From the figure, you should notice that it is made up of a right triangle and a trapezoid. The lower base of the trapezoid is also the hypotenuse of the right triangle.

First, find the length of the hypotenuse of the right triangle using the Pythagorean Theorem.

\(\displaystyle \text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}\)

\(\displaystyle \text{Hypotenuse}=\sqrt{7^2+12^2}=\sqrt{193}\)

Next, use this value to find the area of the trapezoid.

\(\displaystyle \text{Area of Trapezoid}=\frac{1}{2}(\text{base 1}+\text{base 2})(\text{height})\)

Plug in the given and found values to find the area.

\(\displaystyle \text{Area of Trapezoid}=\frac{1}{2}(16+\sqrt{193})(8)=64+4\sqrt{193}\)

Next, find the area of the triangle.

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(\text{base})(\text{height})\)

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(7)(12)=42\)

To find the area of the figure, add the two areas together.

\(\displaystyle \text{Area of Figure}=\text{Area of Trapezoid}+\text{Area of Triangle}\)

\(\displaystyle \text{Area of Figure}=64+4\sqrt{193}+42=161.57\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

From the figure, you should notice that it is made up of a right triangle and a trapezoid. The lower base of the trapezoid is also the hypotenuse of the right triangle.

First, find the length of the hypotenuse of the right triangle using the Pythagorean Theorem.

\(\displaystyle \text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}\)

\(\displaystyle \text{Hypotenuse}=\sqrt{5^2+11^2}=\sqrt{146}\)

Next, use this value to find the area of the trapezoid.

\(\displaystyle \text{Area of Trapezoid}=\frac{1}{2}(\text{base 1}+\text{base 2})(\text{height})\)

Plug in the given and found values to find the area.

\(\displaystyle \text{Area of Trapezoid}=\frac{1}{2}(11+\sqrt{146})(3)=\frac{33+3\sqrt{146}}{2}\)

Next, find the area of the triangle.

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(\text{base})(\text{height})\)

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(5)(11)=\frac{55}{2}\)

To find the area of the figure, add the two areas together.

\(\displaystyle \text{Area of Figure}=\text{Area of Trapezoid}+\text{Area of Triangle}\)

\(\displaystyle \text{Area of Figure}=\frac{33+3\sqrt{146}}{2}+\frac{55}{2}=62.12\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

From the figure, you should notice that it is made up of a right triangle and a trapezoid. The lower base of the trapezoid is also the hypotenuse of the right triangle.

First, find the length of the hypotenuse of the right triangle using the Pythagorean Theorem.

\(\displaystyle \text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}\)

\(\displaystyle \text{Hypotenuse}=\sqrt{22^2+26^2}=\sqrt{1160}\)

Next, use this value to find the area of the trapezoid.

\(\displaystyle \text{Area of Trapezoid}=\frac{1}{2}(\text{base 1}+\text{base 2})(\text{height})\)

Plug in the given and found values to find the area.

\(\displaystyle \text{Area of Trapezoid}=\frac{1}{2}(34+\sqrt{1160})(18)=306+9\sqrt{1160}\)

Next, find the area of the triangle.

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(\text{base})(\text{height})\)

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(22)(26)=286\)

To find the area of the figure, add the two areas together.

\(\displaystyle \text{Area of Figure}=\text{Area of Trapezoid}+\text{Area of Triangle}\)

\(\displaystyle \text{Area of Figure}=306+9\sqrt{1160}+286=898.53\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

From the figure, you should notice that it is made up of a right triangle and a trapezoid. The lower base of the trapezoid is also the hypotenuse of the right triangle.

First, find the length of the hypotenuse of the right triangle using the Pythagorean Theorem.

\(\displaystyle \text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}\)

\(\displaystyle \text{Hypotenuse}=\sqrt{9^2+10^2}=\sqrt{181}\)

Next, use this value to find the area of the trapezoid.

\(\displaystyle \text{Area of Trapezoid}=\frac{1}{2}(\text{base 1}+\text{base 2})(\text{height})\)

Plug in the given and found values to find the area.

\(\displaystyle \text{Area of Trapezoid}=\frac{1}{2}(13+\sqrt{181})(8)=52+4\sqrt{181}\)

Next, find the area of the triangle.

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(\text{base})(\text{height})\)

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(9)(10)=45\)

To find the area of the figure, add the two areas together.

\(\displaystyle \text{Area of Figure}=\text{Area of Trapezoid}+\text{Area of Triangle}\)

\(\displaystyle \text{Area of Figure}=52+4\sqrt{181}+45=150.81\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

From the figure, you should notice that it is made up of a right triangle and a trapezoid. The lower base of the trapezoid is also the hypotenuse of the right triangle.

First, find the length of the hypotenuse of the right triangle using the Pythagorean Theorem.

\(\displaystyle \text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}\)

\(\displaystyle \text{Hypotenuse}=\sqrt{7^2+12^2}=\sqrt{193}\)

Next, use this value to find the area of the trapezoid.

\(\displaystyle \text{Area of Trapezoid}=\frac{1}{2}(\text{base 1}+\text{base 2})(\text{height})\)

Plug in the given and found values to find the area.

