Diagonal \(\displaystyle d\) forms a triangle with adjacent sides \(\displaystyle a\). Since this is a square we know this is a right triangle and we can use the Pythagorean Theorem to determine the length of \(\displaystyle d\). Sides of length \(\displaystyle a\) form each of the legs and \(\displaystyle d\) is the hypotenuse. So the equation looks like this:

\(\displaystyle 12^{2}+12^{2}=d^{2}\)

Solve for \(\displaystyle d\)

\(\displaystyle 144+144=d^{2}\)

\(\displaystyle 288=d^{2}\)

\(\displaystyle d=\sqrt{288}\)

We can simplify this to

\(\displaystyle \sqrt{144\cdot 2}=\sqrt{144}\cdot \sqrt{2}=12\sqrt{2}\)

The path from home plate to first base is a side of a perfect square; the path from home plate to second base is a diagonal. As two sides and a diagonal form a \(\displaystyle 45^{\circ } -45^{\circ } -90^{\circ }\) triangle, the diagonal measures \(\displaystyle \sqrt{2}\) as long as a side.

The distance to second base from home is \(\displaystyle \sqrt{2}\) times the distance to first base:

\(\displaystyle d = 44 \cdot \sqrt{2} \approx 44 \cdot 1.414 \approx 62\)

A square is also a rhombus, so its area can be calculated as one half the product of its diagonals:

\(\displaystyle A= \frac{1 }{2}d^{2}\),

where \(\displaystyle d\) is the common diagonal length.

Since \(\displaystyle A=1200\), \(\displaystyle 1200 = \frac{1 }{2}d^{2}\).

\(\displaystyle 2400 = d^{2}\)

\(\displaystyle d = \sqrt{2400} \approx 49\)

The distance between opposite corners is about 49 meters.

The diagonal line cuts the square into two equal triangles.  Their hypotenuse is the diagonal of the square, so we can solve for the hypotenuse.

We need to use the Pythagorean Theorem: \(\displaystyle c^2=a^2+b^2\), where a and b are the legs and c is the hypotenuse.

The two legs have lengths of 8.  Plug this in and solve for c:

\(\displaystyle c^2=8^2+8^2=64+64=128\)

\(\displaystyle c=\sqrt{128}=\sqrt{2\cdot 64}=8\sqrt2\)

You can do this problem in two different ways that lead to the final answer:

1. Pythagorean Theorem

2. Special Triangles (45-45-90)

1. For the first idea, use the Pythagorean Theorem: \(\displaystyle a^2 +b^2 =c^2\), where a and b are the side lengths of the square and c is the length of the diagonal.

\(\displaystyle 4^2+4^2=c^2\)

\(\displaystyle c^2=32\)

\(\displaystyle c=\sqrt{32}\)

\(\displaystyle c=\sqrt{16*2}=\sqrt{16}\, \times \sqrt{2} = 4\sqrt{2}\)

2. If you know that ALL squares can be made into two special right triangles such that their angles are 45-45-90, then there's a formula you could use:

Let's say that your side length of the square is "a". Then the diagonal of the square (or the hypotenuse of the right triangle) will be \(\displaystyle a\sqrt{2}\).

So using this with a=4:

\(\displaystyle 4\sqrt{2}\)

Perimeter = side * 4

48 = side * 4

Side = 12

We can break up the square into two equal right triangles. The diagonal of the sqaure is then the hypotenuse of these two triangles.

Therefore, we can use the Pythagorean Theorem to solve for the diagonal:

\(\displaystyle side^2 + side^2 = hypotenuse^2\)

\(\displaystyle 12^2+12^2=hypotenuse^2\)

\(\displaystyle 144 + 144=hypotenuse^2\)

\(\displaystyle \sqrt{144+144}=\sqrt{hypotenuse^2}\)

\(\displaystyle hypotenuse=\sqrt{2\times 144}=\sqrt{2\times 12\times 12}=12\sqrt{2}\)

From the perimeter, we can find the length of each side of the square. The side lengths of a square are equal by definition therefore, the perimeter can be rewritten as,

\(\displaystyle 4s=28\)

\(\displaystyle s=7\)

Then we use the Pythagorean Theorme to find the diagonal, which is the hypotenuse of a right triangle with each leg being a side of the square.

\(\displaystyle a^2+b^2=c^2\)

\(\displaystyle 7^2+7^2=c^2\)

\(\displaystyle c=\sqrt{49+49}=\sqrt{98}=\sqrt{2\cdot49}=7\sqrt2\)

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,

\(\displaystyle \text{Diagonal}=4\sqrt2\)

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,

\(\displaystyle \text{Diagonal}=10\sqrt2\)

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,

\(\displaystyle \text{Diagonal}=15\sqrt2\)