Can You Figure It Out? The *Almost* Impossible Geometry Puzzle Solved. [Not Clickbait]

I solved this problem by comparing triangles' ratio.

Similare triangels

Little funny you didn't just find the integral of a piecewise function, instead of doing a triangle with coordinate geometry and another area with calc.

I found the 4 vertices and used the shoelaces algorithm.

I did it mostly similarly, but left the squares unrotated with the origin on the left corner. Calculated the formulas for the lines, and calculated the points on the squares, and the area as the area of the small square minus the areas of the two triangles.

I hope he used similarity of triangles

Thank god he didnt say gougo theorum anywhere , i wouldve lost it

just make the two lines y(x)=3x/5 and g(x)=3x/2. evaluate the integral of g(x)-y(x) from 0 to 4, and subtract the small triangle area formed in the L of the two squares to do the whole thing in about 5 lines of working

Will you please make videos on integral Inequalitys... please,

I think Dr Peyam could do it better honestly. Steve BPRP better as well. I like fights btw Great Mathematicians, this kind looks like the one btw Leibnitz and Newt .... Ah, no.

I am suddenly thirsty for some fresh tapwater

Can this problem be solved by elementary school students?

There are much simpler methods available using similar triangles

Coord-bash

I think it's very rude to say, "How you play with your girlfriend at home." Some of us would prefer to play with our boyfriends, well, if I had one. Until then I'll just continue to play with myself.

4*4 - (10*6/2)*(4/10)^2 - (6*4/2)*(4/6)^2 = 16 - 24/5 - 16/3 = 88/15

We could easily do it with Pythagoras theorem and similarity of triangles concept

Is there a joke I’m missing? He says “times” and “sides” with the same vowel, like in “eye”

But he says “five” and “minus” with a different vowel, like the vowel in “odd”

Is this on purpose?

This isn’t hard if you know the concepts of similar triangle, you doesn’t need integrals to solve this problem

Mirrored the 2 squares in the x-axis, calculate length of sides grayed area, then calculate the white triangles areas in 4x4 square.
16 - 4.8 - 5.33... = 5.866...

I sAw thIs oN FrEsh ToADwalkEr

Pffff, how to make something very easy very difficult, right?
Just add a rectangle of 4x2 to the left square so you make a rectangle of 4x6 (area=24). In this rectangle you now have 3 triangles and the area we're after. Area of the top-left triangle is 1/2*4*6=12. For the area of the bottom-right triangle you need the height (where the bottom diagonal meets the 4x4 square). This you can get with tan A=6/10=height/4 so 10*height=4*6=24 so height is 24/10=12/5. Its area then is 1/2*4*(12/5)=24/5. Now we need to know the are of the small top-right triangle and for this we need to know its base (where the top diagonal meets the 4x4 square). This you can get with tan B=4/6=x/4 so 6x=16 so x= 16/6=8/3. The base of the triangle is 4-x=4-8/3=4/3. Its area then is 1/2*4/3**2=4/3.
Now all we need to do is deduct the areas of the three triangles from the area of the rectangle so 24-12-24/5-4/3=180/15-72/15-20/15=180/15-92/15=88/15. Right then, time to go play with the girlfriend.

You got us used to way more complicated problems ahah

Remember lads God loves you all

I simply turned the two diagonals into linear equations with y intercept at 0. y = mx = rise/run.
- So you get y1 = 3x/5, y2 = 3x/2. Integrate y2-y1 from 0 to 4 = 36/5. That is the triangle area (the shaded region + the small triangle on top). {You could alternatively use similar triangles to find the area of the smaller triangle on the bottom then subtract that from the right triangle = 4*6/2 - A(small triangle on bottom).}
- Then the horizontal line which is the base of the smaller triangle is simply y = 4.
- So find the intersection point of y2 and 4 == x=8/3. So integrate from 8/3 to 4 on y2-4 to get the small triangle area = 4/3.
- Now do the triangle - smaller triangle = 36/5-4/3 = 88/15.
Done. This way you don't have y intercepts.

I just used analytic geometry (without integrals). But yeah analytic geometry is pretty fun!

Sorry, but the problem is trivial. The area required is inside a square of side 4 along with two other triangles with areas A (lower one) and B (upper one): so the answer is 16 - A - B.

Now A has (horizontal) base 4 and perpendicular height h, say and one sees that h:4 is the same as ratio 6 : 10 therefore h=12/5 and so A= 24/5.

Similarly B has a vertical base 4 and a perpendicular height l, say, and one sees that l$ is the same as 4:6 leaving l=8/3 and so B=16/3.

Therefore the area is 16 - 24/5 - 16/3 = 88/15.

Much much easier with similar triangles for students who wish for another say to solve this.

There's nothing better than an overly complicated but absolutely elegant answer to a math problem

There is more simpler method available to solve the problem. Just use principles of similar triangles as taught to 7th/8th standard students here in India and get the solution. The result is 88/15 sq units.

Thus literally solvable in 2 minutes by high school students

100% Clickbait

You can also solve the area of trapezium

How can i support your channel without money? Im still 15, not enough money in my pocket 😂

I reckon I managed to solve this madness in my head (got 5.86666...)

I feel like this puzzle is a sort of test to see if you fully understand similar triangles

He made it so messy. It could be done just by basic geometry.

I for one applied pythagoras, triangle area formulae, and cosine law to solve this. This felt like a Presh Talwalker style problem so I knew it didn't need more than high school trigonometry to figure out

Edit: Wait was I not supposed to use trigonometry

Eso se resuelve con dos homotecias. No hace falta ninguna integral ni ninguna función.

This problem is probably too hard to do entirely in your head!

I think the videos could be shorter and more concise.

This took me 5 mins using the traditional triangle area formula!

dude just make the corner where liness originate the origin
becomes way more easy

This is much easier with similar triangles. Why you are doing it in a long and complicated way.
And what is your problem with MYD?
Presh make videos for all ages. So obviously there will be some easy problems too.

Not that tough.

This prob is not hard. I’m 11 years old and i solved it simple by similar triangles. Your way is very long.