Area of SAS & ASA Triangles - Video Tutorials & Practice Problems

Calculating Area of SAS Triangles

Everyone. So you may remember that to calculate the area of any triangle, you'll always need the height. You may recognize this one half base times height formula. It works even for non right triangles. As long as you're given height and base, you can find the area. But some questions will give you angles and sides and you won't actually be given the height. But don't worry because in these types of problems, what I'm gonna show you is that you can find it by using the sign of a known angle. So what I'm gonna show you in this video is we're gonna work out the areas of these two triangles and they're exactly the same. We're just gonna use different information to get there. And then I'll show you three new equations that you'll need to solve for the area. Let's get started here. So in this first example, we will have the one half uh if we have the base and the height of this triangle, and we could just use the familiar one half base times height formula to find the area. So really what this becomes here is one half the base of this triangle is eight and the height of this triangle is just equal to three. When you work this out, this ends up being 12, pretty straightforward. That's the area of this triangle, by the way, doesn't necessarily have to be B as the base. You could have A or C, all these things could be potentially different letters. But the area of this is 12. What about this triangle over here? The key difference is that we actually don't have that height that's already drawn for us. But the principle is still the same, we're still just gonna use one half of base times height. What's the base of this triangle? Well, actually, it's the same thing as the, as the triangle on the left, the base really is just B, so I'm just gonna write that over here. What about the height of this triangle? Well, here's where it's a little bit different because whenever you wanna draw the height of a triangle, you're just gonna take the base that you've drawn and you're basically gonna draw up a perpendicular line up to one of the corners. And this effectively turn sort of turns this into a right triangle. So what is the value of this height over here? Which I'll write as h well, the idea here is that if you have the sides, then basically you can do is turn this into a right triangle and we have an angle and the hypotenuse of this triangle. So we can use sokoto up. So the sign of this over here, we're gonna use the sign of angle A is equal to uh C which is the hypotenuse or sorry. This is gonna be the opposite side which is H over the hypotenuse, which is C if I rearrange for this, what I can find is that H is equal to C times the sign of A, all right. So basically what happens here. So if I have the hypotenuse in angle, I can figure out the heights and this effectively comes becomes the height of my triangle. So really what happens is that the base becomes the letter B and the heights of this triangle actually becomes the C times the S of A B is the base and then C times sign of A is the height, all right. So let's see actually what this works out to be. This is gonna be area equals one half and I'm done, my base is still eight exactly like it was on the left. And then now what happens is instead of three, I'm just gonna plug in C times the sign of A, which is six times the sign of 30 degrees when you work this out, which you're actually gonna get is that this actually equals 12 exactly like what it did on the left side. So the angles or sorry, the areas of these two triangles are exactly the same. This actually totally makes sense because if you sol for the height of this triangle, this is really just six times the sign of 30 which is actually just equal to three. So using different information, we found the height of this triangle is still three and therefore the areas are gonna be exactly the same. That's really all there is to it. All right. Now remember that the positions of these variables will be in different places throughout your triangles. So the sort of three different variations of this formula kind of like how there was for the law of cosines. I'm just gonna go ahead and show you those. Sometimes you'll have BC sin A. Another variation that you might see is ac times the sign of big B and the last one you'll see here is A B times the sign of C. So what you'll notice here is that the angle is always gonna be the sort of the last, the third letter that is not these other two over here. And it, so these two lowercase letters and the sign of an angle over here. All right. Now, I remember that all of these things will be multiplied. So you can sort of shuffle these letters around. But this is the idea behind finding the area of an, of a non right triangle. Thanks for watching and let's get some practice.

