1) A triangle could possibly intersect a circle at the following number of points:
- I only
- I & II only
- I & III only
- II & III only
- 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?
4) Given that ABCDE is a regular pentagon, what is the measure of ∠ACE?
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?
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 =
Answers and explanations will come at the end of this article.
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 far-reaching implications, while other situations can encourage naïve assumptions. If this topic is rusty, you may find some refreshers in those posts.
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.
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 8-15-17 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 n-sided 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 5-12-13 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.
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.