# Solutions

# Chapter 1 – Numbers

## 1.1

1.1

2.1

Answers may vary. Some of the patterns are:

The **left column** can be divided in 4 groups, with each group possessing 19, 17, 13, and 11 notches. The sum of these being 60. Those are the 4 successive prime numbers between 10 and 20.

The** ****central column** is divided in groups of 8. By an approximate count, one can find (in the parenthesis, is the maximum number): 7 (8), 5 (7), 5 (9), 10, 8 (14), 4 (6), 6, 3. The minimal sum is 48, while the maximal sum is 63.

The **right column** is divided into 4 groups, where each group has 9, 19, 21, and 11 notches. The sum of these 4 numbers is 60.

The numbers in the left column were compatible with a numeration system based on 10, since 21 = 20 + 1, 19 = 20 – 1, 11 = 10 +1, and 9 = 10 -1. These numbers are also prime numbers between 10 and 20: 11, 13, 17, 19.

## 1.2

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2.1

**1) How many symbols did the Babylonians use to express numbers?**

*2 unique symbols in 60 different arrangements.*

**2) In our decimal system we have 10 digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. How many digits did the Babylonians have?**

*60 digits*

* ***3) Calculate the value of**

a) *2 x 60 = 120, 6 x 1 = 6
126*

b) *2 x 60 = 120, 12 x 1 = 12
132*

c) *10 x 3600 = 36000, 10 x 60 = 600, 2 x 1 = 2
36602*

d) How would Babylonians write 65? 615? 665?

## 1.3

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2.1

**A granary of barley contains 2400 gur, where 1 gur equals 480 sila. If workers are to receive 7 sila of grain for a day’s work, how many men can be paid from this granary?**

[latex]\begin{array}{l}1\ gur\ =\ 480\ sila\\ 2400\ gur\ =\ 2400\ x\ 480\ sila\ =\ 115200\ sila\\ divide\ by\ 7\ sila\ per\ wor\ker\\ \frac{115200\ sila}{7\ \frac{sila}{wrkr}}\approx16457.142\ wrkrs\end{array}[/latex]

* *

## 1.4

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2.1

**A woman weaves a textile that is to be 48 rods long. In one day, she weaves 1/3 rod. In how many days will she cut the textile from the loom?**

*Length = 48 rods*

*Rate = ⅓ rod / day*

*Total time = ? days*

[latex]\begin{array}{l}total\ time\ =\ \frac{total\ length}{rate}\\ =\frac{48\ rods}{\frac{1}{3}rod\ per\ day}=\frac{48\cdot3}{1}=48\cdot3=144\ rods\end{array}[/latex]

## 1.5

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2.1

**a) Find values of the numbers**

**b) Write numbers 85, 121 and 2222 in Mayan numerals**

## 1.6

1.1

2.1

**There is a tree with 100 branches; each branch has 100 nests; each nest, 100 eggs; each egg, 100 birds. How many nests, eggs and birds are there?**

100 branches per tree (100)

100 nests per branch (100 x 100=10000 nests)

100 eggs per nest (100 x 10000 = 1000000 eggs)

100 birds per egg (100 x 1000000 = 100000000 birds)

Total nests, eggs, and birds = 10000 + 1000000 + 100000000 = 101010000 total objects

## 1.7

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2.1

**Three hundred pigs are to be prepared for a feast. They are to be prepared in three batches on three successive days with an odd number of pigs in each batch. Can this be accomplished?**

Odd + Odd = even

Odd + Even = odd

Three batches: Odd + Odd + Odd

Since Odd + Odd = even, then: Odd + Odd + Odd = Even + Odd = Odd.

Since there are 300 pigs (which is even), no this is not possible.

## 1.8

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2.1

**A leech invited a slug for lunch a leuca away. But he could only crawl an inch a day. How long will it take the slug to get his meal? [1 leuca = 1500 paces; 1 pace = 5 feet.]**

1 leuca = 1500 paces

1500 paces = 1500 x 5 feet = 7500 feet

7500 feet = 7500 x 12 inches = 90000 feet

Therefore, it will take 90000 days for the slug to get his meal.

90000 days at 365 days per year = 246 years with a remainder of 210 days.

## 1.9

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2.1

**A certain man had in his trade four weights with which he could weigh integral pounds from 1 up to 40. How many pounds was each weight?**

1, 3, 9 and 27.

