 So here are a couple more problems dealing with similar triangles. We want to find the height of this building given the fact that we are looking across a flagpole. So we're still dealing with similar triangles, and let's trace out those triangles. We have a triangle here, and then a larger triangle that includes the height of the building. So in order to find the height of the building, X, we want to figure out how to set up proportions. So the height of the building is unknown. That's X. The height of the flagpole is known. That's six and a half feet. So let's set that ratio equal to ratios in length. The length from this viewing spot right here to the base of the flagpole is 42 plus seven feet. In other words, it's 49 feet. That corresponding distance in terms of the flagpole is seven feet. So now we have our ratios. Let's cross multiply to solve the problem. Seven times X and six and a half times 49. Dividing by seven on both sides gives us that X is equal to 45 and a half feet. Let's take a look at one more problem. Here our job is to find the length of this lake, kind of a cruddy looking lake, but let's pretend that our job is to find the length of AB. So this fact, the fact that AB is parallel to DE, that tells us some information about about the naming of the triangles. So if AB is parallel to DE, then we know that, for example, angle B would be congruent to angle D because they would be alternate interior angles. If you also consider the vertical angles created by point C, I know that these two triangles are similar by AA. And in particular, let's name the triangles. So triangle ABC would be similar to triangle EDC. And that information will help us set up proportions to find the length of the lake. Let's find the, let's call that length X. So I know that the length AB corresponds with first two letters of the second triangle, ED. There's our unknown ratio, and now we're looking for a known ratio. I see 20 refers to length BC, and that would correspond with length DC. So 50. So for the known ratio, we'll use the length BC, which corresponds with DC. So let's substitute the values of those, those lengths. AB is unknown. Let's call it X. ED is known. It's 40 meters. And that equals BC, which is 200 meters, divided by 50 meters. And so now we'll solve for X using cross multiplication. So the lake is 160 meters across.