanshu* states four methods:

又術曰：周徑相乘，四而一。
又術曰：徑自相乘，三之，四而一。
又術曰：周自相乘，十二而一。
Another method: Multiply the circumference and the diameter, divide by 4.
Another method: Multiply the diameter by itself, multiply by 3, and divide by 4.
Another method: Multiply the circumference by itself, divide by 12.

In modern mathematical notation, the formulas are:

(1)

# A

C
2

⋅

d
2

{\displaystyle A = \frac{C}{2} \cdot \frac{d}{2} }

(2)

# A

C
d

4

{\displaystyle A = \frac{Cd}{4} }

(3)

# A

3

d
2

4

{\displaystyle A = \frac{3d^2}{4} }

(4)

# A

C
2

12

{\displaystyle A = \frac{C^2}{12} }

where

A
,
C
,
d

{\displaystyle A, C, d }

represent the area, circumference, and diameter of the circle respectively.

Methods (1) and (2) are identical formulas and give the exact area if the circumference is indeed

π

{\displaystyle \pi }

 times the diameter. However, a close inspection of the given measurements of the circumference and diameter in each problem indicates that the original authors of the *Jiuzhang Suanshu* took

# π
3

{\displaystyle \pi = 3 }

. This crude approximation for

π

{\displaystyle \pi }

 is then used to express the formulas for methods (3) and (4). Had they expressed the formulas in terms of

π

{\displaystyle \pi }

 where

π
≠
3

{\displaystyle \pi \ne 3 }

, the last two formulas would be exact:

(3)

# A

π

d
2

4

{\displaystyle A = \frac{\pi d^2}{4} }

(4)

# A

C
2

4
π

.

{\displaystyle A = \frac{C^2}{4\pi}. }

### **Problem Study 1: Circular Field**

答曰：七十五步。

Answer: 75 [square] *bu*.

**Solution for *Jiuzhang Suanshu* [1.31]**

Substitute the given circumference and diameter into each area formula.

**Formula 1**

A

=

30

b
u

2

×

10

b
u

2

=
15

b
u
×
5

b
u

=
75

b

u
2

{\displaystyle \begin{align}
A &= \frac{30 \; bu}{2} \times \frac{10 \; bu}{2} \\
&= 15 \; bu \times 5 \; bu \\
& = 75 \; bu^2
\end{align}}

**Formula 2**

A

=

30

b
u
×
10

b
u

4

=

300

b

u
2

4

=
75

b

u
2

{\displaystyle \begin{align}
A &= \frac{30 \; bu \times 10 \; bu}{4} \\
&= \frac{300 \; bu^2}{4} \\
& = 75 \; bu^2
\end{align}}

**Formula 3**

A

=

3
×
(
10

b
u

)
2

4

=

3
×
100

b

u
2

4

=
75

b

u
2

{\displaystyle \begin{align}
A &= \frac{3\times (10 \; bu)^2}{4} \\
&= \frac{3 \times 100 \; bu^2}{4} \\
& = 75 \; bu^2
\end{align}}

**Formula 4**

A

=

(
30

b
u

)
2

12

=

900

b

u
2

12

=
75

b

u
2

{\displaystyle \begin{align}
A &= \frac{(30 \; bu)^2}{12} \\
&= \frac{900 \; bu^2}{12} \\
& = 75 \; bu^2
\end{align}}

Since the formulas equivalent, each method gives the same answer. If only the given circumference is taken to calculate the precise area of this circle, its true area would be, to 2 decimal places,

71.62

b

u
2

{\displaystyle 71.62 \; bu^2 }

. If only the given diameter is used to calculate the precise area of this circle, its true area would be, to 2 decimal places,

78.54

b

u
2

{\displaystyle 78.54 \; bu^2 }

.