Weight of object to be measured |
Left side |
Right side |

1 | O | 1 |

2 | O, 1 | 3 |

3 | O | 3 |

4 | O | 3, 1 |

5 | O, 3, 1 | 9 |

6 | O, 3 | 9 |

7 | O, 3 | 9, 1 |

8 | O, 1 | 9 |

9 | O | 9 |

10 | O | 9, 1 |

11 | O, 1 | 9, 3 |

12 | O | 9, 3 |

13 | O | 9, 3, 1 |

14 | O, 9, 3, 1 | 27 |

15 | O, 9, 3 | 27 |

16 | O, 9, 3 | 27, 1 |

17 | O, 9, 1 | 27 |

18 | O, 9 | 27 |

19 | O, 9 | 27, 1 |

20 | O, 9, 1 | 27, 3 |

21 | O, 9 | 27, 3 |

22 | O, 9 | 27, 3, 1 |

23 | O, 3, 1 | 27 |

24 | O, 3 | 27 |

25 | O, 3 | 27, 1 |

26 | O, 1 | 27 |

27 | O | 27 |

28 | O | 27, 1 |

29 | O, 1 | 27, 3 |

30 | O | 27, 3 |

31 | O | 27, 3, 1 |

32 | O, 3, 1 | 27, 9 |

33 | O, 3 | 27, 9 |

34 | O, 3 | 27, 9, 1 |

35 | O, 1 | 27, 9 |

36 | O | 27, 9 |

37 | O | 27, 9, 1 |

38 | O, 1 | 27, 9, 3 |

39 | O | 27, 9, 3 |

40 | O | 27, 9, 3, 1 |

## 1.10

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2.1

**I have a cloak 100 cubits long and 80 cubits wide. I wish to make small cloaks with it; each small cloak is 5 cubits long, and 4 wide. How many small cloaks can I make?**

Since 5 divides into 100 and 4 divides into 80, we can assume that all of the fabric will be used if we orient the small cloaks such that their 5 cubit sides align along the 100 cubit side of the large cloak, and the 4 cubit sides along the 80 cubit.

Along the long side, we can fit 20 (100/5) small cloaks in a row.

Along the short side, we can fit 20 (80/4) small cloaks in a row.

Therefore we have an array of 20 small cloaks by 20 small cloaks. The total is 400 cloaks.

## 1.11

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2.1

**A father, when dying, gave to his sons 30 glass flasks, of which 10 were full of oil, 10 were half full, and the last 10 were empty. Divide the oil and the flasks so that each of the three sons received equally of both glass and oil.**

Each son needs to receive 10 actual flasks (glass)

There is a total of 10 x 1 + 10 x 0.5 = 15 flasks worth of oil. Therefore each son needs to receive 5 flasks worth of oil.

One possible solution is to give:

Son 1 – 5 full and 5 empty

Son 2 – 5 full and 5 empty

Son 3 – 10 half-empty

## 1.12

1.1

2.1

**A king ordered his servants to collect an army from 30 manors in such a way that from each manor he would take the same number of men he had collected up until then. The servant went to the first manor alone; to the second he went with one other; to the next he took three with him. How many were collected from the 30 manors?**

Manor # |
How many brought |
Collected at this |
New Total |

1 | 0 | 1 | 1 |

2 | 1 | 1+1=2 | 3 |

3 | 3 | 2+3=5 | 8 |

4 | 8 | 5+8=13 | 21 |

5 | 21 | 13+21=34 | 55 |

6 | 55 | 34+55=89 | 144 |

7 | 144 | 89+144=233 | 377 |

8 | 377 | 233+377=610 | 987 |

9 | 987 | 610+987=1597 | 2584 |

10 | 2584 | 1597+2584=4181 | 6765 |

11 | 6765 | 4181+6765=10946 | 17711 |

12 | 17711 | 10946+17711=28657 | 46368 |

13 | 46368 | 28657+46368=75025 | 121393 |

14 | 121393 | 75025+121393=196418 | 317811 |

15 | 317811 | 196418+317811=514229 | 832040 |

16 | 832,040 | 514229+832040=1,346,269 | 2,178,309 |

17 | 2,178,309 | 1,346,269+2,178,309=
3,524,578 |
5,702,887 |

18 | 5,702,887 | 3,524,578+ |

## 1.13

1.1

2.1

**A gentleman has a household of 30 people and orders that they be given 30 measures of grain. He directs that each man should receive 3 measures, each woman 2 measures, and each child 1/2 measure. How many men, women, and children are there?**

The solution can be found by trial and error method.

