The figure represents five concentric quarter-circles. The length of the radius of the largest quarter-circle is x. The length of the radius of each successively smaller quarter-circle is one less than that of the next larger quarter-circle. What is the combined area of the shaded regions (black), in terms of x?

A. \(\pi(x^2-4x+10)\)

B. \(\frac{\pi}{2}(x^2-4x+10)\)

C. \(\frac{\pi}{4}(x^2-4x+10)\)

D. \(\frac{\pi}{8}(x^2-4x+10)\)

E. \(\frac{\pi}{16}(x^2-4x+10)\)


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The radii of the 5 quarter-circles are: x (the largest red), x-1 (the largest white), x-2 (next red), x-3 (next white), and x-4 (the inner red), so 5 radii for 5 quarter-circles.

We want the sum of 3 red (black) regions (so without two white regions in the middle of them).

The area of the largest red region is 1/4 of the are of the largest circle minus 1/4 of the area of the circle with radius of x-1 (the largest white circle) --> \(\frac{\pi{x^2}}{4}-\frac{\pi{(x-1)^2}}{4}=\frac{\pi}{4}(x^2-(x-1)^2)=\frac{\pi}{4}(2x-1)\);

The area of the next red region is 1/4 of the are of the circle with radius of x-2 (the next red circle) minus 1/4 of the area of the circle with radius of x-3 (the next white circle) --> \(\frac{\pi{(x-2)^2}}{4}-\frac{\pi{(x-3)^2}}{4}=\frac{\pi}{4}(2x-5)\);

Finally, the are of the red region in the center with the radius of (x-4) is \(\frac{\pi{(x-4)^2}}{4}=\frac{\pi}{4}(x^2-8x+16)\);

Sum: \(\frac{\pi}{4}(2x-1+2x-5+x^2-8x+16)=\frac{\pi}{4}(x^2-4x+10)\).
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Using x=5, Area of shaded region

= \(\frac{1}{4} [π+(9π-4π)+(25π-16π)]\)
= \(\frac{15}{4}π\)

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Added the answer choices.
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