### Video Transcript

Find the length of the projection of line segment ππΈ on line ππ·.

When weβre considering projections, we need to start by considering the target line, the line onto which the projection will fall. In this case, thatβs the line ππ·, which is here. And we want to see the projection of line segment ππΈ onto this line, so we highlight ππΈ. And when we think about projections, we need to consider perpendicular lines to the target line. In this image, we see that the line segment πΈπ· is perpendicular to the target line ππ·. And when weβre thinking about projections, we want to consider the set of all the lines that are perpendicular to ππ·.

If we consider that these perpendicular lines are a light source, the projection of line segment ππΈ is going to be the shadow that line segment ππΈ would make onto line ππ·. The point πΈ as a shadow would be the point π·. And since π already falls on our target line, it would be itself. The shadow of line segment ππΈ onto line ππ· would be the line segment ππ·. But in order to find the length of ππ·, weβre going to have to think about what we know about right triangles.

First of all, we can clear off our perpendicular lines. We see that all four of these segments are equal in length, so we can label them as 18 centimeters. In order to find a missing side length in a right triangle, we need at least two of the side lengths to use the Pythagorean theorem, which tells us that the two smallest sides squared and added together will equal the square of the longest side. This π-value is the hypotenuse. Itβs the side length opposite the right angle. And that means weβll be working with the triangle π·πΆπ. However, inside triangle π·πΆπ, weβre missing two side lengths, which means weβre not able to solve for a third side length. This is also the case in triangle πΆπ΅π.

However, if we look at triangle π΅π΄π, we do have two side lengths. We can start by finding the hypotenuse of triangle π΅π΄π. We say that 18 squared plus 40 squared will equal the hypotenuse squared, π squared. When we square those values, we get 324 plus 1600 equals π squared. Adding those two values together, we get 1924 equals π squared. And if we take the square root of both sides, we can say that π equals the square root of 1924. Weβre only interested in the positive square root, since weβre dealing with length. Now, instead of trying to estimate or simplify the square root of 1924, I know that we will be squaring this value later, so I can just leave it as is. Side length ππ΅ is equal to the square root of 1924.

And when we look at the next right triangle, we now have two side lengths. For the triangle πΆπ΅π, square root of 1924 squared plus 18 squared will equal π squared. 1924 plus 324 equals π squared, which is 2248. And again, weβll take the square root of both sides of the equation. And we get a hypotenuse value of the square root of 2248. Again, we wonβt worry about simplifying as weβre just using this value to find another value. We just say that side length ππΆ equals the square root of 2248.

And now, we finally have two side lengths in the triangle weβre interested in, π·πΆπ. And we can say the square root of 2248 squared plus 18 squared equals π squared. 2248 plus 324 equals π squared, which is 2572. And then we take the square root of both sides, which means the projection ππ· is equal to the square root of 2572. And if we wanted to simplify this value, we could see that 2572 is divisible by four. Four times 643 equals 2572. And we know the square root of four is two. So, we can rewrite this to say two times the square root of 643.

The length of the projection ππΈ onto ππ· is equal to two times the square root of 643 centimeters.