nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Modern Puzzles/123 - Making an Octagon

== by : $123$ ==
;Making an Octagon
:Can you cut the regular octagon from a square piece of paper without using compasses or ruler,
::or anything but scissors?
:You can fold the paper to make creases.
:
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Paper-Folding Puzzles: $123$. -- Making an Octagon
* : Geometrical Problems: Paper Folding Puzzles: $363$. Folding an Octagon
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/123 - Making an Octagon/Solution

== by : $123$ ==
;Making an Octagon
:Can you cut the regular octagon from a square piece of paper without using compasses or ruler,
::or anything but scissors?
:You can fold the paper to make creases.
:
== Solution ==
Let the square be $\Box ABCD$.
:
Fold $AD$ to $BC$ and $AB$ to $DC$ to obtain the midpoints $E$, $F$, $G$ and $H$ of $AD$, $BC$, $AB$ and $DC$ respectively.
Fold $EG$, $FG$, $EH$ and $FH$.
Fold at $H$ to place $DH$ against $EH$, then at $E$ to place $ED$ against $EH$.
Where those folds cross at $I$ is one of the vertices of the regular octagon we are seeking.
The other three vertices are obtained similarly.
== Sources ==
* : Solutions: $123$. -- Making an Octagon
* : Answers: $363$. Folding an Octagon
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/124 - Drawing a Straight Line

== by : $124$ ==
;Drawing a Straight Line
:If we want to describe a circle we use an instrument that we call a pair of compasses,
::but if we need a straight line we use no such instrument --
::we employ a ruler or other straight edge.
:In other words, we first seek a straight line to produce our required straight line,
::which is equivalent to using a coin, saucer of other circular object to draw a circle.
:Now, imagine yourself in such a position that you cannot obtain a straight edge --
::not even a piece of thread.
:Could you devise a simple instrument that would draw your straight line,
::just as the compasses describe a circle?
:It is an interesting question, but, of course, of no practical value.
:We shall continue to use the straight edge.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $124$. -- Drawing a Straight Line
* : Geometrical Problems: Plane Geometry Puzzles: $303$. Drawing a Straight Line
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/124 - Drawing a Straight Line/Historical Note

== Historical Note on by : $124$ - Drawing a Straight Line ==
, in his $1968$ repackaging , mentions , in which a number of other linkages are described, all doing the same job.
== Sources ==
* : Geometrical Problems: Plane Geometry Puzzles: $303$. Drawing a Straight Line
Category:Historical Notes
Henry Ernest Dudeney/Modern Puzzles/124 - Drawing a Straight Line/Solution

== by : $124$ ==
;Drawing a Straight Line
== Solution ==
:Prime.mover suggests:
:You can always fold a piece of paper.
However, 's suggestion is more elaborate, as follows.
:
The diagram shows a linkage of rods which are free to rotate at the points at which they are joined.
The geometry is such that:
:the longer rods are both the same length
:the shorter rods are all the same length
:the distance $AB$ is the same as the distance $BC$.
The points $A$ and $B$ are pinned to the drawing board so as to be able to rotate around those pins.
A writing instrument is attached to $D$ so as to draw on the surface when it is moved.
When $D$ is moved, a straight line is drawn.
Note that it is not strictly necessary for the rods themselves to be accurately straight, merely that the distances between the hinges are correct.
== Proof ==
It remains to be proved that the line drawn by the writing instrument at $D$ is straight.
== Historical Note ==
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $124$. -- Drawing a Straight Line
* : Geometrical Problems: Plane Geometry Puzzles: $303$. Drawing a Straight Line
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/125 - Making a Pentagon

== by : $125$ ==
;Making a Pentagon
:How do you construct a regular pentagon on a given unit straight line?
's original article was unnecessarily prolix.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $125$. -- Making a Pentagon
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $257$. Making a Pentagon
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/125 - Making a Pentagon/Solution

== by : $125$ ==
;Making a Pentagon
:How do you construct a regular pentagon on a given unit straight line?
's original article was unnecessarily prolix.
== Solution ==
:
Let $AB$ be the unit straight line segment.
Construct $BC$ perpendicular to $AB$ half the length of $AB$.
Draw $AC$ and produce it to $D$ such that $CD = CB$.
Then $BD$ is the radius of the circle which circumscribes a regular pentagon whose side is $1$ unit.
== Proof ==
== Sources ==
* : Solutions: $125$. -- Making a Pentagon
* : Answers: $257$. Making a Pentagon
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/126 - Drawing an Oval

