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A question on area

Since we aren’t told anything about where the triangles meet, it must not matter.

So, we can do the calculation for two triangles that meet at one of the box corners.

That yields 28.

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You are right!

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Ok, that was a little more elegant

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sigh

The average singapore sixth grader must be smarter than my college students

@delete in 24 hours

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I'm impressed by that reasoning. I think I would have tried to calculate the meeting point.

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based on what?

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Don't know. I'd probably try to logic that out, and if not possible eventually maybe realize the trick.

In other words, I'd waste a lot of time brute forcing it before I'd stumble upon a more elegant solution.

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That’s what I started doing. Once I had it sketched out and parameterized, I realized that the intersection point could be anywhere in the middle box.

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Ah, I thought you took one look, saw that you weren't given enough information for the intersection, and reasoned the intersection could be anywhere, including on a straight line between the two corners.

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I did start out with an inclination that there was a cute simple answer like that, but I started working it out to make sure I wasn’t missing something.

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3 sats \ 0 replies \ @WaxLuw 27s

The trick is realizing the meeting point doesn’t matter. Took me a minute to see it :)

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100% I'm not.

Been a long time since I played with these, but it seems I can just work it out and fill in the missing lengths... I used to enjoy these little tricksy exercises in math class.

  • Pythagorean to get the left triangle hypotenuse
  • Pythagorean to get the right-most white triangle hypothenuse
  • white triangles in the middle have same base and hourglass-like height, so must be half the area of a 7x8 rectangle.

....oh wait, I can ignore the sections to the sides and just look at the shades area as part of a twisted rectangle (rhomboids??)

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