3 men, 5 women, 22 children

## 1.14

1.1

2.1

**A man wanting to build a house contracted with six builders, five of whom were master builders, and the sixth an apprentice, to build it for him. He agreed to pay them a total of 25 pence a day, with the apprentice to get half the rate of a master builder. How much did each receive per day?**

A master builder got 4.54 and apprentice got 2.27 pence a day.

## 1.15

1.1

2.1

**A man in the east wanted to buy 100 assorted animals for 100 shillings. He ordered his servant to pay 5 shillings for a camel, 1 shilling for an ass, and 1 shilling for 20 sheep. How many camels, asses, and sheep did he buy?**

The solution can be found by trial and error method or by making a list.

19 camels, 1 ass and 80 sheep

## 1.16

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2.1

**There is an estate that contains 7 houses; each house has 7 cats; each ****cat catches 7 mice; each mouse eats 7 spelt of seeds; each spelt was ****capable of producing 7 hekats of grain. How many things were in the estate?**

7 + 49 + 343 + 2401 + 16,807.= 19,607

## 1.17

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2.1

**Suppose a scribe tells you that four overseers have drawn 100 great quadruple hekats of grain, and their work gangs consist of 12, 8, 6, and 4 men. How much grain does each overseer receive?**

100 / (12 + 8 + 6 + 4)= 3 1/3

12(3 1/3)= 40

8(3 1/3) = 26 2/3

6(3 1/3)= 20

4(3 1/3)= 13 1/3

## 1.18

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2.1

In the middle register we see 835 horned cattle on the left, right behind them are some 220 animals (cows?) and on the right 2235 goats. In the bottom register we see 760 donkeys on the left and 974 goats on the right

## 1.19

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2.1

## 1.20

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2.1

**Given, four whole numbers where, if added together three at a time, their sums are 20, 22, 24, and 27. What are the numbers?**

Suppose the numbers are a, b, c and d. Then:

a+b+c = 20

b+c+d = 22

a+c+d = 24

a+b+d = 27

If we add all of these equations together we find:

3(a+b+c+d) = 22+24+27+20 = 93

Dividing both ends by 3 we find:

a+b+c+d = 93/3 = 31

Then we can find the values of a, b, c and d:

a = (a+b+c+d) – (b+c+d) = 31-22 = 9

b = (a+b+c+d)-(a+c+d) = 31-24=7

c = (a+b+c+d)-(a+b+d) = 31-27 = 4

d = (a+b+c+d)-(a+b+c) = 31-20 = 11

# Chapter 2 – Circles

## 2.1

1.1

2.1

**One of the geometric rules given in the Talmud is “How much is the square greater than the inscribed circle? A quarter” What was the Talmudic value for pi?**

**Case 1 – Assuming we are discussing area, then:**

Let S be the area of the square

Let C be the area of the circle

l (of square) = d = 2r

S = 1.25C

Therefore:

[latex]\begin{array}{l}l^2=\pi r^2\\ substituting\ 2r\ for\ l:\left(2r\right)^2=\pi r^2\\ 4r^2=\pi r^2\\ 4=\pi\end{array}[/latex]

**Case 2 – if we are discussing circumference:**

[latex]\begin{array}{l}S=1.25C\\ l=d\\ 4d=1.25\pi d\\ 1.25\pi=4\\ \pi=\frac{4}{1.25}=\frac{16}{5}=3.2\end{array}[/latex]

Although the wording of the question seems to suggest area as the measurement being discussed in this question, the calculation for pi in the circumference solution is much closer to the modern value of pi than the area solution.

