Sister's math problem X-post

[ihardlyknowher]

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For OOT extra credit:

How many trapezoids are there?

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Same method.

(7 + 6 + 5 + 4 + 3 + 2 + 1) * 2 = 56

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I get more.

[jakethebake]

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jake, you're an argumentative guy, that is obvious. But you should stick to arguing over matters of opinion, not fact.

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Pregnant man gives birth. That's a fact.

[jakethebake]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
For OOT extra credit:

How many trapezoids are there?

[/ QUOTE ]

Same method.

(7 + 6 + 5 + 4 + 3 + 2 + 1) * 2 = 56

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I get more.

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See! I told you!

[bosoxfan]

wouldn't it be 85 if you count the big triangle that is the perimeter of the diagram?

[jakethebake]

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wouldn't it be 85 if you count the big triangle that is the perimeter of the diagram?

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it was already counted.

[Patrick del Poker Grande]

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wouldn't it be 85 if you count the big triangle that is the perimeter of the diagram?

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Yes, but only if you count it twice.

[ChromePony]

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For OOT extra credit:

How many trapezoids are there?


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I get 56.

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Yup.

This is basically the same problemm except the top row cant make trapazoids. 28*2 = 56.

[bosoxfan]

ummm...oops.
this is why I stay out of threads like this

[ihardlyknowher]

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the top row cant make trapazoids.

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This is right, but your answer is wrong.

[BOTW]

This might actually be more interesting than it appears on the surface.

Each vertical line added to a triangle gives us the next triangular number of triangles, ((n+1)*(n+2))/2. Third diagonal of Pascal's triangle.

Ok, maybe not.