Many geometry problems ask you to find the area of a shape or to compare the area of two shapes or to use the area of a shape to deduce some other value. In previous posts, we have already given the standard formulas for finding the area of such shapes (see our appendix for a full list of such formulas). But the GRE will often give you the relevant information in subtle ways, or ask you to do surprising things with the information they’ve given. So in what follows, we give some practice problems intended to help you become used to using these standard formulas in less standard ways.
But first, here are some general steps to follow in trying to solve a geometry problem:
1. Record what the problem tells you.
- The problem may give you certain lengths or angle measurements or areas or perimeters; keep track of this. It is best to write it down somewhere, either on a diagram or just in a list.
2. If you’re stuck, write down some formulas you think may be relevant.
- Thus, if the problem asks you to find the area of some shape, write down the area formula for that shape! Or if the problem gives you a 45-45-90 right triangle, write down the ratio of the side lengths. Sometimes writing such stuff down can help spark connections in your mind.
3. If you’re stuck, look back carefully at the problem and see if there’s any information you missed.
- Generally, these problems are quite parsimonious: they give you exactly what is needed to solve the problem, and no more. So if you find yourself with some unused fact, try and fit it into the problem somewhere. It’s unlikely they would’ve included something totally irrelevant to your problem.
4. If you don’t think you can get the solution quickly, just move on!
- Remember that the math section gives you 35 minutes for 20 questions, so you can only afford to spend an average of one minute and 45 seconds per question. There’s no shame in flagging a question that looks tough so that you can come back to it if time permits. Look at it this way: if you don’t get through the test, there may be easy questions down the road that you’re effectively giving up on. By flagging it and coming back later, you help ensure that you get all the low-hanging fruit in the test.
1. In the following diagram, and are both squares with side lengths of 2 and area of area of area of Find the area of the shaded rhombus.
Note that since both are squares with a side length of 2, we know that This is because all of those line segments must be perpendicular to a line that is parallel to . This means that , and are all rectangles. And as 's area equals the area of , and they together comprise the area of the entire square , we conclude that the area of area of Thus, and
Now, we are given that the shaded figure is a rhombus, so we can apply the formula for the area of a rhombus, namely which equals 2. Thus, the shaded region has an area of 2.
2. In the following diagram, , and are equally spaced. How many scalene triangles can be formed with vertices at those points?
The trick to answering this question is to notice that, no matter which three points you pick, you are really forming what is, in some sense, the same triangle (more precisely, they are all congruent). So, for example, the triangle is really just a rotated version of the triangle . And, of course, both triangles are isosceles (not scalene).
Alternatively, you could just check all the possible triangles. Now, since you need exactly three distinct points to form a triangle, there are four possible triangles (as there are 4 ways to choose three points from four possible points; see our posts on combinatorics here). And you can check each such triangle (namely ) to see that they are all isosceles. So the answer is 0.
3. The following points are equally spaced. How many equilateral triangles can be formed? If three distinct points are randomly chosen, what is the likelihood that they form an equilateral triangle?
To determine how many equilateral triangles can be formed, note that there are only two possible equilateral triangles: and . This can either just be seen based on the diagram, or can be inferred by noting that for an equilateral triangle, the points chosen must be the same distance away from one another. There are six points in total, so one must pick every other point in order to form an equilateral triangle (otherwise some side of the triangle will be longer than some other side). And there are only two ways to pick every other point: you can pick or
Now, to determine the probability that a randomly chosen set of three points will generate an equilateral triangle, we simply need to find how many ways you can choose three points from six points, since any choice of three distinct points will generate a triangle. Thus, we calculate Thus, there is a probability of getting an equilateral triangle.
4. The outer rectangle below is units apart from the smaller rectangle on the top and bottom sides, and unit apart on the left and right sides. The inner rectangle has a height of 4 units and a base of 2 units. What is the area of the outer rectangle?
Let's fill in the above diagram with the information given in the question:
Thus, we can see that the dimensions of rectangle are a base of 4 units and a height of 8 units. So the area of must be 32 units squared.
5. The radius of both circles below is 1. What is the area of triangle ?
The first thing to note is that triangle is an equilateral triangle since all of its sides are radii with length 1. So really, this question asks about the area of an equilateral triangle with sides of length 1. If we split the triangle down the middle, we create two 30-60-90 right triangles, like so:
and you can either use the Pythagorean Theorem or what you know about 30-60-90 right triangles to conclude that the height of the triangle must be . Thus, the area of the triangle is
6. What is the area of ?
The crucial observation here is that triangle is similar to triangle . How do we know this? Well, we know:
And thus, that . So we conclude that and thus and share all the same angles. So they are similar.
We also know the ratio between their sides since the short side of is 3 units long while the short side of is 4 units long. Thus, the length of must be So now we can find the area of the whole triangle as:
7. In the following diagram, What is the area of ?
This problem just requires knowing the ratios of a 30-60-90 right triangle. By our remarks in problem 6 above, we know that and and are all similar triangles. So we know that by the ratios of a 30-60-90 right triangle and we know that is equal to (as is a 30-60-90 right triangle) so we conclude that Thus, we get: as the area of
Challenge Problem: You are not likely to see anything as convoluted as the following on the GRE. Still, it might be a good way to push your understanding of the above material:
8. The radius of the following circle is 5. Find the shaded area.
Note that we can begin by drawing the following right triangle:
There are several ways to conclude that the triangle is a right triangle. For example, note that the diagonal of the square is the diameter of the circle and hence is equal to 10. Then, say that each side of the square is units long. By the Pythagorean Theorem, and so . By the Reverse Pythagorean Theorem, we conclude that the triangle above is a right triangle since it has legs of length 5 and a hypotenuse of length , satisfying the Pythagorean Theorem.
Alternatively, you could look at the picture and argue that since the figure is symmetric across the horizontal and vertical axes, the angle of the triangle at the center of the circle must be the same as the other three angles, so each must be 90 degrees.
Now, we know that the area of that sector of the circle must be the total area of the circle. Now, we just need to subtract the area of the non-highlighted bits.
The large triangle's area is easy to find since we know the base and height are both 5, as radii of the circle. So it has an area of But the smaller triangle is a little trickier. Let's add a little more to our drawing:
Now, we know that since the triangle is isosceles and the line splits it in half. Thus, we know that And we know that and thus as a line has 180 degrees. So we can use the Pythagorean Theorem to find the length of Doing so, we get Thus, So the area of the smaller triangle, in total, must be
We can thus conclude that the area of the shaded region must be: