- Which circle formula uses the radius squared? - Keep \(\pi\) in the answer because an exact value is requested.

1. Use \(A=\pi r^2\). 2. Substitute \(r=5\,\text{cm}\): \(A=\pi(5)^2=25\pi\,\text{cm}^2\).

\(25\pi\,\text{cm}^2\)

- Look at where the segment starts and ends. - How many points on the circle are endpoints of a diameter? - Which name means center-to-edge distance?

1. The segment starts at the circle's center and ends on the circle. 2. Such a segment is a radius. A diameter would reach across the circle through its center from one edge to the other.

A radius. It connects the center to one point on the circle.

- Which measurement does the area formula square? - What is \(4\cdot4\)? - Are you calculating a length or an area?

1. Circle area uses the square of the radius: \(A=\pi r^2\). 2. Here \(r^2=4\cdot4=16\), so the area is \(16\pi\,\text{cm}^2\).

\(16\), because the radius is squared.

- Think about the regions covered by the two shapes. - Does the circle cover each corner of the square?

1. The circle lies completely inside the square. 2. Four corners of the square remain outside the circle, so the square has greater area.

The square has greater area because part of the square lies outside the circle.

- How are diameter and radius related? - Which formula gives the area of a circle? - Subtract the smaller area from the larger area.

1. The radii are \(5\,\text{in.}\) and \(8\,\text{in.}\). 2. The small pizza's area is \(A_s = \pi(5\,\text{in.})^2 = 25\pi\,\text{in.}^2 \approx 78.54\,\text{in.}^2\). 3. The large pizza's area is \(A_l = \pi(8\,\text{in.})^2 = 64\pi\,\text{in.}^2 \approx 201.06\,\text{in.}^2\). 4. The area difference is \(64\pi - 25\pi = 39\pi\,\text{in.}^2 \approx 122.52\,\text{in.}^2\).

Small pizza: approximately \(78.54\,\text{in.}^2\) Large pizza: approximately \(201.06\,\text{in.}^2\) Difference: approximately \(122.52\,\text{in.}^2\)

- Which measurement is squared in the formula for circle area? - Look at the number multiplying \(\pi\) in the area. - What positive number has that square?

1. The area of a circle is \(A=\pi r^2\). 2. From \(\pi r^2=64\pi\,\text{ft}^2\), the radius squared must be \(64\,\text{ft}^2\). 3. Since \(8^2=64\) and a radius is positive, \(r=8\,\text{ft}\).

\(8\,\text{ft}\).

- Which factor of the area is equal to the radius squared? - Match that number to a familiar whole-number square. - Which part asks for twice the radius rather than the radius itself?

1. For a), \(A=\pi r^2\) gives \(r^2=9\,\text{cm}^2\), so the positive radius is \(r=3\,\text{cm}\). 2. For b), \(r^2=100\,\text{m}^2\), so \(r=10\,\text{m}\). 3. The diameter is twice the radius: \(d=2\cdot10\,\text{m}=20\,\text{m}\).

a) \(3\,\text{cm}\). b) \(20\,\text{m}\).

- Which formula relates circumference directly to diameter? - How are radius and diameter related? - Which formula gives circle area?

1. From \(C=\pi d\), \(d=\frac{12}{\pi}\,\text{ft}\approx3.82\,\text{ft}\). 2. The radius is \(r=\frac{6}{\pi}\,\text{ft}\). 3. The area is \(A=\pi r^2=\frac{36}{\pi}\,\text{ft}^2\approx11.46\,\text{ft}^2\).

a) \(\frac{12}{\pi}\,\text{ft} \approx 3.82\,\text{ft}\) b) \(\frac{36}{\pi}\,\text{ft}^2 \approx 11.46\,\text{ft}^2\)

- Which circle formula helps you work backward from circumference? - Why should you keep the radius unrounded when you calculate the area? - In part b), first match the factor multiplying \(\pi\) to a familiar square.

1. For a), \(r=\frac{25.13}{2\pi}\,\text{cm}\approx4.0\,\text{cm}\). 2. Using the unrounded radius, \(A=\frac{25.13^2}{4\pi}\,\text{cm}^2\approx50.3\,\text{cm}^2\). 3. For b), \(A=\pi r^2=9\pi\,\text{cm}^2\), so \(r^2=9\,\text{cm}^2\) and \(r=3\,\text{cm}\). 4. Therefore, \(d=2r=6\,\text{cm}\) and \(C=2\pi r=6\pi\,\text{cm}\approx18.8\,\text{cm}\).

a) \(r\approx4.0\,\text{cm}\), \(A\approx50.3\,\text{cm}^2\). b) \(d=6\,\text{cm}\), \(C=6\pi\,\text{cm}\approx18.8\,\text{cm}\).

