Multiply height \(\displaystyle BD\) by base \(\displaystyle CD\) to get the area.

By the 30-60-90 Theorem:

\(\displaystyle BD = \frac{BC}{2} = \frac{10}{2}} = 5\)

and

\(\displaystyle CD =BD \cdot \sqrt{3} = 5\sqrt{3}\)

The area is therefore

\(\displaystyle BD \cdot CD = 5 \cdot 5\sqrt{3} = 25\sqrt{3}\)

Multiply height \(\displaystyle BD\) by base \(\displaystyle CD\) to get the area.

By the 45-45-90 Theorem, 

\(\displaystyle BD = CD = \frac{BC}{\sqrt{2}} = \frac{10}{\sqrt{2}}\). 

The area is therefore

\(\displaystyle BD \cdot CD = \frac{10}{\sqrt{2}} \cdot \frac{10}{\sqrt{2}} = \frac{100}{2} = 50\)

As can be seen in the diagram, there are three possible locations of the fourth point of the parallelogram:

Regardless of the location of the fourth point, however, the triangle with the given three vertices comprises exactly half the parallelogram. Therefore, the parallelogram has double that of the triangle.

The area of the triangle can be computed by noting that the triangle is actually a part of a 12-by-12 square with three additional right triangles cut out:

The area of the 12 by 12 square is \(\displaystyle 12^{2} = 144\)

The area of the green triangle is \(\displaystyle \frac{1}{2} \times 12 \times 9 = 54\).

The area of the blue triangle is \(\displaystyle \frac{1}{2} \times 9 \times 12 = 54\).

The area of the pink triangle is \(\displaystyle \frac{1}{2} \times 3 \times 3 = 4 \frac{1}{2}\).

The area of the main triangle is therefore

\(\displaystyle 144- 54-54-4\frac{1}{2} = 31\frac{1}{2}\)

The parallelogram has area twice this, or \(\displaystyle 2 \times 31\frac{1}{2} = 63\).

The length of a segment with endpoints \(\displaystyle (0,0)\) and \(\displaystyle (4, 5 )\) can be found using the distance formula with \(\displaystyle x_{1} = y_{1} = 0\), \(\displaystyle x_{2} = 4\), \(\displaystyle y_{2} = 5\):

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

\(\displaystyle = \sqrt{(4-0)^{2}+(5-0 )^{2}}\)

\(\displaystyle = \sqrt{4^{2}+5^{2}}\)

\(\displaystyle = \sqrt{16+25 }\)

\(\displaystyle = \sqrt{41}\)

This is the length of one side of the square, so the area is the square of this, or 41.

The height of the parallelogram is \(\displaystyle BD\), and the base is \(\displaystyle DC\). By the 45-45-90 Theorem, \(\displaystyle BD=CD\). Since the product of the height and the base of a parallelogram is its area, 

\(\displaystyle BD \cdot CD = A\)

\(\displaystyle \left (BD \right )^{2} = X\)

\(\displaystyle BD = \sqrt{X}\)

Also by the 45-45-90 Theorem, 

\(\displaystyle CD = AB = BD = \sqrt{X}\), and

\(\displaystyle AD = BC = \sqrt{X} \cdot \sqrt{2} = \sqrt{2X}\)

The perimeter of the parallelogram is

\(\displaystyle AB + CD + BC + AD = \sqrt{X} + \sqrt{X} + \sqrt{2X}+ \sqrt{2X} = 2 \sqrt{X} + 2 \sqrt{2X}\)

By the 45-45-90 Theorem, 

\(\displaystyle AB= CD = BD = 8\sqrt{2}\)

\(\displaystyle AD= BC = BD \cdot \sqrt{2} = 8\sqrt{2} \cdot \sqrt{2} = 8 \cdot 2 = 16\)

The perimeter of the parallelogram is

\(\displaystyle AB + CD + BC + AD = 8 \sqrt{2} + 8 \sqrt{2} + 16+16 = 32+ 16 \sqrt{2}\)

By the 30-60-90 Theorem:

\(\displaystyle AB = CD= \frac{BD}{\sqrt{3}} = \frac{30 \sqrt{6}}{\sqrt{3}}= 30\sqrt{2}\), and

\(\displaystyle BC = AD = 2\cdot AB = 2 \cdot 30\sqrt{2} =60\sqrt{2}\)

The perimeter of the parallelogram is

\(\displaystyle AB + CD + BC + AD = 30\sqrt{2}+ 30\sqrt{2}+60\sqrt{2}+60\sqrt{2}= 180\sqrt{2}\)

A parallelogram has 4 sides.  A base (where the opposite side is equal) and a side (where the opposite side is equal).  So, we will use the following formula:

\(\displaystyle P = b+b+s+s\)

where b is the base and s is the side of the parallelogram.  

We know the base has a length of 6in.  We also know the side has a length of 8in. 

Knowing this, we can substitute into the formula.  We get

\(\displaystyle P = 6\text{in} + 6\text{in} + 8\text{in} + 8\text{in}\)

\(\displaystyle P = 28\text{in}\)

Find the sum of the interior angles of the polygon using the following equation where n is equal to the number of sides.

\(\displaystyle =180(n-2)\)

\(\displaystyle 180(4-2)=180(2)=360\)

The sum of the angles must equal 360.

\(\displaystyle 360=x+71+x+71=2x+142\)

\(\displaystyle 360-142=2x+142-142\)

\(\displaystyle 218=2x\)

\(\displaystyle 109=x\)