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:

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

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

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

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

By the 30-60-90 Theorem, the length of the short leg of \(\displaystyle \bigtriangleup BDC\) is the length of the long leg divided by \(\displaystyle \sqrt{3}\), so 

\(\displaystyle CD = \frac{BD}{\sqrt{3}} = \frac{15}{\sqrt{3}} = \frac{15\cdot \sqrt{3}}{\sqrt{3}\cdot \sqrt{3}}= \frac{15 \sqrt{3}}{3} = 5\sqrt{3}\)

Its hypotenuse has twice the length of the short leg, so 

\(\displaystyle BC = 2 \cdot CD = 2 \cdot 5\sqrt{3}= 10 \sqrt{3}\)

The perimeter of the parallelogram is

\(\displaystyle AB + BC + CD + DA\)

\(\displaystyle =CD + BC + CD + BC\)

\(\displaystyle =5 \sqrt{3} + 10\sqrt{3} + 5 \sqrt{3} + 10\sqrt{3} = 30 \sqrt{3}\)

By the 45-45-90 Theorem, the lengths of the legs of \(\displaystyle \bigtriangleup BDC\)are equal, so

\(\displaystyle CD = BD = 17\)

Its hypotenuse has measure \(\displaystyle \sqrt{2}\) that of the common measure of its legs, so 

\(\displaystyle BC = CD \cdot \sqrt{2} = 17 \sqrt{2}\)

The perimeter of the parallelogram is

\(\displaystyle AB + BC + CD + DA\)

\(\displaystyle =CD + BC + CD + BC\)

\(\displaystyle =17+ 17 \sqrt{2}+ 17 + 17\sqrt{2}= 34 + 34 \sqrt{2}\)

\(\displaystyle \frac{BD}{CD} = \tan 65^{\circ}\)

\(\displaystyle \frac{CD}{BD} = \frac{1}{\tan 65^{\circ}}\)

\(\displaystyle CD= \frac{BD}{\tan 65^{\circ}} \approx \frac{10}{2.1445} \approx 4.66\)

\(\displaystyle \frac{BD}{BC} = \sin 65^{\circ}\)

\(\displaystyle \frac{BC}{BD} = \frac{1}{\sin 65^{\circ}}\)

\(\displaystyle BC= \frac{BD}{\sin 65^{\circ}} \approx \frac{10}{0.9063} \approx 11.03\)

The perimeter of the parallelogram is

\(\displaystyle AB + BC + CD + DA\)

\(\displaystyle =CD + BC + CD + BC\)

\(\displaystyle \approx 4.66 + 11.03 + 4.66 + 11.03 \approx 31.4\)

By the 45-45-90 Theorem, \(\displaystyle BD = CD\). Since \(\displaystyle BD\) and \(\displaystyle CD\) are its base and height:

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

\(\displaystyle 100 = CD \cdot CD = \left (CD \right ) ^{2}\)

\(\displaystyle CD =10\)

Also by the 45-45-90 Theorem,

\(\displaystyle BC = CD \cdot \sqrt{2} = 10 \cdot 1.4142 \approx 14.14\)

The perimeter of the parallelogram is

\(\displaystyle AB + BC + CD + DA\)

\(\displaystyle =CD + BC + CD + BC\)

\(\displaystyle \approx 10+ 14.14 + 10 + 14.14\)

\(\displaystyle \approx 48.3\)

By the 30-60-90 Theorem, 

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

The area of the parallelogram is the product of height \(\displaystyle BD\) and base \(\displaystyle BC\), so 

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

\(\displaystyle 100 = CD\cdot \sqrt{3}\cdot CD\)

\(\displaystyle 100 = \left (CD\right )^{2} \sqrt{3}\)

\(\displaystyle \left (CD\right )^{2} = \frac{100 }{\sqrt{3}}\)

\(\displaystyle \left (CD\right )^{2} \approx \frac{100 }{1.7321}\)

\(\displaystyle \left (CD\right )^{2} \approx 57.73\)

\(\displaystyle CD \approx \sqrt{57.73} \approx 7.60\)

Also by the 30-60-90 Theorem,

\(\displaystyle BD =2\cdot CD \approx 2 \cdot 7.60 = 15.20\)

The perimeter of the parallelogram is

\(\displaystyle AB + BC + CD + DA\)

\(\displaystyle =CD + BC + CD + BC\)

\(\displaystyle \approx 7.60 + 15.20 + 7.60 + 15.20 \approx 45.6\)

\(\displaystyle \frac{BD}{CD} = \tan 65^{\circ}\)

\(\displaystyle BD=CD \cdot \tan 65^{\circ}\)

The area of the parallelogram is the product of height \(\displaystyle BD\) and base \(\displaystyle BC\), so 

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

\(\displaystyle 100 = CD \cdot \tan 65^{\circ}\cdot CD\)

\(\displaystyle (CD)^{2} \cdot \tan 65^{\circ} = 100\)

\(\displaystyle (CD)^{2} = \frac{100}{\tan 65^{\circ} } = \frac{100}{2.1445} \approx 46.63\)

\(\displaystyle CD \approx \sqrt{46.63} \approx 6.83\)

\(\displaystyle \frac{CD}{BC} = \cos 65^{\circ}\)

\(\displaystyle \frac{BC} {CD}= \frac {1}{ \cos 65^{\circ}}\)

\(\displaystyle BC = \frac{CD}{ \cos 65^{\circ}}\)

\(\displaystyle BC \approx \frac{6.83}{0.4226} \approx 16.16\)

The perimeter of the parallelogram is

\(\displaystyle AB + BC + CD + DA\)

\(\displaystyle =CD + BC + CD + BC\)

\(\displaystyle \approx 6.83 + 16.16 + 6.83 +16.16\)

\(\displaystyle \approx 46.0\)