\(\displaystyle \text{Area of Trapezoid}=\frac{1}{2}(14+\sqrt{193})(5)=35+\frac{5\sqrt{193}}{2}\)

Next, find the area of the triangle.

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(\text{base})(\text{height})\)

\(\displaystyle \text{Area of Triangle}=\frac{1}{2}(7)(12)=42\)

To find the area of the figure, add the two areas together.

\(\displaystyle \text{Area of Figure}=\text{Area of Trapezoid}+\text{Area of Triangle}\)

\(\displaystyle \text{Area of Figure}=35+\frac{5\sqrt{117}}{2}+42=111.73\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

The area of a trapezoid is equal to one half the product of half the height of the trapezoid and the sum of the lengths of the bases. This is 

\(\displaystyle A = \frac{1}{2} h (b_{1}+ b_{2})\)

\(\displaystyle A = \frac{1}{2} \cdot TP \cdot (TR + PA)\)

or, equivalently,

\(\displaystyle A = TP \cdot \frac{TR + PA}{2}\)

The height of the trapezoid is 

\(\displaystyle TP = TM + MP = 5 + 5 = 10\). 

The lengths of bases \(\displaystyle \overline{TR}\) and \(\displaystyle \overline{PA}\) are not given, so it might appear that determining the area of the trapezoid is impossible. 

However, it is given that \(\displaystyle TM = MP\) and \(\displaystyle RM = MA\) - that is, the segment \(\displaystyle \overline{MN}\) bisects both legs of the trapezoid. This makes \(\displaystyle \overline{MN}\) the midsegment of the trapezoid, the length of which is the arithmetic mean of those of the bases:

\(\displaystyle MN = \frac{TR+ PA}{2}\).

Therefore, the formula for the area of the trapezoid can be rewritten as

\(\displaystyle A = TP \cdot MN\),

the product of the height and the length of the midsegment.

\(\displaystyle TP = 10\) and \(\displaystyle MN = 20\), so

\(\displaystyle A =10 \cdot 20 = 200\),

making the statement true.

To find the length of the diagonal, we need to use the Pythagorean Theorem. Therefore, we need to sketch the following triangle within trapezoid \(\displaystyle ABCD\):

We know that the base of the triangle has length \(\displaystyle 9\: m\). By subtracting the top of the trapezoid from the bottom of the trapezoid, we get:

\(\displaystyle 12\: m - 6\: m = 6\: m\)

Dividing by two, we have the length of each additional side on the bottom of the trapezoid:

\(\displaystyle \frac{6\: m}{2} = 3\: m\)

Adding these two values together, we get \(\displaystyle 9\: m\).

The formula for the length of diagonal \(\displaystyle AC\) uses the Pythagoreon Theorem:

\(\displaystyle AC^2 = AE^2 + EC^2\), where \(\displaystyle E\) is the point between \(\displaystyle A\) and \(\displaystyle D\) representing the base of the triangle.

Plugging in our values, we get:

\(\displaystyle AC^2 = (9\: m)^2 + (4\: m)^2\)

\(\displaystyle AC^2 = 81\: m^2 + 16\: m^2\)

\(\displaystyle AC^2 = 97\: m^2\)

\(\displaystyle AC = \sqrt{97}\: m\)

All of the lengths with one mark have length 5, and all of the side lengths with two marks have length 4. With this knowledge, we can add side lengths together to find that one diagonal is the hypotenuse to this right triangle:

Using Pythagorean Theorem gives:

\(\displaystyle 4^2 + 9^2 = C^2\)

\(\displaystyle 16+81=C^2\)

\(\displaystyle 97 = C^2\) take the square root of each side

\(\displaystyle \sqrt{97}= C\)

Similarly, the other diagonal can be found with this right triangle:

Once again using Pythagorean Theorem gives an answer of \(\displaystyle \sqrt{65}\)

To find the length of the diagonals, split the top side into 3 sections as shown below:

The two congruent sections plus 8 adds to 14. \(\displaystyle 14 - 8 = 6\), so the two congruent sections add to 6. They must each be 3. This means that the top of the right triangle with the diagonal as a hypotenuse must be 11, since \(\displaystyle 14-3=11\).

We can solve for the diagonal, now pictured, using Pythagorean Theorem:

\(\displaystyle 11^2 + 8^2 = C^2\)

\(\displaystyle 121 + 64 = C^2\)

\(\displaystyle 185 = C^2\) take the square root of both sides

\(\displaystyle \sqrt{185}=C\)

In order to calculate the length of the diagonal, we first must assume that the height is perpendicular to both the top and bottom of the trapezoid. 

Knowing this, we can draw in the diagonal as shown below and use the Pythagorean Theorem to solve for the diagonal. 

\(\displaystyle Using\ the\ pythagorean\ theorem\ we\ get: 12^2+20^2=diagonal^2\)

\(\displaystyle This\ simplifies\ to\ become\ 144+400=diagonal^2\)

\(\displaystyle 544=diagonal^2\)

We now take the square root of both sides: 

\(\displaystyle \sqrt544=\sqrt diagonal^2\)

\(\displaystyle \sqrt16*\sqrt34=diagonal\)

\(\displaystyle 4\sqrt34=diagonal\)