- Compare the intercepted arc length directly with the radius. - What ratio defines the radian measure of a central angle? - What does that ratio become when the two lengths are equal?

1. Radian measure is the ratio of intercepted arc length to radius: \(\theta=\frac{s}{r}\). 2. Here \(s=r\), so \(\theta=\frac{r}{r}=1\). 3. Therefore, \(\angle AOB\) measures exactly \(1\) radian.

\(1\) radian

- Start with the area of the whole circle. - What fraction of the full circle does the sector occupy? - Keep \(\pi\) in the answer rather than converting to a decimal.

1. The area of the full circle is \(\pi r^2=\pi(8)^2=64\pi\,\text{cm}^2\). 2. The sector is one quarter of the circle, so its area is \(\frac{1}{4}\cdot64\pi=16\pi\,\text{cm}^2\).

\(16\pi\,\text{cm}^2\)

- How many right-angle turns make a full turn? - What fraction of a circle corresponds to that turn?

1. The central angle is \(90^\circ\), which is one-fourth of \(360^\circ\). 2. The intercepted minor arc is therefore \(\frac{1}{4}\) of the circumference.

\(\frac{1}{4}\) of the circumference, because the central angle is \(90^\circ\), which is \(\frac{90}{360}=\frac{1}{4}\) of a full turn.

- Count the equal regions bounded by the radii. - How many of those regions do parts a) and b) cover?

1. The diagram divides the circle into four equal regions. 2. Sector \(AOB\) is one of these regions, so its area is \(\frac{1}{4}\) of the whole. 3. The two sectors from \(A\) through \(B\) to \(C\) make two of the four regions, or \(\frac{1}{2}\) of the whole.

a) \(\frac{1}{4}\), because sector \(AOB\) is one of four equal sectors. b) \(\frac{1}{2}\), because the two sectors cover two of the four equal parts.

- What does one radian describe geometrically? - Compare the given arc length with the radius.

1. One radian is the central angle that intercepts an arc as long as the radius. 2. Here \(s=r=5\,\text{cm}\), so the angle is \(1\) radian.

\(1\) radian, because the arc length equals the radius.

- Identify the given and unknown quantities. - Which formula connects sector area, radius, and arc length? - Rearrange the formula so the arc length is isolated.

1. Use \(A=\frac{1}{2}sr\). 2. Solve for arc length: \(s=\frac{2A}{r}\). 3. Substitute: \(s=\frac{2\cdot24.5\,\text{cm}^2}{7\,\text{cm}}=7\,\text{cm}\).

\(7\,\text{cm}\)

- What fraction of a full circle is represented by each central angle? - Check whether a radius or diameter is given. - Begin with the formula for the full circumference.

1. Use \(s=\frac{\theta}{360^\circ}\cdot2\pi r\). 2. For a), \(s=\frac{48}{360}\cdot2\pi\cdot7.5\,\text{cm}=2\pi\,\text{cm}\approx6.28\,\text{cm}\). 3. For b), the radius is \(r=\frac{1.20\,\text{m}}{2}=0.60\,\text{m}\). 4. Then \(s=\frac{210}{360}\cdot2\pi\cdot0.60\,\text{m}=0.7\pi\,\text{m}\approx2.20\,\text{m}\).

a) \(s\approx6.28\,\text{cm}\) b) \(s\approx2.20\,\text{m}\)

- First find the circumference of the entire pizza. - What fraction of the full circumference belongs to one of \(8\) equal slices? - You could also find the central angle of one slice.

1. The pizza's circumference is \(C=\pi(12\,\text{in.})=12\pi\,\text{in.}\). 2. Each of the \(8\) equal slices has one-eighth of the full circumference as its outer arc. 3. The arc length is \(s=\frac{12\pi}{8}\,\text{in.}=1.5\pi\,\text{in.}\approx4.7\,\text{in.}\).

Approximately \(4.7\,\text{in.}\)

- A half-turn is both \(180^\circ\) and \(\pi\) radians. - For each part, choose a conversion factor that cancels the starting unit. - Keep \(\pi\) exact in part a rather than replacing it with a decimal.

