To square the circle, you must construct a square having the same area as a given circle using compasses and a straightedge. And you encapsulated the difficulty when you said "find a square with the exact area of a given circle". Yes, you could do that by constructing a line segment with length exactly 5\sqrt{\pi}. How do you do that? If you are given a circle with radius exactly 5 you could use that to construct a segment of length exactly 1 (dividing a segment into n equal parts is easy). But then you need to get to \sqrt{\pi}.
The point is that not all numbers are constructible. In fact, there exist relatively few numbers, x, such that given a line segment of length 1, we can construct a line segment of length x. Given a segment of length 1, we can, of course, duplicate it any number of times and so "construct" a segment of length any integer n. And because we can divide a segment into any number of equal parts, we can construct a segment of line 1/n for any postive integer n. It is classic to construct a segment of length \sqrt{2} and that can be extended to constructing any square root of an integer.
To cut this short (perhaps the site dijkarte links to gives the details but they are very deep) the "constructible numbers" are precisely the real numbers that are "algebraic of order a power of 2". A number is said to be "algebraic of order n" if it satisfies a polynomial equation, with integer coefficients, of degree n but no such equation of lower degree. For example, every rational number x= n/m, satisfies mx-n= 0, a polynomial equation of degree 1, with integral coefficients. That is, the numbers that are "algebraic of order 1" are exactly the rational numbers. \sqrt{2} satisfies x^2- 2= 0 and so is "algebraic of order 2" (and so constructible) but \sqrt[3]{2} is algebraic of order 3, not a power of 2, and so is not constructible.
\pi, and so \sqrt{\pi}, is not algebraic of any order (such numbers are called "transcendental"). Since \sqrt{\pi} is not algebraic of order 2, it, and any multiple such as 5\sqrt{\pi} is not "constructible"- you cannot construct a line segment, using compasses and straightedge, of length 5\sqrt{\pi}.