A farmer has 3 square fields shown above, each containing 18, 20, and 26 acres. In order to get a fence around his property, he bought the 4 intervening triangular fields.

Can you figure out the whole area of his estate?

*Will unmask any submitted answers later on this week*

HammyCat| Profile November 18th, 2008 - 12:35 amThe 26 acre square has sides 5.1 (rounded to 1 d.p)

The 20 acre square has sides 4.5

The 18 acre square has sides 4.2

Knowing these sides, all you need to do is find the area of the four triangles, but I havn’t covered these kinds of triangles in class yet (only for right-angled ones) I’d take a wild guess at 84 acres though

bilbao| Profile November 18th, 2008 - 4:41 amWhole area: 104,69 acres (423.670 m2).

Length of the fence: 2.124,35 m

suineg| PUZZLE MASTER | Profile November 18th, 2008 - 8:52 amOk, I think its like this:

Area Squares: 64 acres

Area triangles: I think you need to get the lenght of the sides of the squares, then by trigonometry get the angle between them and get the other sides of the triangles then get the height to calculate the Area of the 8: 90 degree triangles that will be formed, man thats hard,not posting yet just thinking at loud jajajaja there have to be an easiest way to solve this one.

Shawn| PUZZLE GRANDMASTER | Profile November 18th, 2008 - 9:01 am100 acres.

I enjoyed this puzzle. I would not have guessed that the outer triangle areas were all equal to the inner triangle area. This appears to hold true for squares of any size in this configuration – very cool!!

vips| Profile November 18th, 2008 - 4:12 pm128

michaelc| Profile November 18th, 2008 - 6:12 pmYes.

Top left triangle 9 acres.

Top right triangle 9 acres.

Center triangle 9 acres.

Bottom Triangle, what do you know!! 9 acres.

Total acres 100 acres.

What nice numbers. Should be a prequisite in all good puzzles!

Wish I could say I found an easy way to solve it, but not really that lucky. I used Heron’s formula and the Law of Cosines to solve it.

Looking forward for someone who had a better way… and I do believe there is a better way than the one I had…

Falwan| Profile November 18th, 2008 - 10:43 pmtotal area = 112 acres

Mehmet Karatay| Profile November 19th, 2008 - 7:27 amI get a total area of 94.5 acres.

The area of the center triangle is 10.9 acres.

The areas of the three outer triangles are: 6.05, 2.56 and 10.9 acres.

The angles of the inner triangle are: 71.6, 56.3 and 52.1 degrees.

All the figures quoted are to three significant figures. I did the calculations using the full values. I hope I didn’t make any silly mistakes!

Mehmet

Bobo The Bear| PUZZLE MASTER | Profile November 19th, 2008 - 9:35 amEach of the triangular lots has an area of 9 acres, and the total area is 100 acres.

Mashplum| PUZZLE MASTER | Profile November 19th, 2008 - 8:24 pm100 acres

Someone| Profile November 19th, 2008 - 9:05 pm89 acres

glupi_zmaj| Profile November 20th, 2008 - 5:41 amExclude the squares and rotate the triangles to get this picture

Z

\

|\

| \

| \b

| \

| \

| \C

| / \’ .

| / \ .

| / \ .

| /c b\ .

|/ \ .

A/_____a_____\___a___.Y

/ . ‘B

/ . ‘

c/ . ‘

/ . ‘

/. ‘

X

a=sqrt(18), b=sqrt(20), c=sqrt(26)

(angle ABC) + (angle YBC) = 180 because when added with two 90 degrees angles from two squares they add up to 360 (original picture).

This means that triangles ABC and BYC have the same height and basis, therefore they have the same areas.

Same calculation can be used for ABX and ACZ triangles.

This means that the whole area is area of the three squares(64acres) plus four ABC triangle areas.

ABC triangle area can be calculated with Heron’s formula:

sqrt[s*(s-a)*(s-b)*(s-c)], s = (a+b+c)/2

suineg| PUZZLE MASTER | Profile November 20th, 2008 - 10:24 amok, its clear that the sides of each square can be found,

A= square(a)–> A: area; a: side of any of the squares. So you have the sides.

By the tangent rule you could get the inner angle that is form inside the sides that are known( the ones shared with a square)

Now with that angle you could get by the sine law the unknow side of each traingle ( the ones that are not shared with a square)

Until now everything looks easy (beside all those horrible decimals jajaja) right: Here is where the thing gets funny, you draw an imaginary line to make every triangle a rectangle triangle, the height is that imaginary line and by sine law you get:

sin(angle) = height/hypotenuse(side share with a square)

then you have all the height of each triangle, and the base( at first the unknow sides or the non shared with square sides); you calculate the area of the triangles and add that to the 64 inches of the square land, cool. ( Sorry for not working with numbers but I dont like decimals so much)

suineg| PUZZLE MASTER | Profile November 20th, 2008 - 10:33 amFor the one that work with numbers I have one curiosity:

I noticed that the three border triangles share two side with the middle triangle; I search for a law that indicates a relationship between the areas of triangles that share two sides, and did not found it, however it seems to me that the area of the triangles could be the same, the amount that you gain in the base is lost in the height, but maybe( very possibly) I am wrong. Cool.

