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# Circle Problems on the GMAT Here are eight practice problems involving circles.  The difficulty of these problems ranges from medium to very challenging. 1) In the diagram above, O is the center of the circle and angle AOB = 144º.  What is the area of the circle?

Statement #1: The area of sector AOB is 40% of the area of the circle

Statement #2: Arc ACB has a length of . 2) In the diagram above, ABC is an equilateral triangle.  D is the midpoint of AC.  BD is a diameter of the circle.  If AD = 4, what is the area of the circle?  3) In the diagram above, O is the center of the circle.  What is the length of chord AC?

Statement #1: chord BC = 14

Statement #2: the circle has an area of  4) Points P, Q, R, S, and T all lie on the same line.  The larger circle has center S and passes through P and T.  The smaller circle has center R and passes through Q and S.  What is the ratio of the area of the larger circle to the area of the smaller circle?

Statement #1: ST:PQ = 5/2

Statement #2: RT:PR = 13/7

5) A square and a circle intersect at more than one point.  Does the square have more area than the circle?

Statement #1: there are exactly four intersection points

Statement #2: at least two of the intersection points are on vertices of the square 6) In the diagram above, all the points are on a line, and the number of each point indicates how many units that point is from zero.  The points #1 – #6 are the centers of the six circles, and all circles pass through point zero.  What is the total area of the shaded region?  7)  In the diagram above, JKL is an equilateral triangle.  Point M is the midpoint of segment JL, and M is the center of a circle that passes through points J and L.  The shaded regions in the diagram indicate all the regions inside the circle that are outside the triangle.  What fraction of the total area of the circle is outside the triangle?  8) In the diagram above, angle C = 90º and AC = BC.  Point M is the midpoint of AB.  Arc AXB has its center at C, and passes through A and B.  Arc AYB has its center at M and passes through A and B.  The shaded region between the two arcs is called a “lune.”  What is the ratio of the area of the lune to the area of triangle ABC? ## Circles

Ah, the beauty of the circle, the most symmetrical of shapes!  For more information about circles, see these blogs:

If you have any insights while reading those blogs, you may want to give the problems above a second look.  If you find any typos or anything on this page is unclear, please let us know in the comments section. ## Practice problem explanations

1) First of all, we should think about the prompt a bit.  This angle, 144º, is what fraction of a full circle?  Well, both 144 and 360 are divisible by 4: 144 ÷ 4 = 36 and 360 ÷ 4 = 90, so This angle is 2/5 of the whole circle, so the arc is 2/5 of the whole circumference.  Also, remember that if we can find anything about the whole circle, for example, the circumference, then we can find the radius, which would allow us to find the area.

Statement #1: This is a tautological statement.  A tautological statement is a statement that, by definition, has to be true, and because of this, it contains no information.  Statements such as “My car is a car” and “My employer employs me” are verbal tautologies: they contain no useful information.  Much in the same way, we already know from the prompt that the angle takes up 2/5 of the circle, so of course the sector would take up 2/5, or 40%, of the area.  This statement repeats information in the prompt, and contains no new information, so it doesn’t help us at all to figure out anything else.  This statement, alone and by itself, is not sufficient.

Statement #2: We already know this arc is 2/5 of the whole circumference, so we could set up a proportion to find the circumference.  From that, we could find the radius, and that would allow us to find the area.  This statement, alone and by itself, is sufficient.

2) Because ABC is an equilateral triangle, right triangle ABD is a 30-60-90 triangle, one of the GMAT favorite triangles.  From the ratios of that triangle: Half of that is radius of the circle: Use that to find the area: 3) The fact that AB is a diameter guarantees that angle C = 90º.  If we had two sides of right triangle ABC, we could find the third using the Pythagorean Theorem.

Statement #1: this gives us only one side of a right triangle: not helpful.  This statement, alone and by itself, is not sufficient.

Statement #2: this allows us to solve for the radius and, hence, the diameter, so we can determine side AB.  Nevertheless, this gives us only one side of a right triangle: also not helpful.  This statement, alone and by itself, is not sufficient.

Combined statements: We get the length of BC from the first statement, and the length of AB from the second.  Now, we have two sides of the right triangle, so we can use the Pythagorean Theorem to solve for the third side, AC.  Combined, the statements are sufficient.

4) Call the radius of the larger circle y, and y = PS = ST.  Call the radius of the smaller circle x, and x = RS = RQ.  If we took a ratio of the areas, the factors of would cancel and we would be left with the ratio (y/x) squared.  If we could solve for this simpler ratio, y/x, then we could find the ratio of areas.

Statement #1: ST = y and PQ = y – 2x, so This allows us to solve for the ratio y/x, which would allow us to find the ratio of areas.  This statement, alone and by itself, is sufficient.

Statement #2: RT = y + x and PR = y – x, so Cross-multiply. This allows us to solve for the ratio y/x, which would allow us to find the ratio of areas.  This statement, alone and by itself, is sufficient.

5) Statement #1: this information, with nothing more, could mean that the circle is either smaller or larger. This statement, alone and by itself, is insufficient.