**Calculation of

π

## {\displaystyle \pi }**

Prior to the Han dynasty (206 BC - 220 AD), Chinese mathematicians had always taken the ratio of the circumference of a circle to its diameter as

3
:
1

{\displaystyle 3:1 }

. This ratio, denoted using the 16th Greek letter

π

{\displaystyle \pi }

 in modern mathematics, was called the circle rate (圓率 *yuan lü*) or circumference rate (圓周率 *yuan zhou lü*) in ancient China. The earliest attempt to refine the value of

π

{\displaystyle \pi }

 in China took place between the years 1 - 5 AD, right before the brief interregnum of Wang Mang (王莽, c. 45 BC - 23 AD) from 9 AD to 23 AD. This task was undertaken by an eminent mathematician, astronomer, and historian named Liu Xin (劉歆, c. 50 BC - 23 AD).

Although the details of Liu Xin's method are unknown, his approximation for

π

{\displaystyle \pi }

 can be extrapolated from a geometric figure that can be found on Wang Mang's calibrated standard measures of capacity (律嘉量斛 *lü jia liang hu*). In 9 AD, Wang Mang ordered these standard measures to be casted out of bronze. Each cylindrical bronze vessel gives five standard measures, the largest being 1 *hu* (斛) , followed by 1 *dou* (斗), then 1 *sheng* (升), then 1 *ge* (合), and the smallest being 1 yue (龠). At the bottom of the 1 *hu* (斛) measure, there is a square inside a circle (Figure 2). The square shares one center with the circle but it does not touch the circle. Instead, there is a gap between each corner of the square to the circle called *tiao pang* (庣旁). The width of the square measures 1 *chi* (尺) in length, or equivalently 10 *cun* (寸), and each *tiao pang* (庣旁) measures 9 *li* (釐) 5 *hao* (毫), or 0.095 *cun* (寸). The diameter of the circle is thus

10

2

2
×
0.095
≈
14.3321

c
u
n

{\displaystyle 10\sqrt{2}+2\times 0.095 \approx 14.3321 \; cun}

The cylindrical cavity measuring 1 *hu* is 10 *cun* in depth and carries a capacity of 1620 [cubic] *cun*. Thus, the area of the circle is 162 [square] *cun*. Using this area and the above diameter gives Liu Xin's

π

{\displaystyle \pi }

 approximation:

# π

4
×
162

(
14.3321
)

2

≈
3.1547

{\displaystyle \pi = \frac{4\times 162}{{(14.3321)}^{2}} \approx 3.1547 }

The next famous

π

{\displaystyle \pi }

approximation was calculated by the polymath Zhang Heng (張衡, 78 - 139 AD).

In the 3rd century AD, Liu Hui (c. 220 - 280 AD) developed a rigorous algorithm for computing

π

{\displaystyle \pi }

 called the *ge yuan shu* (割圓術, "method of cutting the circle"). Liu Hui was the first Chinese mathematician to base his reasoning on limits and geometry. He employed the theory of limits to approximate the area of the circle using a sequence of inscribed polygons. Liu Hui noticed that the ratio of the perimeter of a regular hexagon (6-gon) inscribed in a circle and the circle's diameter is

3
:
1

{\displaystyle 3:1 }

. Since the area of the inscribed regular hexagon is less than that of the circle, he deduced that

3
:
1

{\displaystyle 3:1 }

 cannot be the true ratio of the circle's circumference to its diameter. Liu Hui realized that by doubling the number of sides to the inscribed polygon, the polygon will be a closer approximation to the circle. In consequence, the area of each successive polygon will approach the true area of the circle.

Unlike Liu Xin, Liu Hui wrote a detailed account of his calculations in his commentary of the *Jiuzhang Suanshu*. From the area of a 96-gon and 192-gon, Liu Hui computed the bound

3.141024
<
π
<
3.142708

{\displaystyle 3.141024 < \pi < 3.142708 }

. He then devised a "quick method" and obtained the approximation

π
≈
3.1416

{\displaystyle \pi \approx 3.1416 }

.

