To find the length of the diagonal, we can consider only the triangle \(\displaystyle ACD\) and use the law of cosines to find the length of the unknown side.

The Law of Cosines:

\(\displaystyle c^2 = a^2 + b^2 - 2ab\cos \left ( C\right )\)

Where \(\displaystyle c\) is the length of the unknown side, \(\displaystyle a\) and \(\displaystyle b\) are the lengths of the known sides, and \(\displaystyle C\) is the angle between \(\displaystyle a\) and \(\displaystyle b\). 

From the problem:

\(\displaystyle \overline{AC}^2 = \overline{CD}^2 + \overline{DA}^2 - 2\left ( \overline{CD} \right )\left (\overline{DA} \right )\cos \left ( \measuredangle D\right )\)

\(\displaystyle \overline{AC}^2 = 35^2 + 82 ^2 - 2\cdot35\cdot82\cdot \cos \left ( 143^{\circ}\right )\)

\(\displaystyle \overline{AC}^2 = 12553\)

\(\displaystyle \overline{AC} = 112\)

To find the length of the diagonal, we can consider only the triangle \(\displaystyle ACD\) and use the law of cosines to find the length of the unknown side.

The Law of Cosines:

\(\displaystyle c^2 = a^2 + b^2 - 2ab\cos \left ( C\right )\)

Where \(\displaystyle c\) is the length of the unknown side, \(\displaystyle a\) and \(\displaystyle b\) are the lengths of the known sides, and \(\displaystyle C\) is the angle between \(\displaystyle a\) and \(\displaystyle b\). 

From the problem:

\(\displaystyle \overline{AC}^2 = \overline{CD}^2 + \overline{DA}^2 - 2\left ( \overline{CD} \right )\left (\overline{DA} \right )\cos \left ( \measuredangle D\right )\)

\(\displaystyle \overline{AC}^2 = 12^2 + 40 ^2 - 2\cdot12\cdot40\cdot \cos \left ( 135^{\circ}\right )\)

\(\displaystyle \overline{AC}^2 = 2423\)

\(\displaystyle \overline{AC} = 49.2\)

To find the length of the diagonal, we can consider only the triangle \(\displaystyle ACD\) and use the law of cosines to find the length of the unknown side.

The Law of Cosines:

\(\displaystyle c^2 = a^2 + b^2 - 2ab\cos \left ( C\right )\)

Where \(\displaystyle c\) is the length of the unknown side, \(\displaystyle a\) and \(\displaystyle b\) are the lengths of the known sides, and \(\displaystyle C\) is the angle between \(\displaystyle a\) and \(\displaystyle b\). 

From the problem:

\(\displaystyle \overline{AC}^2 = \overline{CD}^2 + \overline{DA}^2 - 2\left ( \overline{CD} \right )\left (\overline{DA} \right )\cos \left ( \measuredangle D\right )\)

\(\displaystyle \overline{AC}^2 = 10^2 + 20 ^2 - 2\cdot10\cdot20\cdot \cos \left ( 120^{\circ}\right )\)

\(\displaystyle \overline{AC}^2 = 700\)

\(\displaystyle \overline{AC} = 26.5\)

To find the length of the diagonal, we can consider only the triangle \(\displaystyle ABD\) and use the law of cosines to find the length of the unknown side.

The Law of Cosines:

\(\displaystyle c^2 = a^2 + b^2 - 2ab\cos \left ( C\right )\)

Where \(\displaystyle c\) is the length of the unknown side, \(\displaystyle a\) and \(\displaystyle b\) are the lengths of the known sides, and \(\displaystyle C\) is the angle between \(\displaystyle a\) and \(\displaystyle b\). 

From the problem:

\(\displaystyle \overline{BD}^2 = \overline{AB}^2 + \overline{DA}^2 - 2\left ( \overline{AB} \right )\left (\overline{DA} \right )\cos \left ( \measuredangle A\right )\)

\(\displaystyle \overline{BD}^2 = 12^2 + 40 ^2 - 2\cdot12\cdot40\cdot \cos \left ( 45^{\circ}\right )\)

\(\displaystyle \overline{BD}^2 = 1065\)

\(\displaystyle \overline{BD} = 32.6\)

To find the length of the diagonal, we can consider only the triangle \(\displaystyle ABD\) and use the law of cosines to find the length of the unknown side.

