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## High school geometry

### Course: High school geometry>Unit 8

Derive using similarity the fact that the length of the arc intercepted by an angle is proportional to the radius, and define the radian measure of the angle as the constant of proportionality. Created by Sal Khan.

## Want to join the conversation?

• This is a stupid question, but at one point Sal talks about similarity. This makes me think about it and although this isn't the exact video to ask this, I have a question about similarity.

Say there is a square (2d) with a side length of 2
Now say there is a cube (3d) with a side length of also 2

Would these two shapes be considered similar since you could get it by dilating the square in such a way it makes it 3d? • As David said, not a stupid question. But you've confused two types of transformation: dilation and projection.

A dilation of a shape preserves angles, so if two lines meet at some angle, then they will still meet at that same angle after dilation.

If you draw the main diagonal of your cube, and draw the diagonal of the bottom face, those two lines will meet at 45º. But after collapsing the cube into a square, the lines coincide (meet at 0º). So the "collapse a cube into a square" transformation is not a dilation.

But it is an example of a projection. You'll see a limited version of this when you get to vectors. To project something is to basically consider its shadow on another surface or line. Or you can think of it as taking all of the points in the cube and setting their z-coordinates to 0.
• I don't understand how the measure of that angle is equal to the ratio of the arc to the radius. Can anyone explain this? • You start from the theorem that the measure of the central angle equals the measure of the arc. Further, we note that we can also convert between degrees and radians, 360 degrees = 2π radians. So Sal sets up a simple proportion of part to whole=part to whole. In degrees, you have m<ABC/360 and in lengths, you have m(arc)AC/(2πr). These are proportional, so you get m<ABC/360 = m(arc)AC/(2πr), and r=AB or BC. Then, 360 and 2π cancel on both sides to end up with the equation m<ABC=m(arc)AC/BC. Same logic would apply for the other larger circle, you would just have a different radius, so m<ABC=m(arc)DE/BE.   