Example involving properties of medians. Created by Sal Khan.
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- for any triangle do you need to use these specific letters to describe them?(4 votes)
- At7:00I tried to find the length of FH by using the pythagorean theorem, already knowing the length of GH and FG. So GH is 6, as Sal showed, and I thought that CF=AD=15. FG is "the shorter part" of CF, so it should equal 1/3*15=5, shouldn't it? Now the problem occurs: according to pythagorean theorem, FG^2=GH^2+FH^2 and this is equal to 5^2=6^2+2^2 ! Where did I go wrong...?(5 votes)
- @2:40is Sal making an assumption that AG is the longest part ?(4 votes)
- He may be eyeballing it, but you can also tell by the fact that line AG goes all the way past point F on its side, which is also the point that bisects line AE. That shows that AG is longer by proportion than line GD, or the longer part of the median if that made sense :)(3 votes)
- Around2:06, he says that the area of all the smaller triangles are 18, and we know that the longer leg is also 18. Is just a coincidence?
At5:45, he writes that AF, and HE, both equal 6, but AE equals 12, leaving no room for FH, at8:26, he says that FH is equal to 2. But Af(6), HE(6), and FH(2), does not equal AE(12).
Why is that?(4 votes)
- Yeah, it is just a coincidence.
He actually does not say that HE = 6, he says that FE = 6 and FH = 2 and 6-2 = 4 so really HE = 4. The total is 6 + 2 + 4 = 12. Hope that helps!(5 votes)
- Why is there so many different way's to solve this problem? Wouldn't it be a lot easer if there was just one way to solve it?(4 votes)
- Actually, this is the <b>magical part<b> of mathematics, there are so many ways to prove a single proposition, and all of them are authentic.
Let's suppose that you are willing to meet your friend Alex. And you have 4 options to reach him:
1. You can reach there by foot,
2. Take a taxi
3. Take your own car, (if you have one)
4. Take the bus if available.
see, all these options are legitimate and will take you to your destination. so happy journey.(1 vote)
- Is angle AEB 45*?(8 votes)
- No a median doesn't necessarily bisect the angle.
Also look at <HEG which is the same measure as AEB.
Point G is 4 above and 6 to the right of point E, if it was 45 degrees those two distances would have to be the same and they are not.(5 votes)
- Do you need to divide by six?(3 votes)
- In this video:
Sal proved that the medians of a triangle divide that triangle into six smaller triangles of equal area. So if we know the area of the whole triangle ( 1/2Base*Height, or in this case 1/2 *18*12), and we divide that by six, then we will know the area of the six smaller triangles.(2 votes)
- How could you know that the triangles are similar?6:58(1 vote)
it is wrong because 1. You cannot prove triangles similar by proofs (ASA, AAS, etc); proofs prove triangle congruence. 2. AAA is not a form of triangle congruence.
the two triangles are similar because Sal states that the hypotenuse of triangle AHG (10) is 2/3 of the length of the hypotenuse of triangle AED (15). Same with their bases; HG (6) is 2/3 of the length of HG (9). The same can also be said for their heights; AH (8) is again 2/3 of the length of AE (12). :)(3 votes)
- why is it /6 there are 7 triangles(2 votes)
So we're told that AE is equal to 12. That's this side right over here. And EC is equal to 18. And then they've drawn a bunch of the medians here for us. So we know that they are medians because, when they intersect the opposite side, they're telling us that this length is equal to this length, or that ED is equal to DC, CB is equal to BA, AF is equal to FE, or that F, B, and D are the midpoints and that G, then, would be the centroid where the medians intersect. And so the first thing they ask us is, what is the area of BGC? So BGC right here. That is this triangle right over there. And to figure out that area, we just have to remind ourselves that the three medians of a triangle divide a triangle into six triangles that have equal area. So if we know the area of the entire triangle-- and I think we can figure this out. This is a right triangle. They're telling us that. AE-- this entire distance right over here-- is going to be 12. So this is going to be 12. Let me make sure I have enough space. This entire distance right over here is 18. They tell us that. So the area of AEC is going to be equal to 1/2 times the base-- which is 18-- times the height-- which is 12-- which is equal to 9 times 12, which is 108. That's the area of this entire right triangle, triangle AEC. If we want the area of BGC or any of these smaller of the six triangles-- if we ignore this little altitude