Save Time and Calculations on GRE Triangles

Like the Bermuda Triangle, GRE triangle questions often bring with them a sense of mystery and uncertainty. However, while the Bermuda Triangle will likely remain a mystery, triangles themselves can be understood.

Let’s look at a GRE triangle question:

The first thing you want to do on a problem like this is ask yourself: What do I know? We know the lengths of sides AB, BD and BC.  We also know that angle BDC is a right (90^o) angle.

The next thing you should ask yourself is: What does the question ask me to find? The question above asks us to find the perimeter of triangle ABC.  The perimeter is the distance all the way around.

Finally, you should ask yourself: What do I need to know in order to answer the question? Before we can find the perimeter of triangle ABC, we need to find the missing length, AC. If we can’t find AC independently, we can determine the lengths of AD and DC and add those values together to find AC (this is the method we’ll use for this problem.)

Here is where it gets interesting—and easy. Trust me!

Since triangle BDC is a right triangle, we know that BDA is also a right triangle. (Remember the rule: two angles that form a straight line must add up to 180^o.)  When we have a right triangle and we know the length of any two sides, we can use the Pythagorean Theorem (a² + b² = c², where a and b are the two perpendicular sides and c is the hypotenuse, or longest side) to solve for the length of the missing side.

However, just because we can use the Pythagorean Theorem doesn’t mean we should. Really, who wants to deal with exponents and square roots if you don’t have to? If you know two magical ratios, you will seldom need the Pythagorean Theorem.

Often on GRE triangles, the lengths of the sides of a right triangle will occur in the ratio of 3:4:5 or 5:12:13. It’s important to remember that these ratios do not necessarily give the actual lengths of the sides—because the values are ratios, they represent the side lengths pared down to their simplest form. The actual lengths could be 6:8:10 or 10:24:26 or any other multiple of the basic ratios.

Now, when we look at triangle BDA, we see that we have a multiple of 5 for AB, the hypotenuse. That means we could have a 3:4:5 triangle. Dividing 30 by 5 gives a value of 6, which is going to be the number by which the entire ratio has been multiplied; 6 times 3:4:5 yields 18:24:30. Since 24 and 30 are accounted for, the length of AD must be 18. (If multiplying the ratio by 6 did not give 2 of the 3 lengths the test-maker provided, then we would have to use the Pythagorean Theorem, after all.)

Let’s do the same thing for triangle BDC. The hypotenuse, BC, measures 26, which is a multiple of 13 (13 x 2 = 26.) If we multiply the entire 5:12:13 ratio by 2, we get 10:24:26. Since 24 and 26 are accounted for, the length of DC must be 10. (Again, if multiplying the ratio by 2 did not give 2 of the 3 lengths the test-maker provided, then we would have to use the Pythagorean Theorem.)

If we add 18 and 10, we find that the length of AC is 28.

Now, we have enough information to answer the question: What is the perimeter of triangle ABC? The perimeter will be the length of AB + the length of BC + the length of AC. Adding up the actual numbers, we find that 30 + 26 + 28 = 84.

Be sure that the Pythagorean Theorem and these two classic ratios are in your GRE toolbox for Test Day!