To find the perimeter of a square given its area, take the square root of the area to find its sidelength; then, multiply that sidelength by 4.

$\displaystyle N^{2 } - 6N + 9$ is a perfect square trinomial, since $\displaystyle \left (\frac{-6}{2} \right )^{2} =(-3)^{2}= 9$

so its square root is $\displaystyle \sqrt{N^{2 } - 6N + 9} = N - 3$, the sidelength.

Multiply this by 4 to get the perimeter: $\displaystyle 4 \left (N - 3 \right ) = 4N - 12$

Cutting the triangle in half yields a right triangle with the diagonal becoming the hypotenuse and the other two legs being the sides of the square. Using the Pythagorean Theorem, we can solve for the other legs of the triangle.

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

Since both sides of the square are equal to eachother, $\displaystyle a=b$, therefore:

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

$\displaystyle 2a^2=\left(\frac{x}{2}\right)^2$

$\displaystyle 2a^2= \frac{x^2}{4}$

$\displaystyle a^2=\frac{x^2}{8}$

$\displaystyle a= \frac{x}{\sqrt{8}}$

To find the area of the square:

$\displaystyle A=s^2$ with leg $\displaystyle a$ being one of the sides

$\displaystyle A= \left(\frac{x}{\sqrt{8}}\right)^2$

$\displaystyle A=\frac{x^2}{8}$

The length of one side of the square is $\displaystyle 1,200 \div 4 = 300$ feet, or $\displaystyle 300 \div 3 = 100$ yards. Square this to get the area in square yards:

$\displaystyle A =100^{2} =10,000$ square yards.

The square, the pentagon, and the hexagon have a total of 15 sides, all of which are of equal length; the sum of the lengths is one mile, or 5,280 feet, so the length of one side of any of these polygons is 

$\displaystyle 5,280 \div 15 = 352$ feet.

The square has area equal to the square of this sidelength:

$\displaystyle 352^{2} = 123,904 \textup{ ft} ^{2}$

The regular perimeter has sidelength 60 centimeters and therefore perimeter $\displaystyle 5 \times 60 = 300$ centimeters. The square has as its sidelength $\displaystyle 300 \div 4 = 75$ centimeters and area $\displaystyle 75 ^{2} = 5,625$ square centimeters.

The areas of the squares are the squares of the sidelengths, so add the squares of the sidelengths. Since 1 foot is equal to 12 inches and 2 feet are equal to 24 inches, the sum of the areas is:

$\displaystyle 8^{2} + 12 ^{2} + 15 ^{2} + 20 ^{2} + 24^{2} + 25^{2}$

$\displaystyle = 64 + 144 + 225 + 400 + 576 + 625$

$\displaystyle =2,034$ square inches

As a square, $\displaystyle ABCD$ is also a rhombus. The area of a rhombus is half the product of the lengths of its diagonals, one of which is $\displaystyle \overline{AC }$. Since the diagonals are congruent, this is equal to half the square of $\displaystyle AC$ :

$\displaystyle \frac{1}{2} \left (AC \right) ^{2}$

$\displaystyle = \frac{1}{2} \left (2t + 5 \right) ^{2}$

$\displaystyle = \frac{1}{2}\left [ \left (2t \right) ^{2} + 2\left (2t \right) \cdot 5 + 5^{2} \right ]$

$\displaystyle = \frac{1}{2}\left ( 4t ^{2} +20t + 25 \right )$

$\displaystyle = 2t ^{2} +10t +\frac{ 25 }{2}$

This problem requires us to find the area of a square. Don't let the story behind it distract you, it is simply an area problem. Use the following equation to find our answer:

$\displaystyle Area=s^2=(15\:m)^2=225\:m^2$

$\displaystyle s$ is the length of one side of the square; in this case we are told that it is $\displaystyle 15\:m$, so we can solve accordingly!

Square $\displaystyle ABCD$ has area 49, so each of its sides has as its length the square root of 49, or 7. Each side of Square $\displaystyle WXYZ$ is therefore a hypotenuse of a right triangle with legs 1 and $\displaystyle 7-1 = 6$, so each sidelength, including $\displaystyle WX$, can be found using the Pythagorean Theorem:

$\displaystyle WX = \sqrt{(WB)^{2}+(BX)^{2}}$

$\displaystyle WX = \sqrt{1^{2}+6^{2}} = \sqrt{1 +36} = \sqrt{37}$

The square of this, which is 37, is the area of Square $\displaystyle WXYZ$.

Square $\displaystyle WXYZ$ has area 25, so each side has length the square root of 25, or 5. 

Specifically, $\displaystyle WX = 5$, and, as given, $\displaystyle BW = 1$.

Since $\displaystyle \bigtriangleup WBX$ is a right triangle with hypotenuse $\displaystyle \overline{WX}$ and legs $\displaystyle \overline{BW}$ and $\displaystyle \overline{BX}$, $\displaystyle BX$ can be found using the Pythagorean Theorem:

$\displaystyle BX = \sqrt{(WX)^{2}-(BW)^{2}}$

$\displaystyle = \sqrt{5^{2}-1^{2}}$

$\displaystyle = \sqrt{25-1}$

$\displaystyle = \sqrt{24}$

$\displaystyle =\sqrt{4} \cdot \sqrt{6}$

$\displaystyle = 2\sqrt{6}$

The area of $\displaystyle \bigtriangleup WBX$ is 

$\displaystyle \frac{1}{2} \cdot BW \cdot BX = \frac{1}{2} \cdot 1 \cdot 2 \sqrt{6} = \sqrt{6}$

Since all four triangles, by symmetry, are congruent, all have this area. the area of Square $\displaystyle ABCD$ is the area of Square $\displaystyle WXYZ$ plus the areas of the four triangles, or $\displaystyle 25 + 4\sqrt{6}$.