== by : $126$ ==
;Drawing an Oval
:It is well-known that you can draw an ellipse by sticking two pins into the paper, enclosing them with a loop of thread,
::and keeping the loop taut, running a pencil all the way round till you get back to the starting point.
:Suppose you want an ellipse with a given major axis and minor axis.
:How do you arrange the position of the pins, and what would be the length of the thread?
's original article was unnecessarily prolix.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $126$. -- Drawing an Oval
* : Geometrical Problems: Plane Geometry Puzzles: $304$. Drawing an Ellipse
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/126 - Drawing an Oval/Solution

== by : $126$ ==
;Drawing an Oval
== Solution ==
Place the pins $2 \sqrt {a^2 - b^2}$ apart, where $2 a$ and $2 b$ are the major axis and minor axis respectively.
Then make the string $2 a + 2 \sqrt {a^2 - b^2}$ long.
== Proof ==
Let $L$ denote the length of the thread.
Let $E$ denote the ellipse being drawn.
Let $f_1$ and $f_2$ denote the foci of $E$.
Let $2 a$ and $2 b$ be the major axis and minor axis respectively of $E$.
Let $2 c$ be the distance between $f_1$ and $f_2$.
From the equidistance property of the ellipse, the pins are to be placed at $f_1$ and $f_2$.
From Equidistance of Ellipse equals Major Axis, the sum of the lengths of the sections of thread between the pins and the perimeter of the ellipse equals $2 a$.
From Focus of Ellipse from Major and Minor Axis:
:$c^2 = a^2 - b^2$
Hence the required length of string needed to draw $E$ is given by:
:$L = 2 a + 2 \sqrt {a^2 - b^2}$
and the pins are to be placed a distance $2 \sqrt {a^2 - b^2}$ apart.
 solution focuses on the special case where the major axis is $12$ inches and the minor axis is $8$ inches, and uses an ad hoc semi-geometrical method of solution.
== Sources ==
* : Solutions: $126$. -- Drawing an Oval
* : Answers: $304$. Drawing an Ellipse
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/127 - With Compasses Only

== by : $127$ ==
;With Compasses Only
:Can you show how to mark off the four corners of a square, using the compasses only?
:You simply use a sheet of paper and the compasses,
::and there is no trick, such as folding the paper.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $127$. -- With Compasses Only
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $258$. With Compasses Only
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/127 - With Compasses Only/Solution

== by : $127$ ==
;With Compasses Only
== Solution ==
The solution given by makes the assumption that the compass is stiff enough that when you lift it from the paper, you can put the point down and make another circle with the same opening that you used for the last one.
:
Draw a circle $OA$.
Then, using the same radius $OA$, make:
:the arc at $B$ with center $A$
:the arc at $C$ with center $B$
:the arc at $D$ with center $C$.
Then make:
:an arc at $E$ with center $A$ and radius $AC$
:another arc at $E$ with center $D$ and radius $DC$.
$OE$ is now the length of one side of the square required.
Now, using the length $OE$ as the radius, draw the arcs cutting the original circle at $F$ and $G$.
The square now has vertices $AFDG$.
Some may consider the solution flawed, as you have to draw the arcs with radius $OE$ at a point different from both $O$ and $E$.
You can rescue this by constructing another circle whose center is at $D$ and whose radius is $O$, and constructing a point in the same position relative to it as $E$ is to circle $OA$.
Then one of the arcs used to construct that point intersects the circle $OA$ at $F$, and the job is accomplished thus.
== Sources ==
* : Solutions: $127$. -- With Compasses Only
* : Answers: $258$. With Compasses Only
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/128 - Lines and Squares

== by : $128$ ==
;Lines and Squares
:Here is a simple question.
:With how few straight lines can you make exactly one hundred squares?
:Thus, in the first diagram it will be found that with nine straight lines I have made twenty squares
::(twelve with sides of the length $AB$, six with sides $AC$, and two with sides of the length $AD$).
:
:In the second diagram, although I use one more line, I only get seventeen squares.
:So, you see, everything depends on how the lines are drawn.
:Remember there must be exactly one hundred squares -- neither more nor fewer.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $128$. -- Lines and Squares
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $259$. Lines and Squares
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Henry Ernest Dudeney/Modern Puzzles/128 - Lines and Squares
Henry Ernest Dudeney/Modern Puzzles/128 - Lines and Squares/General Solution

