To find the area of a sector, you need to find a percentage of the total area of the circle.  You do this by dividing the sector angle by the total number of degrees in a full circle (i.e. \(\displaystyle 360\)˚).  Thus, for our circle, which has a sector with an angle of \(\displaystyle 81\)˚, we have a percentage of:

\(\displaystyle \frac{81}{360}\)

Now, we will multiply this by the total area of the circle.  Recall that we find such an area according to the equation:

\(\displaystyle A = \pi * r^2\)

For our problem, \(\displaystyle r=12.3\)

Therefore, our equation is:

 \(\displaystyle \frac{81}{360} * \pi * 12.3^2 = \frac{12254.49\pi}{360}\)

Using your calculator, you can determine that this is approximately \(\displaystyle 106.94\).

To find the area of a sector, you need to find a percentage of the total area of the circle.  You do this by dividing the sector angle by the total number of degrees in a full circle (i.e. \(\displaystyle 360\)˚).  Thus, for our circle, which has a sector with an angle of \(\displaystyle 122\)˚, we have a percentage of:

\(\displaystyle \frac{122}{360}\)

Now, we will multiply this by the total area of the circle.  Recall that we find such an area according to the equation:

\(\displaystyle A = \pi * r^2\)

For our problem, \(\displaystyle r=9.12\)

Therefore, our equation is:

 \(\displaystyle \frac{122}{360} * \pi * 9.12^2 = \frac{10147.2768\pi}{360}\)

Using your calculator, you can determine that this is approximately \(\displaystyle 88.55\).

\(\displaystyle \bigtriangleup ABC\) has two angles of degree measure 45; the third angle must measure 90 degrees, making \(\displaystyle \bigtriangleup ABC\) a right triangle.

For the sake of simplicity, let \(\displaystyle BC = 1\); the reasoning is independent of the actual length. The legs of a 45-45-90 triangle are congruent, so \(\displaystyle AB = 1\). The area of a right triangle is half the product of its legs, so 

\(\displaystyle A = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2} = 0.5\)

Also, if \(\displaystyle BC = 1\), then the orange semicircle has diameter 1 and radius \(\displaystyle \frac{1}{2}\). Its area can be found by substituting \(\displaystyle r = \frac{1}{2}\) in the formula:

\(\displaystyle A = \frac{1}{2} \cdot \pi r^{2}\)

\(\displaystyle = \frac{1}{2} \cdot \pi \left ( \frac{1}{2} \right ) ^{2}\)

\(\displaystyle = \frac{1}{2} \cdot \pi \cdot \left ( \frac{1}{4 } \right )\)

\(\displaystyle = \frac{1}{8} \cdot \pi\)

\(\displaystyle \approx 0. 125 \cdot 3.14\)

\(\displaystyle \approx 0. 3925\)

\(\displaystyle \bigtriangleup ABC\) has a greater area than the orange semicircle.

\(\displaystyle \bigtriangleup ABC\) has two angles of degree measure 60; its third angle must also have measure 60, making \(\displaystyle \bigtriangleup ABC\) an equilateral triangle 

For the sake of simplicity, let \(\displaystyle BC = 1\); the reasoning is independent of the actual length. The area of equilateral \(\displaystyle \bigtriangleup ABC\) can be found by substituting \(\displaystyle s = 1\) in the formula

\(\displaystyle A = \frac{s^{2}\sqrt{3}}{4}\)

\(\displaystyle = \frac{1^{2}\sqrt{3}}{4}\)

\(\displaystyle = \frac{1 \cdot \sqrt{3}}{4}\)

\(\displaystyle \approx \frac{1.7}{4}\)

\(\displaystyle \approx 0.425\)

Also, if \(\displaystyle BC = 1\), then the orange semicircle has diameter 1 and radius \(\displaystyle \frac{1}{2}\). Its area can be found by substituting \(\displaystyle r = \frac{1}{2}\) in the formula:

\(\displaystyle A = \frac{1}{2} \cdot \pi r^{2}\)

\(\displaystyle = \frac{1}{2} \cdot \pi \left ( \frac{1}{2} \right ) ^{2}\)

\(\displaystyle = \frac{1}{2} \cdot \pi \cdot \left ( \frac{1}{4 } \right )\)

\(\displaystyle = \frac{1}{8} \cdot \pi\)

\(\displaystyle \approx 0. 125 \cdot 3.14\)

\(\displaystyle \approx 0. 3925\)

\(\displaystyle \bigtriangleup ABC\) has a greater area than the orange semicircle.