1. To convert degrees to radians, multiply by \(\frac{\pi}{180^\circ}\): \(150^\circ\cdot\frac{\pi}{180^\circ}=\frac{5\pi}{6}\). 2. To convert radians to degrees, multiply by \(\frac{180^\circ}{\pi}\): \(\frac{7\pi}{6}\cdot\frac{180^\circ}{\pi}=210^\circ\).

a) \(\frac{5\pi}{6}\) radians b) \(210^\circ\)

- Which quantities are given: arc length, radius, or angle? - Recall the direct relationship among \(s\), \(r\), and \(\theta\) when \(\theta\) is measured in radians. - Solve that relationship for the unknown angle before substituting values.

1. For an angle measured in radians, arc length satisfies \(s=r\theta\). 2. Solve for the angle: \(\theta=\frac{s}{r}=\frac{18}{12}=\frac{3}{2}\). 3. The wheel turns through \(\frac{3}{2}\) radians.

\(\frac{3}{2}\) radians

- What fraction of a full turn is the marked angle? - Find the whole circumference before taking that fraction. - Keep \(\pi\) in the exact answer.

1. The arc corresponds to \(\frac{120}{360}=\frac{1}{3}\) of the circumference. 2. With radius \(6\,\text{cm}\), the full circumference is \(2\pi\cdot6=12\pi\,\text{cm}\). 3. One-third of it is \(4\pi\,\text{cm}\).

\(4\pi\,\text{cm}\)

- Does the question ask for curved length or enclosed area? - What fraction of the whole circle does the marked central angle give? - Multiply the whole circle's area by that fraction.

1. The sector's central angle is \(135^\circ\), or \(\frac{135}{360}=\frac{3}{8}\) of a full turn. 2. The circle's area is \(\pi\cdot4^2=16\pi\,\text{cm}^2\). 3. The sector's area is \(\frac{3}{8}\cdot16\pi=6\pi\,\text{cm}^2\).

\(6\pi\,\text{cm}^2\)

- Keep the central-angle fraction unchanged in both sectors. - How does circumference depend on radius? - How does circle area depend on radius?

1. At a fixed central angle, arc length is a fixed fraction of \(2\pi r\), so doubling \(r\) doubles the arc length. 2. Sector area is the same fraction of \(\pi r^2\), so doubling \(r\) multiplies the area by \(2^2=4\). 3. Length depends on one factor of radius; area depends on two.

Arc length doubles because it is proportional to the radius when the central angle stays fixed. Sector area becomes four times as large because it is proportional to the radius squared.

- What portion of the full \(360^\circ\) turn is represented by the arc? - Does the matching sector cover the same part of a full circle?

1. The arc's central angle is \(\frac{3}{8}\cdot360^\circ=135^\circ\). 2. The sector uses the same central-angle fraction of the whole circle, so it has \(\frac{3}{8}\) of the area.

The central angle is \(\frac{3}{8}\cdot360^\circ=135^\circ\), and the sector contains \(\frac{3}{8}\) of the circle's area. Both fractions match because they come from the same fraction of a full turn.

- Compare the angle with a straight angle. - How many radians make a half-turn?

1. A half-turn is \(180^\circ=\pi\) radians. 2. The marked \(60^\circ\) angle is one-third of a half-turn, so it measures \(\frac{\pi}{3}\) radians.

\(\frac{\pi}{3}\) radians.

- What portion of the whole circle does each central angle cover? - Find the whole circumference and area before taking a portion. - Check whether the area and the radius-times-arc-length product change in the same way.

1. The circle's circumference is \(12\pi\,\text{cm}\), and its area is \(36\pi\,\text{cm}^2\). 2. A \(60^\circ\) sector takes one-sixth of both: \(s=2\pi\,\text{cm}\) and \(A=6\pi\,\text{cm}^2\). 3. A \(120^\circ\) sector takes one-third: \(s=4\pi\,\text{cm}\) and \(A=12\pi\,\text{cm}^2\). 4. In the first case \(rs=12\pi\,\text{cm}^2\), and in the second \(rs=24\pi\,\text{cm}^2\). The area is half of \(rs\) in both cases.

a) \(s=2\pi\,\text{cm}\), \(A=6\pi\,\text{cm}^2\) b) \(s=4\pi\,\text{cm}\), \(A=12\pi\,\text{cm}^2\) c) In both cases, \(A=\frac{1}{2}rs\).

- Which formula relates radius, arc length, and central angle? - Is there a sector-area formula that uses arc length directly? - What fraction of the full circumference is the given arc?