Calculating Area of SAS Triangles Example 1

Welcome back everyone. So in this problem, we're gonna try to calculate the area of this triangle. Let's get started here. Remember at the end of the day, an area is just one half times a base times a height. So area is equal to one half times the base times the height and the base of this triangle appears to be B equals eight. What about the height of this triangle? Remember we're gonna try to draw from one of the corners down to where it intersects the base. The problem is if I try to do this here and try to figure out this height, there's a piece of this hypotenuse of this triangle that still keeps going up and I'm losing all of this area if I try to plug it into one half, base times height. All right. So it turns out that we can't really do this. So I kind of want to explain what's going on in these triangles. But first of the actually just take a look at the variables. We know that little B is eight, little A is six and the big C is equal to 100 and 20. We know that there's sort of three versions of the area formula of a triangle. And usually what happens to these problems is that as you look through your variables, two of them won't make sense. Let's take a look here. We have one half B times C times sine of A. The problem is, I don't know what A is in this problem or little C. So I have two unknowns. That equation is not gonna work. The second one is gonna be one out of A times CS B. Same problem. I don't know what C little C is or the big B, both of these variables or angles are unknown. The only one that makes any sense at all is using this one over here because I actually know what all three of these values are A B and C. But I wanna show you why this works in this equation. So we're gonna have one half of B oh Sorry of A times B times the sine of C. Remember area is always a base times the height. What's the base in this problem? It's really just that little B. So what I'm gonna do is I'm actually just gonna switch the variables around in this equation. This is gonna be one half of base of B which this little B here is the base. So then what's the height of this triangle? Well, it's really just gonna be this other piece over here a times the sign of C. All right. But I want to show you why that ends up being height in this problem. All right. So we already know that we can't sort of draw an, a, a little dash line straight up and have that as the base. But what we can do is we can go all the way to the corner over here of B and then basically just draw this line straight down even if it sticks out past the base of my triangle. Because what I can do is I could basically just pretend that the base of this triangle sort of keeps on moving on, going on and sort of ongoing straight. So really this height over here is actually the height of my triangle and notice that what we've done is we've made a smaller right triangle. That sort of is an extension of my obtuse triangle. And the heights of this right triangle is the same as the heights of the obtuse triangle that I'm trying to find the area of. What is the height of this triangle related to the variables that I know. Well, I've got the hypotenuse which is A and just using so Katoa, this H would just be a times the sign of whatever this angle is over here. What is this angle? Well, it's basically just if we, assuming that this line here is gonna be supplemental, it's gonna be 100 and 80 degrees, then this is really just gonna be 100 and 80 degrees minus 100 and 20 which ends up just being 60 degrees. So this angle over here is 60 degrees. Therefore, the heights would just be a times the sign of 60. In other words, it would really just be six times the sign of 60. All right. Now, it might seem like you have to do all this work to figure out this angle over here. But actually you don't because what I'm gonna show you here is that you could have done all this work to figure out that this angle was 60 or you can actually just use six times the sign of the original angle C that you were given, which is 100 and 20. What I want you to do is remember an old identity that you may have learned uh back in a previous chapter, which is the sign of any angle is equal to the sign of 180 minus that angle. So an example of this in this problem would just be. So for example, the sign of 120 degrees is the same thing as the sign of 180 minus 100 and 20 which is the sign of 60. So you don't have to actually do all of this work to figure out what the sign of this sort of smaller right, right triangle is. Because what I'm gonna show you whatever, what I've shown you here is that the sign of this 120 which was your original angle inside of your two triangle is actually the same as the sign of that supplementary angle. All right. In fact, you can actually plug these things into your calculator. And what you'll see is that six times, the sign of 60 ends up being 5.2 and six times 120 also ends up being 5.2. So what I'm trying to tell you in these problems here, what I'm really, what I'm trying to say is that you actually, it might seem like you have to do all of this extra work to figure out this supplementary angle. But you actually don't, we could have just gone straight into using this equation A B times the sign of C because taking the sign of this angle with this hypotenuse is the same as taking the sign of this angle to figure out the height of this triangle. OK. That's really what's going on here. So really, I'm just gonna take the area over here and this is gonna be one half. I know the base is eight. And what's the height I just calculated the height of this triangle is 5.2. So this is gonna be uh sorry, this is gonna be a sc which ends up being 5.2 and the area of this triangle ends up being 20.8 units. All right. And that is the final answer. That is the area included inside of this obtuse triangle. All right. Thanks for watching. And I'll see you in the next one.

