#1

ok heres a problem on a take home test i had... se if anyone can get it.

Start with an equilateral triangle with sides of 10cm. Connect the midpoints of the sides to make a smaller triangle inside. Repeat infinite times and find the sum of the areas of all the triangles.

good luck

Start with an equilateral triangle with sides of 10cm. Connect the midpoints of the sides to make a smaller triangle inside. Repeat infinite times and find the sum of the areas of all the triangles.

good luck

#2

42?

#3

100 cm squared?

#4

Tell the asshole math teacher that it can't be done... Does he expect you to repeat it infinite times? It's infinite.

#5

Uh.... "

*Repeat infinite times*"?
#6

Umm, it doesnt converge to a point so you cant work out the exact area? Im not sure

#7

Its impossible. If you do something infinite, its going to keep going. Theres no possible way you could get a sum from something

*Infinite*.
#8

Won't it just be the area of the original equilateral triangle? If you divide a square, the area of the divisions still add up to be the total area of the original square.

#9

^Only if you don't include the triangles within. If you don't then you're right but I thought you had to include them and that would make it an infinitely long decimal.

EDIT\/:

EDIT\/:

*Last edited by Meths at Jan 7, 2007,*

#10

ok heres a problem on a take home test i had... se if anyone can get it.

Start with an equilateral triangle with sides of 10cm. Connect the midpoints of the sides to make a smaller triangle inside.Repeat infinite timesand find the sum of the areas of all the triangles.

good luck

You idiot.

#11

well simply its one half base time height and seeing as how making smaller triangles out of the same triangle isn't going to change the AREA of the triangle it should be 1/2 bh which you would need sin and cos for and all that good stuff, so I don't wanna figure it out

#12

Won't it just be the area of the original equilateral triangle?

We did this as an example of doing limits, and yes, it approaches (never quite reaches) the area of the original.

#13

ok heres a problem on a take home test i had... se if anyone can get it.

Start with an equilateral triangle with sides of 10cm. Connect the midpoints of the sides to make a smaller triangle inside. Repeat infinite times and find the sum of the areas of all the triangles.

good luck

do you do it just for the internal triangle or for the three surrounding it? because after you connect the midpoints of the sides you have 4 small triangles and one big triangle. if you were to sum the areas of all non-overlapping triangles you get the area of the original triangle.

#14

50cm^2 ?

#15

the infinte times thing was a joke.... anyways you have to add up the areas of all the triangles you make (even though technically its inside of the original one)

#16

i dont think anyone has gotten it yet but who knows if i did it right or not....maybe i should say nobody has gotten the answer i got

#17

May aswell throw this one in. I think I found it in some book that was trying far too hard to be clever:

When drawing, say a 10cm line with a ruler you begin at 0cm, then get to 1cm and towards the end reach 9cm. You carry on going towards 10cm; 9.1cm, 9.2cm and so on. Until 9.9cm, you keep going till you reach 9.99cm, then you get to 9.999cm, then 9.9999cm. You keep getting infinitley closer to 10cm but when do actually reach it?

Stupid, really.

When drawing, say a 10cm line with a ruler you begin at 0cm, then get to 1cm and towards the end reach 9cm. You carry on going towards 10cm; 9.1cm, 9.2cm and so on. Until 9.9cm, you keep going till you reach 9.99cm, then you get to 9.999cm, then 9.9999cm. You keep getting infinitley closer to 10cm but when do actually reach it?

Stupid, really.

#18

this is so obv its the area in the triangle you had to begin with so its 10 times 10 then halfed so the answer is 50 cm sqaured

#19

When it's infinite... i think it can be done to a point. Because the area will get so small it will only add an incredbily small amount so if it's to 2 decimal points then it will reach a number which wont change.

So just deal with the most important values... maybe to the 10

Area of equilateral triangle = 0.5(A x B) x 2 (i know the 0.5 and 2 cancel but just to make it obvious)

10x8.66 = 86.6

5x...

2.5...

1.25... so on

and so on... then add them all up otherwise... you will never be able to write enough numbers.

So just deal with the most important values... maybe to the 10

Area of equilateral triangle = 0.5(A x B) x 2 (i know the 0.5 and 2 cancel but just to make it obvious)

10x8.66 = 86.6

5x...

2.5...

1.25... so on

and so on... then add them all up otherwise... you will never be able to write enough numbers.

#20

See, the area of the first smaller circle is 1/2 the original, then the second smaller is half of the half, or 1/4, then the third would be half of that, so 1/8 and so on.

When you add them all up, you'd end up with 1/2area+1/4area+1/8area+1/16area+1/32area+1/64area and so on, which comes out to .99999999999area. Safe enough to say its the same as the original area.

When you add them all up, you'd end up with 1/2area+1/4area+1/8area+1/16area+1/32area+1/64area and so on, which comes out to .99999999999area. Safe enough to say its the same as the original area.

