Mensa Triangle Problem - Seeking Answers

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The Mensa Triangle Problem involves analyzing the slopes of the triangles presented. Participants noted that one of the shapes does not qualify as a true triangle due to differing slopes. Specifically, the hypotenuse of the red triangle has a slope of 3/8, while the green triangle's slope is 2/5, indicating that the overall shape is not straight. This discrepancy leads to an additional area that is often overlooked when counting squares. Understanding these slope differences is crucial to solving the problem correctly.
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Ok, I've just signed up to this forum and i see there are a lot of very smart people and i just wondered if anyone knows the answer to the mensa triangle problem. I've sat and stared at it for hours and still i have no answer. Any theories.
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Use the search function. There have been several threads about this already.

Bottom line: Look at the slopes, one of the shapes isn't even a real triangle.

For example, the second triangle's green part has a steeper slope than the red part.

EDIT: The green triangle slopes don't equal the red triangle slope.

EDIT2: To see this, line up a ruler to the first triangle, connect the bottom left to the top right. Notice how the actual "triangle" is slightly below this. Do the same to the second one, and you will see the "triangle" is slightly bent upwards. So the change in that area is that extra square.
 
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Damn you make it all sound easy but i see what you mean. I've just been counting the number of squares per triangle. Thats where I've been going wrong. Cheers
 
Oh, that thing! Check the slopes of the hypotenuse of the red right triangle: it has base 8 and height 3; slope 3/8. Now check the slope on the green right triangle: it has base 5 and height 2; slope 2/5. Since those are not the same, that "hypotenuse" of the entire "triangle" is NOT a straight line and it isn't really a triangle at all.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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