To solve, simply realize that the radius is half the diameter. Thus, our answer is 12.

If two chords of a circle intersect, the measure of the angles they form is equal to half the sum of the measures of the angles they intercept that is, 

\(\displaystyle \frac{1}{2} \left ( m \overarc {AD}+ m \overarc {BC} \right ) = m \angle AXD\)

The ratio of the measures of arcs on the same circle is equal to that of their lengths, so

\(\displaystyle \frac{ m \overarc {BC} }{m \overarc {AD} } = \frac{18 \pi}{9 \pi} = 2\)

and 

\(\displaystyle m \overarc {BC} = 2 \cdot m \overarc {AD}\)

Substituting:

\(\displaystyle \frac{1}{2} \left ( m \overarc {AD}+ 2 \cdot m \overarc {AD} \right ) =72^{\circ }\)

\(\displaystyle \frac{1}{2} \left ( 3 \cdot m \overarc {AD} \right ) =72^{\circ }\)

\(\displaystyle \frac{3}{2} \cdot m \overarc {AD} =72^{\circ }\)

\(\displaystyle \frac{2} {3}\cdot \frac{3}{2} \cdot m \overarc {AD} =\frac{2} {3}\cdot 72^{\circ }\)

\(\displaystyle m \overarc {AD} =48^{\circ }\)

 if \(\displaystyle C\) is the circumference of the circle, and the length of the arc \(\displaystyle \overarc {AD}\) is \(\displaystyle l\),

\(\displaystyle l = \frac{m \overarc{AC}}{360 ^{\circ }} \cdot C\)

\(\displaystyle \overarc {AD}\) has length \(\displaystyle 9 \pi\) and measure \(\displaystyle 48^{\circ }\) so

\(\displaystyle 9 \pi = \frac{48^{\circ }}{360 ^{\circ }} \cdot C\)

or

\(\displaystyle 9 \pi = \frac{2 }{15 } \cdot C\)

Since, if the radius is \(\displaystyle r\), \(\displaystyle C = 2 \pi r\),

\(\displaystyle 9 \pi = \frac{2 }{15 } \cdot 2 \pi r\)

\(\displaystyle 9 \pi = \frac{4 \pi }{15 } r\)

Solve for \(\displaystyle r\):

\(\displaystyle \frac{4 \pi }{15 } r = 9 \pi\)

\(\displaystyle \frac{15 } {4 \pi } \cdot \frac{4 \pi }{15 } r =\frac{15 } {4 \pi } \cdot 9 \pi\)

\(\displaystyle r =\frac{135} {4 } = 33\frac{3}{4}\)

The ratio of the degree measures of the arcs is the same as that of their lengths. Therefore, 

\(\displaystyle \frac{m \overarc {BT}}{m \overarc {AT}} = \frac{75 \pi}{20 \pi} = \frac{15}{4}\)

and 

\(\displaystyle m \overarc {BT}= \frac{15}{4} \cdot m \overarc {AT}\)

If a secant and a tangent are drawn to a circle from a point outside the circle, the measure of the angle they form is equal to half the difference of the measures of their intercepted arcs; therefore, 

\(\displaystyle \frac{1}{2} \left ( m\overarc {BT} - m \overarc {AT} \right ) = m \angle BNT\)

Substituting:

\(\displaystyle \frac{1}{2} \left ( m\overarc {BT} - m \overarc {AT} \right ) = 50 ^{\circ }\)

\(\displaystyle \frac{1}{2} \left ( \frac{15}{4} \cdot m \overarc {AT} - m \overarc {AT} \right ) = 50 ^{\circ }\)

\(\displaystyle \frac{11}{8} \cdot m \overarc {AT} = 50 ^{\circ }\)

\(\displaystyle \frac{8}{11} \cdot \frac{11}{8} \cdot m \overarc {AT} =\frac{8}{11} \cdot 50 ^{\circ }\)

\(\displaystyle m \overarc {AT} = 36 \frac{4}{11}^{\circ }\)

Since \(\displaystyle \overarc {AT}\) has length \(\displaystyle 20 \pi\), then, if we let \(\displaystyle C\) be the circumference of the circle,

\(\displaystyle \frac{C}{360^{\circ }} = \frac{20 \pi} {36\frac{4}{11}^{\circ } }\)

\(\displaystyle \frac{C}{360^{\circ }} \cdot 360^{\circ }= \frac{20 \pi} {36\frac{4}{11}^{\circ } }\cdot 360^{\circ }\)

\(\displaystyle C=198 \pi\)

Divide the circumference by \(\displaystyle 2 \pi\) to obtain the radius:

\(\displaystyle r = \frac{C}{2 \pi} = \frac{198 \pi }{2 \pi} = 99\)

