The side length is half of the base length:

\(\displaystyle h=\frac{w}{2}=\frac{4t+6}{2}=\frac{4t}{2}+\frac{6}{2}=2t+3\)

The perimeter of a parallelogram is:

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

Where:

  \(\displaystyle w\) is the base length of the parallelogram and \(\displaystyle h\) is the side length

\(\displaystyle =2[(4t+6)+(2t+3)]\)

\(\displaystyle =2(6t+9)\)

\(\displaystyle =12t+18\)

The base length is three times more than the side length, so we have:

Base length \(\displaystyle =3(t^2+1)=3t^2+3\)

The perimeter of a parallelogram is:

\(\displaystyle 2(w+h)\)

Where:

\(\displaystyle w\) is the base length of the parallelogram and \(\displaystyle h\) is the side length. So we get:

\(\displaystyle Perimeter=2(w+h)=2\left [ (3t^2+3)+(t^2+1) \right ]\)

\(\displaystyle =2(4t^2+4)\)

\(\displaystyle =8t^2+8\)

Like any polygon, the perimeter of a parallelogram is the total distance around the outside, which can be found by adding together the length of each side. In case of a parallelogram, each pair of opposite sides is the same length, so the perimeter is twice the base plus twice the side length. Or as a formula we can write:

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

Where:

\(\displaystyle w\) is the base length of the parallelogram and \(\displaystyle h\) is the side length. So we can write:

\(\displaystyle Perimeter=2(w+h)=2(10+6)=32\)

Let:

Side length \(\displaystyle =x\).

The perimeter of a parallelogram is:

\(\displaystyle 2w+2h\)

where:

\(\displaystyle w\) is the base length of the parallelogram and \(\displaystyle h\) is the side length. The perimeter is known, so we can write:

\(\displaystyle Perimeter=24=2(t+2)+2x\)

Now we solve the equation for \(\displaystyle x\):

\(\displaystyle 24=2(t+2)+2x\)

\(\displaystyle \Rightarrow 24=2t+4+2x\)

\(\displaystyle \Rightarrow 24-2t-4=2x\)

\(\displaystyle \Rightarrow 20-2t=2x\)

\(\displaystyle \Rightarrow 10-t=x\)

The perimeter of a parallelogram is:

\(\displaystyle 2(w+h)\)

Where:

\(\displaystyle w\) is the base length of the parallelogram and \(\displaystyle h\) is the side length. In this problem the base length and side length are identical, that means:

\(\displaystyle w=h\)

So we can write:

\(\displaystyle Perimeter=40=2(w+w)\Rightarrow 20=2w\Rightarrow w=10\)

The perimeter of a parallelogram is:

\(\displaystyle 2(w+h)\)

where:

\(\displaystyle w\) is the base length of the parallelogram and \(\displaystyle h\) is the side length. So we have:

\(\displaystyle Perimeter=2(w+h)=2\left [ (2t+3)+(2t-3) \right ]=2(4t)=8t\)

and:

\(\displaystyle t=1\Rightarrow 8t=8\times 1=8\)

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

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

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

\(\displaystyle BD = 10\)

By the \(\displaystyle 45^{\circ}-45^{\circ}-90^{\circ}\) Theorem, 

\(\displaystyle CD = AB = BD = 10\), and

\(\displaystyle AD = BC = BD \cdot \sqrt{2} = 10\sqrt{2}\)

The perimeter of the parallelogram is

\(\displaystyle AB + CD + BC + AD = 10 + 10 + 10\sqrt{2}+ 10\sqrt{2} = 20+20\sqrt{2}\)

By the \(\displaystyle 30^{\circ}-60^{\circ}-90^{\circ}\) Theorem:

\(\displaystyle AB = CD= BD\sqrt{3}= 10\sqrt{3}\), and

\(\displaystyle AD=BC= 2\cdot BD = 2\cdot 10 = 20\)

The perimeter of the parallelogram is

\(\displaystyle AB + CD + BC + AD = 20+20+ 10\sqrt{3}+ 10\sqrt{3} = 40+20\sqrt{3}\)