== by : $128$ ==
;Lines and Squares
== General Solution ==
With $n$ straight lines we can make as many as:
:$\dfrac {\paren {n - 3} \paren {n - 1} \paren {n + 1} } {24}$ squares if $n$ is odd
:$\dfrac {\paren {n - 2} \paren {n - 1} n } {24}$ squares if $n$ is even.
== Proof ==
Suppose with $n$ lines, we create a rectangular grid with $a$ columns and $b$ rows.
Further suppose that $a < b$.
We can add an extra column and remove a row simultaneouly without changing the total number of lines $n$.
By adding one more column, we create:
:$b$ squares with side length $1$;
:$b - 1$ squares with side length $2$;
:...
:$b - a + 1$ squares with side length $a$.
The number of new squares this process adds will be:
By removing a row now, we remove:
:$a + 1$ squares with side length $1$;
:$a$ squares with side length $2$;
:...
:$1$ square with side length $a + 1$.
The number of squares this process removes will be:
The net change of the number of squares is:
Therefore for the same $n$, the closer the number of rows and columns are, the more squares we would create.
For even $n$, we would have $\dfrac n 2$ vertical lines and horizontal lines each.
This creates a square grid with side length $\dfrac n 2 - 1$.
The number of squares with side length $1$ is $\paren {\dfrac n 2 - 1}^2$.
The number of squares with side length $2$ is $\paren {\dfrac n 2 - 2}^2$.
...
The number of squares with side length $\dfrac n 2 - 1$ is $1^2$. 
Hence there would be:
Now suppose we create a rectangle with $n + 1$ lines. 
By adding an extra column (or row), we would create:
:$\dfrac n 2 - 1$ more squares with side length $1$;
:$\dfrac n 2 - 2$ more squares with side length $2$;
:...
:$1$ more square with side length $\dfrac n 2 - 1$.
The total number of squares will now become:
By doing the substitution $n \to n - 1$, we obtain the formula for odd $n$:
:$\dfrac {\paren {n - 3} \paren {n - 1} \paren {n + 1} } {24}$
== Sources ==
* : Solutions: $128$. -- Lines and Squares
* : Answers: $259$. Lines and Squares
Category:Henry Ernest Dudeney/Modern Puzzles/128 - Lines and Squares
Henry Ernest Dudeney/Modern Puzzles/128 - Lines and Squares/Solution

== by : $128$ ==
;Lines and Squares
== Solution ==
We use $15$ straight lines as follows:
:
There are:
:$40$ squares with sides of length $AB$
:$28$ squares with sides of length $AC$
:$18$ squares with sides of length $AD$
:$10$ squares with sides of length $AE$
:$4$ squares with sides of length $AF$
making $100$ altogether.
We could make as many as $112$ squares with $15$ straight lines, but we were asked for exactly $100$.
With $14$ straight lines we can make only $91$ squares.
=== General Solution ===
== Sources ==
* : Solutions: $128$. -- Lines and Squares
* : Answers: $259$. Lines and Squares
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Category:Henry Ernest Dudeney/Modern Puzzles/128 - Lines and Squares
Henry Ernest Dudeney/Modern Puzzles/129 - The Circle and Discs

== by : $129$ ==
;The Circle and Discs
:During a recent visit to a fair we saw a man with a table,
::on the oilcloth covering of which was painted a large red circle,
::and he invited the public to cover this circle entirely with five tin discs which he provided,
::and offered a substantial prize to anyone who was successful.
:The circular discs were all of the same size, and each, of course, smaller than the red circle.
:he showed that it was "quite easy when you know how," by covering up the circle himself without any apparent difficulty,
::but many tried over and over again and failed every time.
:It was a condition that when once you had placed any disc you were not allowed to shift it,
::otherwise, by sliding them about after they had been placed, it might be tolerably easy to do.
:Let us assume that the red circle is six units in diameter.
:Now, what is the smallest possible diameter for the five discs in order to make a solution possible?
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $129$. -- The Circle and Discs
* : Geometrical Problems: Circle Puzzles: $287$. The Circle and Discs
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/129 - The Circle and Discs/Solution

