What is the area when these two rectangles overlap?

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[spicemasterii]

"Dad, I didn't get 100 in math"
Presh: why not?
"What the hell is Pythagoras Theorem?"

[knetknecht]

Gougu Theorem: never heard of that
Seeing the equation: Oh, it's just Pythagoras

[adrianshum]

Am I the only one that feel the last step is unnecessarily complicated?
The base of the parallelogram is 2-3/4= 5/4. Height =1. Hence area = 5/4

[Wecoc]

I've heard people mispronouncing 'Pythagoras' in many ways but this one is the most creative

[genius]

Actually, we could have removed a step. Since the area of a parallelogram is
base x height,
once we figured out that the other leg of the triangle was 3/4, we could easily figure out the base of the parallelogram: 2 - 3/4, that is, 5/4.
base x height =
5/4 x 1 =
5/4.

[xongi]

I didn't know what the solution will be, but I knew that he will use the Gougu Theorem.

[MaxMathGames]

You can always judge the level of the question by the time duration. 😁

[UnbreakablePickaxe]

No one:
Mind your decision: "Gougu Theorem"

[jaiskreno]

I love how this is the first one I actually solved myself

[emiltonklinga]

- HE USED THE PYTHAGOREAN THEOREM!! -

(He just couldn't pronounce it.)

[thomashughes]

Presh, what I like even more than solving these fantastic puzzles is their applications in so many areas of engineering. I build clocks, metronomes, other solid figures, etc., and your puzzles have lead me to "discover" general formulas for future reference. Thanks a million!

[hurktang]

Trigonometry solution : The diagonal angle of the rectangle is tan-1(1/2)~=26.56degree so the angle of the parallelogram is 2x = 53.13 degree. The remnant of the 90 degree is 36.86 degree so X=tan(36.76). From there solve with the big rectangle and the small rectangle as shown in the solution 2*1-tan(36.76)*1=1.25

[muhammadzidanealhalita]

What Gougu theorem is? It's just a Pythagorean theorem

[jeffreytao]

The formula for calculating the area of such an overlap using any quadrangle with 4 right angles comes to (x^3)/2y + xy/2 where x<y and x and y are base and height. This equation was obtained by multiplying the longest diagonal of the area (the hypotenuse) by its perpendicular line of longest length that fits within that area (the base) divided by 2. Presh, can you explain where those two components of the formula arise from? The other version of the formula is x(x^2+y^2)/2y

[popogast]

I suggest a different approach:
The longer diagonal of the rhombus has the length of sqr(5). The rectangles' side length ratio is 1:2. Thus the shorter diagonal of the rhombus has the length of sqr(5)/2. The area of the rhombus can be considered as two mirrored triangles with widths and heights of sqr(5)/2. Thus the total area of the blue rhombus is A = (sqr(2)/2)^2 = 5/4.

[gemlacambra]

Everyone: Pythagorean Theorem
MindYourDecisions: GoUgU THeOreM

[rutgerdekok]

I divided the rhombus by two lines from opposite corner to oposite corner, forming four congruent triangles. Looking at one triangle's sides that meet in the middle, I label the long side a and the short one b. The Area of the rhombus is 2ab. If you draw a line dividing the rectangle in two squares you can observe, by similar triangles that a:b = 1:1/2, so a = 2b -> Area = 2ab = a^2. And we can use Pythagoras, Area = a^2 = 1^2 + 1/2^2 = 5/4

[loglnlg]

I solved it in another way: firstly show that that quadilateral is rhombus, then if we draw diagonals of that rhombus, we know tangent of one angle which is 1/2 because that angle is also in big rectangle

Then smaller diagonal of rhombus is 1/2 of big diagonal which is sqrt(5), and its area is 1/2*d1*d2=5/4 where d1 and d2 are its diagonals

[aadityasehrawat]

What happened to me....
I am watching this for entertainment.

[ameyanrd]

My first reaction on Gougu Theorem, "Let's check the comments!".XD