A problem by Ajay Sambhriya
Level
2
A square is inscribed in a quarter circle as shown above.
Given that AF meets CB at E, and the radius of this quarter circle is 12. Find the length of AE to 2 decimal places.
The answer is 10.39.
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DB is a radius and so = 12. It is also the diagonal of the square ABCD, which means each side of the square is 6*sqrt(2). We can coordinize the problem. Let D be the origin. D = (0,0(. F = (12,0) A =(0,6sqrt(2) The equation of the line AF is y = (-sqrt(2)/2)x + 6sqrt(2). The equation of the line CB is x = 6sqrt(2). The intersection is the coordinates of E. E = (6sqrt(2), -6 + 6sqrt(2). The distance AE is given by sqrt((6sqrt(2) - 0)^2 + (-6 + 6sqrt(2) -6sqrt(2))^2) = sqrt(72 + 36) = sqrt(108) = 6sqrt(3) = 10.392. Ed Gray