#Rectangle
44 messages · Page 1 of 1 (latest)
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what are ABCD?
i tried to do like this. I picked triangles with same areas
arbitrary points?
I think those are arbitrary.
but i assume AB=CD right
Yeah, that's the correct approach.
That is guaranteed due to O being the intersection of diagonals, I believe.
these are same. So from 2 of this kind of triangles both are shaded. so 2/2.
from 4 triangles that got same area 2 are shaded. which is 2/4
Right. Those two triangles are half of the rectangle.
So, what's the shaded fraction?
but this is not correct
. this one isnt correct idk why
These ones, I mean.
Well, 1/4, yeah.
So, their total area is 1/2.
Now, let's get rid of them both. What's the total remaining area?
oh i got your point
I think your mistake was that here you considered it as the fraction of the whole rectangle instead of the remaining area.
ohhh my god exactly
So, it's not 1/2 + 2/4, but 1/2 + (2/4)(1/2), rather.
The approach you can use here is just to use symmetry.
Here's the rectangle with the side triangles cut out.
Can you see that the white and gray parts are the same, just inverted around the center?
Yeah, exactly!
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