- The circle area tells you what the square of the radius must be. - How do you obtain the diameter from the radius? - Which formula uses the radius to find the length around the circle?

1. The area formula \(A=\pi r^2\) gives \(r^2=81\,\text{cm}^2\), so \(r=9\,\text{cm}\). 2. The diameter is \(d=2r=18\,\text{cm}\). 3. The circumference is \(C=2\pi r=18\pi\,\text{cm}\approx56.5\,\text{cm}\).

\(r=9\,\text{cm}\), \(d=18\,\text{cm}\), \(C=18\pi\,\text{cm}\approx56.5\,\text{cm}\).

- Find the square's area first. - Convert the circle's diameter to a radius. - Compare the two areas using the requested approximation only at the end.

1. The square pizza has area \((10\,\text{in.})^2=100\,\text{in.}^2\). 2. The circular pizza has radius \(6\,\text{in.}\), so its area is \(36\pi\,\text{in.}^2\approx113.10\,\text{in.}^2\). 3. Since \(113.10>100\), the circular pizza has the greater area.

The circular pizza has the greater area: \(36\pi\,\text{in.}^2\approx113.10\,\text{in.}^2\), compared with \(100\,\text{in.}^2\) for the square pizza.

- How are radius and diameter related? - Which formulas connect radius with circumference and area? - When circumference is given, find diameter or radius before area.

1. Circle A: \(d=9\,\text{cm}\), \(C\approx2\cdot3.14\cdot4.5\,\text{cm}=28.26\,\text{cm}\), and \(A\approx3.14(4.5)^2\,\text{cm}^2=63.59\,\text{cm}^2\). 2. Circle B: \(r=6\,\text{dm}\), \(C\approx3.14\cdot12\,\text{dm}=37.68\,\text{dm}\), and \(A\approx3.14(6)^2\,\text{dm}^2=113.04\,\text{dm}^2\). 3. Circle C: \(d\approx15.7\div3.14=5\,\text{m}\), so \(r=2.5\,\text{m}\). Then \(A\approx3.14(2.5)^2\,\text{m}^2=19.63\,\text{m}^2\).

a) \(d=9\,\text{cm}\), \(C\approx28.26\,\text{cm}\), \(A\approx63.59\,\text{cm}^2\) b) \(r=6\,\text{dm}\), \(C\approx37.68\,\text{dm}\), \(A\approx113.04\,\text{dm}^2\) c) \(r=2.5\,\text{m}\), \(d=5\,\text{m}\), \(A\approx19.63\,\text{m}^2\)

- How does the area formula connect an area to the radius squared? - Look for a familiar whole-number square after removing the factor \(\pi\). - Which formula gives the distance around the bed once you know its radius?

1. Use \(A=\pi r^2\). Since \(16\pi=\pi r^2\), the radius squared is \(16\). 2. The positive number whose square is \(16\) is \(4\), so \(r=4\,\text{ft}\). 3. The length around the bed is \(C=2\pi r=8\pi\,\text{ft}\approx25.13\,\text{ft}\).

Radius: \(4\,\text{ft}\). Edging length: \(8\pi\,\text{ft}\approx25.13\,\text{ft}\).

- For a), begin with the circumference formula to find the radius. - For b), which number has the square shown in the area after dividing by \(\pi\)? - Leave the radius unrounded until the last step of a calculation.

1. For a), first find the radius: \(r=\frac{C}{2\pi}=\frac{31.4}{2\pi}\,\text{cm}\approx4.997\,\text{cm}\). 2. Without rounding the radius first, \(A=\pi\left(\frac{31.4}{2\pi}\right)^2\,\text{cm}^2\approx78.5\,\text{cm}^2\). 3. For b), \(A=\pi r^2=36\pi\,\text{cm}^2\) means \(r^2=36\,\text{cm}^2\), so \(r=6\,\text{cm}\). 4. The circumference is \(C=2\pi r=12\pi\,\text{cm}\approx37.7\,\text{cm}\).

a) \(A\approx78.5\,\text{cm}^2\). b) \(C=12\pi\,\text{cm}\approx37.7\,\text{cm}\).