The Law of Cosines:

\(\displaystyle c^2 = a^2 + b^2 - 2ab\cos \left ( C\right )\)

Where \(\displaystyle c\) is the length of the unknown side, \(\displaystyle a\) and \(\displaystyle b\) are the lengths of the known sides, and \(\displaystyle C\) is the angle between \(\displaystyle a\) and \(\displaystyle b\). 

From the problem:

\(\displaystyle \overline{BD}^2 = \overline{AB}^2 + \overline{DA}^2 - 2\left ( \overline{AB} \right )\left (\overline{DA} \right )\cos \left ( \measuredangle A\right )\)

\(\displaystyle \overline{BD}^2 = 10^2 + 20 ^2 - 2\cdot10\cdot20\cdot \cos \left ( 60^{\circ}\right )\)

\(\displaystyle \overline{BD}^2 = 300\)

\(\displaystyle \overline{BD} = \sqrt{300}\)

The formula to find the area of a parallelogram is

\(\displaystyle base \times height\)

The base, \(\displaystyle DC\), is given by the question.

\(\displaystyle DE+EC=DC\)

\(\displaystyle 12+6=18\)

You should recognize that \(\displaystyle AE\) is not only the height of parallelogram \(\displaystyle ABCD\), but it is also a leg of the right triangle \(\displaystyle AED\).

Use the Pythagorean Theorem to find the length of \(\displaystyle AE\).

\(\displaystyle {AE}^2+{DE}^2={AD}^2\)

\(\displaystyle {AE}^2+12^2=20^\)

\(\displaystyle {AE}^2=256\)

\(\displaystyle AE=16\)

Now that we have the height, multiply it by the base to find the area of the parallelogram.

The formula for the area of a parallelogram is given by the equation \(\displaystyle A =b \cdot h\), where \(\displaystyle b\) is the base and \(\displaystyle h\) is the height of the parallelogram. 

The only given information is that the base is \(\displaystyle 12\:cm\), the side is \(\displaystyle 5\:cm\), and the base of the right triangle in the parallelogram (the triangle formed between the edge of the top base and the bottom base) is \(\displaystyle 3\:cm\) because \(\displaystyle 12\:cm\div4=3\:cm\).

The last part of information that is required to fulfill the needs of the area formula is the parallelogram's height, \(\displaystyle h\). The parallelogram's height is given by the mystery side of the right triangle described in the question. In order to solve for the triangle's third side, we can use the Pythagorean Theorem, \(\displaystyle a^2 + b^2 =c^2\).

In this case, the unknown side is one of the legs of the triangle, so we will label it \(\displaystyle a\). The given side of the triangle that is part of the base we will call \(\displaystyle b\), and the side of the parallelogram is also the hypotenuse of the triangle, so in the Pythagorean Formula its length will be represented by \(\displaystyle c\). At this point, we can substitute in these values and solve for \(\displaystyle a\):

\(\displaystyle a^2 + 3^2=5^2\)

\(\displaystyle a^2 + 9 = 25\)

\(\displaystyle a^2 = 25 - 9\)

\(\displaystyle a^2 = 16\)

\(\displaystyle a=\sqrt{16}\)

\(\displaystyle a=\pm 4\) , but because we're finding a length, the answer must be 4. The negative option can be negated. 

Remembering that we temporarily called \(\displaystyle h\) "\(\displaystyle a\)" for the pythagorean theorem, this means that \(\displaystyle h=4\:cm\).

Now all the necessary parts for the area of a parallelogram equation are available to be used:

\(\displaystyle A = (12\:cm)(4\:cm)\)

\(\displaystyle A = 48\:cm^2\)

The length is 5 x 3 = 15 inches. Multiplied by the width of 3 inches, yields 45 in².

Rectangle

B = 2H

B = 8”

H = B/2 = 8/2 = 4”

Area = B x H = 8” X 4” = 32 in²

For a rectangle, \(\displaystyle P = 2l + 2w\) and \(\displaystyle A = lw\), where \(\displaystyle l\) is the length and \(\displaystyle w\) is the width.

Let \(\displaystyle w\) be equal to the width. We know that the length is equal to "two more than twice with width."

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

The equation to solve for the perimeter becomes \(\displaystyle 58 = 2(2w + 2) + 2w\).

\(\displaystyle 58 = 4w + 4+2w=6w+4\)

\(\displaystyle 54=6w\)

\(\displaystyle 9=w\)

Now that we know the width, we can solve for the length.

\(\displaystyle l=2w + 2 = 20\)

Now we can find the area using \(\displaystyle A = lw\).

\(\displaystyle A=(20ft)(9ft)=180ft^2\)