\(\displaystyle \bigtriangleup AXB\) is a right triangle whose hypotenuse \(\displaystyle \overline{AB}\) has length \(\displaystyle \sqrt{2}\) times that of leg \(\displaystyle \overline{AX}\). This is characteristic of a triangle whose acute angles both have measure \(\displaystyle 45 ^{\circ }\) -and consequently, whose acute angles are congruent. Therefore,

\(\displaystyle \angle A \cong \angle B\)

These inscribed angles being congruent, the arcs they intercept, \(\displaystyle \overarc{AD}\) and \(\displaystyle \overarc{BC}\), are also congruent.

\(\displaystyle \angle A\) is an inscribed angle, so its degree measure is half that of the arc it intercepts, \(\displaystyle \overarc {BC }\):

\(\displaystyle m \angle A = \frac{1}{2} \cdot m \overarc {BC } = \frac{1}{2} \cdot 100^{\circ } = 50^{\circ }\).

\(\displaystyle \angle A\) and \(\displaystyle \angle B\) are acute angles of right triangle \(\displaystyle \bigtriangleup AXB\). They are therefore complimentary - that is, their degree measures total \(\displaystyle 90^{\circ }\). Consequently,

\(\displaystyle m \angle B+ m \angle A = 90^{\circ }\)

\(\displaystyle m \angle B+ 50 ^{\circ } = 90^{\circ }\)

\(\displaystyle m \angle B+ 50 ^{\circ }- 50 ^{\circ } = 90^{\circ } - 50 ^{\circ }\)

\(\displaystyle m \angle B = 40 ^{\circ }\)

\(\displaystyle m \angle A > m \angle B\).

The measures of the angles of a quadrilateral have sum \(\displaystyle 360^{\circ }\). If \(\displaystyle x\) is the measure of the unknown angle, then:

\(\displaystyle x + 100 + 105 + 110 = 360\)

\(\displaystyle x + 315= 360\)

\(\displaystyle x + 315-315= 360-315\)

\(\displaystyle x = 45\)

The measure of the fourth angle is \(\displaystyle 45^{\circ }\).

A quadrilateral has four angles totalling \(\displaystyle 360^{\circ}\). So, first add up the three angles given. The sum is \(\displaystyle 219^{\circ}\). Then, subtract that from 360. This gives you the missing angle, which is \(\displaystyle 141^{\circ}\).

The sum of the measures of the angles of a quadrilateral is \(\displaystyle 360^{\circ }\); the sum of the measures if a pentagon is \(\displaystyle 540^{\circ }\). Therefore, 

\(\displaystyle x + 80 + 80 + 80 = 360\)

\(\displaystyle x + 240 = 360\)

\(\displaystyle x = 120\)

and 

\(\displaystyle y + 100+ 100+ 100+ 100 = 540\)

\(\displaystyle y + 4 00 = 540\)

\(\displaystyle y = 140\)

\(\displaystyle y > x\), so (B) is greater.

The measures of the angles of a quadrilateral have sum \(\displaystyle 360^{\circ }\). If \(\displaystyle x\) is the measure of the unknown angle, then:

\(\displaystyle x + 100 + 105 + 110 = 360\)

\(\displaystyle x + 315= 360\)

\(\displaystyle x + 315-315= 360-315\)

\(\displaystyle x = 45\)

The measure of the fourth angle is \(\displaystyle 45^{\circ }\).

(a) One meter is equal to 100 centimeters; a square with this sidelength has perimeter \(\displaystyle 100 \times 4 = 400\) centimeters.

(b) A regular pentagon has five congruent sides; since the sidelength is 75 centimeters, the perimeter is \(\displaystyle 75 \times 5 = 375\) centimeters.

This makes (a) greater.

Let \(\displaystyle s\) be the sidelength of Square 1. Then the length of a diagonal of this square - which is \(\displaystyle \sqrt{2}\) times this sidelength, or \(\displaystyle s \sqrt{2}\) by the \(\displaystyle 45^{\circ }-45^{\circ }-90^{\circ }\) Theorem - is the same as the diameter of this circle, which, in turn, is equal to the sidelength of Square 2. 

Since the perimeter of a square is four times its sidelength, Square 1 has perimeter \(\displaystyle 4s\); Square 2 has perimeter \(\displaystyle 4 s \sqrt{2}\), which is \(\displaystyle \sqrt{2}\) times the perimeter of Square 1. \(\displaystyle \sqrt{2} < 2\), making the perimeter of Square 2 less than twice than the perimeter of Square 1.

The perimeters of the squares are 

\(\displaystyle 4 \times 1 = 4\) feet

\(\displaystyle 4 \times 2 = 8\) feet

\(\displaystyle 4 \times 3 = 12\) feet

\(\displaystyle 4 \times 4 = 16\) feet

\(\displaystyle 4 \times 5 = 20\) feet

The mean of the perimeters is their sum divided by five;

\(\displaystyle (4+8+12+16+20) \div 5 = 60 \div 5 = 12\) feet.

The median of the perimeters is the value in the middle when they are arranged in ascending order; this can be seen to also be 12 feet.

The quantities are equal.

First find the perimeters of the squares:

\(\displaystyle 4 \times 100 = 400\) centimeters (one meter being 100 centimeters)

\(\displaystyle 4 \times 100 = 400\) centimeters

\(\displaystyle 4 \times 120 = 480\) centimeters

\(\displaystyle 4 \times 140 = 560\) centimeters

The mean of the perimeters is their sum divided by four:

\(\displaystyle (400+400+480+560) \div 4 = 1,840 \div 4 = 460\) feet.

The median of the perimeters is the mean of the two values in the middle, assuming the values are in numerical order:

\(\displaystyle (400 + 480) \div 2 = 880 \div 2 = 440\)

The mean, (A), is greater.