Statement #2: this information, with nothing more, could mean that the circle is either smaller or larger. This statement, alone and by itself, is insufficient.

Combined Statements: One possibility is the circle that intersects the square four times by passing through all four vertices: That circle is clearly bigger than the square.  The circle absolutely cannot pass through exactly three vertices.  If it pass through two vertices, it would have to intersect the side two more times.  Possibilities include the following (point C is the center of the circle). Notice that, as point C approaches the top side of the square, it gets closer and closer to the circle that has this top side as a diameter, equivalent to the first circle in the statement #1 diagram.  That circle is clearly has less area than the square.  Well, that circle won’t work here, because it intersects at only two points, but because point C could get closer and closer to the top side without touching it, which means the area of the circle in this diagram could get closer and closer to the area of the first circle in the statement #1 diagram.  This means that we could make the circle in this diagram have less area than the square has.

Thus, even with the constraints of both statements, we can construct a circle that has an area that is either greater than or less than that of the square.  Even with both statements, we cannot give a definitive answer to the prompt question.  Both statements combined are insufficient.

6) First, let’s look at the outer “lobe,” the one between 10 and 12.  The circle through point 12 has a center a 6 and radius of 6, so its area is .  The circle through point 10 has a center a 5 and radius of 5, so its area is .  If we subtract the latter from the former, we an area of for this lobe.

Now, let’s look the middle lobe, the one between 6 and 8.  The circle through point 6 has a center a 3 and radius of 3, so its area is .  The circle through point 8 has a center a 4 and radius of 4, so its area is .  If we subtract the latter from the former, we an area of for this lobe.

Now, let’s look the smallest lobe, the one between 2 and 4.  The circle through point 2 has a center a 1 and radius of 1, so its area is .  The circle through point 4 has a center a 2 and radius of 2, so its area is .  If we subtract the latter from the former, we an area of for this lobe.

Add the areas of the three separate lobes: .

7) This is a difficult one.  For the sake of argument, let’s say that JM = ML = 1.  Then the area of the total circle would be .  That’s the denominator of our ratio.

Clearly, part of the shaded region is the semicircle beneath JL.  That part has an area of .  That’s the easy part.

The circle intersections sides JK and KL: call these points A & B. Point A must be the midpoint of JK, and point B, the midpoint of KL.  Triangle JAM must be an equilateral triangle, with sides equal to 1.  First, we will compute the area of the sector: The angle at M is 60º, so this is one sixth of the area of the circle. To get the area of that small shaded part, known as a circular segment, we need to subtract the area of equilateral triangle JAM from the area of the sector.  See this blog for the area of an equilateral triangle. area of segment = (area of sector) – (area of triangle JAM) We have two circular segments, so we double this to get both: Now, add the area of the semicircle to get the area of the whole shaded region: Divide this by the area of the circle, , to get the ratio: 8) This problem is based on a famous theorem of Hippocrates of Chios (c. 470  – c. 410 BCE), a predecessor of Euclid.   The easiest way to see this is as follows.

Let AC = BC = 2S.  This is the radius of the larger circle, which has an area of , so that the quarter circle: must have an area of .  Hold that thought.

Now, notice that triangle ABC is a 45-45-90 triangle, so that the length of the hypotenuse must be and the radius of the smaller circle is The area of that whole circle would be , and the area of the semicircle: would be half of that, .  Thus, the quarter circle of the larger circle and semicircle of the smaller circle have the same area.   That’s a very big idea!

Now, look at the circular segment: We don’t need to know that area: simply call it J.

If we subtract the circular segment, J, from the quarter circle of the bigger circle, we get the triangle: – J = area of triangle ABC

If we subtract the circular segment, J, from the semicircle of the smaller circle, we get the lune: – J = area of the lune

Because those two areas equal the same thing, they must equal each other.

area of triangle ABC = area of the lune

The two are equal, so their ratio is 1.  Answer = (A)  ### 18 Responses to Circle Problems on the GMAT

1. Milind August 28, 2018 at 7:03 am #

train is moving on a circular track whose Centre is o let A and B are two consecutive points on the track then angle aob is same as angle in equilateral triangle of the distance from Centre to respective position is 12 CM find the area of sector AOB and triangle AOB …please send answer

• Magoosh Test Prep Expert August 31, 2018 at 9:35 am #

Hi Milind,

This doesn’t appear to be a Magoosh question or an Official GMAT question, but please let me know if I’m mistaken! Unfortunately, we aren’t able to answer questions from outside sources. We are a test prep program and must prioritize our materials and Official GMAT materials. Thanks for your understanding!

2. Hari January 10, 2015 at 9:20 am #

Hi Mike,

Let me assume CB =1
Area of CAB(quarter of circle) = pi *1^2/4 = pi/4
AB = sq.rt(2)
Area of AYB (semi-circle) = pi*(sq.rt(2)/2)^2 =pi/2

Is this ratio not 2:1

Regards,
Hari

Sorry i could not express this in a better as I am unable to put symbols here. Thanks in advance

• Mike January 10, 2015 at 1:49 pm #

Dear Hari,
My friend, there’s a tiny flaw in your work. It’s true that the larger circle (center C, radius AC) has a total area of (pi), and therefore a quarter of it has an area of [(pi/4]. On that, we agree. Now, consider the whole circle with center M and radius AM: its area is (pi)/2, which you calculated correctly, but you misinterpreted this as the area of the total circle, not the semicircle. You have to divide by 2 to get the semicircle: (pi)/4, which is equal the area of the larger quarter circle. That’s why you wound up with a 2:1 ratio — you were comparing the whole of the smaller circle to a quarter of the larger circle, and that indeed would be a 2:1 ratio. The semicircle of the smaller circle to the quarter of the larger circle is the 1:1 ratio that’s germane to this solution.
Does all this make sense?
Mike 🙂

3. Arun Panda December 10, 2014 at 9:26 am #

The blog not only brings out some of the very important properties of Circle but also provides a good guidance on the topic, Undoubtedly, questions on this simplest closed curve appear complex in the test, especially, when other figures are added to it. I find that each questions digs deep into some of the logical relations that exist between circle and other regular figures such as square, equilateral and right angle triangle.

Thanks a lot- these questions are fabulous.

Arun.

• Mike December 10, 2014 at 10:38 am #

Dear Arun,
You are quite welcome, my friend. 🙂 Thank you for your kind words. Best of luck to you!
Mike 🙂

4. Rodrigo August 30, 2014 at 11:39 am #

Hi Mike, thanks for this post!

In the question 3, why can we assume that AB is a diameter? We do not have this information in the diagram nor in the text in statements A and B. Visually it seems that AB is a diameter, but I read in other post named ““DRAWN AS ACCURATELY AS POSSIBLE” that we cannot assume facts like these.

In case we cannot guarantee that AB is a diameter, the correct answer would be E because A or B are not vertices of a right triangle, and then we cannot calculate the length of chord AC.

To clarify… angle AOB could be 179° (instead of 180°) and then we could not calculate the length of chord CA because we would not have a right triangle

Please, could you clarify this point?

Thank you!

• Mike August 30, 2014 at 12:45 pm #

Dear Rodrigo,
That’s a great question. 🙂 In #3, we are told explicitly that O is the center of the circle, and any chord that passes through the center is a diameter. One thing you always can assume from a diagram is “straightness” — if a line from A to O to B looks straight, it is. Yes, it’s true we could draw a 179.9° angle there, it it would visually “look” straight, but the test is not going to do that to you. You absolutely can assume that lines are straight if they appear straight. If points A & O & B appear collinear, then they are.
We can’t assume parallel. We can’t assume that something that looks perpendicular is. But we can assume that something that looks straight is.
Does this distinction make sense?
Mike 🙂

• Rodrigo August 30, 2014 at 12:59 pm #

That is perfect Mike! Thank you very much for the clarification!

• Mike August 30, 2014 at 1:00 pm #

Dear Rodrigo,
You are quite welcome! Best of luck to you!
Mike 🙂

5. Sheriar Irani July 20, 2014 at 1:32 pm #

Hi Mike,

in the first question, on what basis do we assume that arc ACB corresponds to angle AOB? Since this is a DS question (and the figure need not adhere to the 2 conditions) couldn’t arc ACB also be the longer arc of the circle?

Also, riveting post! 🙂

• Mike July 20, 2014 at 2:54 pm #

Dear Sheiari,
I’m happy to help. 🙂 On GMAT DS, it’s true, we can’t assume, for example, that lines are parallel or perpendicular — anything in which it could be a degree or two off and look right and be close but be not exact. We can assume, though, fundamental topological qualities such as “betweenness.” From the diagram, we absolutely know that points A & B & C are on the circle, and we absolutely know that C is between A & B on the smaller arc.
Does all this make sense?
Mike 🙂

6. Skr July 19, 2014 at 9:22 pm #

Yes, it does Mike. 🙂 Thank you so much!! 🙂

• Mike July 20, 2014 at 2:47 pm #

Dear Skr,
You are quite welcome, my friend. Best of luck to you.
Mike 🙂

7. Skr July 18, 2014 at 10:07 pm #

Hi Mike,

For problem 7, how did you conclude that “Point A must be the midpoint of JK, and point B, the midpoint of KL”?

Skr

• Mike July 18, 2014 at 11:07 pm #

Dear Skr,
Look at triangle JAM. We know that JM = AM, because they are both radii of the same circle. Because those two sides are equal, the triangle is Isosceles, so we can apply the Isosceles Triangle Theorem. See:
https://magoosh.com/gmat/2012/isosceles-triangles-on-the-gmat/
The angle at J is 60, because it’s part of the larger equilateral triangle, JKL. The Isosceles Triangle Theorem tells us that the angle at A must be equal — that angle also must be 60. Well, if two of the angles in a triangle are 60, the third must be 60 as well. This means that JAM is an equilateral triangle. Since JM is obviously half the length of JL, it means that AJ must be half the length of JK. Does all this make sense?
Mike 🙂

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