Calculating Area of ASA Triangles

What we've seen in the last couple of videos is that they calculate the area of a triangle. When you're not given the heights, we find it by using the sign function. Now in some triangles, like what one we're gonna work out here. We actually are not given all of the information that we need to plug into our area formulas. For example, this is an A S A triangle. We, we have two angles and a side between them. But all of the area formulas require that we know two sides out of three. So it seems like we're stuck here. But what I'm gonna show you this video is that there's nothing to worry about because whenever this happens, we can always use the other formulas that we've learned a lot of signs and cosigns to find any missing information to plug into our area equations. Let's just jump into this problem here. So I can show you it's really not so bad. All right, let's get started. So we have a triangle over here. We have big A is 40 degrees, big C is 60 degrees and little B is four. All right, I've already drawn this here just to make it a little bit simpler. But basically, the idea here is that I want to sort of orient this triangle. So I've got my base over here and then I can draw the heights, which is gonna be over here on this side. We've seen something similar to this. All right. So what is the height of this triangle? Basically, if I make this little right triangle like this, the heights using? So Katoa, it's just gonna be the hypotenuse of this triangle times the angle over here. In other words, it's just gonna be C times the sign of A. The problem is I actually don't know what C is. All right. So what happens is in, in order to go my area formula, my area formula is gonna be one half the base of this triangle here is gonna be B which I know and then the height of this triangle is gonna be C times the sign of A. When you start looking at these variables, you know the B and you know the big A but you don't know C. So in order to calculate the area, I need to actually go find this little C first, how do I do that? Well, this is a triangle in which I know, I know I, I have a si I have some angles and sides here. So in order to find C, what I can do is I could just use the law of signs, right. That's what we saw from Missing sides. So, what I'm gonna do here is I'm gonna set up my Law of Signs. I've sign A over A equals sign B over B the sequel Sine C over C. All right. Remember I'm looking for this little C over here. So what I'm gonna have to do is I'm gonna have to pick two out of these three ratios in which I know three out of four variables. Let's go ahead and sort of, you know, uh sort of uh check out which variables. I know, I know little B and I know big A and I also know big C. So it might look here like I'm kind of stuck again because I only have three out of the four out of the six possible variables. But remember here that whenever you have two angles like A and C, you can always solve for that missing angle by using the angle sum formula. So what I'm gonna do here is I'm just gonna go right up here and I'm gonna say that A plus B plus C is equal to 100 and 80 degrees. So therefore, B is just equal to 100 and 80 minus the other two angles. Whenever you know, two out of three angles you can always easily solve for that other one. So this is gonna be 100 and 80 minus 40 60 this ends up equaling 80 degrees. So this angle over here is 80. And the reason we need that is we need this to basically figure out a another variable so we can set up our law of science. So now that I know what this B is, if you take a look here, the only sort of two ratios I can use are these last two ones over here. This is where I have three out of four variables. So I'm gonna pick these two out of three ratios. And we've seen how to solve for something like C before I would do is I'm gonna flip these variables. I've got C over sine C is equal to B over sine B. So therefore, C is equal to, I'm just gonna plug in everything. Now, I've got B is equal to four and then, then sign of Big B which I just figured out that angle is gonna be the sign of 80. And I'm gonna multiply this by Big C which goes up here, which is gonna be the sign of 60 degrees. When you go ahead and plug this in. What you should get for little C is you should get 3.5. Now, remember that's not your answer. That's not the area of this triangle. That's really just this third s this one of these sides over here that you were missing the C here equals 3.5. And the reason you need that is because you need that to figure out the height of this triangle. So now that we figured out c we have everything we need to solve the problem because we figured this out. OK. So this is just gonna be one half of B, which is four times C, which I just figured it was 3.5 times the sign of angle A which is 40 degrees when you plug all this stuff into your calculator carefully, what you should find for the area is that the area is equal to about uh 4.5. So this is gonna be 3.5. And I'm sorry, this is gonna be 4.5 and that is the area of this triangle. All right. So thanks for watching and let me know if you have any questions.

Find the area of the triangle: $A=30\degree$, $b=10m$, $B=80\degree$.

• A. $44.8m^2$
• B. $11.9m^2$
• C. $26.2m^2$
• D. $23.9m^2$