Difference between revisions of "Talk:2007 AMC 10B Problems/Problem 11"

(another solution)
 
 
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Another possible solution is to plot the circle and triangle on a graph with the circle having center (0,0).
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Another possible solution is to plot the circle and triangle on a graph with the circle having center (0,0).                      
Radius of circle = r
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Radius of circle = r                              
distance between origin and base of triangle = a
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distance between origin and base of triangle = a                                                      
1 + a^2 = r^2
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1 + a^2 = r^2                          
r + a = 2sqrt(2)
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r + a = 2sqrt(2)                      
a =  (2)sqrt(2) - r
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a =  (2)sqrt(2) - r                  
9 - (4r)sqrt(2) = 0
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9 - (4r)sqrt(2) = 0                
r = ((9)sqrt(2))/8
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r = ((9)sqrt(2))/8                    
 
πr^2 = 81π/32
 
πr^2 = 81π/32

Latest revision as of 21:12, 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