Difference between revisions of "Talk:2007 AMC 10B Problems/Problem 11"
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− | Another possible solution is to plot the circle and triangle on a graph with the circle having center (0,0). | + | Another possible solution is to plot the circle and triangle on a graph with the circle having center (0,0). |
− | Radius of circle = r | + | Radius of circle = r |
− | distance between origin and base of triangle = a | + | distance between origin and base of triangle = a |
− | 1 + a^2 = r^2 | + | 1 + a^2 = r^2 |
− | r + a = 2sqrt(2) | + | r + a = 2sqrt(2) |
− | a = (2)sqrt(2) - r | + | a = (2)sqrt(2) - r |
− | 9 - (4r)sqrt(2) = 0 | + | 9 - (4r)sqrt(2) = 0 |
− | r = ((9)sqrt(2))/8 | + | r = ((9)sqrt(2))/8 |
πr^2 = 81π/32 | πr^2 = 81π/32 |
Latest revision as of 22:12, 11 November 2017
Another possible solution is to plot the circle and triangle on a graph with the circle having center (0,0). Radius of circle = r distance between origin and base of triangle = a 1 + a^2 = r^2 r + a = 2sqrt(2) a = (2)sqrt(2) - r 9 - (4r)sqrt(2) = 0 r = ((9)sqrt(2))/8 πr^2 = 81π/32