1. Use the arc-length formula \(s=\frac{\theta}{360^\circ}\cdot 2\pi r\). 2. Solve for the central angle: \(\theta=\frac{s\cdot360^\circ}{2\pi r}=\frac{12\cdot360^\circ}{2\pi\cdot15}=\frac{144^\circ}{\pi}\approx45.84^\circ\). 3. Use \(A=\frac{1}{2}sr\): \(A=\frac{1}{2}\cdot12\,\text{cm}\cdot15\,\text{cm}=90\,\text{cm}^2\).

\(\theta=\frac{144^\circ}{\pi}\approx45.84^\circ\) and \(A=90\,\text{cm}^2\)

- Compare how \(\theta\) and \(r\) appear in the two formulas. - Which variable is squared? - What happens to a product when one factor is multiplied by a constant?

1. The formulas are \(s=\frac{\theta}{360^\circ}\cdot2\pi r\) and \(A=\frac{\theta}{360^\circ}\cdot\pi r^2\). 2. In part a), \(\theta\) is a linear factor in both formulas. Tripling \(\theta\) triples both \(s\) and \(A\). 3. In part b), \(r\) is a linear factor in the arc-length formula, so doubling \(r\) doubles \(s\). 4. Radius is squared in the area formula, so doubling \(r\) multiplies \(A\) by \(2^2=4\).

a) Both \(s\) and \(A\) are multiplied by \(3\). b) \(s\) is multiplied by \(2\), and \(A\) is multiplied by \(4\).

- Rearrange the sector-area formula to isolate \(r\). - What pieces make up the boundary of a sector? - Keep the radius unrounded until the final perimeter calculation.

1. From \(A=\frac{\theta}{360^\circ}\pi r^2\), solve for radius: \(r=\sqrt{\frac{A\cdot360^\circ}{\theta\pi}}\). 2. Substitute the given values: \(r=\sqrt{\frac{62.83\cdot360}{72\pi}}\,\text{cm}\approx10.00\,\text{cm}\). 3. Using the unrounded radius, the arc length is \(s=\frac{72}{360}\cdot2\pi r\approx12.57\,\text{cm}\). 4. The sector perimeter is \(P=s+2r\approx32.57\,\text{cm}\).

a) \(r\approx10.00\,\text{cm}\) b) \(P\approx32.57\,\text{cm}\)

- First use the given radius and arc length to find the current area. - In \(A=\frac{1}{2}sr\), what happens when \(r\) doubles and \(s\) stays fixed? - Check the relationship by calculating the new area.

1. Use \(A=\frac{1}{2}sr\). 2. The current area is \(A=\frac{1}{2}\cdot15\,\text{ft}\cdot12\,\text{ft}=90\,\text{ft}^2\). 3. With the arc length fixed, area is proportional to radius in this formula. Doubling the radius doubles the area. 4. The new area is \(A=\frac{1}{2}\cdot15\,\text{ft}\cdot24\,\text{ft}=180\,\text{ft}^2\).

The original area is \(90\,\text{ft}^2\). With a \(24\,\text{ft}\) reach and the same arc length, the area doubles to \(180\,\text{ft}^2\).

- Compare the arc length with the radius. - How do radius and arc length determine a central angle in radians? - What formula relates a sector's area to its radius and arc length?

1. The radius and arc length are both \(10\,\text{cm}\). 2. The sector area is \(A=\frac{1}{2}rs=\frac{1}{2}\cdot10\cdot10=50\,\text{cm}^2\). 3. The angle in radians is \(\theta=\frac{s}{r}=\frac{10}{10}=1\). 4. In degrees, \(\theta=\frac{180^\circ}{\pi}\approx57.3^\circ\).

The area is \(50\,\text{cm}^2\). The angle is exactly \(1\) radian, or approximately \(57.3^\circ\).

- Rearrange the arc-length formula to isolate radius. - At a fixed radius, what happens to arc length when the angle increases? - How must two factors change inversely to keep their product constant?

1. From \(s=\frac{\theta}{360^\circ}\cdot2\pi r\), solve for radius: \(r=\frac{s\cdot360^\circ}{2\pi\theta}\). 2. For part a, \(r=\frac{15.7\cdot360}{2\pi\cdot90}\,\text{cm}=\frac{31.4}{\pi}\,\text{cm}\approx10.0\,\text{cm}\). 3. Arc length is proportional to the product \(r\theta\). To keep that product constant when \(\theta\) doubles, \(r\) must be divided by \(2\). 4. The new radius is approximately \(5.0\,\text{cm}\).

a) \(r\approx10.0\,\text{cm}\) b) The radius must be halved to approximately \(5.0\,\text{cm}\).

