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08 Feb 2012, 17:13
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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70% (03:14) correct
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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. \(\frac{\pi}{16}(x^2-4x+10)\)
Attachment:
Circles+.PNG [ 2.4 KiB | Viewed 26144 times ]
Area of the larger circle = pi \(x^2\)
Area of middle circle = pi (x-1)\(^2\)
Area of outer circle = pi (x-2)\(^2\)
Combined area should be the sum of the above 3. So when I do the maths I get combined area as
pi (3\(x^2\) - 4x +5). But I am stuck after this.
09 Feb 2012, 02:27
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enigma123
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 \(\frac{\pi(5^2 - 4*5 + 10)}{4} = 15\pi/4\)
08 Feb 2012, 17:46
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Attachment:
circles.PNG [ 25.38 KiB | Viewed 25327 times ]
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: \(\frac{\pi}{4}(2x-1+2x-5+x^2-8x+16)=\frac{\pi}{4}(x^2-4x+10)\).