## 2.2

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2.1

**In 4000 year old Egyptian papyrus we read “A circular field has diameter 9 chet. What is its area?” The Ahmes, the scribe of the papyrus, gives the solution: “Subtract 1/9 of the diameter namely 1 chet. The remainder is 8 chet. Multiply 8 by 8; it makes 64. Therefore, it contains 64 square chet of land”**

**For the Egyptians a square with the side length 8/9 of the diameter of a circle was a good enough approximation to calculate the area of a circle.**

**Calculate the ancient Egyptian value of pi.**

[latex]A=\ \left(\frac{8}{9}d\right)^2[/latex]

If we substitute the modern formula for area of a circle, we can solve for pi:

[latex]\begin{array}{l}A=\ \left(\frac{8}{9}d\right)^2\\ \pi r^2=\left(\frac{8}{9}d\right)^2\\ d=2r\\ \pi r^2=\left(\frac{16r}{9}\right)^2\\ \pi r^2=\frac{256r^2}{81}\\ \pi=\frac{256r^2}{81r^2}=\frac{256}{81}\end{array}[/latex]

Therefore π = 256/81≈ 3.16

## 2.3

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2.1

**In Babylonian mathematics a circle was equated to a regular hexagon which side was equal to the radius of a circle. Calculate the Babylonian value of pi.**

The perimeter of the hexagon is P≅6r. The circumference C=2πr. 2π=6 π=3

## 2.4

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2.1

**There is a round field which contains 400 yards in its circumference. How many square yards will its area be? Alcuin obtains 10000 square yards. How does he get that, you might ask? Calculate the value of pi he used.**

We begin with the formulas for Area of a Circle and Circumference of a Circle:

[latex]\begin{array}{l}A=\pi r^2\\ C=2\pi r\end{array}[/latex]

(Note the use of the radius-centred circumference formula, rather than the diameter formula. This is done to preserve variables)

We know that C = 400, and A = 10000. Therefore, we can solve for pi.

[latex]\begin{array}{l}10000=\pi r^2\\ 400=2\pi r\\ r=\frac{400}{2\pi}\\ \therefore10000=\pi\left(\frac{400}{2\pi}\right)^2\\ 10000=\frac{160000\pi}{4\pi^2}\\ 10000=\frac{40000}{\pi}\\ 10000\pi=40000\\ \pi=4\end{array}[/latex]

## 2.5

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2.1

**There is a city which is 8000 feet in circumference. How many houses could the city contain if each house is 30 feet long and 20 feet wide?**

There are two scenarios here. The first, which is the simplistic scenario, is the assumption that the shape of the houses can be squished into a shape to meet the curve of the circle of the city. Assuming unrealistic city design where each house was connected to each other like bricks and there were no additional buildings or spaces within the city, we can calculate the theoretical maximum number of houses. Each house has a constant area of 600 square feet, notwithstanding some slight variance in the shape of each house to accommodate the curved edges of the city. Therefore the theoretical maximum (TM) is the area of the circle in square feet divided by 600 square feet per house.

[latex]\begin{array}{l}TM=\frac{A}{600}\\ A=\pi r^2\\ C=2\pi r\\ If\ C=8000\\ then\ 8000=2\pi r\\ r=\frac{8000}{2\pi}=\frac{4000}{\pi}\\ TM=\frac{\pi r^2}{600}\\ =\frac{\pi\left(\frac{4000}{\pi}\right)^2}{600}\\ =\frac{16000000}{600\pi}\\ =\frac{80000}{3\pi}\\ If\ we\ take\ \pi\ \approx3.1415\\ TM\approx8488.26\end{array}[/latex]

Therefore the theoretical maximum if every square foot of the city was house (unrealistic) is 8488 houses. Answers will vary, and the solutions need to be justified based on realistic considerations, such as:

- Other buildings (businesses, markets, schools, community centres, religious buildings, government buildings, utility buildings, etc.)
- Open spaces (paths, roads, squares, etc.)
- Green space (parks, plazas, hiking trails, etc.)
- Attractions (fairs, sports centres, arts centres, etc.)
- Topography (whether there are hills, valleys, rivers, etc.)
- Transportation requirements (stables, bus facilities, trains, airport, etc.)
- How much space people take up in a city when outside their homes
- And perhaps most importantly, the fact that buildings need to have separation from each other, and not be squished together in all directions.

