First of all, a bit of theory: since the area of a square is given by

where s is the length of the square. So, if we invert this function we have
.
Moreover, the diagonal of a square cuts the square in two isosceles right triangles, whose legs are the sides, so the diagonal is the hypothenuse and it can be found by

So, the diagonal is the side length, multiplied by the square root of 2.
With that being said, your function could be something like this:
double diagonalFromArea(double area) {
double side = Math.sqrt(area);
double diagonal = side * Math.sqrt(2);
return diagonal;
}
Answer:

5 StartRoot 10 EndRoot
Step-by-step explanation:
we know that
The legs of a 45°-45°-90° triangle are congruent
Let
x ----> the length of one leg of the triangle
Applying the Pythagorean Theorem

where
c is the hypotenuse
a and b are the legs
we have


substitute


Simplify

5 StartRoot 10 EndRoot
Answer:
Step-by-step explanation:
2/5
do the run over rise