Free Math Worksheets

- 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 formula relates circle area and radius? - What operation undoes squaring? - How are diameter and radius related?

1. From \(A=\pi r^2\), solve for radius: \(r=\sqrt{\frac{A}{\pi}}\). 2. For a), \(r=\sqrt{\frac{28.27\,\text{cm}^2}{\pi}}\approx3.00\,\text{cm}\). 3. For b), \(r=\sqrt{\frac{314.16\,\text{m}^2}{\pi}}\approx10.00\,\text{m}\). 4. The diameter is \(d=2r\approx20.00\,\text{m}\).

a) \(r \approx 3.00\,\text{cm}\) b) \(d \approx 20.00\,\text{m}\)

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

1. From \(C=\pi d\), \(d=\frac{12.00\,\text{ft}}{\pi}\approx3.82\,\text{ft}\). 2. The radius is \(r=\frac{d}{2}=\frac{6}{\pi}\,\text{ft}\). 3. The area is \(A=\pi r^2=\pi\left(\frac{6}{\pi}\right)^2\,\text{ft}^2=\frac{36}{\pi}\,\text{ft}^2\approx11.46\,\text{ft}^2\).

a) Approximately \(3.82\,\text{ft}\) b) Approximately \(11.46\,\text{ft}^2\)

- How are circumference and area related to radius? - Which operation reverses squaring? - How are radius and diameter related?

1. For a), \(r=\frac{C}{2\pi}=\frac{25.13\,\text{cm}}{2\pi}\approx4.0\,\text{cm}\). 2. Without rounding the radius first, \(A=\pi\left(\frac{25.13}{2\pi}\right)^2\,\text{cm}^2\approx50.3\,\text{cm}^2\). 3. For b), \(r=\sqrt{\frac{28.27\,\text{cm}^2}{\pi}}\approx3.0\,\text{cm}\). 4. Then \(d=2r\approx6.0\,\text{cm}\) and \(C=2\pi r\approx18.8\,\text{cm}\).

a) \(r\approx4.0\,\text{cm}\); \(A\approx50.3\,\text{cm}^2\) b) \(d\approx6.0\,\text{cm}\); \(C\approx18.8\,\text{cm}\)

- Which formula relates area and radius? - How are radius and diameter related? - How do you find circumference from radius? - Round only after completing the calculations.

1. Solve \(A=\pi r^2\) for radius: \(r=\sqrt{\frac{50}{\pi}}\,\text{cm}\approx4.0\,\text{cm}\). 2. The diameter is \(d=2r\approx8.0\,\text{cm}\). 3. Without using the rounded radius, the circumference is \(C=2\pi\sqrt{\frac{50}{\pi}}\,\text{cm}\approx25.1\,\text{cm}\).

\(r\approx4.0\,\text{cm}\) \(d\approx8.0\,\text{cm}\) \(C\approx25.1\,\text{cm}\)

- Find the square's area. - Convert the circle's diameter to a radius. - Use the circle-area formula and compare the results.

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

The circular pizza has the greater area: approximately \(113.10\,\text{in.}^2\), 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 first.

1. Circle A: \(d=2\cdot4.5\,\text{cm}=9\,\text{cm}\). Also, \(C\approx2\cdot3.14\cdot4.5\,\text{cm}=28.26\,\text{cm}\), and \(A\approx3.14\cdot(4.5\,\text{cm})^2=63.585\,\text{cm}^2\approx63.59\,\text{cm}^2\). 2. Circle B: \(r=\frac{12\,\text{dm}}{2}=6\,\text{dm}\). Also, \(C\approx3.14\cdot12\,\text{dm}=37.68\,\text{dm}\), and \(A\approx3.14\cdot(6\,\text{dm})^2=113.04\,\text{dm}^2\). 3. Circle C: \(d\approx\frac{15.7\,\text{m}}{3.14}=5\,\text{m}\), so \(r=2.5\,\text{m}\). Then \(A\approx3.14\cdot(2.5\,\text{m})^2=19.625\,\text{m}^2\approx19.63\,\text{m}^2\).