## 2.6

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2.1

**A circular road A that is 48 km in circumference touches at point P another circular ****road B of circumference 36 km. A cow and a horse start walking from ****the point P along the road A and B, respectively. The cow walks 6 km per day and the ****horse walks 12 km per day. How many days later days later do the cow and horse meet ****again at P?**

The cow walks the full circle A in 48/6 = 8 days. The horse walks the full circle B in 36/12 = 3 days. Finding the lowest common multiple, which is 3×8=24, we find that 24 days later the cow and horse meet again at P

## 2.7

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2.1

**A circle of radius r is inscribed in an isosceles triangle with sides 10 and 12. Find r.**

Draw the diagram with some auxiliary lines

Label the vertices. By the Pythagorean theorem BN=8.

We can see now that triangles ABN and ODB are similar (the angles are equal). For better visual we overlap them. Label the sides in the diagram by their length.

Now we can write the ratio AN : AB = OD : DB or 6 : 10 = r : (8-2r).

r=3.

## 2.8

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2.1

**The centers of a loop of the circles of radius r form the vertices of a polygon, as shown in the figure below. Let S1 be the sum of the shaded areas of the circles, and S2 the sum of the unshaded areas of the circles. Find [latex]S2 S1.[/latex]**

Match the shaded and unshaded pieces. Then count the rest of unshaded pieces. If you combine them you get two full circles. Therefore [latex]S2 S1 = 2πr2[/latex]

# Chapter 5 – Pythagorean Theorem

## 5.1

1.1

2.1

**In the center of a river whose width is 10 chi grows a reed whose top reaches 1 chi above the water level. If we pull the reed towards the bank, its top is even with the water’s surface. What is the depth of the river?**

Let’s set up a model showing the reed with height h that, when growing vertically, extends 1 above the water. Therefore the depth of the river is equal to h – 1.

Since the width of the river is 10, the width of half the river is 5. Therefore, we are solving a Pythagorean triangle with side lengths 5 and h-1, and hypotenuse h

[latex]\begin{array}{l}5^2+\left(h-1\right)^2=h^2\\ 5^2+h^2-2h+1=h^2\\ 25+h^2-2h+1=h^2\\ 25-2h+1=0\\ 26=2h\\ h=13\end{array}[/latex]

Therefore, the depth, or h-1, equals 12 chi.

## 5.2

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2.1

**Given a pyramid 300 cubits high, with the square base 500 cubits on a side, determine the distance from the center of any side to the apex.**

Given a pyramid 300 cubits high, with a square base 500 cubits on a side, determine the distance from the centre of any side to the apex

When we consider a vertical cross-section of the pyramid that intersects with the top vertex, we notice that it is an isosceles triangle with base of 500 cubits and height of 300 cubits. If we draw a height line down from the top vertex, we have a line that bisects the base, creating 2 congruent right triangles with base 250 cubits and height 300 cubits.

The point at which the cross-section intersects the edge of the base is the centre of the side. Therefore, the distance being asked is the hypotenuse of the right triangles in the diagram below:

[latex]\begin{array}{l}a^2+b^2=hyp^2\\ 250^2+300^2=hyp^2\\ 62500+90000=hyp^2\\ hyp^2=152500\\ hyp=\sqrt{152500}=50\sqrt{61}\approx390.5125\end{array}[/latex]

Therefore, the distance from the centre of each side to the apex is approximately 390.5125 cubits.

## 5.3

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2.1

**A pyramid has a base of 360 cubits and a height of 250 cubits. What is its seqt (the ratio of vertical and horizontal dimensions)?**

The ratio of vertical and horizontal dimensions or seqt is [latex]250/(360/2)=25/18[/latex]

## 5.4

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2.1

**Find the height of a square pyramid with a seqt (the ratio of vertical and horizontal dimensions) of 21 fingers per cubit (cubit equals 28 fingers) and a base of 140 cubits on one side. Find the length of the slope.**

The seqt or the ratio of vertical and horizontal dimensions is [latex]21/28=3/4[/latex]

Thus if the horizontal distance is [latex](140/2)=70[/latex] cubits then the vertical is [latex]70/4×3=52.5[/latex] cubits or 52 cubits 14 fingers.

You can calculate the slope by two methods:

1) Since the legs of the right triangle in the diagram are in 3 to 4 ratio then the hypotenuse is in 5 to 4 to 3 ratio (a Pythagorean triplet). The length of the slope is [latex]70/4×5=87.5[/latex] cubits or 87 cubits and 14 fingers

2) By the Pythagorean theorem the length of the slope is a square root of [latex]70^2+52.5^2=87.5[/latex]

## 5.5

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2.1

**A rectangular plot is 60 cubit square; the diagonal is 13 cubits. How many cubits does it take to make the sides?**

You can solve the problem by three methods.