#21

this is so obv its the area in the triangle you had to begin with so its 10 times 10 then halfed so the answer is 50 cm sqaured

wow... you just did so many stupid things i dont know how to begin mocking you

#22

wow... you just did so many stupid things i dont know how to begin mocking you

no he just did one thing wrong, the fact is the height is not 10, but he was doing 1/2bh formula which is right

#23

this is so obv its the area in the triangle you had to begin with so its 10 times 10 then halfed so the answer is 50 cm sqaured

But astonishingly, it isn't 50 cm squared!

Learn your basic geometry

Area = Half of the HEIGHT X the Base

The height of an equalateral triangle with 10 cms for each side is 5 radical 3.

The area of the triangle is 25 radical 3.

#24

Um ... the same as the area of the triangle. 25 root 3 i think, can't be bothered getting a calculator.

#25

no he just did one thing wrong, the fact is the height is not 10, but he was doing 1/2bh formula which is right

well he didnt even read the question correctly.... you have to add up all the triangles you form inside the original ones even though their areas technically overlap

#26

...maybe i should say nobody has gotten the answer i got

And the answer you got was...spaghetti?

#27

This is definitely possible. Eventually the area of the triangle formed becomes infinitely small. OK.

Using the triangle area equation:

area = (1/2)bcsinA

Triangle 1:

(1/2)(10)(10)sin(60) = 43.30127019

Triangle 2:

(1/2)(5)(5)sin(60) = 10.82531755

Triangle 3:

(1/2)(2.5)(2.5)sin(60) = 2.706329387

Etc. Just keep halving the b and c values. Eventually the numbers become infinitely small.

The total comes out to about 57.678 or so.

Using the triangle area equation:

area = (1/2)bcsinA

Triangle 1:

(1/2)(10)(10)sin(60) = 43.30127019

Triangle 2:

(1/2)(5)(5)sin(60) = 10.82531755

Triangle 3:

(1/2)(2.5)(2.5)sin(60) = 2.706329387

Etc. Just keep halving the b and c values. Eventually the numbers become infinitely small.

The total comes out to about 57.678 or so.

*Last edited by JoHNNERz at Jan 7, 2007,*

#28

the answer is an equation since it goes on infinitely. but i dont feel like thinking of it

#29

This is definitely possible. Eventually the area of the triangle formed becomes infinitely small. OK.

Using the triangle area equation:

area = (1/2)bcsinA

Triangle 1:

(1/2)(10)(10)sin(60) = 43.30127019

Triangle 2:

(1/2)(5)(5)sin(60) = 10.82531755

Triangle 3:

(1/2)(2.5)(2.5)sin(60) = 2.706329387

Etc. Just keep halving the b and c values. Eventually the numbers become infinitely small.

The total comes out to about 57.678 or so.

wow finally someone who is not a complete idot(or at least didnt just guess)

nice job im pretty sure thats correct

#30

Why are we still getting answers? The case is closed.

/Thread

/Thread

#31

Okay, seems like this is all cleared up.

*CLOSED*

...

*CLOSED*

...

#32

Dude, I just ****ing owned this thread.

Just wanted to say that.

Just wanted to say that.

#33

57.73502691896258... to be a little more exact...

#34

57.73502691896258... to be a little more exact...

25 root 3 would be the most exact answer.

Result? I > you.

#35

THE BULGE is right, it doesn't matter how many triangles you make, the internal area of the original triangle remains constant, so the answer is 50cm2

#36

THE BULGE is right, it doesn't matter how many triangles you make, the internal area of the original triangle remains constant, so the answer is 50cm2

No it isn't! Learn your basic geometry! The height of the triangle is not 5 cms. It is 5 radical 3! That makes the area 25 radical 3!!!!!

25 root 3 would be the most exact answer.

Result? I > you.

25 root 3 = 43.30127019 not 57.73502691896258

*Last edited by DorkusMalorkus at Jan 7, 2007,*

#37

I didn't actually do the math but bxh should give you the area because all the smaller triangles are fractions of the largest one, i excluded the /2 because you want the area of ALL the triangles that includes the total of the fraction triangles and the largest one

#38

The area of a triangle is 1/2 x base x PERPENDICULAR HEIGHT).

You can use pythagoras' theorem (a^2 + b^2 = c^2) to find the perpendicular height. It's the square root of 75. 8.6 or something.

EDIT: My brother's friend saw a t-shirt with this on on eBay.

Pythagoras meets Einstein:

E= m(a^2 + b^2).

Haha, geek jokes ftw.

#39

25 root 3 = 43.30127019 not 57.73502691896258

Ah well, so Almer was wrong, not me. Assuming we're agreed that the answer is 25 root 3.

#40

43.30127019 is the area of just the original triangle, isnt it?