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) \(r=2.5\,\text{m}\), \(d=5\,\text{m}\), \(A\approx19.63\,\text{m}^2\)

- Which formula relates area and radius? - Rearrange that formula to isolate the radius. - Which measurement do you need before finding circumference?

1. The radius is \(r=\sqrt{\frac{50.27\,\text{ft}^2}{\pi}}\approx4.00\,\text{ft}\). 2. Without rounding the radius first, the circumference is \(C=2\pi\sqrt{\frac{50.27}{\pi}}\,\text{ft}\approx25.13\,\text{ft}\).

Radius: approximately \(4.00\,\text{ft}\) Edging length: approximately \(25.13\,\text{ft}\)

- How are the radius and circumference related? How are the radius and area related? - Rearrange the appropriate formula to find the radius first. - What intermediate measure connects circumference and area?

1. For part 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, \(A=\pi\left(\frac{31.4}{2\pi}\right)^2\,\text{cm}^2\approx78.460\,\text{cm}^2\), so \(A\approx78.5\,\text{cm}^2\). 3. For part b, first find the radius: \(r=\sqrt{\frac{A}{\pi}}=\sqrt{\frac{50.27}{\pi}}\,\text{cm}\approx4.000\,\text{cm}\). 4. Without rounding the radius first, \(C=2\pi\sqrt{\frac{50.27}{\pi}}\,\text{cm}\approx25.134\,\text{cm}\), so \(C\approx25.1\,\text{cm}\).

a) \(A\approx78.5\,\text{cm}^2\) b) \(C\approx25.1\,\text{cm}\)

- How are the radius and diameter related? - Which formula gives the distance around a circle? - Which formula gives the area inside a circle?

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

a) About \(50.27\,\text{ft}\) b) About \(201.06\,\text{ft}^2\)

- Which formula relates area and radius? - Can you use the area to find the radius first? - Which operation reverses squaring? - Which formula gives circumference from radius?

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

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

- 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 \(1519.8\,\text{cm}^2\).

- 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 \(C_2 = 2\pi r_2\), \(r_2 = \frac{30\pi}{2\pi}\,\text{cm} = 15\,\text{cm}\). 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) \(C_1 \approx 37.70\,\text{cm}\); \(A_1 \approx 113.10\,\text{cm}^2\) 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 formula gives the area of a circle? - What inverse operation undoes squaring when you solve for radius? - Use the given value in each part to decide which quantity to find first.

1. For a), \(C = 2\pi\cdot4\,\text{cm} = 8\pi\,\text{cm} \approx 25.13\,\text{cm}\). Also, \(A = \pi\cdot4^2\,\text{cm}^2 = 16\pi\,\text{cm}^2 \approx 50.27\,\text{cm}^2\). 2. For b), \(r = \frac{31.42\,\text{m}}{2\pi} \approx 5.00\,\text{m}\). Without rounding the radius first, \(A = \pi r^2 = \frac{(31.42\,\text{m})^2}{4\pi} \approx 78.56\,\text{m}^2\). 3. For c), \(r = \sqrt{\frac{153.94\,\text{mm}^2}{\pi}} \approx 7.00\,\text{mm}\). Without rounding the radius first, \(C = 2\pi\sqrt{\frac{153.94}{\pi}}\,\text{mm} \approx 43.98\,\text{mm}\).

a) \(C \approx 25.13\,\text{cm}\), \(A \approx 50.27\,\text{cm}^2\) b) \(r \approx 5.00\,\text{m}\), \(A \approx 78.56\,\text{m}^2\) c) \(r \approx 7.00\,\text{mm}\), \(C \approx 43.98\,\text{mm}\)

- 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 \(A_1 = \pi(6\,\text{cm})^2 = 36\pi\,\text{cm}^2\). 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 \(C_2 = \pi(36\,\text{cm}) = 36\pi\,\text{cm} \approx 113.10\,\text{cm}\). 4. The second area is \(A_2 = \pi(18\,\text{cm})^2 = 324\pi\,\text{cm}^2 \approx 1017.88\,\text{cm}^2\). 5. The area ratio is \(\frac{A_2}{A_1}=\frac{324\pi}{36\pi}=9\).

