+817 Let R be the circle centered at (0,0) with radius 10. The lines x=6 and y=4 divide R into four regions R1, R2, R3 , and R4. Let R_i denote the area of region R_i If R1>R2>R3>R4, then find R1 + R2 + R3 + R4.
+1537 Since R1, R2, R3, R4 are the four regions into which the unit circle is divided by the lines x=6 and y=4, we see that regions R1 and R3 are congruent, and regions R2 and R4 are congruent. Therefore, R1+R2+R3+R4=2(R1+R2).
To compute the area of R1, we can construct sector OAC, where O is the center of the circle and A is the point at which the circle intersects the line y=4. Sector OAC has central angle 90 degrees, and its radius is 10, so its area is \frac{1}{4}(10^2)\pi = \frac{25}{2}\pi.
Since regions R1 and R3 are congruent, and region R1 is the upper right quadrant of sector OAC, the area of R1 is \boxed{\frac{25\pi}{4}}.