- Which formula gives the distance around a circle from its diameter? - How are radius and diameter related? - Keep \(\pi\) symbolic until the requested approximation.

1. The edging length is the circumference: \(C=16\pi\,\text{ft}\approx50.27\,\text{ft}\). 2. The radius is \(8\,\text{ft}\). 3. The area is \(A=64\pi\,\text{ft}^2\approx201.06\,\text{ft}^2\).

a) \(16\pi\,\text{ft}\approx50.27\,\text{ft}\) b) \(64\pi\,\text{ft}^2\approx201.06\,\text{ft}^2\)

- Which formula connects circle area and radius? - Divide by the approximation for \(\pi\) to find the radius squared. - Which positive whole number has the square you found? - How can that radius give the circumference?

1. Use the area formula: \(A=\pi r^2\). 2. Substitute the given values: \(200.96=3.14r^2\). 3. Divide to get \(r^2=200.96\div3.14=64\). 4. Because \(8^2=64\) and a radius is positive, \(r=8\,\text{cm}\). 5. Then \(C=2\pi r\approx2\cdot3.14\cdot8\,\text{cm}=50.24\,\text{cm}\).

\(50.24\,\text{cm}\).

- According to the formula, which operation should you perform first? - Square each diameter before continuing. - Use the hundredths digit to decide how to round to the nearest tenth. - Substitute each diameter into the formula separately.

1. For \(d=8\,\text{cm}\), \(A=\frac{3.14\cdot8^2}{4}=50.24\,\text{cm}^2\approx50.2\,\text{cm}^2\). 2. For \(d=20\,\text{cm}\), \(A=\frac{3.14\cdot20^2}{4}=314.0\,\text{cm}^2\). 3. For \(d=30\,\text{cm}\), \(A=\frac{3.14\cdot30^2}{4}=706.5\,\text{cm}^2\). 4. For \(d=44\,\text{cm}\), \(A=\frac{3.14\cdot44^2}{4}=1519.76\,\text{cm}^2\approx1519.8\,\text{cm}^2\).

The areas are \(50.2\,\text{cm}^2\), \(314.0\,\text{cm}^2\), \(706.5\,\text{cm}^2\), and \(1519.8\,\text{cm}^2\).

- Identify whether the labeled segment in the diagram is a radius or a diameter. - Use the circle-area formula with the radius read from the diagram.

1. The diagram shows a radius of \(6\,\text{cm}\). 2. Use \(A=\pi r^2\): \(A=\pi(6)^2=36\pi\,\text{cm}^2\). 3. Numerically, \(36\pi\approx113.1\,\text{cm}^2\).

a) \(36\pi\,\text{cm}^2\) b) Approximately \(113.1\,\text{cm}^2\)

- What does the exponent in \(A=\pi r^2\) mean? - Check Mia's number against the area of a square with side equal to the radius.

1. The area is \(A=\pi r^2=\pi\cdot5^2=25\pi\,\text{cm}^2\). 2. Mia doubled the radius instead of squaring it. Doubling gives \(10\), but squaring gives \(25\).

\(25\pi\,\text{cm}^2\). Mia used twice the radius instead of its square.

- Does the labeled line measure a radius or a diameter? - Which measurement belongs in the circle-area formula? - Leave \(\pi\) in your final answer.

1. The segment through the center is the diameter, \(10\,\text{cm}\). 2. The radius is \(10\div2=5\,\text{cm}\). 3. The area is \(A=\pi\cdot5^2=25\pi\,\text{cm}^2\).

\(25\pi\,\text{cm}^2\).

- Start with the area of the whole circle. - What fraction of the circle is one semicircle?

1. The full circle has area \(\pi\cdot6^2=36\pi\,\text{cm}^2\). 2. A semicircle covers half that region, so its area is \(18\pi\,\text{cm}^2\).

\(18\pi\,\text{cm}^2\).

- Compare the squares of the two radii. - Is area proportional to the radius or its square?

1. A's area is \(\pi\cdot2^2=4\pi\,\text{cm}^2\). 2. B's area is \(\pi\cdot4^2=16\pi\,\text{cm}^2\). 3. The ratio is \(16\pi\div4\pi=4\), not \(2\).

Four times as large. Doubling the radius multiplies the area by \(4\).

- Which circle covers the path and flower bed together? - Does the flower bed belong in the path's area? - How can you use both circle areas to count only the path?

