As anyone who’s ever chewed on a pencil knows, it doesn’t take much force to put a dent in a regular old #2 pencil. You might have an opportunity to use this to your advantage on the SAT. Occasionally, a geometry question will appear that asks you to figure out the length of a segment based on information about the lengths of other segments. If the figure is drawn to scale, you and your thumb nail are in business.

I don’t want to oversell this. You’re not going to see a huge score increase because of it. But if you desperately need to compare the lengths of a few segments, you can make notches in your pencil with your thumb nail, and then use your new makeshift ruler to solve the problem. Let’s look at an example where this might come in handy.

  1. In the figure above, PTACis the midpoint of perpendicular segments BD and PT, and D is the midpoint of AC. If AC = 4, BD = 8, and PQ = 3, what is the length of PT?(A) 6
    (B) 8
    (C) 10
    (D) 10.5
    (E) 11

So there’s a mathematical solution to this, of course, but since it’s drawn to scale and all we’re asked to do is compare the lengths of a couple line segments, I’m going to show you how to do this one the quick and dirty way. (Commenters, feel free to post the geometric solution if you like.)

Buckle up, things are about to get meta:

That’s right. I printed my own blog, and took pictures of it, and now I’m posting them on my blog. But seriously though, do you see what I’ve done here? I know BD = 8, and since the figure is drawn to scale, I can make thumbnail dents in my pencil and use them to approximate other lengths. Let’s look at PT:

OMG. It looks like PT = 8 too! In fact, there aren’t any other choices that are close enough to 8 for me to do any more work. I’m going to pick (B) and move right along.

Now, let me repeat what I said above: this trick is not going to win you beaucoup points; you might go through an entire test without an opportunity to use it. But it’s one more way to attack a geometry question that’s drawn to scale, and it just might help extricate you from a bind. Keep it in the back of your mind.

Comments (6)

I’ve been thinking of an alternate solution. If AC is 4, then AD is 2. PQ is 3 and seems to overlap with the AD if you draw a vertical line up from A. With guestimate, There looks to be the 1/2 of AD left, so that is 1. That makes 4, which becomes 8 when doubled. Does that make any sense?

I actually haven’t figured out any sort of algebra I could use to solve this. I was trying right triangles though. Will this sort of question ever come up as a not drawn to scale?

I figured it out with the similar triangles. They can be similar by AAA or just by looking at it I suppose. The big triangle’s height is 8 and the small’s height is 4. The big triangle’s base is 4. So the small triangle’s base is 2. Point R is the midpoint, so each side’s base is 1 in the small triangle.

Sorry to bump an old post:

I figured that connecting points B, T, D, P into a square would do the trick.

Although I have an explanation, I don’t think I can type it down concisely in a way that makes sense. But here goes something anyway.

Points P and T are at equal distances away from point B because point R is the midpoint of line PT, and also because point B lies in the same line as point R.

The distance from point D to points P and T is the same with the above, because point R is the midpoint of line BD. In other words, point D is just like a reflection of point B over the “imaginary” x-axis.

So our square BTDP has four equal sides, because the distances/lengths of lines BT, TD, DP, and PB are equal.

Notice that BD and PT are really diagonals of our square BTDP. And since diagonals of squares have the same length, whatever the length of line BD is the length of line PT.

With our given length of line BD = 8, we can tell that line PT is 8 as well.

Hopefully my approach made sense.

P.S. The pencil ruler trick is brilliant. It’s really mind-blowing.

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