If each of the five congruent sides has measure \(\displaystyle x\), then the other two sides have measures \(\displaystyle x + 250\) and \(\displaystyle \left (x + 250 \right ) +500 = x + 750\). Add the sides to get the perimeter, which is equal to \(\displaystyle 5,280\) feet, the solve for \(\displaystyle x\):

\(\displaystyle x + x + x + (x+250) + (x+ 750) = 5,280\)

\(\displaystyle 5x + 1,000 = 5,280\)

\(\displaystyle 5x + 1,000 - 1,000 = 5,280- 1,000\)

\(\displaystyle 5x = 4,280\)

\(\displaystyle 5x\div 5 = 4,280 \div 5\)

\(\displaystyle x = 856\) feet

Now we can compare (a) and (b).

(a) The longest side has measure \(\displaystyle x + 750= 856 + 750 = 1,606\) feet.

(b) The three shortest sides each have length 856 feet; twice this is \(\displaystyle 856 \times 2 = 1,712\) feet.

(b) is greater.

One yard is equal to 36 inches. A regular pentagon has five sides of equal length, so one side of the pentagon has length

\(\displaystyle 36 \div 5 = 7 \frac{1}{5}\) inches.

Since \(\displaystyle 7 \frac{1}{5} > 7\), (A) is greater.

A regular pentagon has five sides of the same length.

One foot is equal to twelve inches; since the sum of the lengths of three of the congruent sides is twelve inches, each side measures

\(\displaystyle 12 \div 3 = 4\) inches. 

The perimeter is 

\(\displaystyle 4 \times 5 = 20\) inches.

Let \(\displaystyle s\) be the length of one side of the hexagon. Then its perimeter is \(\displaystyle 6s\).

Each side of the pentagon is 20% greater than this length, or 

\(\displaystyle s + 0.20s = 1.20 s\).

The perimeter is five times this, or \(\displaystyle 5 \cdot 1.2s = 6s\).

The perimeters are the same.

Let \(\displaystyle s\) be the length of one side of the regular pentagon. Then its perimeter is \(\displaystyle 5s\).

The length of one side of the regular octagon is 60% of \(\displaystyle s\), or \(\displaystyle 0.6s\), so its perimeter is \(\displaystyle 8 \cdot 0.6s = 4.8 s\).The answer is therefore the percent \(\displaystyle 4.8s\) is of \(\displaystyle 5s\), which is

\(\displaystyle \frac{4.8s}{5s} \times 100 = 0.96 \times 100 = 96 \%\)

Let \(\displaystyle s\) be the length of one side of the pentagon. Then its perimeter is \(\displaystyle 5s\).

Each side of the hexagon is 20% less than this length, or 

\(\displaystyle s - 0.20s = 0.80 s\).

The perimeter is five times this, or \(\displaystyle 6 \cdot 0.8s = 4.8s\).

Since \(\displaystyle 5 > 4.8\) and \(\displaystyle s\) is positive, \(\displaystyle 5s > 4.8s\), so the pentagon has greater perimeter, and (A) is greater.

The angles of a pentagon measure a total of \(\displaystyle 180 (5-2) = 540\). From the information, we know that:

\(\displaystyle x + x + x + y + 2y = 540\)

\(\displaystyle 3x + 3y = 540\)

\(\displaystyle 3\left ( x + y \right )= 540\)

\(\displaystyle 3\left ( x + y \right )\div 3= 540\div 3\)

\(\displaystyle x + y = 180\)

making the two quantities equal.

The angles of a pentagon measure a total of \(\displaystyle 180 (5-2) = 540\). From the information given, we know that:

\(\displaystyle x + x + x + y + y = 540\)

\(\displaystyle 3x + 2y = 540\)

However, we cannot tell whether \(\displaystyle x\) or \(\displaystyle y\) is greater. For example, if \(\displaystyle x = 90\), then \(\displaystyle y = 135\); if \(\displaystyle x = 140\), then \(\displaystyle y = 60\). 

In both situations, the two adjacent sides and the diagonal form an isosceles triangle.

By the Isosceles Triangle Theorem,  \(\displaystyle m \angle BAC = m \angle BCA\) and  \(\displaystyle m \angle VUW = m \angle VWU\). Also, since the measures of the angles of a triangle total \(\displaystyle 180^{\circ }\), we know that 

\(\displaystyle m \angle BAC + m \angle BCA + m \angle ABC = 180\)

and 

\(\displaystyle m \angle VWU + m \angle VUW + m \angle UVW = 180\).

We can use these equations to compare \(\displaystyle m \angle BAC\) and \(\displaystyle m \angle VUW\).

(a) \(\displaystyle m \angle ABC = \frac{ 180 (5-2)}{5} = 108 ^{\circ }\)

\(\displaystyle m \angle BAC + m \angle BCA + m \angle ABC = 180 ^{\circ }\)

\(\displaystyle m \angle BAC + m \angle BAC + 108 ^{\circ } = 180 ^{\circ }\)

\(\displaystyle 2\cdot m \angle BAC + 108 ^{\circ }= 180 ^{\circ }\)

\(\displaystyle 2\cdot m \angle BAC + 108 ^{\circ } -108 ^{\circ } = 180 ^{\circ }-108 ^{\circ }\)

\(\displaystyle 2\cdot m \angle BAC = 72 ^{\circ }\)

\(\displaystyle 2 \cdot m \angle BAC \div 2 = 72 ^{\circ } \div 2\)

\(\displaystyle m \angle BAC = 36 ^{\circ }\)

(b) \(\displaystyle m \angle UVW = \frac{ 180 (6-2)}{6} = 120 ^{\circ }\)

\(\displaystyle m \angle VWU + m \angle VUW + m \angle UVW = 180 ^{\circ }\)

\(\displaystyle m \angle VUW + m \angle VUW + 120^{\circ } = 180 ^{\circ }\)

\(\displaystyle 2 \cdot m \angle VUW + 120^{\circ } = 180 ^{\circ }\)

\(\displaystyle 2 \cdot m \angle VUW + 120^{\circ } -120^{\circ } = 180 ^{\circ }-120^{\circ }\)

\(\displaystyle 2 \cdot m \angle VUW = 60^{\circ }\)

\(\displaystyle 2 \cdot m \angle VUW \div 2 = 60^{\circ }\div 2\)

\(\displaystyle m \angle VUW = 30^{\circ }\)

\(\displaystyle m \angle BAC > m \angle VUW\)

Each diagonal, along with two consecutive sides of its polygon, forms a triangle. All of the sides of the pentagon and the hexagon are congruent to one another, so between the two triangles, there are two pairs of two congruent corresponding sides:

\(\displaystyle \overline{AB} \cong \overline{ UV }\)

\(\displaystyle \overline{BC} \cong \overline{ VW }\)

Their included angles, \(\displaystyle \angle B\) and \(\displaystyle \angle V\), are interior angles of the pentagon and hexagon, respectively. The angle with greater measure will be opposite the longer side. We can use the Interior Angles Theorem to calculate the measures:

\(\displaystyle m\angle B =\frac{ 180 (5-2)}{5} =108 ^{\circ }\)

\(\displaystyle m\angle V =\frac{ 180 (6-2)}{6} =120 ^{\circ }\)

\(\displaystyle m\angle V > m\angle B \Rightarrow UW > AC\)