The triangle on the left side of the figure has a \(\displaystyle 30^{\circ}\) and a \(\displaystyle 90^{\circ}\) angle. Since all of the angles of a triangle must add up to \(\displaystyle 180^{\circ}\), we can find the angle measure of the third angle:

\(\displaystyle 30+90=120\)

\(\displaystyle 180-120=60^{\circ}\)

Our third angle is \(\displaystyle 60^{\circ}\) and we have a \(\displaystyle 30-60-90\) triangle.

A \(\displaystyle 30-60-90\) triangle has sides that are in the corresponding ratio of \(\displaystyle x-x\sqrt{3}-2x\). In this case, the side opposite our \(\displaystyle 30^{\circ}\) angle is \(\displaystyle 6\), so

\(\displaystyle x=6\)

We also now know that

\(\displaystyle x\sqrt{3}=6\sqrt{3}\)

\(\displaystyle 2x=12\)

Now we know all of our missing side lengths.  The right and left side of the parallelogram will each be \(\displaystyle 12\). The bottom and top will each be \(\displaystyle 20+6\sqrt{3}\). Let's combine them to find the perimeter:

\(\displaystyle Perimeter=2w+2l\)

\(\displaystyle 2\cdot (12)+2\cdot (20+6\sqrt{3})\)

\(\displaystyle 24+40+12\sqrt{3}\)

\(\displaystyle 64+12\sqrt{3}\)

The formula for the perimeter of a trapezoid is:

\(\displaystyle P=b_1+b_2+e_1+e_2\),

where \(\displaystyle b\) is the length of the base and \(\displaystyle e\) is the length of the edge.

Opposite sides of a parallelogram have the same length. Therefore, both edges are \(\displaystyle 8m\) and both bases are \(\displaystyle 12m\).

Plugging in our values, we get:

\(\displaystyle P=12m+12m+8m+8m=40m\)

The formula for the perimeter of a parallelogram is:

\(\displaystyle P = 2(s_{long}) + 2(s_{short})\)

where \(\displaystyle s_{long}\) is the length of the longer side and \(\displaystyle s_{short}\) is the length of the shorter side.

Plugging in our values, we get:

\(\displaystyle P = 2(16m) + 2(9m) = 50m\)

The formula for the perimeter of a parallelogram is

\(\displaystyle P = 2(side)+2(base)\).

Plugging in our values, we get:

\(\displaystyle P = 2(4\sqrt{2}\ m)+2(19\ m)\)

\(\displaystyle P = 38 + 8\sqrt{2}\ m\)

If all of the angles in triangle ABD are equal and line BD divides the parallelogram, then all angles in triangle BDC must be equal as well.

We now have two equilateral triangles, so all sides of the triangles will be equal.

All sides therefore equal 5.

5+5+5+5 = 20

Opposite angles are equal, and adjacent angles must sum to 180.

Therefore, we can set up an equation to solve for z:

(z – 15) + 2z = 180

3z - 15 = 180

3z = 195

z = 65

Now solve for x:

2z = x = 130°

To solve this question you must know the formula for the area of a parallelogram.

In this equation, \(\displaystyle B\) is the length of the base and \(\displaystyle H\) is the length of the height. We can plug in the side length for both base and height, as given in the question.

\(\displaystyle A=8*7\)

\(\displaystyle A=56\)

To solve this question you must know the formula for the area of a parallelogram.

The formula is 

So we can plug in the side length for both base and height to yield \(\displaystyle A=12*8\)

Perform the multiplication to arrive at the answer of \(\displaystyle A=96\).

The formula for the area of a parallelogram is:

\(\displaystyle A=(b)(h)\),

where \(\displaystyle b\) is the length of the base and \(\displaystyle h\) is the length of the height.

In order to the find the height of the parallelogram, use the formula for a \(\displaystyle 30-60-90\) triangle:

\(\displaystyle a-a\sqrt{3}-2a\), where \(\displaystyle a\) is the side opposite the \(\displaystyle \measuredangle30\).

The left side of the parallelogram forms the following \(\displaystyle 30-60-90\) triangle:

\(\displaystyle 4m-4\sqrt{3}m-8m\), where \(\displaystyle 4\sqrt{3}m\) is the length of the height.

Plugging in our values, we get:

\(\displaystyle A=(12m)(4\sqrt{3}m)=48\sqrt{3}m^2\)

Use the Pythagorean Theorem to determine the length of the diagonal:

\(\displaystyle A^2 + B^2 = C^2\)

\(\displaystyle (9m)^2 + B^2 = (16m^2)\)

\(\displaystyle B^2 = 175m^2\)

\(\displaystyle B = 5 \sqrt{7}m\)

The area of the parallelogram is twice the area of the right triangles:

\(\displaystyle A_{Triangle} = \frac{1}{2} (b)(h)\)

\(\displaystyle A_{Triangle} = \frac{1}{2} (9m)(5m\sqrt{7}) = \frac{45m^2\sqrt{7}}{2}\)

\(\displaystyle A_{Parallelogram} = 2 (A_{Triangle}) = 2\left(\frac{45m^2\sqrt{7}}{2}\right)=45m^2\sqrt{7}\)