What is the area of the region enclosed by lines y=x, x=y,

The lines y=x, x=−y intersecting at (0,0) and they are perpendicular to each other.
The circle y^2+x^2=4 is centered at (0,0) and has a radius of 2.
These two lines are dividing the circle y^2+x^2=4 into four equal segments.
Area of each segment = (PI*r^2) / 4 = (PI*2^2) / 4 = PI

Answer is D.

Hi Bunuel,
in the above problem, i understand that the circle is centered at the origin and has radius 2 but lines y=x, and x = -1 represents the half of the circle so the area enclosed should be half of the total area, I mean 4pi/2. i know, i am missing something, please clarify!
and How do I know that question asks the area of upper 1/4 of the area or How do i determine the 90 degree portion??

kamalahmmad1



Two lines are perpendicular if their slopes are -ve reciprocal of each other. E.g. line 1 : y = mx and line 2: y = (-1/m)x then line 1 and line two are perpendicular

Here y = x and y = -x meet the above stated criteria for perpendicular lines

//kudos please, if the above explanation is good.

The two lines are y=x and y=-x (x=-y), not x = -1. Both lines are shown on the diagram in my post.

Lines y=x and y=-x make 90 degrees.

To find the Area of a sector

Arc angle Area of sector
___________ = ______________
360 Area of circle


90 Area of sector
___ = _____________
360 4 Pi

Area of sector = 4 pi x 90/ 360 = pi

...
My solution(with detail explanation):
This diagram is indispensable:

All the 3 parts i.e x=y; x=-y; and x^2+y^2=4 makes one forth of the circular region x^2+y^2=4 area of which is 4pi so the required area is pi. hope this one makes it clear.

Question corrected!

Circle is y^2+x^2=4 has radius 2. Area of circle = pi * 4
y=x is the line that passes through 1 and 3 quadrants which has 45 degree to x-axis.
y=-x is the line that passes through 2 and 4 quadrants which has 45 degree to x-axis.

The two lines, divides the circle into 4 parts.

Combining all the three, (question to find the area of the upper crescent) Area = 4 pi/4 = pi

Why or How does Circle is y^2+x^2=4 has radius 2 ?

In the equation x^2+y^2=4

all the points (2,2)(-2,2)(0,2)(2,0) are on the circle. And these have the center as (0,0)

Hence the radius is 2.

Another explanation:
Equation Of A Circle
(x - a)^2 + (y - b)^2 = r^2
where
a is the x co-ordinate of the centre of the circle
b is the y co-ordinate of the centre of the circle
r is the radius of the circle

Comparing the above equation with the equation (x-0)^2+(y-0)^2=4
r^2 = 4
r=2

center is (0,0)

hi

the area of the circle is represented by the equation, x^2 + y^2 = 4, means that the area of the circle is x^2 + y^2 = 4

now, as 2 sides are equal to each other, it can be said that area is (x^2 + y^2) OR 4

since area is 4, the radius is 2

I don't know whether this reasoning is okay, but I have seen the scenario this way
Expert's reply on this issue is highly desired

thanks
???

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

Yes, I think it's a perfectly valid question. Notice that the average time it took users to answer it correctly is just 1:07 minutes.

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!

Thanks, gmatcracker2018, I will keep your explanation in mind next time I see such question.

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