RK| Founder | Profile November 20th, 2008 - 11:28 amwill post the answer later on today…

michaelc| Profile November 20th, 2008 - 12:54 pmAs Shawn states above, this appears to hold true for all types of these configurations.

Area of triangle = 1/2 *ab Sin C

where a and b are the corresponding lengths of the adjacent sides of angle C in a triangle.

In the above case, the top left triangle and center triangle intersect with supplementary angles and equal adjacent sides, sqrt(26) and sqrt(18).

The Sine function of supplementary angles is equal. Therefore the area of the triangles are equal.

So it not just appears to be true, it it true. Anytime 3 squares meet and form a triangle as such, all triangle will have equivalent areas!

That’s like poetry man.

suineg| PUZZLE MASTER | Profile November 20th, 2008 - 9:18 pmCool, as Shawn and MichaelC points out turn out that my guess about the areas of the triangles was right, nice, however, my solution use a similar, well maybe exact approach that the method that MichaelC describe, but how the numbers where so beautiful, I only look at the square root of each square area and I did not have an integer result ( the sides like hammy-cat) jajaja well dont know how the result turn to be 100, but I belive in those guys. Well at least I think my solution was compatible algebraically talking.

LadyInsomnia| Profile November 20th, 2008 - 9:47 pmSay the 26-acre square has side-length X, the 20-acre square has side-length Y, and the 18-acre square has side-length Z. We must convert acres to square feet in order to have a measure of length available. The numbers for X, Y, and Z are messy in feet, but luckily we’ll only need them as X^2, Y^2, and Z^2, which are nice integers. We can use Heron’s formula to determine the area of the center triangle, call it C. Heron’s formula:

A = sqrt[s(s-X)(s-Y)(s-Z)]

where s = (X+Y+Z)/2

With a little algebra we get

A = sqrt[4Y^2*Z^2 -(X^2-Y^2-Z^2)^2]/4

Substituting gives A=9 acres for triangle C.

Now consider the upper-left triangle, call it L. Triangle L shares two sides with triangle C, of lengths X and Y. Say the angle between X and Y in triangle C is phi. Then the area of C can also be expressed by A = XY*sin(phi)/2. But the angle between X and Y in triangle L, call it theta, is supplementary to phi, so sin(theta)=sin(phi). Thus

A = XY*sin(phi)/2 = XY*sin(theta)/2

so the area of L = the area of C = 9 acres. The same argument holds for the two remaining triangles. Thus the total area is

T = 26+20+18+4(9) = 100 acres.

suineg| PUZZLE MASTER | Profile November 20th, 2008 - 9:55 pmSo the curiosity kill me and I had to do the math, yes the result by my method gives me 99.97 ok thats 100, cause Someone put me to rethink, man I wish i knew before the Heron formula, with that it was a ride on the park, Herons formula is great, new knowledge nice so I also get the 100 result when putting the numbers in my equation, now thats was something I would never have guessed, jaja cool.

RK| Founder | Profile November 20th, 2008 - 10:23 pmyes, 100 is correct.

Here is one way to view the problem:

http://www.smart-kit.com/wp-co.....lution.jpg

Every little dotted square in the diagram represents an acre.

The complete area of the squared diagram is 12 × 12 = 144 acres, and the portions 1, 2, 3, 4, not included in the estate, have the respective areas of 12½, 17½, 9½, and 4½. These added together make 44, which, deducted from 144, leaves 100 as the required area of the complete estate

Another way to approach: We can also see that the area of triangle E is 2½, F is 4½, and G is 4. These added together make 11 acres, which we deduct from the area of the rectangle, 20 acres, and we find that the field A contains exactly 9 acres. If you want to prove that B, C, and D are equal in size to A, divide them in two by a line from the middle of the longest side to the opposite angle, and you will find that the two pieces in every case, if cut out, will exactly fit together and form A.

turner| Profile November 22nd, 2008 - 9:14 pmI agree the area is 100. An easy way to compute the area of the inner triangle is the following. Drop a perpendicular line to the long side (i.e. sqrt(26)), this produces two simple triangles with a 90 deg angle. One can find the height of the triangle and the sides by solving the pair of equations

20 = h^2 + (sqrt(26)-a)^2 & 18 = h^2 + a^2

where h denotes the height and a is missing length of the triangle with side = sqrt(18). The solution yields

[h = ((9 * sqrt(2) * sqrt(13))/13), a = ((6 * sqrt(26))/13)]

Now from the law of sines one computes the angles

sin(ang1)/h = sin(pi/2)/a

=> ang1 = asin( h/a )

sin(ang2)/h = sin(pi/2)/(sqrt(26)-a)

=> ang2 = asin( h/(sqrt(26-a) )

The missing angle must be defined by: pi = ang1+ang2+ang3

=> ang3 = pi-ang1-ang2

At each “point” of the triangle solve for the missing angle

=> 2*pi = angI + pi/2 + pi/2 + C

=> C = pi – angI

Then use the area relation identified by others: AI = s1*s2*sinC

Then for Suieng and others the “ugly” decimals all work their magic.