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

- Work backward from the circle area formula. - Which operation undoes squaring? - Keep \(\pi\) unrounded until the final step.

1. Start with \(A = \pi r^2\) and substitute the given area: \(200 = \pi r^2\). 2. Divide by \(\pi\): \(r^2 = \frac{200}{\pi}\). 3. Take the positive square root: \(r = \sqrt{\frac{200}{\pi}} \approx 7.98\,\text{ft}\).

Approximately \(7.98\,\text{ft}\)

- How are diameter and radius related? - How do you calculate area per dollar? - Compare the two unit rates.

1. The radii are \(5\,\text{in.}\) and \(6\,\text{in.}\). 2. The areas are \(A_1 = \pi(5\,\text{in.})^2 \approx 78.54\,\text{in.}^2\) and \(A_2 = \pi(6\,\text{in.})^2 \approx 113.10\,\text{in.}^2\). 3. The \(10\,\text{in.}\) pizza provides \(\frac{78.54}{7.50} \approx 10.47\,\text{in.}^2/\text{dollar}\). 4. The \(12\,\text{in.}\) pizza provides \(\frac{113.10}{11.00} \approx 10.28\,\text{in.}^2/\text{dollar}\). 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 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 \(30^2\,\text{cm}^2=900\,\text{cm}^2\). 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: approximately \(1145.92\,\text{cm}^2\) 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 square on the grid paper? - Which formula relates the area of a circle, its radius, and \(\pi\)? - How do you calculate percent error between an experimental value and an accepted 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 \(A = \pi r^2\), solve for \(\pi\): \(\pi \approx \frac{A}{r^2}\). 4. Substitute the estimated area and radius: \(\pi \approx \frac{113}{6^2} = \frac{113}{36} \approx 3.1389\). 5. 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 circular disk? - Why does the mass ratio equal the area ratio? - Write an equation involving the square's mass, the disk's mass, and \(\pi\).

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 \(\frac{A_{\text{circle}}}{A_{\text{square}}} = \frac{\pi r^2}{(2r)^2} = \frac{\pi}{4}\). 2. The theoretical disk mass is \(720\,\text{g} \cdot \frac{3.1416}{4} = 565.488\,\text{g}\), or about \(565.49\,\text{g}\). 3. For the measured mass, solve \(\frac{565}{720} \approx \frac{\pi}{4}\) for \(\pi\). 4. Then \(\pi \approx 4 \cdot \frac{565}{720} = \frac{113}{36} \approx 3.1389\).

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

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

1. For Group A, use \(C = \pi d\): \(\pi \approx \frac{C}{d} = \frac{157}{50} = 3.14\). 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 for \(\pi\): \(\pi \approx 4 \cdot \frac{19}{24} = \frac{19}{6} \approx 3.1667\). 4. Group A's absolute error is \(|3.14 - 3.14159| = 0.00159\). Group B's absolute error is \(|3.166666\ldots - 3.14159| \approx 0.02508\). 5. Because \(0.00159 < 0.02508\), Group A's estimate is closer.

Group A: \(\pi \approx 3.14\) Group B: \(\pi \approx 3.1667\) Group A's estimate is closer to \(3.14159\).

- What shape does the rotating arm sweep out? - Which formulas give the area and the distance traveled by the tip? - Convert \(10\) hours to minutes before calculating the requested speed.

