Let $\displaystyle b$ be the length of the shorter base. Then $\displaystyle 3b$ is the length of its longer base, and $\displaystyle 2 (3b)= 6b$ is the height. Substitute $\displaystyle B = 3b, h = 6b , A = 6,666$ in the area formula:

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

$\displaystyle \frac{1}{2} ( 3b + b) \cdot 6b= 6,000$

$\displaystyle \frac{1}{2} \cdot 4b \cdot 6b= 6,000$

$\displaystyle 12b^{2}= 6,000$

$\displaystyle 12b^{2} \div 12 = 6,000 \div 12$

$\displaystyle b^{2} = 500$

$\displaystyle b = \sqrt{500} = \sqrt{100}\cdot \sqrt{5 } =10 \sqrt{5 }$

The height is six times this, or $\displaystyle h = 6 \cdot 10 \sqrt{5 } = 60 \sqrt{5 }$ inches.

The length of the midsegment of a trapezoid is the mean of the lengths of its bases; the height is irrelevant. The bases are of length 4,000 feet and 5,280 feet; their mean is 

$\displaystyle \frac{1}{2} \left ( 5,280 + 4,000 \right ) = 4,640$ feet, the length of the midsegment.

The area $\displaystyle A$ of a trapezoid with a larger base $\displaystyle B$, a shorter base $\displaystyle b$, and a height $\displaystyle h$ is defined by the equation $\displaystyle A=\frac{1}{2}(B+b)h$. Restructuring to solve for the shorter base $\displaystyle b$:

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

$\displaystyle 2A=(B+b)h$

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

$\displaystyle \frac{2A}{h}-B=b$

Plugging in our values for $\displaystyle A$, $\displaystyle B$, and $\displaystyle h$: 

$\displaystyle \frac{2(2500)}{50}-70=b$

$\displaystyle 30=b$ 

The area $\displaystyle A$ of a trapezoid with a larger base $\displaystyle B$, a shorter base $\displaystyle b$, and a height $\displaystyle h$ is defined by the equation $\displaystyle A=\frac{1}{2}(B+b)h$. Restructuring to solve for the shorter base $\displaystyle b$:

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

 $\displaystyle 2A=(B+b)h$

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

$\displaystyle \frac{2A}{h}-B=b$

Plugging in our values for $\displaystyle A$, $\displaystyle B$, and $\displaystyle h$: 

$\displaystyle \frac{2(1000)}{20}-60=b$

$\displaystyle 40=b$

The area $\displaystyle A$ of a trapezoid with a larger base $\displaystyle B$, a shorter base $\displaystyle b$, and a height $\displaystyle h$ is defined by the equation $\displaystyle A=\frac{1}{2}(B+b)h$. Restructuring to solve for the shorter base $\displaystyle b$:

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

 $\displaystyle 2A=(B+b)h$

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

$\displaystyle \frac{2A}{h}-B=b$

Plugging in our values for $\displaystyle A$, $\displaystyle B$, and $\displaystyle h$: 

$\displaystyle \frac{2(4000)}{40}-120=b$

$\displaystyle \frac{8000}{40}-120=b$

$\displaystyle 200-120=b$

$\displaystyle 80=b$

Opposite sides of a parallelogram are congruent; if $\displaystyle CD = 12$, then $\displaystyle CD \neq AB$, violating this condition.

Consecutive angles of a parallelogram are supplementary; if $\displaystyle m \angle A = 114^{\circ}$, then $\displaystyle m \angle A + m \angle B= 114 + 76 = 190$, violating this condition.

Opposite angles of a parallelogram are congruent; if $\displaystyle m \angle D < 76^{\circ}$, then $\displaystyle m \angle D \neq m \angle B$, violating this condition.

Adjacent sides of a parallelogram, however, may or may not be congruent, so the condition that $\displaystyle BC > 12$ would not by itself prove that the quadrilateral is not a parallelogram.

The four interior angles of a quadrilateral measure a total of $\displaystyle 360^{\circ }$, so we test each group of numbers to see if they have this sum.

$\displaystyle 75+85+95+105 = 360$

$\displaystyle 80+80+100+100 = 360$

$\displaystyle 60+100+100+100 = 360$

$\displaystyle 75 + 75 + 75 + 125 = 350$

This last group does not have the correct sum, so it is the correct choice.

A circle can be circumscribed about any triangle regardless of its sidelengths or angle measures, so we can eliminate the two triangle choices. 

A circle can be circumscribed about a quadrilateral if and only if both pairs of its opposite angles are supplementary. An isosceles trapezoid has this characteristic; this can be proved by the fact that base angles are congruent, and by the Same-Side Interior Angles Statement. For a parallelogram to have this characteristic, since opposite angles are congruent also, all angles must measure $\displaystyle 90^{\circ }$; the rectangle fits this description, but the rhombus does not.

The diagonals of a rhombus always intersect at right angles, so $\displaystyle m \angle HPO = 90 ^{\circ }$. The measures of the interior angles of the rhombus are irrelevant.

Two opposite angles of a quadrilateral inscribed inside a circle are supplementary, so 

$\displaystyle m \angle Q = 180^{\circ } - m \angle A = 180^{\circ } - 100^{\circ } = 80^{\circ }$