Does this go here? Illusion, I suppose

[Fatstep]

Can someone please explain this to me.

 

 

I'm sorry if it doesn't belong here, but it involves math? maybe.

[Queso]

It's more comfortable arranged in a certain way.

No magic there.

[freeztar]

The second triangle is simply bad display of Tetris skills. :esmoking:

 

But seriously, it would seem like there could be a relatively easy mathematical solution to this. I'm sure it would involve the relationship of the three triangles.

[Turtle]

Can someone please explain this to me.

 

 

I'm sorry if it doesn't belong here, but it involves math? maybe.

 

it's an illusion perpetrated by thickening the lines in the first drawering. this is not math; it's parlor magic. :esmoking:

 

PS been there, done that. >> http://hypography.com/forums/physics-mathematics/1407-triangle-mystery.html?highlight=triangle+mystery

[CraigD]

Can someone please explain this to me.

It’s a famous “trick” paradox. It’s appeared in hypography threads before – but there’s fun, not harm, in posting it again.

 

I’d be depriving you and other’s who haven’t seen it before of the fun if I gave the trick away, but I’ll provide a one-word hint - “slope?” - and the following: if you laid a straightedge on it, you’d find something not-quite-right about the whole “triangle”…

[Mercedes Benzene]

AHA! I see it now. Tricky little benz...

...i mean, err, bends.

[Farsight]

Look at them triangles from a sidelong angle to see it.

[freeztar]

Ok, I finally spotted it.

:shrug:

 

I was beginning to feel like that one person who didn't get the joke.

[CraigD]

As most of us have now discovered, the illusion is due to the 2 triangle pieces having slightly different slopes, 3/8 vs. 2/5. In geometric terms, they're not similar. This causes the “missing square” assembly of them to have a very skinny parallelogram (see attached sketch) with area 1 that the original does not.

 

So we're not guilty of having a thread in our Physics & Math forum with not actual numbers in it, here's a math challenge:

Given: the red triangle is right, with short sides length 8 and 3; the green triangle is right with short sides 5 and 2; the “hole” is a square with side length 1.

 

Find find the length of the sides and diagonals of the parallelogram.

 

Finding the lengths of the sides, and one of the diagonals, is trivial. Finding the length of the other diagonal is not difficult, but not quite too trivial to be challenging. :doh:

[New-ideas]

The triangles cannot be the same area, the seond one has to be larger than the first one, maybe the seperating lines are thicker, the hypotense is possible not a straight line, it may be curved, making the shape have a bigger area.

[New-ideas]

Ops, I only just read the comments, everyone already got that, sorry :beer:

[freeztar]

As most of us have now discovered, the illusion is due to the 2 triangle pieces having slightly different slopes, 3/8 vs. 2/5. In geometric terms, they're not similar. This causes the “missing square” assembly of them to have a very skinny parallelogram (see attached sketch) with area 1 that the original does not.

 

So we're not guilty of having a thread in our Physics & Math forum with not actual numbers in it, here's a math challenge:

Given: the red triangle is right, with short sides length 8 and 3; the green triangle is right with short sides 5 and 2; the “hole” is a square with side length 1.

 

Find find the length of the sides and diagonals of the parallelogram.

 

Finding the lengths of the sides, and one of the diagonals, is trivial. Finding the length of the other diagonal is not difficult, but not quite too trivial to be challenging. :)

 

I'd like to solve the puzzle....

 

All four sides are of length 1 and both diagonals are of length [math]{\sqrt{2}[/math].

 

Hence the square "hole"/parallelogram.

 

I came to this conclusion by applying Pythagorean Theory [math]a^2 + b^2=c^2[/math].

 

[math]1+1=c^2[/math]

 

[math]2=c^2[/math]

 

[math]{sqrt{2}=c[/math]

[CraigD]

Another way of describing the before and after (“missing square”) “triangles” is that they are actually irregular quadrilaterals with sides [math]13[/math], [math]5[/math], [math]sqrt{29}[/math], and [math]sqrt{73}[/math].

 

It’s possible to alter the proportions of the pieces to make the dissimilarity between the 2 triangles more pronounced – see the attached thumbnail. The parallelogram I refer to in post #9 is formed by joining the 2 thin triangles formed by the piece edges and the true hypotenuse of the big triangle

I'd like to solve the puzzle....

 

All four sides are of length 1 and both diagonals are of length [math]{\sqrt{2}[/math].

 

Hence the square "hole"/parallelogram.

Not the parallelogram I had in mind, freeztar – you’ve given the sides and diagonals of the missing square. I’m looking for the very-flattened parallelogram with sides the length of the 2 triangles’ hypotenuses, [math]\sqrt{8^2 +3^2} = \sqrt{73}[/math] and [math]\sqrt{5^2 +2^2} = \sqrt{29}[/math]. One diagonal is just the true hypotenuse of the big triangle, [math]\sqrt{13^2 +5^2} = \sqrt{194}[/math]. The problem is to find the other, short diagonal, and show that the area of the skinny parallelogram is 1, the same as the missing square.

[freeztar]

That picture definitely clears things up. :)

 

I'll give this a go tomorrow when I have some free time (if that happens).

[addicted-]

It's actually really simple. All it has to do with is the angles and alignment in the parallelogram, Most peices can fit together because of the angle they are while left over peices are just simply put into the place they fit. :) really hard to explain

[september_13]

good to see something like this... agiant .