1. The larger circle has area \(\pi\cdot5^2=25\pi\,\text{m}^2\). 2. The flower bed has area \(\pi\cdot3^2=9\pi\,\text{m}^2\). 3. The path has area \(25\pi-9\pi=16\pi\,\text{m}^2\).

\(16\pi\,\text{m}^2\).

- If circumference is multiplied by a factor, what happens to radius? - Use the circumference and area formulas for a circle. - Compare the two area expressions as a ratio. - How do lengths and areas scale when a figure is enlarged?

1. For the first circle, \(C_1 = \pi(12\,\text{cm}) = 12\pi\,\text{cm} \approx 37.70\,\text{cm}\). 2. Its radius is \(6\,\text{cm}\), so \(A_1 = \pi(6\,\text{cm})^2 = 36\pi\,\text{cm}^2 \approx 113.10\,\text{cm}^2\). 3. Since \(C_2 = 2.5C_1\), the second circumference is \(30\pi\,\text{cm}\). 4. From \(C_2 = 2\pi r_2\), \(r_2 = 15\,\text{cm}\). 5. The second area is \(A_2 = \pi(15\,\text{cm})^2 = 225\pi\,\text{cm}^2\). 6. Therefore, \(\frac{A_2}{A_1} = \frac{225\pi}{36\pi} = 6.25\). 7. A circumference scale factor of \(k\) gives the same radius scale factor \(k\), while area changes by \(k^2\). Here, \(2.5^2 = 6.25\).

a) \(C_1 = 12\pi\,\text{cm} \approx 37.70\,\text{cm}\); \(A_1 = 36\pi\,\text{cm}^2 \approx 113.10\,\text{cm}^2\) b) \(15\,\text{cm}\) c) The area scale factor is \(6.25\), which is the square of the circumference scale factor: \(2.5^2 = 6.25\).

- Which circle formula gives circumference, and which gives area? - In part b), start from the given circumference and work toward the radius. - In part c), what familiar whole number has a square matching the factor multiplied by \(\pi\)?

1. For a), \(C=2\pi\cdot4\,\text{cm}=8\pi\,\text{cm}\approx25.13\,\text{cm}\). Also, \(A=\pi\cdot4^2\,\text{cm}^2=16\pi\,\text{cm}^2\approx50.27\,\text{cm}^2\). 2. For b), \(r=\frac{31.42}{2\pi}\,\text{m}\approx5.00\,\text{m}\). Using the unrounded radius, \(A=\frac{31.42^2}{4\pi}\,\text{m}^2\approx78.56\,\text{m}^2\). 3. For c), \(A=\pi r^2=49\pi\,\text{mm}^2\) means \(r^2=49\,\text{mm}^2\), so \(r=7\,\text{mm}\). Then \(C=2\pi\cdot7\,\text{mm}=14\pi\,\text{mm}\approx43.98\,\text{mm}\).

a) \(C\approx25.13\,\text{cm}\), \(A\approx50.27\,\text{cm}^2\). b) \(r\approx5.00\,\text{m}\), \(A\approx78.56\,\text{m}^2\). c) \(r=7\,\text{mm}\), \(C=14\pi\,\text{mm}\approx43.98\,\text{mm}\).

- First find the first circle's diameter. - How large is the second diameter when it is \(3\) times the first? - Compare the two areas by dividing. - What happens to area when radius is multiplied by \(3\)?

1. The first diameter is \(d_1 = 2 \cdot 6\,\text{cm} = 12\,\text{cm}\), and the first area is \(A_1 = 36\pi\,\text{cm}^2\). 2. The second diameter is \(d_2 = 3 \cdot 12\,\text{cm} = 36\,\text{cm}\), so \(r_2 = 18\,\text{cm}\). 3. The second circumference is \(C_2 = 36\pi\,\text{cm} \approx 113.10\,\text{cm}\). 4. The second area is \(A_2 = 324\pi\,\text{cm}^2 \approx 1017.88\,\text{cm}^2\). 5. The area ratio is \(\frac{A_2}{A_1} = \frac{324\pi}{36\pi} = 9\).

a) \(C_2 = 36\pi\,\text{cm} \approx 113.10\,\text{cm}\); \(A_2 = 324\pi\,\text{cm}^2 \approx 1017.88\,\text{cm}^2\) b) \(\frac{A_2}{A_1} = 9\), so the second area is \(9\) times the first area.

