Free Math Worksheets

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

- 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 \(s=\frac{\theta}{360^\circ}\cdot2\pi r\). 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 \(A=\frac{1}{2}sr\): \(A=\frac{1}{2}\cdot12\,\text{cm}\cdot15\,\text{cm}=90\,\text{cm}^2\).

\(\theta \approx 45.84^\circ\) 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? - Do not omit the two straight radii when finding perimeter.

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) \(r \approx 10.00\,\text{cm}\) b) \(P \approx 32.57\,\text{cm}\)

- 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 \(180\,\text{ft}^2\).

- Use the radius and arc length directly in the sector-area formula. - For the angle, compare the arc length with the full circumference. - Rearrange the arc-length formula to isolate the central angle.

1. The radius and arc length are both \(10\,\text{cm}\). 2. The sector area is \(A=\frac{1}{2}sr=\frac{1}{2}\cdot10\,\text{cm}\cdot10\,\text{cm}=50\,\text{cm}^2\). 3. Use \(s=\frac{\theta}{360^\circ}\cdot2\pi r\) and solve for \(\theta\). 4. Then \(\theta=\frac{s\cdot360^\circ}{2\pi r}=\frac{10\cdot360^\circ}{2\pi\cdot10}=\frac{180^\circ}{\pi}\approx57.3^\circ\).

\(A=50\,\text{cm}^2\) and \(\theta\approx57.3^\circ\)

- 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 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 \(5.0\,\text{cm}\).

- Start with the sector-area formula that uses the central angle. - How does the sector's fraction of the circle relate to its angle? - Use the area, arc length, and radius in the direct relationship to check your result.

1. Use \(A=\frac{\theta}{360^\circ}\pi r^2\): \(25=\frac{\theta}{360^\circ}\pi(5)^2\). 2. Solving gives \(\theta=\frac{360^\circ}{\pi}\approx114.6^\circ\). 3. The arc length is \(s=\frac{\theta}{360^\circ}\cdot2\pi r\). Using the exact angle, \(s=\frac{360/\pi}{360}\cdot2\pi(5\,\text{cm})=10\,\text{cm}\). 4. Check: \(\frac{1}{2}sr=\frac{1}{2}\cdot10\,\text{cm}\cdot5\,\text{cm}=25\,\text{cm}^2\), which matches the given area.

a) \(\theta\approx114.6^\circ\) 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 formula to find the angle. - Rearrange each formula before substituting.

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 \(\theta\approx45.84^\circ\)

- Think of the swept region as starting away from the pivot. - Model the cleaned area as the difference of two sectors. - For part c, how is sector area related to the angle when the radii stay fixed?

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

a) \(A\approx605.98\,\text{in.}^2\) b) \(s\approx59.52\,\text{in.}\) c) Approximately \(16.1\%\)

- 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 \(r = \frac{3}{\sin(25^{\circ})} \approx 7.10\,\text{cm}\). 2. Use the unrounded radius in the arc-length formula: \(s = \frac{50}{360}\cdot 2\pi r \approx \frac{50}{360}\cdot 2\pi \cdot 7.0986 \approx 6.19\,\text{cm}\).

Radius: \(r \approx 7.10\,\text{cm}\) Minor arc length: \(s \approx 6.19\,\text{cm}\)