The inner rectangle, which represents the garden, has length and width \(\displaystyle (100 - 2\times 6) = 88\) feet and \(\displaystyle (60 - 2\times 6) = 48\) feet, respectively, so its perimeter is

\(\displaystyle P = 88+ 88 + 48 + 48 = 272\)  feet.

The four sides of a rhombus have equal length, so we can eliminate three choices by demonstrating that at least two sidelengths are not equal.

\(\displaystyle 0.1\textrm{ km}, \textup{1,000}\textrm{ m}, \textup{1,000}\textrm{ m}, 1\textrm{ hm}\):

1,000 meters is, by definition, equal to 1 kilometer, not 0.1 kilometers. Therefore,

\(\displaystyle 0.1\textrm{ km} \neq \textup{1,000}\textrm{ m}\)

and this choice is incorrect.

\(\displaystyle 1\textrm{ mi}, \textrm{1 mi}, 1,760 \textrm{ ft}, 586 \frac{2}{3} \textrm{ yd}\):

1 mile is, by definition, equal to 5,280 feet, not 1,760 feet. Therefore,

\(\displaystyle 1\textrm{ mi} \neq 1,760\)

and this choice is incorrect.

\(\displaystyle 10 \textrm{ cm}, 1\textrm{ dm}, 0.01 \textrm{ m}, 0.01 \textrm{ m}\)

By definition, 1 decimeter, not 0.1 decimeter, is equal to 1 meter. Therefore,

\(\displaystyle 1 \textrm{ dm} \neq 0.01 \textrm{ m}\)

and this choice is incorrect.

\(\displaystyle 18\textrm{ in}, 18\textrm{ in}, 1\frac{1}{2} \textrm{ ft}, \frac{1}{2} \textrm{ yd}\):

\(\displaystyle \frac{1}{2}\) yard is equal to \(\displaystyle \frac{1}{2} \times 36 = 18\) inches and, also, \(\displaystyle \frac{1}{2} \times 3 = 1\frac{1}{2}\) feet. Therefore,

\(\displaystyle 18\textrm{ in} = 18\textrm{ in} = 1\frac{1}{2} \textrm{ ft}= \frac{1}{2} \textrm{ yd}\)

All four sides have equal length so this is the rhombus. This is the correct choice.

A polygon with six sides is called a hexagon.

By definition, any polygon with four sides is called a quadrilateral. All three figures fit this description.

A four-sided figure, or quadrilateral, with one pair of parallel sides and its other sides nonparallel is called a trapezoid.

Polygon \(\displaystyle EFCD\) has four sides and is therefore a quadrilateral. \(\displaystyle \overleftrightarrow{AE} ||\overleftrightarrow{BD}\), so \(\displaystyle \overline{FE }|| \overline{CD}\). Also, since \(\displaystyle \angle FCD\) is acute and \(\displaystyle \angle CDE\) is right, \(\displaystyle \angle FCD \ncong \angle CDE\), so \(\displaystyle \overline{FC } \nparallel \overline{DE}\). 

The quadrilateral has one pair of parallel sides, and the other two sides are not parallel. Therefore, it is a trapezoid.

A parallelogram, by definition, has two pairs of parallel sides. Figures A and B fit that criterion, but Figure C does not.

The length of the garden is \(\displaystyle 2x\) than that of the entire lot, or 

\(\displaystyle L = 100 - 2x\).

The width of the garden is \(\displaystyle 2x\) than that of the entire lot, or 

\(\displaystyle W = 60 - 2x\).

The perimeter is twice the sum of the two:

\(\displaystyle P = 2 (L + W)\)

\(\displaystyle = 2 (100-2x+60-2x) = 2\left ( 160-4x\right )= 320-8x\)

The four sides of a rhombus are congruent. Also, the diagonals of a rhombus are perpendicular bisectors to each other, so the four triangles they form are right triangles. Therefore, the Pythagorean theorem can be used to determine the common sidelength of Quadrilateral \(\displaystyle ABCD\).

We focus on \(\displaystyle \Delta AXB\). The diagonals of a rhombus, as is the case with any parallelogram, are each the other's bisector, so 

\(\displaystyle AX = \frac{1}{2} (AC) = \frac{1}{2} \cdot 24 = 12\)

\(\displaystyle XB = \frac{1}{2} (BD) = \frac{1}{2} \cdot 10 = 5\)

By the Pythagorean Theorem, 

\(\displaystyle AB = \sqrt{(AX)^{2}+ (XB)^{2}} = \sqrt{12^{2}+5^{2}} = \sqrt{144+25} = \sqrt{169} = 13\)

13 is the common length of the four sides of Quadrilateral \(\displaystyle ABCD\), so its perimeter is \(\displaystyle 4 \times 13 = 52\).

The perimeter of Rectangle \(\displaystyle ABED\) is

\(\displaystyle P = AB +BE + ED + AD\).

Opposite sides of a rectangle are congruent, so 

\(\displaystyle AB= ED = 14, AD = BE = 7\)

and 

\(\displaystyle P = 14 + 7 + 14 + 7 = 42\).

The perimeter of Rectangle \(\displaystyle ACGF\) is 

\(\displaystyle P = AC + CG + GF + AF\).

Opposite sides of a rectangle are congruent, so

\(\displaystyle GF = AC = AB + BC = 14 + 6 = 20\),

\(\displaystyle CG = AF = AD + DF = 7 + 3 = 10\),

and

\(\displaystyle P = 20 + 10 + 20 + 10 = 60\).

The ratio of the perimeters is

\(\displaystyle \frac{60}{42} = \frac{60\div 6}{42\div 6} =\frac{10}{7}\) - that is, 10 to 7.