- How are diameter and radius related? - How do you calculate area per dollar? - Compare the two unit rates after rounding them to the requested place.

1. The radii are \(5\,\text{in.}\) and \(6\,\text{in.}\). 2. The areas are \(25\pi\,\text{in.}^2 \approx 78.54\,\text{in.}^2\) and \(36\pi\,\text{in.}^2 \approx 113.10\,\text{in.}^2\). 3. The \(10\,\text{in.}\) pizza provides \(\frac{25\pi}{7.50} \approx 10.47\,\text{in.}^2/\text{dollar}\). 4. The \(12\,\text{in.}\) pizza provides \(\frac{36\pi}{11.00} \approx 10.28\,\text{in.}^2/\text{dollar}\). 5. Since \(10.47 > 10.28\), the \(10\,\text{in.}\) pizza provides more area per dollar.

The \(10\,\text{in.}\) pizza provides more area per dollar: approximately \(10.47\,\text{in.}^2/\text{dollar}\), compared with \(10.28\,\text{in.}^2/\text{dollar}\) for the \(12\,\text{in.}\) pizza.

- The wire length is the perimeter of each shape. - Find the square's side length and the circle's radius first. - For percent increase, compare the area difference with the square's area.

1. For the square, each side is \(\frac{120\,\text{cm}}{4}=30\,\text{cm}\), so its area is \(900\,\text{cm}^2\). 2. For the circle, \(120\,\text{cm}=2\pi r\), so \(r=\frac{60}{\pi}\,\text{cm}\). 3. The circle's area is \(A=\pi\left(\frac{60}{\pi}\right)^2\,\text{cm}^2=\frac{3600}{\pi}\,\text{cm}^2\approx1145.92\,\text{cm}^2\). 4. The area difference is approximately \(1145.92-900=245.92\,\text{cm}^2\). 5. Relative to the square, the percent increase is \(\frac{245.92}{900}\cdot100\%\approx27.3\%\).

a) Square: \(900\,\text{cm}^2\); circle: \(\frac{3600}{\pi}\,\text{cm}^2 \approx 1145.92\,\text{cm}^2\) b) Approximately \(27.3\%\) greater

- What happens to a circle's area when its diameter is doubled? - Compare the area scale factor with the price scale factor. - You do not need the exact value of either area.

1. The large pizza's diameter and radius are \(2\) times the small pizza's corresponding measurements. 2. Circle area is proportional to the square of the radius, so the large pizza has \(2^2=4\) times the area. 3. The price scale factor is \(\frac{18}{6}=3\). 4. The large pizza provides \(4\) times the area for \(3\) times the price, so it provides more area per dollar.

The large pizza provides more area per dollar. Its area is \(4\) times as great, while its price is only \(3\) times as great.

- What is the area of one grid square? - Which formula relates the area of a circle, its radius, and \(\pi\)? - For percent error, compare the experimental estimate with the stated reference value.

1. Each grid square has area \(0.5\,\text{cm} \cdot 0.5\,\text{cm} = 0.25\,\text{cm}^2\). 2. The estimated circle area is \(A = 452 \cdot 0.25\,\text{cm}^2 = 113\,\text{cm}^2\). 3. From \(A = \pi r^2\), \(\pi \approx \frac{A}{r^2} = \frac{113}{36} \approx 3.1389\). 4. The percent error is \(\frac{|3.138888\ldots - 3.14159|}{3.14159} \cdot 100\% \approx 0.08598\%\), which rounds to \(0.09\%\).

a) \(113\,\text{cm}^2\) b) \(\pi \approx 3.1389\) c) \(0.09\%\)

- What fraction of the square sheet's area is occupied by the inscribed circular disk? - Why does the mass ratio equal the area ratio here? - Use the requested rounding place only after setting up each ratio.

1. Because the sheets have the same material and thickness, the mass ratio equals the area ratio. The circle fits exactly inside the square, so \(\frac{A_{\text{circle}}}{A_{\text{square}}} = \frac{\pi}{4}\). 2. The theoretical disk mass is \(720\,\text{g} \cdot \frac{3.1416}{4} = 565.488\,\text{g} \approx 565.49\,\text{g}\). 3. For the measured mass, \(\frac{565}{720} \approx \frac{\pi}{4}\). 4. Thus \(\pi \approx 4 \cdot \frac{565}{720} = \frac{113}{36} \approx 3.1389\).

a) Approximately \(565.49\,\text{g}\) b) \(\pi \approx 3.1389\)

- Which equation relates circumference and diameter? - For pieces made from the same uniform material and thickness, how are mass and area related? - Compare each estimate's absolute error from \(3.14159\).