Method 1. One of the Pythagorean triples is 5, 12, 13. Thus the sides are 5 and 12 cubits.

Method 2. Quadratic equation,

Since we know that d = 13 and A = 60

[latex]\begin{array}{l}bh=60\\ b=\frac{60}{h}\\ \therefore13=\left(\frac{60}{h}\right)^2\end{array}By\ substitution:\begin{array}{l}13^2=b^2+h^2\\ 13^2=\left(\frac{60}{h}\right)^2+h^2\\ 169=\frac{3600}{h^2}+h^2\\ 169h^2=3600+h^4\\ h^4-169h^2+3600=0\\ \left(h^2\right)^2-169\left(h^2\right)+3600=0\end{array}Use\ Quadratic\ Formula\ with\ h^2\ as\ the\ variable:\begin{array}{l}h^2=\frac{169\pm\sqrt{\left(-169\right)^2-4\left(3600\right)}}{2}\\ \ \ =\frac{169\pm\sqrt{28561-14400}}{2}\\ \ \ =\frac{169\pm\sqrt{14161}}{2}\\ \ \ =\frac{169\pm119}{2}\\ h^2=144\ or\ 25\\ h=\pm12\ or\ \pm5\end{array}[/latex]

Since the side lengths must be positive values, these values must be 12 cubits and 5 cubits respectively.

Method 3. Trial and error

Students may try a trial and error with different side lengths that result in an area of 60, as laid out in the following table (as well as the opposite order):

[latex]\begin{array}{l}1\times60=60,\ d\approx60.01&&4\times15=60,\ d\approx15.52\\ 2\times30=60,\ d\approx30.07&&5\times12=60,\ d=13\\ 3\times20=60,\ d\approx20.22&&6\times10=60,\ d\approx11.66\end{array}[/latex]

## 5.6

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2.1

**An erect pole of 10 cubits has its base moved 6 cubits. Determine the new height and the distance the top of the pole is lowered.**

An erect pole, 10 cubits in length, has its base moved outwards, 6 cubits. Determine the new height and the distance the top of the pole has been lowered.

Consider that the original height is the length of the pole, or 10 cubits. With this consideration, we can model the problem:

Animation:

Diagram:

Method 1 – Pythagorean theorem

The hypotenuse (the pole) is known. Therefore:

[latex]\begin{array}{l}x^2+6^2=10^2\\ x=\sqrt{10^2-6^2}\\ x=\sqrt{100-36}\\ x=\sqrt{64}\\ x=8\end{array}[/latex]

Method 2 – Pythagorean triples

The lengths of the sides form a Pythagorean Triple 6-8-10. The two sides are 6 and 10 (with 10 as the hypotenuse), therefore the missing side is 8 cubits.

## 5.7

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2.1

**There is a hole at the foot of a pillar 9 hastas high, and a pet peacock standing on top of it. Seeing a snake returning to its hole at a distance from the pillar equal to three times its height, the peacock swoops down upon the snake slantwise. Say quickly, how far from the pole does the meeting of their paths occur?**

Let the peacock and snake meet at x m from the pillar as shown in the above figure.

Again we can solve the question by the Pythagorean triples. In this case it is 9, 12, 15 triple. So the peacock and snake are going to meet at a distance of 12 feet from the pillar. Or we can write and solve the quadratic equation.

## 5.8

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2.1

**A fish is resting at the northeast corner of a rectangular pool. A heron standing at the northwest corner spies the fish. When the fish sees the heron looking at him, he quickly swims towards the south. When he reaches the south side of the pool, he has the unwelcome surprise of meeting the heron who has calmly walked due south along the side and turned at the southwest corner of the pool and proceeded due east, to arrive simultaneously with the fish on the south side. Given that the pool measures 12 units by 6 units, and that the heron walks as quickly as the fish swims, find the distance the fish swam.**

We begin by drawing out a model. There are two possible orientations for the 12×6 pool, one in which the 12-unit length is oriented North-South and the second that is oriented east-west. We will call these Scenario 1 and Scenario 2, respectively. These are indicated below:

In all cases:

f is the distance travelled by the fish and h is the distance travelled by the heron

h can be separated into the southbound portion (labelled s) and the eastbound portion (labelled e). Therefore h = f = s + e.