1. The watered region is a circle with radius \(120\,\text{m}\), so \(A=\pi(120\,\text{m})^2=14{,}400\pi\,\text{m}^2\approx45{,}238.93\,\text{m}^2\). 2. The outer nozzle travels one circumference per rotation: \(C=2\pi(120\,\text{m})=240\pi\,\text{m}\approx753.98\,\text{m}\). 3. Convert the time: \(10\,\text{hr}=600\,\text{min}\). 4. The speed is \(v=\frac{240\pi\,\text{m}}{600\,\text{min}}\approx1.26\,\text{m/min}\).

a) Approximately \(45{,}238.93\,\text{m}^2\) b) Approximately \(1.26\,\text{m/min}\)

- 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. - How does multiplying the radius by \(2\) affect the area? - Use the area scale factor to find the new diameter.

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 \(\frac{\$8.50}{113.10/10}\approx\$0.75\). 4. The family price per \(10\,\text{in.}^2\) is \(\frac{\$11.00}{153.94/10}\approx\$0.71\), so the family pizza is less expensive per unit area. 5. Quadrupling the standard area requires doubling the radius, so the new diameter is \(24\,\text{in.}\).

a) Standard: approximately \(113.10\,\text{in.}^2\); family: approximately \(153.94\,\text{in.}^2\) 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) \(24\,\text{in.}\)

- Rearrange the area and circumference formulas to isolate the radius. - How are radius and diameter related? - Compare the calculated diameters directly.

1. For pane A, \(r_A=\sqrt{\frac{700}{\pi}}\,\text{cm}\approx14.93\,\text{cm}\). Its diameter is \(d_A=2\sqrt{\frac{700}{\pi}}\,\text{cm}\approx29.85\,\text{cm}\). 2. For pane B, \(r_B=\frac{95}{2\pi}\,\text{cm}\approx15.12\,\text{cm}\). Its diameter is \(d_B=\frac{95}{\pi}\,\text{cm}\approx30.24\,\text{cm}\). 3. Since \(30.24\,\text{cm}>29.85\,\text{cm}\), pane B has the greater diameter.

Pane A has radius about \(14.93\,\text{cm}\) and diameter about \(29.85\,\text{cm}\). Pane B has radius about \(15.12\,\text{cm}\) and diameter about \(30.24\,\text{cm}\). Pane B has the greater diameter.

- How can you rearrange the formula so that the diameter is isolated? - Which operation reverses squaring? - For area calculations, check the order in which you square and multiply. - Use the tenths digit to round to the nearest whole number.

1. For circle \(1\), \(A=\frac{3.14\cdot12^2}{4}=113.04\,\text{cm}^2\approx113\,\text{cm}^2\). 2. To find a diameter from an area, rearrange the formula: \(d=\sqrt{\frac{4A}{\pi}}\). 3. For circle \(2\), \(d=\sqrt{\frac{4\cdot1256}{3.14}}\,\text{cm}=\sqrt{1600}\,\text{cm}=40\,\text{cm}\). 4. For circle \(3\), \(A=\frac{3.14\cdot50^2}{4}=1962.5\,\text{cm}^2\approx1963\,\text{cm}^2\). 5. For circle \(4\), \(d=\sqrt{\frac{4\cdot2826}{3.14}}\,\text{cm}=\sqrt{3600}\,\text{cm}=60\,\text{cm}\).

a) \(A=113\,\text{cm}^2\) b) \(d=40\,\text{cm}\) c) \(A=1963\,\text{cm}^2\) d) \(d=60\,\text{cm}\)

- Translate the relationship into an equation. - Substitute the circle formulas for area and circumference. - Simplify the equation to isolate the radius.

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. Simplify: \(\pi r^2=8\pi r\). 4. Since the circle has positive radius, divide by \(\pi r\) to get \(r=8\,\text{cm}\). 5. The area is \(A=\pi\cdot8^2\,\text{cm}^2=64\pi\,\text{cm}^2\approx201.1\,\text{cm}^2\).

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