To find the area of a quadrilateral, multiply length times width. In a square, since all sides are equal, $\displaystyle a$ is both the length and width.

$\displaystyle a^{2}=21\times21=441$

To find the area of the floor, multiply the length of the room by the width (which is the same forumla used to find the area of a square).  The equation can be written: $\displaystyle A=l\cdot w$

Substitute $\displaystyle 10$ feet for $\displaystyle l$ and $\displaystyle 12$ feet for $\displaystyle w$:

$\displaystyle A=10\cdot 12$

Amy will need $\displaystyle 120 ft^{2}$ of carpet.

The perimeter of the rectangle is

$\displaystyle P = 2L + 2W = 2\cdot100 +2\cdot40 = 200 + 80 = 280$ centimeters.

This is also the perimeter of the square, so divide this by $\displaystyle 4$ to get its sidelength:

$\displaystyle s = \frac{P}{4} = 280 \div 4 = 70$ centimeters.

The area is the square of this, or $\displaystyle A = s^{2} = 70^{2} = 4,900$ square centimeters.

The areas of the four squares can be calculated by squaring their sidelengths. Add these areas, then divide by 4:

$\displaystyle \frac{4^{2}+8^{2}+12^{2}+16^{2}}{4} = \frac{16+64+144+256}{4} = \frac{480}{4} = 120$ square inches

Multiply the sidelength by 36 to convert from yards to inches:

$\displaystyle 1 \frac{1}{6} \times36 = \frac{7}{6} \times \frac{36}{1} = \frac{7}{1} \times \frac{6}{1} = 42\ inches$

Square this to get the area:

$\displaystyle 42^{2} = 1,764$ square inches

The area of a square is equal to the product of one side multiplied by another side. Therefore, the area will be equal to:

$\displaystyle \sqrt{\frac{1}{4}+\frac{1}{3}}\cdot \sqrt{\frac{1}{4}+\frac{1}{3}}=\frac{1}{4}+\frac{1}{3}$

The next step is to convert the fractions being added together to a form in which they have a common denominator. This gives us:

$\displaystyle \frac{3}{12}+\frac{4}{12}=\frac{7}{12}$

The length of a segment with endpoints $\displaystyle (6, 2)$ and $\displaystyle (-5, 1)$ can be found using the distance formula with $\displaystyle x_{1} = 6$, $\displaystyle y_{1} = 2$, $\displaystyle x_{2} = -5$, $\displaystyle y_{2} = 1$:

$\displaystyle d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$\displaystyle = \sqrt{(6 - (-5))^{2}+(2-1)^{2}}$

$\displaystyle = \sqrt{11^{2}+1^{2}}$

$\displaystyle = \sqrt{121+1}$

$\displaystyle = \sqrt{122}$

This is the length of one side of the square, so the area is the square of this, or 122.

The length of a segment with endpoints $\displaystyle (A, B)$ and $\displaystyle (-B, A)$ can be found using the distance formula as follows:

$\displaystyle d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$\displaystyle d = \sqrt{((-B)-A)^{2}+(A-B)^{2}}$

$\displaystyle d = \sqrt{(-B-A)^{2}+(A-B)^{2}}$

$\displaystyle d = \sqrt{(-1)^{2}(B+A)^{2}+(A-B)^{2}}$

$\displaystyle d = \sqrt{1 \cdot (B^{2} +2AB +A^{2})+(A^{2}-2AB+B^{2})}$

$\displaystyle d = \sqrt{ (B^{2} +2AB +A^{2})+(A^{2}-2AB+B^{2})}$

$\displaystyle d = \sqrt{ 2A^{2} +2 B^{2}}$

This is the length of one side of the square, so the area is the square of this, or $\displaystyle 2A^{2} +2 B^{2}$.

Since a square is a rhombus, one way to calculate the area of a square is to take half the square of the length of a diagonal. If we let $\displaystyle d$ be the length of each diagonal, then 

$\displaystyle \frac{1}{2} d^{2} = A$

$\displaystyle \frac{1}{2} d^{2} = 13$

$\displaystyle d^{2} =26$

$\displaystyle d = \sqrt{26}$

Therefore, we want to choose the point that is $\displaystyle \sqrt{26}$ units from the origin. Using the distance formula, we see that $\displaystyle (5, 1)$ is such a point:

$\displaystyle d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$\displaystyle = \sqrt{(5-0)^{2}+(1-0)^{2}}$

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

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

$\displaystyle = \sqrt{26}$

Of the other points:

$\displaystyle (6,0)$:

$\displaystyle d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$\displaystyle = \sqrt{(6-0)^{2}+(0-0)^{2}}$

$\displaystyle = \sqrt{6^{2}+0^{2}}$

$\displaystyle = \sqrt{36+0}$

$\displaystyle = \sqrt{36 }$

$\displaystyle (4, 2)$:

$\displaystyle d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$\displaystyle = \sqrt{(4-0)^{2}+(2-0)^{2}}$

$\displaystyle = \sqrt{4^{2}+2^{2}}$

$\displaystyle = \sqrt{16+4}$

$\displaystyle = \sqrt{20}$

$\displaystyle (3, 3)$:

$\displaystyle d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$\displaystyle = \sqrt{(3-0)^{2}+(3-0)^{2}}$

$\displaystyle = \sqrt{3^{2}+3^{2}}$

$\displaystyle = \sqrt{9+9}$

$\displaystyle = \sqrt{18}$

One of your holiday gifts is wrapped in a cube-shaped box. 

If one of the edges has a length of 6 inches, what is the area of one side of the box?

We are asked to find the area of one side of a cube, in other words, the area of a square.

We can find the area of a square by squaring the length of the side.

$\displaystyle A_{square}=s^2=(6in)^2=36in^2$