- Which sector-area formula uses a central angle measured in radians directly? - Once the angle is in radians, how are radius, angle, and arc length related? - Use the area, arc length, and radius in the direct relationship to check your result.

1. For a radian angle, use \(A=\frac{1}{2}r^2\theta\): \(25=\frac{1}{2}(5)^2\theta\). 2. Solving gives \(\theta=2\) radians. Converting to degrees gives \(2\cdot\frac{180^\circ}{\pi}=\frac{360^\circ}{\pi}\approx114.6^\circ\). 3. Use \(s=r\theta\): \(s=5\,\text{cm}\cdot2=10\,\text{cm}\). 4. Check: \(\frac{1}{2}sr=\frac{1}{2}\cdot10\,\text{cm}\cdot5\,\text{cm}=25\,\text{cm}^2\), matching the given area.

a) \(\theta=2\) radians \(=\frac{360^\circ}{\pi}\approx114.6^\circ\) b) \(s=10\,\text{cm}\) c) \(\frac{1}{2}\cdot10\,\text{cm}\cdot5\,\text{cm}=25\,\text{cm}^2\), so the relationship is verified.

- Use the central angle to determine each sector's fraction of a full circle. - Calculate area and arc length separately. - Compare the two results for each measurement.

1. Slice A has area \(A_A=\frac{45}{360}\pi(6\,\text{in.})^2=4.5\pi\,\text{in.}^2\approx14.14\,\text{in.}^2\). 2. Slice B has area \(A_B=\frac{30}{360}\pi(7\,\text{in.})^2=\frac{49\pi}{12}\,\text{in.}^2\approx12.83\,\text{in.}^2\). Therefore, Slice A has the greater area. 3. Slice A's arc length is \(s_A=\frac{45}{360}\cdot2\pi(6\,\text{in.})=1.5\pi\,\text{in.}\approx4.71\,\text{in.}\). 4. Slice B's arc length is \(s_B=\frac{30}{360}\cdot2\pi(7\,\text{in.})=\frac{7\pi}{6}\,\text{in.}\approx3.67\,\text{in.}\). Therefore, Slice A also has the longer crust.

a) Slice A, with approximately \(14.14\,\text{in.}^2\), has the greater area. b) Slice A, with an arc length of approximately \(4.71\,\text{in.}\), has the longer crust.

- Which formula relates sector area directly to arc length and radius? - After finding the radius, use the arc-length relationship to connect the arc to the central angle. - Keep the angle in exact form before finding its decimal approximation.

1. Use \(A=\frac{1}{2}sr\) and solve for radius: \(r=\frac{2A}{s}\). 2. Then \(r=\frac{2\cdot40\,\text{ft}^2}{8\,\text{ft}}=10\,\text{ft}\). 3. Use \(s=\frac{\theta}{360^\circ}\cdot2\pi r\) and solve for \(\theta\). 4. Thus \(\theta=\frac{8\cdot360^\circ}{2\pi\cdot10}=\frac{144^\circ}{\pi}\approx45.84^\circ\).

\(r=10\,\text{ft}\) and \(\theta=\frac{144^\circ}{\pi}\approx45.84^\circ\)

- Which two radii mark the inner and outer edges of the cleaned region? - What region near the pivot does the blade never touch? - For the percent increase, compare the change in sweep angle with the original angle.

1. The inner radius is \(6\,\text{in.}\), and the outer radius is \(6+16=22\,\text{in.}\). 2. The cleaned area is the difference between two sectors: \(A=\frac{155}{360}\pi(22^2-6^2)\approx605.98\,\text{in.}^2\). 3. The outer arc length is \(s=\frac{155}{360}\cdot2\pi\cdot22\approx59.52\,\text{in.}\). 4. With the radii fixed, area is proportional to the sweep angle. The percent increase is \(\frac{180-155}{155}\cdot100\%\approx16.1\%\).

a) \(605.98\,\text{in.}^2\) b) \(59.52\,\text{in.}\) c) \(16.1\%\)

- Connect the chord’s endpoints to the center to form an isosceles triangle. - Bisect the triangle and use right-triangle trigonometry to find the radius. - Then use the central angle as a fraction of \(360^\circ\) to find the arc length. - Use the unrounded radius in the arc-length calculation.

