[Guest]

I made this on my Inventor software so I don't know if it's possible to do this by hand.

I made two hexagons that have center points 4 inches away from the corners. They also shared a side. Then I rotated them, so that one pair of corners were touching. I want to know what the area of the triangles above and below this corner are. See picture for understanding.

riddle_drawing.bmp

You can do this by hand. I just checked.

Edited April 22, 2009 by hartleydogg33

[Prof. Templeton]

The triangles would be 1/6 of the area of the hexagon or 6.9282 inches for each one.

[Guest]

The triangles would be 1/6 of the area of the hexagon or 6.9282 inches for each one.

Yes you are correct. Did you do this by hand or did you use a drawing software. If by hand, I figured out the base of the triangle. Then, I divided it in half and used the a squared+b squared=c squared to get the height. Then, base times height. Did you like it or was it too easy?

[Pickett]

Since you say the center of the hexagons are 4 inches away from the corners, I assume you mean away from ALL the corners...in which case they are regular hexagons...

That makes the triangles formed equilateral triangles with sides of length 4 inches...which means that the area of each triangle would be 2 * sqrt(12)...or about 6.93 sq. in.

Edited April 22, 2009 by Pickett

[Guest]

Since you say the center of the hexagons are 4 inches away from the corners, I assume you mean away from ALL the corners...in which case they are regular hexagons...

That makes the triangles formed equilateral triangles with sides of length 4 inches...which means that the area of each triangle would be 2 * sqrt(12)...or about 6.93 sq. in.

Yep, another one right. Congratulations!! Here's your free virtual zoo ticket. Just imagine you're in a zoo. By the way, was this too easy?

[Prof. Templeton]

Yes you are correct. Did you do this by hand or did you use a drawing software. If by hand, I figured out the base of the triangle. Then, I divided it in half and used the a squared+b squared=c squared to get the height. Then, base times height. Did you like it or was it too easy?

I assumed regular hexagons whose area is 6 times the area of the equilateral triangles that make them up. If two hexagons are touching at the corners as you've shown the two triangle whose area you are looking for would be equilateral of the same size as the six that make up the hexagons. So I just determined the area of a equilateral triangle with a side length of 4 inches.

[Guest]

I assumed regular hexagons whose area is 6 times the area of the equilateral triangles that make them up. If two hexagons are touching at the corners as you've shown the two triangle whose area you are looking for would be equilateral of the same size as the six that make up the hexagons. So I just determined the area of a equilateral triangle with a side length of 4 inches.

Wow! I never thought of that. Hmmm. At least I learned something while I was at school today!!

[Pickett]

Yep, another one right. Congratulations!! Here's your free virtual zoo ticket. Just imagine you're in a zoo. By the way, was this too easy?

In my opinion, it was too easy. I would say that it was just a basic geometry problem (so, I would guess around 8th/9th grade math level, depending on where you go to school). So, it may have been a good challenge/exercise for some people on here...To make it a bit harder, you could make the hexagons different sizes or irregular, or something else (we could get really crazy with it, but let's just leave it at that)...

[Guest]

In my opinion, it was too easy. I would say that it was just a basic geometry problem (so, I would guess around 8th/9th grade math level, depending on where you go to school). So, it may have been a good challenge/exercise for some people on here...To make it a bit harder, you could make the hexagons different sizes or irregular, or something else (we could get really crazy with it, but let's just leave it at that)...

Thanx for the info. I'm in tenth grade so it's just a little under my level I guess. Thanx for the input though. I'll use it to make a better riddle next time.

[Prof. Templeton]

Yep, another one right. Congratulations!! Here's your free virtual zoo ticket. Just imagine you're in a zoo. By the way, was this too easy?

Perhaps too easy as is. I like to form a basic math problems into a word problem so that it's up to the reader to form his own drawings or determinations and then try to derive to math answer. Alot of times this added step will trip people up. Ex. An ant is walking on a floor tiled with hexagons and squares, if he starts at the center of a hexagon and walks straight to a corner 4 inches away, then proceeds to graze on some juice that was spilled on 1 whole square, how much area does he cover while grazing? Or something like that. You get the idea.

[Guest]

Perhaps too easy as is. I like to form a basic math problems into a word problem so that it's up to the reader to form his own drawings or determinations and then try to derive to math answer. Alot of times this added step will trip people up. Ex. An ant is walking on a floor tiled with hexagons and squares, if he starts at the center of a hexagon and walks straight to a corner 4 inches away, then proceeds to graze on some juice that was spilled on 1 whole square, how much area does he cover while grazing? Or something like that. You get the idea.

Yeah sort of. Ok, I'll try to work on something like that. Thanks for the help guys. I appreciate it!!