== by : $129$ ==
;The Circle and Discs
== Solution ==
:
The dotted lines represent the red circle and a regular pentagon inscribed within it.
The center of this circle is the point $C$.
Find the point $D$ which is equidistant from $A$, $B$ and $C$.
With radius $AD$, draw the circle $ABC$.
Similarly construct the points $E$, $F$, $G$ and $H$.
These points are the centers of the circles representing the discs with which the red circle is covered.
We are given that the red circle is $6$ units in diameter.
The diameter of the discs is then just under $4$ units in diameter.
Covering is possible if the ratio of the two diameters is greater than approximately $0.6094185$.
In the case above, where all five discs reach the center of the red circle, the ratio is the golden mean $\approx 0.6180340$.
== Proof ==
== Sources ==
* 
* : Geometrical Problems: Various Geometrical Puzzles: $129$. -- The Circle and Discs
* : Geometrical Problems: Circle Puzzles: $287$. The Circle and Discs
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/12 - A Weird Game

== by : $12$ ==
;A Weird Game
:Seven men engaged in play.
:Whenever a player won a game he doubled the money of each of the other players.
:That is, he gave each player just as much money as each had in his pocket.
:They played $7$ games and, strange to say, each won a game in turn in the order of their names,
::which began with the letters $\text A$, $\text B$, $\text C$, $\text D$, $\text E$, $\text F$, and $\text G$.
:When they had finished it was found that each man had exactly $2$ shillings and $8$ pence in his pocket.
:How much had each man had in his pocket before play?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $12$. -- A Weird Game
* : Arithmetical and Algebraical Problems: Money Puzzles: $8$. A Weird Game
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/130 - Mr. Grindle's Garden

== by : $130$ ==
;Mr. Grindle's Garden
:"My neighbour," said Mr. Grindle, "generously offered me, for a garden,
::as much land as I could enclose with four straight walls measuring $7$, $8$, $9$ and $10$ rods in length respectively."
:"And what was the largest area you were able to enclose?" asked his friend.
:Perhaps the reader can discover Mr. Grindle's correct answer.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $130$. -- Mr. Grindle's Garden
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $260$. Mr. Grindle's Garden
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Henry Ernest Dudeney/Modern Puzzles/130 - Mr. Grindle's Garden
Henry Ernest Dudeney/Modern Puzzles/130 - Mr. Grindle's Garden/Solution

== by : $130$ ==
;Mr. Grindle's Garden
== Solution ==
Just under $71$ square rods.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* $202$ - A Garden Puzzle in 's posthumous collection of $1932$, which is similar to this.
== Sources ==
* : Solutions: $130$. -- Mr. Grindle's Garden
* : Answers: $260$. Mr. Grindle's Garden
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Category:Henry Ernest Dudeney/Modern Puzzles/130 - Mr. Grindle's Garden
Henry Ernest Dudeney/Modern Puzzles/130 - Mr. Grindle's Garden/Solution/Proof 1

== by : $130$ ==
;Mr. Grindle's Garden
== Solution ==
== Proof ==
Let $\AA$ square rods be the area of the garden.
We are given that $\AA$ is the greatest possible for the given sides.
We also have that the sides of the garden are in arithmetic sequence.
Thus:
== Sources ==
* : Solutions: $130$. -- Mr. Grindle's Garden
* : Answers: $260$. Mr. Grindle's Garden
Category:Henry Ernest Dudeney/Modern Puzzles/130 - Mr. Grindle's Garden
Henry Ernest Dudeney/Modern Puzzles/130 - Mr. Grindle's Garden/Solution/Proof 2

== by : $130$ ==
;Mr. Grindle's Garden
== Solution ==
== Proof ==
Let $\AA$ square rods be the area of the garden.
We are given that $\AA$ is the greatest possible for the given sides.
From Area of Quadrilateral with Given Sides is Greatest when Quadrilateral is Cyclic, the quadrilateral formed by the sides of the garden is cyclic.
Hence we can apply Brahmagupta's Formula:
:$\AA = \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} }$
where $s$ is the semiperimeter:
:$s = \dfrac {a + b + c + d} 2$
Thus:
and so:
Category:Henry Ernest Dudeney/Modern Puzzles/130 - Mr. Grindle's Garden
Henry Ernest Dudeney/Modern Puzzles/131 - The Garden Path

