Mind you, if you do have the options (as you will in GMAT), just plug in a value for x and solve. Working with numbers is much easier.
Radius of smallest circle is 1 and of largest is 5.

Required Area =\(\frac{1}{4}(\pi*25 - \pi*16 + \pi*9 - \pi*4 +\pi)\)
\(= 15\pi/4\)

In the options, just plug x = 5 and get your answer.
In the answer that Bunuel got above, if we plug x = 5, we get \(\frac{\pi(5^2 - 4*5 + 10)}{4} = 15\pi/4\)
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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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The radius of the largest circle is x, so the radius of next is x-1 and so on so the area is
pi[(x^2-{x-1}^2)+({x-2}^2-{x-3}^2)+(x-4)^2]
on solving we get pi(x^2-4x+10)
since this is a quarter of a circle so the area is
1/4*pi(x^2-4x+10)

Using x=5, Area of shaded region

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

Hi Bunuel,

Would be great if you can add answer choices and OA to this question. You may add choices of your wish for the 4 incorrect choices in case the official ones are not available.

Warm Regards,
Pritishd
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Added the answer choices.
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