The diagonal of a square is also a hypotenuse of a right triangle with the side lengths as legs of the triangle.

Use the Pythagorean Theorem to find the length of the diagonal.

$\displaystyle \text{Diagonal}=\sqrt{\text{side length}^2+\text{side length}^2}$

$\displaystyle \text{Diagonal}=\sqrt{2(\text{side length})^2}$

$\displaystyle \text{Diagonal}=(\text{side length})\sqrt2$

For the square given in the question,

$\displaystyle \text{Diagonal}=5\sqrt2$

The diagonal of a square is also a hypotenuse of a right triangle with the side lengths as legs of the triangle.

Use the Pythagorean Theorem to find the length of the diagonal.

$\displaystyle \text{Diagonal}=\sqrt{\text{side length}^2+\text{side length}^2}$

$\displaystyle \text{Diagonal}=\sqrt{2(\text{side length})^2}$

$\displaystyle \text{Diagonal}=(\text{side length})\sqrt2$

For the square given in the question,

$\displaystyle \text{Diagonal}=159\sqrt2$

The diagonal of a square is also a hypotenuse of a right triangle with the side lengths as legs of the triangle.

Use the Pythagorean Theorem to find the length of the diagonal.

$\displaystyle \text{Diagonal}=\sqrt{\text{side length}^2+\text{side length}^2}$

$\displaystyle \text{Diagonal}=\sqrt{2(\text{side length})^2}$

$\displaystyle \text{Diagonal}=(\text{side length})\sqrt2$

For the square given in the question,

$\displaystyle \text{Diagonal}=27\sqrt2$

The diagonal of a square is also a hypotenuse of a right triangle with the side lengths as legs of the triangle.

Use the Pythagorean Theorem to find the length of the diagonal.

$\displaystyle \text{Diagonal}=\sqrt{\text{side length}^2+\text{side length}^2}$

$\displaystyle \text{Diagonal}=\sqrt{2(\text{side length})^2}$

$\displaystyle \text{Diagonal}=(\text{side length})\sqrt2$

For the square given in the question,

$\displaystyle \text{Diagonal}=23\sqrt2$

The diagonal of a square is also a hypotenuse of a right triangle with the side lengths as legs of the triangle.

Use the Pythagorean Theorem to find the length of the diagonal.

$\displaystyle \text{Diagonal}=\sqrt{\text{side length}^2+\text{side length}^2}$

$\displaystyle \text{Diagonal}=\sqrt{2(\text{side length})^2}$

$\displaystyle \text{Diagonal}=(\text{side length})\sqrt2$

For the square given in the question,

$\displaystyle \text{Diagonal}=122\sqrt2$

The diagonal of a square is also a hypotenuse of a right triangle with the side lengths as legs of the triangle.

Use the Pythagorean Theorem to find the length of the diagonal.

$\displaystyle \text{Diagonal}=\sqrt{\text{side length}^2+\text{side length}^2}$

$\displaystyle \text{Diagonal}=\sqrt{2(\text{side length})^2}$

$\displaystyle \text{Diagonal}=(\text{side length})\sqrt2$

For the square given in the question,

$\displaystyle \text{Diagonal}=35\sqrt2$

To find the diagonal, you case use the pythagorean theorem or realize that this in isosceles triangle, and therefore the hypotenuse is

$\displaystyle s\sqrt{2}=8\sqrt{2}$

Finding the diagonal of a square is the same as finding the hypotenuse of a triangle, and uses the Pythagorean Theorem. (Imagine the square being sliced diagonally into two triangles, for you visual learners.) Within this theorem, both a and b are the same number, since the sides of a square are equal. 

$\displaystyle a^2 + b^2 = c^2$

$\displaystyle 15^2 + 15^2 = c^2$

$\displaystyle 225 + 225 = c^2$

$\displaystyle 450 = c^2$

$\displaystyle \sqrt{}450 = c$

$\displaystyle 21.2 ft = c$

To find a diagonal of a square recall that the diagonal will create a triangle in the square for which it is the hypotenuse and the side lengths will be the other two lengths of the triangle.

To solve, simply use the Pythagorean Theorem to solve.

Thus,

$\displaystyle c=\sqrt{a^2+b^2}=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt2$

To solve, simply use the Pythagorean Theorem. Thus,

$\displaystyle c=\sqrt{a^2+b^2}=\sqrt{2^2+2^2}=\sqrt{2(2^2)}=2\sqrt{2}$

Remember, that when simplifying square roots, you can only pull a number out if you have two factors of it. That is why I grouped the end of this answer the way I did so you could see that since I had two squared, I could pull one out, one disappears, and the two on the outside of the parenthesis remains under the radical.