Liu Hui's *ge yuan shu* is often compared to the "method of exhaustion" employed by Archimedes (c. 287 - 212 BC), for both techniques involve the approximation of the circle by regular polygons. Despite the geometric similarities, Archimedes and Liu Hui differed in their approach. Archimedes inscribed and circumscribed the circle with regular polygons (starting with the regular hexagon) to compute a lower bound and upper bound for

π

{\displaystyle \pi }

. He then calculated the length of one side of each polygon so that he could compare the perimeter of the regular polygon to the circle. Archimedes determined the bound

3

10
71

<
π
<
3

1
7

{\displaystyle 3\frac{10}{71} < \pi < 3\frac{1}{7}}

also using a 96-gon.

Two centuries later Zu Chongzhi (429 – 500 AD) applied the *ge yuan shu* to achieve a more accurate bound

3.1415926
<
π
<
3.1415927

{\displaystyle 3.1415926 < \pi < 3.1415927 }

.

古之九數，圓周率三，圓徑率一，其術疏舛[1]。 自劉歆[2]、張衡[3]、劉徽[4]、王蕃[5]、皮延宗[6]之徒，各設新率，未臻折衷。
宋[7]末，南徐州從事史[8]祖沖之，更開密法，以圓徑一億為一丈， 圓週盈數[9]三丈一尺四寸一分五厘九毫二秒七 忽，朒數[10]三丈一尺四寸一分五厘九毫二秒六忽，正數在盈、朒二限之間。 密率，圓徑一百一十三，圓週三百五十五。 約率，圓徑七，週二十二。
According to the nine computational methods of antiquity, the ratio of the circumference to the diameter is 3 to 1. This method is inaccurate. Liu Xin, Zhang Heng, Liu Hui, Wang Fan, Pi Yanzong each found a new ratio, yet could not reach a consensus.
Towards the end of Song, the assistant prefectural governor of southern Xu Zhou, Zu Chongzhi, further devised a precise method. Taking a circle of diameter 10000000 to be 1 *zhang*, the excess value of the circumference is 3 *zhang* 1 *chi* 4 *cun* 1 *fen* 5 *li* 9 *hao* 2 *miao* 7 *hu*, and the deficit value is 3 *zhang* 1 *chi* 4 *cun* 1 *fen* 5 *li* 9 *hao* 2 *miao* 6 *hu*. The actual number lies between the excess and deficit values. The precise ratio is a diameter of 113 to a circumference of 355. The approximate ratio is a diameter of 7 to a circumference of 22.

### **Problem Study 3: Circular Segment**

答曰：一畝九十七步半。
[1.36] 又有弧田，弦七十八步、二分步之一，矢十三步、九分步之七。問：為田幾何？
答曰：二畝一百五十五步、八十一分步之五十六。
術曰：以弦乘矢，矢又自乘，并之，二而一。
Answer: 1 *mu* 97 1/2 [square] *bu*.
[1.28] Again, suppose there is a right trapezoidal field of width 65 *bu*, a vertical base 100 *bu*, and another vertical base 72 *bu*. Question: what is the area of the field?
Answer: 23 *mu* 70 [square] *bu*.
Method: Multiply the base by the altitude, then multiply the altitude by itself. Add the results and divide by 2.

**Solution for *Jiuzhang Suanshu* [1.35]**

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* [1.36]**

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*.

## **Area of the Annulus**

In the *Jiuzhang Suanshu*, the formula for calculating the area of the ring-shaped field (環田 *huan tian*) uses three quantities: the inner circumference (中周 *zhong zhou*), the outer circumference (外周 *wai zhou*), and the width (徑 *jing*). One should note that the term *jing* (徑) is usually defined as the diameter of a full circle. However, for the ring-shaped field, the *jing* is defined as the distance between the two circles, more simply the width of the ring-shaped field.

According to the solution method found in the *Jiuzhang Suanshu*, the formula for calculating the area of a ring-shaped field is analogous to the trapezoid.

# A

C
1

C
2

2

w

{\displaystyle A = \frac{C\_1 + C\_2}{2}w }

### **Problem Study 4: Ring-Shaped Fields**

答曰：二畝五十五步。

[1.38] 又有環田，中周六十二步、四分步之三，外周一百一十三步、二分步之一，徑十二步、三分步之二。問：為田幾何？

答曰：四畝一百五十六步、四分步之一。

Answer: 2 *mu* 55 [square] *bu*.

