Question: Intersection and Arc

Comment on Intersection and Arc

Do we need to memorize these formulas for the GRE Exam? Or will they provide it?
My Exam is less than 10 days and I have no time to memorize ALL of them :(
greenlight-admin's picture

You are not provided with any formulas when taking the GRE.

My exam is in 12 days and I am concerned with how to avoid silly mistakes on the GRE Quant Section. I know the formular and I have the math skills but due to the time constraint and all I end up making some mistakes that I would not have made if given more time; which may have led me to attempt the questions in a more relaxed state of mind. Please what do I do?

Great! For everything you a solution!!

We have been asked to find the length and not the area but we have used the area of arc formula , why ?
greenlight-admin's picture

The formula used in the solution is the formula for finding arc length.

There is no such thing as the area of an arc. If you're thinking about the formula for finding the area of a sector, that formula is:

Area = (central angle/360)(pi)(radius²)

More here:

This question is from Kaplan quantitative reasoning practice sets. And I have found a tedious way to answer this difficulty level hard question. Is there any better approach to solve the question, the one mentioned below. Thanks in advance!
Qn.: Circle A and circle B have their centers at points A and B, respectively. If point A lies on circle B, point B lies on circle A, and line segment AB has a length of 6, what is the area of region that lies within both circle A and circle B?
A) 12π - 6√2
B) 12π - 12√3
C) 24π - 6√3
D) 24π - 18√3
E) 24π - 24√2
greenlight-admin's picture

Very interesting. That's pretty much the same question as ours:

As you'll see, it's a long solution. I'm not sure if there's a faster way.

I was able to find Y no problem, put it into the equation and I get 80/360 (60 pie). Simplified this down to 2/9 (60pie) then you lost me. Where to go after that step is beyond me
greenlight-admin's picture

If you had no problem getting to the point where the length of arc length of CDE = (2/9)(60pi), then here's where we need to go from there:

arc length of CDE = (2/9)(60pi)
= (2/9)(60pi/1)
= (120pi)/9 [I multiplied numerators and denominators]
= (40pi)/3 [I divided top and bottom by 3]
= Answer choice D

I didn't know the formula, so I used proportion.

80 / 360 = s / 60π Where s is the arc length

4800π = 360s

40π / 3 = s

Answer is D

greenlight-admin's picture

Perfect approach!!

Please if O is the center doesn’t it make line AO and OC equal and hence angle x- 90 degrees, if we add an imaginary line from A to C
greenlight-admin's picture

Hi Angel,

Which angle are you saying should equal x-90 degrees?


Please I meant Angle X. should be equal to 90 degrees
greenlight-admin's picture

You're correct to conclude that, since O is the center, AO = OC. However, this does not mean that ∠AOC = 90°

If we add an imaginary line from A to C, we can see that ∆AOC is an ISOSCELES triangle, in which ∠OAC = ∠OCA, but we can't make any conclusions about the values of any of the angles inside ∆AOC.

To make conclusions about the angles inside ∆AOC, we need to use additional information.

Does that help?


I didn't notice the line, I used the given information about x and y, solve it using algebra:
x-y = 20, means x = y +20. Opposite angles are equal which means we have 2(y+20) + 2 (y) = 360 ==> 40 +4y = 360 ==> y = 80.
then use the length of arc formula and the given radius to get the required result.
greenlight-admin's picture

That's a perfectly-logical approach. Nice work!!

I did it like this:

So w + x + y + O = 360
w = y (opposite angles)
x = 20 + w --> x = 20 + y
O = x = 20 + y

So we have:
y + (20 + y) + y + (20 + y) = 360
4y + 40 = 360
4y = 320
y = 80

Then plug this into the formula!
greenlight-admin's picture

That works perfectly. Nice work!!!

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