1. The two radii and the chord form an isosceles triangle. Bisecting it gives \(\sin(25^\circ)=\frac{3}{r}\), so \(r=\frac{3}{\sin(25^\circ)}\approx7.10\,\text{cm}\). 2. Use the unrounded radius in the arc-length formula: \(s=\frac{50}{360}\cdot2\pi r\approx\frac{50}{360}\cdot2\pi\cdot7.0986\approx6.19\,\text{cm}\).

Radius: \(r\approx7.10\,\text{cm}\) Minor arc length: \(s\approx6.19\,\text{cm}\)

- What is the dilation scale factor from the inner circle to the outer circle? - Under a dilation, how do lengths such as radii and arc lengths change? - Compare each arc length with its own radius before interpreting the common ratio.

1. The two sectors are related by a dilation centered at \(O\) with scale factor \(\frac{10}{4}=\frac{5}{2}\). 2. Arc length scales by the same factor, so the outer arc length is \(\frac{5}{2}\cdot\frac{5\pi}{3}=\frac{25\pi}{6}\,\text{cm}\). 3. For the inner arc, \(\frac{s}{r}=\frac{5\pi/3}{4}=\frac{5\pi}{12}\). For the outer arc, \(\frac{s}{r}=\frac{25\pi/6}{10}=\frac{5\pi}{12}\). 4. The ratio \(\frac{s}{r}\) is unchanged by the dilation. By definition, that common ratio is the angle measure in radians, so the marked angle is \(\frac{5\pi}{12}\) radians.

a) \(\frac{25\pi}{6}\,\text{cm}\) b) Both ratios equal \(\frac{5\pi}{12}\). c) \(\frac{5\pi}{12}\) radians

- Check what unit the angle must have in the formula \(A=\frac{1}{2}r^2\theta\). - Convert the given angle before substituting it. - Keep the converted angle exact so the final area can be written exactly in terms of \(\pi\).

1. The formula \(A=\frac{1}{2}r^2\theta\) requires \(\theta\) to be measured in radians, but Jordan substituted \(120\) as though the degree measure were a radian measure. 2. Convert the angle: \(120^\circ\cdot\frac{\pi}{180^\circ}=\frac{2\pi}{3}\) radians. 3. Substitute the radian measure: \(A=\frac{1}{2}(6)^2\left(\frac{2\pi}{3}\right)=12\pi\,\text{cm}^2\).

Jordan used a degree measure in a formula that requires radians. Since \(120^\circ=\frac{2\pi}{3}\) radians, the correct area is \(12\pi\,\text{cm}^2\).

- How are an arc and its sector related to the whole circle? - What stays the same when you compare the sector's share of the circumference and of the area? - Can you use one relationship to simplify the other?

1. Write \(f\) for the fraction of the circle used by the sector. 2. The arc length is \(s=f(2\pi r)\), so \(f=\frac{s}{2\pi r}\). 3. The sector area is \(A=f(\pi r^2)\). 4. Substitution gives \(A=\frac{s}{2\pi r}\cdot\pi r^2=\frac{1}{2}rs\).

Both the arc and sector use the same fraction \(f\) of the circle. Since \(s=f(2\pi r)\) and \(A=f(\pi r^2)\), replacing \(f\) gives \(A=\frac{1}{2}rs\).

- What fraction of the full circumference is represented by an arc of length \(s\)? - The sector occupies that same fraction of the circle's area. - After deriving the area formula, use the sector's two radii and arc to express its perimeter. - Check every algebraic solution against the geometric conditions for a sector.

1. The arc is the fraction \(\frac{s}{2\pi r}\) of the full circumference, so the sector is the same fraction of the circle's area. 2. Therefore, \(A=\frac{s}{2\pi r}\cdot\pi r^2=\frac{1}{2}rs\). 3. The sector perimeter is \(2r+s=26\), so \(s=26-2r\). 4. Use the area condition: \(\frac{1}{2}r(26-2r)=40\). This simplifies to \(r^2-13r+40=0\). 5. Factor: \((r-5)(r-8)=0\), so \(r=5\) or \(r=8\). 6. If \(r=5\), then \(s=16\). If \(r=8\), then \(s=10\). Both arc lengths are positive and less than the corresponding full circumferences, so both sectors are possible.

a) \(A=\frac{1}{2}rs\) b) \((r,s)=(5\,\text{cm},16\,\text{cm})\) or \((8\,\text{cm},10\,\text{cm})\)