[1.38] Again, suppose there is a ring-shaped field (環田 *huan tian*). The inner circumference is

62

3
4

{\displaystyle 62 \frac{3}{4} }

 *bu*, the outer circumference is

113

1
2

{\displaystyle 113 \frac{1}{2} }

 *bu*, and the width (徑 *jing*) is

12

2
3

{\displaystyle 12\frac{2}{3} }

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

Answer: 4 *mu*

156

1
4

{\displaystyle 156 \frac{1}{4} }

 [square] *bu*.

**Solution for *Jiuzhang Suanshu* [1.37]**

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* [1.38]**

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}}

## **Additional Problems**

**Problem 1**
Read the following passage. Illustrate the multiplication process using rod arithmetic (Arabic numbers may be used).

答曰：十一畝九十步、十二分步之一。

60

1
3

{\displaystyle 60\frac{1}{3} }

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

Answer: 11 *mu*

90

1
12

{\displaystyle 90\frac{1}{12} }

[square] *bu*.

**Solution for *Jiuzhang Suanshu* [1.32]**

Substitute the given circumference and diameter into each area formula.

**Formula 1**

A

=

181

b
u

2

×

60

1
3

b
u

2

=
90

1
2

b
u
×
30

1
6

b
u

=
2730

1
12

b

u
2

{\displaystyle \begin{align}
A &= \frac{181 \; bu}{2} \times \frac{60\frac{1}{3} \; bu}{2} \\
&= 90\frac{1}{2} \; bu \times 30\frac{1}{6} \; bu \\
& = 2730\frac{1}{12} \; bu^2
\end{align}}

**Formula 2**

A

=

181

b
u
×
60

1
3

b
u

4

=

10920

1
3

b

u
2

4

=
2730

1
12

b

u
2

{\displaystyle \begin{align}
A &= \frac{181 \; bu \times 60\frac{1}{3} \; bu}{4} \\
&= \frac{10920\frac{1}{3} \; bu^2} {4}\\
& = 2730\frac{1}{12} \; bu^2
\end{align}}

**Formula 3**

A

=

3
×

(

60

1
3

b
u

)

2

4

=

3
×
3640

1
9

b

u
2

4

=
2730

1
12

b

u
2

{\displaystyle \begin{align}
A &= \frac{3\times {\left(60\frac{1}{3} \; bu \right)}^{2}}{4} \\
&= \frac{3 \times 3640 \frac{1}{9} \; bu^2}{4} \\
& = 2730\frac{1}{12} \; bu^2
\end{align}}

**Formula 4**

A

=

(

181

b
u

)

2

12

=

32761

b

u
2

12

=
2730

1
12

b

u
2

{\displaystyle \begin{align}
A &= \frac{{\left(181 \; bu \right)}^{2}}{12} \\
&= \frac{32761 \; bu^2}{12} \\
& = 2730\frac{1}{12} \; bu^2
\end{align}}

Knowing that

1

m
# u
240

b

u
2

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

, convert

2730

b

u
2

{\displaystyle 2730 \; bu^2 }

 from *bu* (步) to *mu* (畝)*.*

2730

b

u
2

240

b

u
2

/

m
u

=
11

90
240

m
u

{\displaystyle \frac{2730 \; bu^2}{240 \; bu^2/mu} = 11\frac{90}{240} \; mu }

Therefore, the final answer is

11

m
u

90

1
12

b

u
2

{\displaystyle 11 \; mu \; 90\frac{1}{12} \; bu^2 }

.

If only the given circumference is taken to calculate the precise area of this circle, its true area would be, to 2 decimal places,

2730.08

b

u
2

{\displaystyle 2730.08 \; bu^2 }

. If only the given diameter is used to calculate the precise area of this circle, its true area would be, to 2 decimal places,

2858.94

b

u
2

{\displaystyle 2858.94 \; bu^2 }

.