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The circumference of a circle is 6.28. What is the area of the circle
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\[\large C = 2\pi r\] \[\large A = \pi r^2\] Notice there is an 'r' in each equation...lets solve the top one for 'r' \[\large r = \frac{C}{2\pi}\] Now lets substitute that into the 2nd equation for 'r' \[\large A = \pi(\frac{C}{2\pi})^2\] \[\large A = \pi(\frac{C^2}{4\pi^2})\] \[\large A=\frac{\cancel{\pi} C^2}{4\pi\cancel{^2}}\] \[\large A = \frac{C^2}{4\pi}\] Now lets substitute the fact that we know C = 6.28 \[\large A = \frac{(6.28)^2}{4\pi}\] \[\large A = \frac{39.4384}{12.56}\] \[\large A = 3.14 = \pi \]
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