1. For Group A, \(\pi \approx \frac{C}{d} = \frac{157}{50} = 3.1400\). 2. For Group B, the circle's diameter equals the square's side length. Because the cardboard is uniform, the mass ratio equals the area ratio, so \(\frac{19}{24} \approx \frac{\pi}{4}\). 3. Solve: \(\pi \approx 4 \cdot \frac{19}{24} = \frac{19}{6} \approx 3.1667\). 4. Group A's absolute error is \(|3.14 - 3.14159| = 0.00159\). Group B's absolute error is approximately \(0.02508\). 5. Group A's estimate is closer.

Group A: \(\pi \approx 3.1400\) Group B: \(\pi \approx 3.1667\) Group A's estimate is closer to \(3.14159\).

- What shape does the rotating arm sweep out? - What distance does the tip travel in one full rotation? - Convert the rotation time to minutes before finding speed.

1. The watered region is a circle with radius \(120\,\text{m}\), so \(A=14{,}400\pi\,\text{m}^2\approx45{,}238.93\,\text{m}^2\). 2. The outer nozzle travels one circumference per rotation: \(C=240\pi\,\text{m}\). 3. Convert the time: \(10\,\text{hr}=600\,\text{min}\). 4. The speed is \(v=\frac{240\pi\,\text{m}}{600\,\text{min}}=\frac{2\pi}{5}\,\text{m/min}\approx1.26\,\text{m/min}\).

a) \(14{,}400\pi\,\text{m}^2 \approx 45{,}238.93\,\text{m}^2\) b) \(\frac{2\pi}{5}\,\text{m/min} \approx 1.26\,\text{m/min}\)

- Rearrange the circle-area formula to find the radius. - What circle measurement represents the fence length? - In which formula is the radius squared?

1. From \(A=\pi r^2\), \(r^2\approx\frac{28.26}{3.14}\,\text{ft}^2=9\,\text{ft}^2\), so \(r=3\,\text{ft}\). 2. The fence length is the circumference: \(C\approx2\cdot3.14\cdot3\,\text{ft}=18.84\,\text{ft}\). 3. Circumference is proportional to radius, so doubling the radius doubles the fence length. 4. Area depends on the square of the radius, so doubling the radius multiplies the area by \(2^2=4\).

a) \(3\,\text{ft}\) b) \(18.84\,\text{ft}\) c) The fence length doubles, and the area is multiplied by \(4\).

- Use the perimeter to find the square's side length. - Use the same perimeter to find the circle's radius. - Calculate and compare the two areas.

1. The square's side length is \(\frac{20\,\text{cm}}{4}=5\,\text{cm}\), so its area is \(25\,\text{cm}^2\). 2. The circle's radius is \(r\approx\frac{20\,\text{cm}}{2\cdot3.14}\approx3.1847\,\text{cm}\). 3. The circle's area is \(A\approx3.14\cdot(3.1847\,\text{cm})^2\approx31.85\,\text{cm}^2\). 4. Since \(31.85>25\), the circle encloses the greater area.

The square encloses \(25\,\text{cm}^2\). The circle encloses approximately \(31.85\,\text{cm}^2\), so the circle has the greater area.

- Use radius, not diameter, in the circle-area formula. - Divide price by the number of \(10\,\text{in.}^2\) units. - What radius scale factor gives an area scale factor of \(4\)?

1. The standard pizza has radius \(6\,\text{in.}\), so \(A_s=36\pi\,\text{in.}^2\approx113.10\,\text{in.}^2\). 2. The family pizza has radius \(7\,\text{in.}\), so \(A_f=49\pi\,\text{in.}^2\approx153.94\,\text{in.}^2\). 3. The standard price per \(10\,\text{in.}^2\) is approximately \(\$0.75\). 4. The family price per \(10\,\text{in.}^2\) is approximately \(\$0.71\), so the family pizza is the better value. 5. Quadrupling area requires doubling the radius, so the new diameter is \(24\,\text{in.}\).

a) Standard: \(36\pi\,\text{in.}^2\approx113.10\,\text{in.}^2\); family: \(49\pi\,\text{in.}^2\approx153.94\,\text{in.}^2\) b) Standard: approximately \(\$0.75\) per \(10\,\text{in.}^2\); family: approximately \(\$0.71\) per \(10\,\text{in.}^2\). The family pizza is the better value. c) \(24\,\text{in.}\)

- Which familiar positive number, when squared, matches pane A's area factor? - How do you recover diameter directly from circumference for pane B? - Compare the diameters, not the areas.