Convert fifteen yards to inches by multiplying by 36:

\(\displaystyle 15 \textrm{ yd} \times 36 \textrm{ in/yd} = 540 \textrm{ in}\)

The radius of a circle is one half its diameter, so multiply this by \(\displaystyle \frac{1}{2}\):

\(\displaystyle \frac{1}{2} \times 540 \textrm{ in} = 270 \textrm{ in}\)

Call \(\displaystyle t = m \overarc {TU}\). The measure of the corresponding major arc is  \(\displaystyle 360 ^{\circ } - m \overarc {TU}= 360 ^{\circ } -t\)

If two tangents are drawn to a circle from a point outside the circle, the measure of the angle they form is equal to half the difference of the measures of their intercepted arcs; therefore

\(\displaystyle \frac{1}{2} \left (360 ^{\circ } -t -t \right ) = m \angle BNT\)

Substituting:

\(\displaystyle \frac{1}{2} \left (360 ^{\circ } -2t\right ) = 66 ^{\circ }\)

\(\displaystyle 180 ^{\circ } -t = 66 ^{\circ }\)

\(\displaystyle 180 ^{\circ } -t + t - 66 ^{\circ } = 66 ^{\circ } + t - 66 ^{\circ }\)

\(\displaystyle 114 ^{\circ } = t\)

Therefore, \(\displaystyle m \overarc {TU} = 114 ^{\circ }\). Since \(\displaystyle \overarc {TU}\) has length \(\displaystyle 30 \pi\), it follows that if \(\displaystyle C\) is the circumference of the circle, 

\(\displaystyle \frac{C}{360^{\circ }} = \frac{30 \pi }{m \overarc{TU}}\)

\(\displaystyle \frac{C}{360^{\circ }} = \frac{30 \pi }{114^{\circ }}\)

\(\displaystyle C= \frac{30 \pi }{114^{\circ }} \cdot 360^{\circ }\)

\(\displaystyle C = 94\frac{14}{19} \pi\)

Divide the circumference by \(\displaystyle 2 \pi\) to obtain the radius:

\(\displaystyle r = \frac{C}{2 \pi} = \frac{94 \frac{14}{19}\pi }{2 \pi} = 47 \frac{7}{19}\).

This makes 47 the correct choice.

IF you draw the longest diagonal across the shape, the length of it is 13.4. This means the radius must be at least 6.7. The answer is 7.

Julie runs for nine minutes, or \(\displaystyle \frac{9}{60} = \frac{3}{20}\) hour; she runs nine miles per hour. Setting \(\displaystyle t = \frac{3}{20}\) and \(\displaystyle r = 9\) in the rate formula, we can evaluate distance in miles:

\(\displaystyle d = rt = 9 \cdot \frac{3}{20} = \frac{27}{20 }\)

Julie runs \(\displaystyle \frac{27}{20 }\) miles, which converts to feet by multiplication by 5,280 feet per mile:

\(\displaystyle \frac{27}{20 } \times 5,280 = 7,128\) feet.

Each side of the octagonal track measures 264 feet, so Julie runs

\(\displaystyle 7,128 \div 264 = 27\)

sides of the track; this is equivalent to running the entire track three times, then three more sides. She is running clockwise, so three more sides from Point A puts her at Point D. This is the correct response.

 One mile comprises 5,280 feet; the perimeter of the track, two-fifths of a mile, is equal to

\(\displaystyle \frac{2}{5} \times 5,280 = 2,112\) feet.

Each (congruent) side of the octagonal track measures one-eighth of this,

\(\displaystyle \frac{1}{8} \times 2,112 = 264\) feet.

By running clockwise from Point A to halfway between Point E and Point F, Boris runs along four and one half sides, each of which has this length, for a total running distance of

\(\displaystyle 4 \frac{1}{2} \times 264 = 1,188\) feet. 

Of the five responses, 1,200 comes closest.

This question is very misleading, because while each answer COULD be true, none of them MUST be true. Between angle A and C, onne of the angles could be very small (0.001 degrees) and the other one could be very large. For instance, if A = 89.9999 and C = 0.0001, AC = 0.009. On the other hand, the two angles could be very siimilar. If B = 90 and D = 90 then BD = 8100 and BD > AC. If we use these same values we disprove AD = BC as 8100 ≠ .009. Finally, if B is a very small value, then B/C will be very small and smaller than A/D.

If angle A is one of the base angles, then the other base angle must measure 50 degrees. Since 50 + 50 + x = 180 means x = 80, the vertex angle must measure 80 degrees.

If angle A is the vertex angle, the two base angles must be equal. Since 50 + x + x = 180 means x = 65, the two base angles must measure 65 degrees.

The only number given that is not possible is 95 degrees.