how come two 3x8 right triangles and two trapezoids with bases of 3 and 5 and a height of 5 fit into a 13 x 5?

Question

rectangle. the area of each triangle is 12. the area of each trapezoid is 20. the total area of the shapes is 64 units squared. how does it fit into a rectangle thats 65 units squared.

Answer 1

I've seen this "paradox". Here's the picture, for reference.
http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/jigsaw.gif

If we look at the top-left piece, the slope of the line under it is 2/5 (the line stretches up 2 boxes and 5 across). But if we look at the bottom-left piece, the slope of the line above it is 3/8.

This isn't exactly the same - the first is 0.4 and the second is 0.375. The difference of 0.025 is just 1 box up in 40 across, so it's too small to notice here. However, it does mean the pieces don't match up exactly. There's a small gap between them, and an equal gap on the right. The area of these gaps add up to exactly 1, which is why it seems to "fit" into 65 squares.

If we imagined they were hiding this missing area in the thickness of the diagonal line itself, then since the box has a diagonal of almost 14 units, the line would only have to be as thick as 1/14 a box - which it's probably close to.

Answer 2

you take 1/2 of b(h), which is 8(3) (1/2 of 24), which is 12. you then double it because you have 2 triangles and you get 24.
(you could just use b(h)) then you take the trapezoids (b1 + b2)(h)and we wont divide by two because there are two of them. that is (3+5)(5)=(8)(5)=40
then you add the total area of the triangles and the trapezoids, and you get 64. 13*5 is 65 65 is clearly greater than 64, so it will have no trouble fitting in the 13 by 5 space

Answer 3

Thanks for the graphic, geofft. I was having trouble visualizing this.

3/5 ≠ 8/13, so when you put the 3x8 triangle on top of the 5x5x3 trapezoid, you do not make a 5x8 right triangle.

The base angle of the trapezoid is
arctan(5/2) = 63.435 deg.

The base angle of the 3x8 triangle is
arctan(8/3) = 71.565 deg.

The angle between the triangle and the trapezoid is
(90-63.435) - (90-71.565) = 8.13 deg

So between the 4 shapes you have a parallelepiped with sides of 5.196 and 8.544, having an area of exactly 1

Answer 4

You say the total area of the shapes is 64 right? And the area of the rectangle is 65? Well, it makes sense... 64 can fit into 65. So that's how it'd fit. Think about... (doesn't take much thinking).

Answer 5

regrettably i won't insert a drawing right here, so words will could do. a million. First divide the pentagon of section A into 5 triangles. 2. next draw the vertical top line fore each and each triangle. Now you have 10 little ideal perspective triangles, each and each of section a, so A = 10a 3. the backside of each and every little triangle is 4 cm. in case you knew the top, you may calculate the section. 4. the proper perspective of each and every little ideal triangle could be 360°/10 = 36°. Now, tan 36° = 0.7265 5. Calculate the top as 4/0.7265 = 5.5 (rounded) 6. a = (5.5 x 4)/2 and A = a hundred and ten cm^2 (greater precisely a hundred and ten.11055.....) i did no longer use the sine rule for this, yet once you utilize the comparable physique of strategies you will discover a fashion employing it.