1. For pane A, \(\pi r_A^2=225\pi\,\text{cm}^2\), so \(r_A=15\,\text{cm}\) because \(15^2=225\). Thus \(d_A=30\,\text{cm}\). 2. For pane B, \(r_B=\frac{95}{2\pi}\,\text{cm}\approx15.12\,\text{cm}\) and \(d_B=\frac{95}{\pi}\,\text{cm}\approx30.24\,\text{cm}\). 3. Because \(\frac{95}{\pi}>30\), pane B has the greater diameter.

Pane A: \(r=15.00\,\text{cm}\), \(d=30.00\,\text{cm}\). Pane B: \(r\approx15.12\,\text{cm}\), \(d\approx30.24\,\text{cm}\). Pane B has the greater diameter.

- In the reverse cases, rearrange the formula to identify the diameter squared. - Match a square value to a familiar whole-number square. - In the forward cases, square the given diameter before multiplying. - Round only the final area, not the intermediate calculation.

1. For a), \(A\approx\frac{3.14\cdot12^2}{4}=113.04\,\text{cm}^2\approx113\,\text{cm}^2\). 2. For b), \(d^2=\frac{4A}{\pi}\approx\frac{4\cdot1256}{3.14}=1600\,\text{cm}^2\). Since \(40^2=1600\), the positive diameter is \(40\,\text{cm}\). 3. For c), \(A\approx\frac{3.14\cdot50^2}{4}=1962.5\,\text{cm}^2\approx1963\,\text{cm}^2\). 4. For d), \(d^2\approx\frac{4\cdot2826}{3.14}=3600\,\text{cm}^2\). Since \(60^2=3600\), the diameter is \(60\,\text{cm}\).

a) \(A\approx113\,\text{cm}^2\). b) \(d=40\,\text{cm}\). c) \(A\approx1963\,\text{cm}^2\). d) \(d=60\,\text{cm}\).

- The labeled segment is the diameter, not the radius. - Convert the diameter to a radius before using the circle-area formula. - For the cost, multiply the unrounded area by the price per square inch and round only at the end.

1. The diameter is \(18\,\text{in.}\), so the radius is \(9\,\text{in.}\). 2. The area is \(A=\pi(9)^2=81\pi\,\text{in.}^2\approx254.5\,\text{in.}^2\). 3. Use the unrounded area for the cost: \(81\pi\cdot\$0.08\approx\$20.36\).

a) Approximately \(254.5\,\text{in.}^2\) b) Approximately \(\$20.36\)

- Track where all of the original circular arcs end up after the sectors alternate. - Think about the center-to-circle length carried by every sector. - Only after identifying the new base and height should you substitute the circumference formula.

1. The curved sector edges together make the entire circumference \(C\). Alternating the sectors places half of those arcs along one long side and half along the other, so each long side approaches \(\frac{C}{2}\). 2. Each sector extends from the center to the circle, so the distance between the two long sides approaches the radius \(r\). 3. The rearrangement preserves area. Its area therefore approaches \(\frac{C}{2}\cdot r\). 4. Substitute \(C=2\pi r\): \(A=\frac{1}{2}(2\pi r)r=\pi r^2\).

a) The circumference; each long side approaches \(\frac{C}{2}\). b) The radius \(r\). c) \(A=\frac{C}{2}r=\frac{1}{2}(2\pi r)r=\pi r^2\).

- Translate the stated numerical relationship into an equation involving \(A\) and \(C\). - Substitute the circle formulas before simplifying. - Keep the area exact until the requested final approximation.

1. Represent the condition by \(A=4C\), comparing the stated numerical values. 2. Substitute \(A=\pi r^2\) and \(C=2\pi r\): \(\pi r^2=4(2\pi r)\). 3. Since the radius is positive, divide by \(\pi r\) to get \(r=8\,\text{cm}\). 4. The area is \(A=64\pi\,\text{cm}^2\approx201.1\,\text{cm}^2\).

\(r=8\,\text{cm}\) \(A=64\pi\,\text{cm}^2\approx201.1\,\text{cm}^2\)