== by : $131$ ==
;The Garden Path
:This is an old puzzle that I find frequently cropping up.
:Many find it perplexing, but it is easier than it looks.
:A man has a rectangular garden, $55$ yards by $40$ yards,
::and he makes a diagonal path, one yard wide, exactly in the manner indicated in the diagram.
:
:What is the area of the path?
:Dimensions for the garden are generally given that only admit of an approximate answer,
:but I select figures that will give an answer that is quite exact.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $131$. -- The Garden Path
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $261$. The Garden Path
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/131 - The Garden Path/Solution

== by : $131$ ==
;The Garden Path
== Solution ==
:$66 \tfrac 2 3$ square yards.
== Proof ==
We solve the general case.
Let $L$ yards be the length of the garden.
Let $B$ yards be the breadth of the garden.
Let $C$ yards be the width of the path.
Let $x$ yards be the length of the boundary between the path and the rest of the garden.
Let $A$ square yards be the area of the path.
Let $y$ be the length of one of the sides of the path.
The path is a parallelogram bounded by two sides of length $x$ and two sides of length $y$.
From Area of Parallelogram:
:$A = C x$
Then we have:
Having done that, we now use the numbers given.
We have:
Inserting these into $(3)$ above:
Finally:
:$A = C x = 66 \tfrac 2 3$
Hence the area of the path is $66 \tfrac 2 3$ square yards.
== Sources ==
* : Solutions: $131$. -- The Garden Path
* : Answers: $261$. The Garden Path
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/132 - The Garden Bed

== by : $132$ ==
;The Garden Bed
:Here is quite a simple little puzzle.
:A man has a triangular lawn of the proportions shown,
::and he wants to make the largest possible rectangular flower-bed without enclosing the tree.
:
:This will serve to teach the uninitiated a simple rule that may prove useful on occasion.
:For example, it would equally apply to the case of a carpenter who had a triangular board
::and wished to cut out the largest possible rectangular table-top without including a bad knot in the wood.
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $132$. -- The Garden Bed
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $262$. The Garden Bed
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/132 - The Garden Bed/Solution

== by : $132$ ==
;The Garden Bed
== Solution ==
From Largest Rectangle Contained in Triangle, the largest rectangle that can be inscribed within a triangle is situated such that:
:Two of its vertices lie on one side of the triangle
:Its other two vertices bisect the other two sides of the triangle.
:
This solution works only if the tree is sufficiently close to the corner that it is not enclosed inside at least one of those largest rectangles.
== Sources ==
* : Solutions: $132$. -- The Garden Bed
* : Answers: $262$. The Garden Bed
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/133 - A Problem for Surveyors

== by : $133$ ==
;A Problem for Surveyors
:There are tricks in every trade, and the science of numbers contains an infinite number of them.
:In nearly every vocation of life there are little wrinkles and short cuts that are most useful when known.
:For example:
:A man bought a little field, and here is a scale map that was given to me.
:I asked my surveyor to tell me the area of the field,
::but he said it was impossible without some further measurements;
::the mere length of one side, $7$ rods, was insufficient.
:What was his surprise when I showed him in about two minutes what was the area!
:Can you tell how it is to be done?
:
=== Click here for solution ===
== Sources ==
* : Geometrical Problems: Various Geometrical Puzzles: $133$. -- A Problem for Surveyors
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $263$. A Problem for Surveyors
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/133 - A Problem for Surveyors/Solution

== by : $133$ ==
;A Problem for Surveyors
== Solution ==
The construction given below takes advantage of the fact that $\angle AFG$ is a right angle.
:
Produce $GA$.
Construct $AC$.
Construct a straight line through $B$ parallel to $AC$ to meet $GA$ at $1$.
Construct $1D$.
Construct a straight line through $C$ parallel to $1D$ to meet $GA$ at $2$.
Construct $2E$.
Construct a straight line through $D$ parallel to $2E$ to meet $GA$ at $3$.
Construct $3F$.
Construct a straight line through $E$ parallel to $3F$ to meet $GA$ at $4$.
Construct $4G$.
Then $\triangle G4F$ is equal in area to the field $ABCDEFG$.
== Proof ==
== Sources ==
* : Solutions: $133$. -- A Problem for Surveyors
* : Answers: $263$. A Problem for Surveyors
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions