1) A triangle could possibly intersect a circle at the following number of points:
I. 3
II. 4
III. 5

(A) I only
(B) I & II only
(C) I & III only
(D) II & III only
(E) I, II, III
2) Each circle in the diagram above has a radius of r = 6. What is the total area of the shaded regions?
3) Triangle STV has sides ST = TV = 17, and SV = 16. What is the area?
(A) 85
(B) 100
(C) 120
(D) 136
(E) 165
4) Given that ABCDE is a regular pentagon, what is the measure of ∠ACE?
(A) 24°
(B) 30°
(C) 36°
(D) 40°
(E) 45°
5) In the diagram above, ED is parallel to GH, and the circle has a diameter of 13. If ED = 5 and GH = 15, what is the area of triangle FGH?
(A) 240
(B) 270
(C) 300
(D) 330
(E) 360
6) In the diagram above, A & B are the centers of the two circles, each with radius r = 6, and ∠A = ∠B = 60°. What is the area of the shaded region?
7) In the diagram above, point B is the center of Circle #1 and point D is the center of Circle #2. If the ratio of the area of Circle #2 to the area of Circle #1 is 3:2, what is the ratio CE:BC?
8) In the diagram above, AB is parallel to EH, and BD is parallel to FH. Also, AB = BC, and EF = FH. If ∠EGC = 70°, then ∠D =
(A) 65°
(B) 70°
(C) 75°
(D) 80°
(E) 85°
Answers and explanations will come at the end of this article.
Geometry
On the GMAT quantitative section, you need be familiar with the basics of Geometry: parallel lines & angles; all ordinary triangles, isosceles triangles, equilateral triangles, the two special triangles, as well as the Pythagorean Theorem and the triplets that satisfy it; quadrilaterals; polygons, including regular polygons; circles — their areas, their arcs, their tangent lines, and their inscribed polygons; and 3D solids. It’s important to know about similarity and scale factors. Some ideas in Geometry have farreaching implications, while other situations can encourage naïve assumptions. If this topic is rusty, you may find some refreshers in those posts.
Summary
If you read some of those linked blogs, you may want to give the practice problems another look before reading the solutions below. Here’s another Geometry question, from inside the Magoosh product.
9) http://gmat.magoosh.com/questions/1024
If you having anything you would like to share about Geometry, or if you have a question about something appearing in this blog, please let us know in the Comments sections.
Practice problem explanations
1) Here are diagrams showing all three cases:
All three are possible. Answer = (E)
2) It would be a huge mistake to do any advanced calculations to solve this. The three lower oval shapes can be inserted into the three hollow spaces adjacent to the upper shaded region, and together, the four shapes will neatly form a single circle. The circle has r = 6, so area = 36(pi). Answer = (B)
3) Let’s think about this triangle:
Triangle STV is isosceles, so the perpendicular line from vertex T is bisects base SV. Thus, SW = 8. Now, look at right triangle ATW: it has leg = 8 and hypotenuse = 17. It will save you a tremendous amount of calculations here if you already have memorized the 81517 Pythagorean Triplet. Thus, TW = 15, and that’s the height. Area = (0.5)bh = (0.5)(16)(15) = 8*15 = 120. Answer = (C)
4) First of all, all the angles in any nsided polygon add up to (n – 2)*180°. For any pentagon, that would be 3*180° = 540°. For a regular pentagon, the five angles are equal, so each one is 540/5 = 108°.
Now, look at isosceles triangle ABC, with an angle of 108° at B. The other two angles are equal: call each x. 108 + x + x = 180, which leads to x = 36°.
So, ∠BACA = 36° and ∠ECD = 36°, which means that ∠ACE = 36° Answer = (C)
5) Let’s begin by focusing on triangle FED. The angle ∠E spans a diameter, so ∠E = 90°. Thus, triangle FED is a right triangle with hypotenuse FD = 13 and leg ED = 5. It will save you a tremendous amount of calculations here if you already have memorized the 51213 Pythagorean Triplet. Thus, FE = 12. Area = (0.5)bh = (0.5)(12)(5) = 30.
Because ED and GH are parallel, all the angles are equal, and the two triangles are similar. From ED = 5 to GH = 15 we scale up by a scale factor of k = 3. Lengths are multiplied by the scale factor, and areas are multiplied by the scale factor squared, k^2 = 9. 30*9 = 270 is the area of FGH. Answer = (B)
6) Step #1: Each circle has an area of
Step #2: One sector, one “slice of pie”, occupies 60°, one sixth of the circle.
Each sector has an area of
Step #3: Now look at the equilateral triangle.
This has a side of s = 6, so it’s area is
Step #4: A circular segment is the name for that little leftover piece, the part of the sector that’s beyond the triangle:
As may be clear, the (area of segment) = (area of sector) – (area of triangle), so the sector has an area of
Step #5: Now, notice that the shaded region in the diagram is two equilateral triangles minus two circular segments. That’s exactly as we’ll calculate it.
Answer = (D)
7) This is tricky. We are given the ratio of areas, and we want to know a ratio involving the radii. Let R be the radius of the bigger circle, and r be the radius of the smaller circle.
Take the square root and rationalize the denominator:
Well, CE = 2R, and BC = r, so CE/BC = 2R/r, which is twice this ratio
Answer = (D)
8) First of all, to solve this, we need to know the properties of parallel lines & angles. Since ∠EGC = 70°, we know that an angle formed by the same line intersecting both EH & AB will make the same angle: ∠A = 70°.
Now, we know that, because AB = BC, triangle ABC is isosceles, which means that ∠ACB = 70°. The three angles have to add up to 180°, so this tells us that ∠B = 40°.
At this point, we reach a very tricky move: both ∠B and ∠H are angles formed by the pairs of parallel lines — the sides of each one are parallel to the corresponding sides of the others. These means ∠B = ∠H = 40°.
Now, we know that, because EF = EH, triangle AFH is also isosceles, which means ∠GEF = 40°. The three angles have to add up to 180°, so this tells us that ∠F = 100°.
Finally, ∠F and ∠D are two angles on the same side of the same line between two parallel lines (“same side interior angles”). These angles must be supplementary, which means they have a sum of 180°. That means ∠D = 80°.
Answer = (D)
More on why ∠B and ∠H are congruent: For some people, this is intuitive, but for others this is a really trick one. Both of those angles are formed by two pairs of parallel lines, so they have to be congruent. Why? One way to think about this — suppose segment DEF were extended to the left of D, and suppose AB were extended above A, so that it intersected the extension of DEF — call that point Q. Then, we would have two triangles, HFE and BDQ — the pairs of angles at F & D and at E & Q would be equal, because the extended version of QDEF would a common transversal crossing both sets of parallel lines. Then, the two triangles would be similar, HFE ~ BDQ, by AA Similarity, and that would mean the angles at B & H would have to be congruent.
Mike:
In problem 6, step 3, based on the given data, how do you determine that triangle is equilateral? In other words, how do you determine CD=6?
Dear Srini,
I’m happy to respond. Look at triangle ACD. We know A is the center of the circle, so AC = AD are radii, and therefore of equal length. This means that triangle ACD is isosceles, with at least two equal sides, AC & AD. Now, angle A = 60, so the other two angles have a sum of 120 degrees. Well, by the Isosceles Angle Theorem, those other two angles must be equal, 60 degrees each. Three 60 degree angles: that’s an equilateral triangle.
This is definitely an example of the kind of geometric reasoning the GMAT could expect you to do in a problem.
Does all this make sense?
Mike
Yes, it does Mike. Thank you!
Dear Srini,
You are more than welcome, my friend. Best of luck to you in the future!
Mike
Hi Mike,
I am confused as to why triangle ABC and EFG in the 8th problem not similar triangles?
AB and BH are parallel aren’t they then by AA rule aren’t the two triangles Similar?
Thanks
Dear Sads,
I’m happy to help.
It’s true that AB and EH are parallel, but one set of parallel lines is not enough to determine more than one angle if the triangles are not between the two parallel lines. It’s true that angle B = angle H, because of the double parallel lines. Because EF = FH, by the Isosceles Triangle theorem, we know that angle H = angle EFG, Then, by transitivity, we know angle B = angle EFG. That’s one equal angle the two triangles have in common. In triangle ABC, we know AB = BC, so the angles at B & C would have to be equal. We have no guarantee that EF and EG are the same length — in fact, in this “drawn to scale” diagram, it very much looks as if EF is considerable longer than EG. Therefore, ABC is an isosceles triangle, and EFG is not, so they can’t possible be similar.
Does all this make sense?
Mike
Hi mangoosh,
I am stuck @ 6th question. I am able to follow till the step where we subtract the area of triangle from the area of the sector.
Area of circular segment=6*Pi* 9 *sqrt{3}.
We have two circular segments so shouldn’t the answer be 2*(6*Pi* 9 *sqrt{3})
= 12*Pi* 18 *sqrt{3}.
Can you please elaborate step 5 too??
thanks
Tensor
Dear Tensor,
I save all the “doubing” for the last step, step #5. In step #4, I simply calculate the area of ONE circular segment. That’s precisely what I intend to do there, because I wait until step #5 to put everything together and double — TWO triangles minus TWO circular segments. Those are the factors of 2 in front of the two parentheses in that step.
Does all this make sense?
Mike
Thanks mike. My bad, I thought question asked us to find the area of two circular segments. Now it makes sense.
In question number 7. can you elaborate how angle B and angle H are corresponding angles? I don’t see any traversal between line segment AB and GH.
Thanks
tensor
Tensor,
That’s a really trick one — both of those angles are formed by two pairs of parallel lines, so they have to be congruent. One way to think about this — suppose segment DEF were extended to the left of D, and suppose AB were extended above A, so that it intersected the extension of DEF — call that point Q. Then, we would have two triangles, HFE and BDQ — the pairs of angles at F & D and at E & Q would be equal, because the extended version of QDEF would a common transversal crossing both sets of parallel lines. Then, the two triangles would be similar, HFE ~ BDQ, by AA Similarity, and that would mean the angles at B & H would have to be congruent.
Does all this make sense?
Mike
At this point, we reach a very tricky move: both ∠B and ∠H are angles formed by the pairs of parallel lines — the sides of each one are parallel to the corresponding sides of the others. These means ∠B = ∠H = 40°.
Sorry Mike Im still not able to grasp this point!
Pls help
Thanks
Dear Sads,
I explained one way to think about this in this response to Tensor, and because you asked, I went ahead and copied that text and added it to the text explanation above. If you cannot visualize all that in your head, I suggest that you write it out on paper and look at it. Often, the best way to understand geometry is visually. Draw out the diagram, add the extended lines as I suggest, and that should make everything clear.
Mike
Hi Mike,
Yes now I understand that the two angles are corresponding angles between parallel lines BD and FH!
Thanks a lot
Dear Sads,
You are quite welcome. Best of luck to you!
Mike