Can you find the length of X?

[@ناصريناصر-س4ب]

It would have been better if the lengths of the sides of the triangle were random and not from Pythagorean triples.

[@lovesharma1699]

In accounting, we symbolize X for wrong answer or solution

[@Maths_physicshubs]

😀☺️

[@EqremKamerri-ch2sp]

4

[@URMBOT]

You did not finish because your denominator is not rational. 😊☝️

[@paparmar]

That's one reason most problems of this kind ask for the area of the square, thereby avoiding having to deal with radicals at the end.

[@Grizzly01-vr4pn]

@@paparmar Well, the area is still irrational, involving a radical. URMBOT means that the final answer should have been presented as (16√17)/17 not 16/√17

[@paparmar]

@@Grizzly01-vr4pn I totally agree with you as to the preferred expression of the side length of the square. However, the area of the square is exactly 256/17, so definitely rational. In fact, you can generalize the set-up to conclude the area of the square is a function of only the lengths of the sides of the right triangle (a & b, where b >=a): Area of square = (b^4) / (b^2 + (b-a)^2). If a and b are integers, the area is going to be rational. In this case, a = 3, b = 4, so area is 4^4 / (4^2 + 1) = 256/17. Note the relation gives the correct result of b^2 for the limiting case of the isosceles right triangle (b =a), where the legs of the triangle are the same as the sides of the square.

[@Grizzly01-vr4pn]

@ Ah yes, apologies. I was mistakenly referring to the side length of the square, rather than the area, which as you say is indeed rational.

[@elliotlambert3817]

if a 3 4 5 triangle is a right triangle then side f b must be 3 and side b c must be 4.your answer seems a crock.

[@Grizzly01-vr4pn]

No. Not all right triangles with hypotenuse = 5 have legs = 3 and 4. It is a false assumption to believe otherwise. △FBC is most certainly not 3-4-5

[@louf7178]

@@Grizzly01-vr4pn Are you sure about that?

[@Grizzly01-vr4pn]

@ You betcha. If anyone can show that △FBC is indeed a 3-4-5 triangle, let's see it. Edit to add: I mean, just by inspection it can easily be seen that the side of the square is < 4 (CE > CD), and obviously > 3.

[@elliotlambert3817]

@ rubbish

[@elliotlambert3817]

@ yes

[@samyyakoub7312]

Is not right. X≈3,3

[@Grizzly01-vr4pn]

Nope. x ≈ 3.88 units²

[@cannotsay5505]

LOL, another non - answer

[@Grizzly01-vr4pn]

How is it a non-answer?