SCENARIO 1

\begin{array}{l}f^2=12^2+\left(6-e\right)^2\\

h=s+e=12+e\\

Since\ h=f,\ then\ f=h=12+e\\

Substituting:\\

\left(12+e\right)^2=12^2+\left(6-e\right)^2\\

144+24e+e^2=144+36-2e+e^2\\

24e=36-2e\\

26e=36\\

e=\frac{36}{26}=1\ \frac{10}{26}=1\ \frac{5}{13}\\

f=12+e\\

f=12+1\ \frac{5}{13}\\

f=13\frac{5}{13}\end{array}

Therefore, each animal travelled [latex]13\frac{5}{13}[/latex] units

SCENARIO 2

\begin{array}{l}f^2=6^2+\left(12-e\right)^2\\

h=s+e=12+e\\

Since\ h=f,\ then\ f=h=12+e\\

Substituting:\\

\left(12+e\right)^2=6^2+\left(12-e\right)^2\\

144+24e+e^2=36+144-24e+e^2\\

24e=36-24e\\

48e=36\\

e=\frac{36}{48}=\frac{3}{4}\\

h=12+e\\

h=12+\frac{3}{4}\\

h=12\frac{3}{4}\end{array}

Therefore, each animal travelled [latex]12\frac{3}{4}[/latex] units.

## 5.9

1.1

2.1

**One monkey came down a tree of height 100 and went to a pond a distance of 200. Another monkey, leaping some distance above the tree, went diagonally to the same place. If their total distances traveled are equal, tell me quickly, learned one —if you have a thorough understanding of calculation- how much is the height of the leap?**

Let h be the height of the leap above the tree. Since the diagonal length of the leap is equal to the height of the tree plus the ground distance (100+200), which is 300 units. Therefore, the model looks like:

This creates a right triangle with side lengths 200, 300 and h + 100

[latex]\begin{array}{l}\left(h+100\right)^2+200^2=300^2\\ \left(h+100\right)^2+40000=90000\\ \left(h+100\right)^2=50000\\ h+100=\sqrt{50000}\\ h=\sqrt{50000}-100\approx123.6\end{array}[/latex]

Therefore, the height of the jump is approximately 123.6

## 5.10

1.1

2.1

**In a certain lake swarming with red geese, the tip of a lotus bud was seen to extend a span [9 inches] above the surface of the water. Forced by the wind, it gradually advanced and was submerged at a distance of 2 cubits [40 inches]. Compute quickly, mathematician, the depth of the pond.**

Let d be the depth of the river. Assuming that the lotus is rooted in a consistent place in the riverbed, then the length of the stem is d+9.

[latex]\begin{array}{l}d^2+40^2=\left(d+9\right)^2\\ 40^2=\left(d+9\right)^2-d^2\\ 1600=d^2+18d+81-d^2\\ 1600=18d+81\\ 1600-81=18d\\ 1519=18d\\ 84.3\overline{8}=d\end{array}[/latex]

## 5.11

1.1

2.1

**Two towers, the heights of which are 30 paces and 40 paces, have a 50 paces distance. Between the two towers there is a font where two birds, flying down from the two towers at the same speed will arrive at the same time. What is the distance of the font from the two towers?**

Let d be the distance that each bird flies. If the speed is the same, and the time of departure and arrival is the same, then the distance must also be the same.

Let t be the distance of the font from the thirty-pace tower.

Let f be the distance of the font from the forty-pace tower.

Using the Pythagorean Theorem, and the given information that the towers are 50 paces apart:

[latex]\begin{array}{l}t^2+30^2=d^2\\ f^2+40^2=d^2\\ t+f=50\end{array} \begin{array}{l}f=50-t\\ Substitute:\\ t^2+30^2=d^2\\ \left(50-t\right)^2+40^2=d^2\\ Substituting\ again:\\ t^2+30^2=\left(50-t\right)^2+40^2\\ t^2+900=2500-100t+t^2+1600\\ 900=2500-100t+1600\\ 100t=2500+1600-900\\ 100t=3200\\ t=32\end{array}[/latex]

Therefore, the font is 32 paces from the thirty-pace tower and 50-32 or 18 paces from the 40-pace tower.