Summary: There are two large right triangles, each composed of several smaller shapes. When the shapes are rearranged in the second large triangle, which appears to be the same overall size / shape, there’s an extra space.
I keep doing the math on the individual figures and the areas don’t add up to the correct area anyway - they add up to 32 squares, and the large triangle is 13 x 5 (using 3rd grade arithmetic, that’s 13x5 = 65, divided by 2, = 32.5).
The ratios also don’t make sense - for them to be able to move the triangles around as they do, they’d all have to have the same slope (ratio between vertic and horizontal axes) - that’d be the sine, trigonometrically speaking. I’m guessing that’s the key to the whole conundrum.
But it sure looks like the reorganization works! I just can’t quite get my head around how they did it and made it look correct.
It’s a classic illusion. The apparently straight hypoteneuse in the upper right triangle isn’t, really – it’s two separate lines that meet at the point where the red and green sub-triangles meet, IIRC, but they aren’t really parallel and don’t form a continuous line. The difference in area account for the missing block.
I mean, look at the two triangles – the red one is a rise of three blocks over a run of 8, while the green one is a rise of 2 over a run of 5 – those aren’t similar triangles, and they can’t have the same slopes on the hypoteneuses.
Either the overall shape isn’t a true triangle or the angles of the red and green triangles are changed between the two layouts. The red triangle appears to have a 3 in 8 slope, with angles of roughly 22 and 68 degrees. The green triangle seems to have a 2 in 5 slope, with angles of roughly 23.6 and 66.4 degrees.
While I was measuring these out in Paint someone else beat me to it.
Copy these two triangles to Paint or Powerpoint. Draw a line along the big hypoteneuse of the top one. Note that there is a slight gap between your line and the hypoteneuse. Over the length of the triangle this amounts to about 1/2 a square.
Draw a similar diagonal line on the bottom image. Note that the line you drew is actually slightly inside the triagle. THis amounts to about 1/2 square over the whole picture.
All of which is why my geometry teacher used to tell us “if the diagram doesn’t say it’s (whatever) you can’t assume it’s (whatever).” Notice that nowhere does the diagram say these are triangles, much less that they are similar.
I tell my SAT/ACT kids the same thing-never trust a figure. Well, on the SAT, they’ll warn you if it is not drawn to scale-on the ACT they won’t, but in practice the ACT has just as many accurate figures as the SAT does-it’s just an additional challenge on the students’ part to determine which ones are, or are not. More to the point, you have to prove it is what it appears to be, based on the givens, else you can’t conclude much of anything at all.
Here’s an exaggerated example of what’s going on that I made in paint. Notice the top triangle is bent inward. On the bottom I cut and pasted the two parts and it bends outward instead. The remaining box in the bottom right is now larger.
I’m not sure why you need to discuss the component shapes.
You can see in the illustration that the the upper boundary of the lower figure is slightly raised.
Notice that two grid squares above the “hole” has more area filled in the bottom figure than the top. A close inspection of all of the squares along the upper boundary shows a difference in area filled.
So the area from the “hole” is spread out and transferred to each of the upper boundary squares.
cornflakes - I think you’ve got the angles wrong - here’s what I get from drawing it up in CAD and comfirming with trig
red triangle - 20.6 and 69.4
green triangle - 21.8 and 68.2
Sigene - the extra/missing area of the slim triangles along the not-really-a-hypotenuse is not about half a sqaure each - it’s exactly half each. I wasn’t sure if it was or wasn’t til I drew it up.
John diFool - I’m not sure if you’re implying the figures aren’t drawn to scale or just making a related comment, but these figures are drawn exactly to scale. I pulled the image into CAD and it lines up with my drawing down to the pixel.
Sorry if using CAD is considered cheating - but I did use some trig!
I printed the examples and then cut out the pieces of the bottom example. When layed on top of the first example, they fit perfectly, exactly the same configuation as the first example.
While this particular type of problem wouldn’t necessarily show up on said tests, the object lesson remains the same-you have to prove that what you see is what you get.
Wow, all of these scientific solutions. All I did to realize that the hypotenuse was different was look at the upper right hand triangle (green). It’s 5 squares wide by 2 squares tall.
The green triangle hypotenuse should go through a point 5 squares over, and 2 squares up. It doesn’t.
Wow… this is one of those things that’s difficult to put into words. A long time ago in honors trig we had to write our answers out like this as our solutions to prove that we fully understood the problem. It was kind of neat actually.
The easiest way for me has always been by looking at the empty sections of the grid right above the triangle. In this pic, the middle one is the most obvious. http://img694.imageshack.us/img694/1634/snap5i.jpg
