Overkill1

Square Root in AutoCAD – Solution 2

In All, Architecture and Engineering by Brian HaileyLeave a Comment

This is the final post that I’ll be doing on the really cool idea that you can calculate the square root of a number by drawing a few lines and a circle. You can see the original post HERE as well as the first solution HERE.

In the first solution, we took an algebraic approach to solving this (using the Pythagoras theorem) but in this example, we’ll take more of a geometric approach. First thing we must understand is that if you draw two lines from the ends of a diameter of the circle to any point on the circle, those two lines will always be perpendicular.

Diameter Right Angle 1

As you can see in the above image, the magenta line is a diameter to the magenta circle. Drawing the two white lines from the ends of the diameter to a point on the circle results in a right angle between them. Don’t believe me? Try it yourself in AutoCAD. I’m not going to get into the details of why this is, we’re just going to assume that this is indeed the case.

Overkill

Now that we have that out of the way, let’s take a look at our situation. The red line X has the length that we want to find the square root of, the green line is one unit length long, and the white line Y is the value we want to find relative to X.

Problem

So, what can we determine from this diagram? Well, there are three different triangles; ACD, ABD, and BCD. The nice thing about thees three triangles is they are all similar. One of the easiest ways to determine if two triangles are similar is to show that they each have two angles that are the same. In the following image, the three triangles have been broken out individually.

Similar

First, let’s take a look at the top two triangles, ACD and BCD. We know that because line AC is a diameter of the circle and point D is on the circle, then we can see that the angle ADC is a right angle. We also know that angle DBC is a right angle because we constructed it that way. The second angle that is the same between them is angle a. Therefor we can conclude that triangles ACD and BCD are similar. We can do the same for the left two triangles ABD and ACD. There’s the one common angle b that they share and we’ve already seen that angle ADC is a right angle and angle ABD is a right angle so these are also similar. We can then see that all three triangles are similar. If triangle A is similar to triangles B and D, then triangles B and D must be similar.

So, how does this help us out? One of the properties of similar triangles is the ratio of the common side between two similar triangles is a constant. Basically, it’s the scaling factor between the two triangles. If triangle A has sides A, B, and C, and it similar to triangle A’ with sides A’, B’, and C’, then A/A’=B/B’=C/C’. To help you visualize it a bit better which sides correspond to each other, imagine taking the bottom left triangle and rotating it ninety degrees clockwise.

In the following image, I’ve gone ahead and marked the common sides between the two triablges and also labeled the lengths of the lines we know or want to know.

Similarity

We know that the ratios of the sides are the same between two similar triangle so side AB on the left divided by side BD on the right will equal AD on the left divided by CD on the right which will also equal BD on the left divided by BC on the right. Since AB=X, BD=Y, and BC=1, we can create an equation using just those values.

Ratios

As you can see, Y is the square root of X.

I hope you’ve enjoyed learning about this as much as I have. I think this is a really cool bit of information. Now go out there and be awesome!

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