Figure(s) Drawn to Scale

This tip for improving your SAT score was provided by Samhita Noone at Veritas Prep.

Most figures on the SAT Math are drawn to scale, and you should take advantage of that. Students often miss the fact that if the question does not state “Figure(s) Not Drawn to Scale” underneath or above the figure, then the figure has been drawn accurately, meaning you have free rein to estimate. You don’t have to do the math explicitly for every single problem.

When should you estimate? If you can crunch the numbers, go ahead. On many problems, the answer choices will be too similar for an estimate to give you a meaningful method of elimination. Estimation will allow you to affirm your answer choice and make sure you didn’t make an error in arithmetic. Furthermore, a couple of options are usually clear outliers, and estimating can allow you to eliminate these. Statistically, if you can eliminate even one answer choice, you should go ahead and guess.

Essentially, if you have no idea how to go about a problem, and the problem comes with a figure drawn to scale, go ahead and check if you can eliminate even one answer and then guess. And even if you have crunched the numbers, quickly check the figure to make sure your answer makes sense.

Let’s take a look at a sample question to see how we can practice this strategy. The following appears on the 2013-14 Official College Board Pretest. First, I’ll crunch the numbers. Second, I’ll solve the problem by estimating.

Method 1
The crucial word in this problem is “semicircle,” which is defined as exactly half a circle. You may not know what to do with that strange curve, but you do have formulas to deal with circles. Knowing that, you can break RS into the two radii of the semicircles. If I were to label the point of intersection of line l and the curve as X, I could say that RS = RX + XS. Since RS = 12, we know that RX + XS = 12, as well.

Next, I’m going to think about what I’m actually trying to solve. I need the length of the curve (which I am going to call C), meaning I need half the circumference of the first circle added to half the circumference of the second circle. The formula for circumference is 2πr. I can substitute RS and XS as the radii(r) in this formula, since the problem gives to us that points R and S are in the middle of the semicircles.

Thus C = 1/2 2πRX + 1/2 2πXS.

I can simplify to C = πRX + πXS
Then I can factor out π from both terms: C = π(RX + XS)

Finally, we already know that RX + XS = 12. We can substitute that in, and we end up with: C = 12π.