The area of a trapezoid is given by

$\displaystyle Area=(\frac{b_{1}+b_{2}}{2})h$,

where $\displaystyle b_{}1$, $\displaystyle b_{}2$ are the lengths of each base and $\displaystyle h$ is the altitude (height) of the trapezoid.

$\displaystyle Area=(\frac{b_{1}+b_{2}}{2})h=(\frac{3t+t}{2})\times h=8t^2\Rightarrow (2t)h=8t^2$

$\displaystyle \Rightarrow h=\frac{8t^2}{2t}=4t$

The area of a trapezoid is given by

$\displaystyle Area=(\frac{b_{1}+b_{2}}{2})h$,

where $\displaystyle b_{}1$, $\displaystyle b_{}2$ are the lengths of each base and $\displaystyle h$ is the altitude (height) of the trapezoid. 

$\displaystyle b_{1}=a$

$\displaystyle b_{2}=b=2a$

$\displaystyle h=a$

Substitute these values into the area formula:

$\displaystyle Area=(\frac{b_{1}+b_{2}}{2})h=(\frac{a+2a}{2})a=54\Rightarrow \frac{3a^2}{2}=54\Rightarrow a^2=36\Rightarrow a=6\ in$

Quadrilateral $\displaystyle SQUR$ - the shaded region - is a trapezoid with bases $\displaystyle \overline{SR}$ and $\displaystyle \overline{QU }$, and altitude $\displaystyle \overline{SQ}$. The area of the trapezoid can be calculated using the formula 

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

where $\displaystyle b_{1} = SR = 24$ and  $\displaystyle b_{2} = QU$, and $\displaystyle h = SQ = 25$.

The length of $\displaystyle \overline{UA}$ can be found by setting $\displaystyle c = 25$ and $\displaystyle b = 24$ and applying the Pythagorean Theorem:

$\displaystyle a = \sqrt{c^{2} - b^{2}} = \sqrt{25^{2} - 24^{2}} =\sqrt{625- 576} = \sqrt{49} = 7$

Therefore, 

$\displaystyle QU = QA - UA = 24 - 7 = 17$.

Substituting:

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

$\displaystyle A = \frac{1}{2} ( 24 + 17) 24$

$\displaystyle A = \frac{1}{2} \cdot 41 \cdot 24$

$\displaystyle A = 492$

Substitute $\displaystyle A = 465, b = 25, h = 15$ into the formula for the area of a trapezoid:

$\displaystyle \frac{1}{2} (B + b)h = A$

$\displaystyle \frac{1}{2} (B + 25) \cdot 15 = 465$

$\displaystyle 2 \cdot \frac{1}{2} (B + 25) \cdot 15 = 2 \cdot 465$

$\displaystyle (B + 25) \cdot 15 = 930$

$\displaystyle 15 B + 375 = 930$

$\displaystyle 15 B + 375 - 375 = 930 - 375$

$\displaystyle 15 B = 555$

$\displaystyle 15 B \div 15 = 555\div 15$

$\displaystyle B = 37$

To find the area of a kite, use the formula $\displaystyle \frac{1}{2}ab$, where $\displaystyle a$ represents one diagnal and $\displaystyle b$ represents the other.

Since Cassie has one tube measuring $\displaystyle 7$ inches, we can substitute $\displaystyle 7$ for $\displaystyle a$. We can also substitute the other tube that measures $\displaystyle 12$ inches in for $\displaystyle b$.

$\displaystyle \frac{1}{2}(7)(12)=42\ in^{2}$

The area of a kite is half the product of the diagonals, i.e.

 $\displaystyle Area=\frac{d_{1}d_{2}}{2}$,

where $\displaystyle d_{1}$ and $\displaystyle d_{2}$ are the lengths of the diagonals. 

$\displaystyle Area=\frac{d_{1}d_{2}}{2}=\frac{6\times 10}{2}=30\ in^2$

When you know the length of two unequal sides of a kite and their included angle, the following formula can be used to find the area of a kite:

$\displaystyle Area=ab\times sinC$,

where $\displaystyle a,b$are the lengths of two unequal sides, $\displaystyle C$ is the angle between them and $\displaystyle sin$ is the sine function.

$\displaystyle Area=ab\times sinC=10\times 15\times sin120^{\circ}=10\times 15\times \frac{\sqrt{3}}{2}\Rightarrow Area=75\sqrt{3}\ cm^2$

To find the area of a kite, we will use the following formula:

$\displaystyle A = \frac{pq}{2}$

where p and q are the lengths of the diagonals of the kite.

Now, we know the length of one diagonal is 18in.  We also know the other diagonal is half of the first diagonal.  Therefore, the second diagonal has a length of 9in.

Knowing this, we can substitute into the formula.  We get

$\displaystyle A = \frac{18\text{in} \cdot 9\text{in}}{2}$

$\displaystyle A = \frac{162\text{in}^2}{2}$

$\displaystyle A = 81\text{in}^2$

Multiply height $\displaystyle BD$ by base $\displaystyle CD$ to get the area.

By the 30-60-90 Theorem:

$\displaystyle BD = \frac{BC}{2} = \frac{10}{2}} = 5$

and

$\displaystyle CD =BD \cdot \sqrt{3} = 5\sqrt{3}$

The area is therefore

$\displaystyle BD \cdot CD = 5 \cdot 5\sqrt{3} = 25\sqrt{3}$