ujjwal80 wrote:
Bunuel wrote:
m22 q20What is the area of the region enclosed by lines \(y=x\), \(x=-y\), and the upper crescent of the circle \(y^2+x^2=4\) ?A. \(\frac{\pi}{4}\)
B. \(\frac{\pi}{2}\)
C. \(\frac{3\pi}{4}\)
D. \(\pi\)
E. \(4\pi\)
The circle represented by the equation \(x^2+y^2 = 4\) is centered at the origin and has the radius of \(r=\sqrt{4}=2\).
Look at the diagram below:
Attachment:
m22-20.png
We need to find the area of the upper crescent, so the area of the yellow sector. Since the central angle of this sector is 90 degrees then its area would be 1/4 of that of the circle (since circle is 360 degrees).
The area of the circle is \({\pi}{r^2}=4\pi\), 1/4 of this value is \(\pi\).
Answer: D.
Hi
Bunuel, I was able to solve this question but it took me a while to visualise the diagram. So just asking by curiosity, can we expect this type of questions in GMAT? As always appreciate your efforts in advance! Thanks
hi
as far as time is concerned, you can see the problem this way
(y =x), means slope is 1 and the line passes through the origin
(y = -x), means slope is -1, and the line passes through the origin
now it is worth noticing that a slope of 1 or -1 creates an angle of 45 degree, so 2 slopes jointly cover a 90 degree
also, when 2 lines are perpendicular to each other, their slopes are negative reciprocal to each other
here 1 is negative reciprocal to -1, so the slopes are perpendicular to each other, creating an angle 90 degree
thus, 2 slopes together add to 90 degrees, which is 1/4 of pi (2)^2 , as the radius is 2 we get from (x^ + Y^ = 4)
OR
you can simply draw the 2 slopes to see the area covered by the upper crescent
hope this helps and is clear!
thanks and cheers!