The radius of a circle is half its diameter; the radius of the circle in the diagram is half of 20, or 10.

To find the area of the circle, set \(\displaystyle r = 10\) in the area formula:

\(\displaystyle A = \pi r ^{2} = \pi \cdot 10 ^{2} = 100 \pi\)

The circle is divided into sixteen sectors of equal size, five of which are shaded; the shaded portion is

\(\displaystyle \frac{5}{16} \cdot 100 \pi = \frac{500}{16} \pi = \frac{125 \pi }{4 }\).

This question is asking for the length of an arc corresponding to \(\displaystyle \frac{58}{60}\) of a circle with radius eight feet. The question can be answered by evaluating for \(\displaystyle r=8\):

\(\displaystyle \frac{58}{60} \cdot 2 \pi r=\frac{58}{60} \cdot 2 \pi \cdot 8 \approx 48.6\)

To compare \(\displaystyle m \widehat{ABC}\) and \(\displaystyle m \widehat{CAB}\), we note that

\(\displaystyle m \widehat{ABC} = m \widehat{AB} + m \widehat{BC}\)

and 

\(\displaystyle m \widehat{CAB} = m \widehat{AB} + m \widehat{AC}\)

We need to be able to compare \(\displaystyle m \widehat{BC}\) and \(\displaystyle m \widehat{AC}\). If we know which of the intercepting angles is the greater, then we know which of the arcs is greater. The intercepting angles are \(\displaystyle \angle A , \angle B\), respectively. However, we are not given this relationship. 

Every hour, the tip of the minute hand travels the circumference of a circle with radius six feet, which is

\(\displaystyle C = 2 \pi r= 2 \pi \cdot 6 = 12 \pi\) feet.

Since it is now 3:50 PM, the minute hand made three complete revolutions since noon, plus \(\displaystyle \frac{50}{60 } = \frac{5}{6}\) of a fourth, so its tip has traveled this circumference \(\displaystyle 3\frac{5}{6}\) times. 

This is

\(\displaystyle 3\frac{5}{6} \times 12 \pi = \frac{23}{6} \times 12 \pi = 46 \pi\) feet. This is 

\(\displaystyle 46 \pi \times 12 = 552 \pi\) inches.

The tip of a minute hand travels a circle whose radius is equal to the length of that minute hand, which, in this question, is seven feet long. The circumference of this circle is \(\displaystyle 2 \pi\) times the radius, or \(\displaystyle 2 \pi \times 7 = 14\pi\) feet; over the course of thrity minutes (or one-half of an hour) the tip of the minute hand covers half this distance, or \(\displaystyle \frac{1}{2} \times 14 \pi = 7 \pi\) feet.

The circumference of a circle seven feet in diameter is \(\displaystyle \pi\) times this diameter, or \(\displaystyle 7 \pi\) feet.

The quantities are equal.

Every hour, the tip of the minute hand travels the circumference of a circle, which here is 

\(\displaystyle C = 2 \pi r = 2 \pi \times 4 \frac{1}{2} = 9 \pi\) feet.

The minute hand has traveled \(\displaystyle 110 \pi\) feet since noon, so it has traveled the circumference of the circle

\(\displaystyle \frac{110 \pi}{9 \pi} = \frac{110 }{9 } =12 \frac{2}{9}\) times.

Since \(\displaystyle 12 \leq 12 \frac{2}{9} \leq 12 \frac{1}{2}\), between 12 and \(\displaystyle 12 \frac{1}{2}\) hours have elapsed since noon, and the time is between 12:00 midnight and 12:30 AM.