Red

Answer: it’s going to fall.

No, it's wedged in that box pretty tight.

Is there really enough information provided to solve this?

du=dv+1v−u=10209(u+v)=d
P = base + height + hypotenuse

Average of left side of disassociated rectangle within the confines of a square:

P=a+b+a2+b2

Using the Pythagorean theorem, distributed to the lateral angle of the unattached side, minus the base, gets you the conclusion, which is an imperfect number due to the inability of the mind to work with less than whole numbers and you get a simple answer of 7

9

X= the red line (looks less than 7)

X = 4.4 ish (or pi x 1.4) for [bleep] & giggles

Correction…. I will go with 5.5 Alex.

🤔

I'm gonna have to check wit my neighbor, Mr. Milton Pythagoras. I bet he's got a theory on this.

5

Two rectangles and a square go into a bar.

The bartender says, "Not these guys again. Last time they showed up it turned into a real circle jerk".

5

5.335

Since they are congruent, I'm going with 5.

Spain

who cares????

If x is the short leg of the triangle, then x = 6 but that is based on the assumption that the triangle is a 30-60-90.

CiK's control statement "2 Congruent Rectangles in a Square" says that it cannot be 5 and it cannot be 6.

Well that sucks.

Then it cannot be a 3-4-5 triangle.

Correct, not with Caveman Math.....Alien Math maybe......Nah

This help?

Correct

Correct

What?

"The man that chooses not to use his brain... will surely loose it." Joe Biden

Apologies. I was incorrect. it can be 5..... or 4.695, or 5.299, or 5.335 or 6. I assumed the columns represented 4 equal rectangles with a width of 2.5 in the square.

The same that you have to assume the angles are 30, 60, and 90 to come up with X=5.

It depends on the angles that you assume (Not provided in OP)

Maybe this will help:
It didn't help me.

No Sir...

All Right Angle Triangles have two legs and a Hypotenuse.

The “congruent” rectangles make use of two Legs and two Hypoteni (SIC?) to equal 10.

See the diagram above.

The Hypotenus legs are more than 2.5.

The Adjacent Legs are less than 2.5.

The red highlights are a true statement.

The blue highlights were in no way quantified in the OP

You can change the length of 2 sides in a right triangle by changing the degrees in angle's a & b. It will still be a right triangle, the "Hypotenuse" would still be 10.

I don't have a clue on what you are trying to say...

An example showing the Hypotenuse C, when you change the angle's and it changes the length of the sides.

Who cares? In my 60 years on this earth I have never needed any math like this and have functioned just perfectly. Most math is total BS for the average person.

I did…
There are ways to use angles and distance to check your landing pattern and get to a good distance abeam. Specifically, a 30/60 right triangle.

Also useful in determining how much lateral separation you have on a bogey in performing a stern conversion.

30-60-90 triangle

Sides are:

X
2X = 10
and X*sqrt(3)

Of course...

But to alter the angles will defeat the congruent rectangles... OR the square they reside in.



Please feel free to render support for your precept.

Mac...

Part of the reason I put up these simple math problems is to watch the railing of folks that can't sort them out... and are determined to tell everyone about it.

The one on the left will become a wrecktangle.

3.33

3:33 AM is The Witching Hour