right over here, the ones that are bounded by the medians-- then we just have to divide this by 6. Because they all have equal area. We've proven that in a previous video. So the area of BGC is equal to the area of AEC, the entire triangle, divided by 6, which is 108 divided by 6. Which is what? It's 60-- let's see. You get 10 and then 48. Looks like it would be 18. It would be 18. And that's right because it would be-- 108 is the same thing as 18 times 6. So we did our first part. The area of that right over there is 18. And if we wanted, we could say, hey, the area of any of these triangles-- the ones that are bounded by the medians-- this is going to be 18. This is going to be 18. This entire FGE triangle is going to be 18, but we did this first part right over there. Now they ask us, what is the length of AG? So AG is the distance. It's the longer part of this median right over here. And to figure out what AG is, we just have to remind ourselves that the centroid is always 2/3 along the way of the medians, or it divides the median into two segments that have a ratio of 2 to 1. So if we know the entire length of this median, we could just take 2/3 of that. And that'll give us the length of AG. And lucky for us, this is a right triangle. And we know that F and D are the midpoints. So for example, we know this AE is 12. That was given. We know that ED is half of this 18. So ED right over here-- I'll do this in a new color. ED is going to be 9. So then we could just use the Pythagorean theorem to figure out what AD is. AD is the hypotenuse of this right triangle. So we're looking at triangle AED right now. Let me write this down. We know that 12 squared plus 9 squared is going to be equal to AD squared. 12 squared is 144. 144 plus 81. And so this is going to be equal to AD squared. So this is what? This is 225. So we have 225 is equal to AD squared. And 225, you may or may not recognize, is 15 squared. So AD is equal to 15. You want to take the principal root, the positive root, because we're talking about distances or lengths of sides. We don't care about the negatives. So AD is equal to 15. So this whole thing right over here is going to be equal to 15. And AG is going to be 2/3 of AD. We proved that in a previous video, that the centroid is 2/3 along the way of any of these medians. And we could do it for any of the medians. So it's equal to 2/3 times 15, which is equal to 10. So AG right over here is equal to 10. So we did the second part. Now, this third part, what is the area of FGH? So let me color it in-- FGH. So if we knew this length-- if we knew HG and if we knew FH-- we could easily figure out what that area is. And there's actually multiple ways of figuring out either one of those things. So one way that we can think about finding what HG is is to remind ourselves that HG is the altitude of either triangle FGE or triangle AFG. And both of them have a base of 6. So this is 6, and this is 6 over here. And then they have a height equal to GH. And we know what the area is. We know that the area is already equal to 18. So let's take this triangle up here. So we're talking about the area of triangle AFG. So we know it's 1/2 times its base, which is 6, times its height, which is GH-- times GH. That's just 1/2 base times height is equal to the area of this triangle, which is going to be equal to 18. And so then, we just have to tell ourselves, well, this is-- 3 times GH is equal to 18. If we divide both sides of this by 3, GH is equal to 6. So that is one way to do it. GH is equal to 6. You could have also made the similarity argument. Then you could have said, look, this triangle up here is similar to this larger triangle over here. This hypotenuse is 2/3 of the length of this entire thing. So this is going to be 2/3 of this 9. So that's another way that you could have gotten 6 there. But either way, we got this length. Now we just have to figure out what FH is. And we could figure out what FH is if we know what AH is. Because we know A to F is 6. So FH is going to be AH minus AF. So let's figure out what AH is. Well, once again, we can make a similarity argument. And if we want to do it formally, we see that both this larger right triangle and the smaller right triangle, both have a 90-degree angle there. They both have this angle in common. So they have two angles in common. They are definitely similar triangles. And so we know the ratio of AH-- let me do it in orange. We know that the ratio of AH to AE-- which is 12-- is equal to the ratio of AG-- which is 10-- to the ratio of AD-- which we already figured out was 15. So one way to think about it is AH is going to be 2/3 of 12. Or we can just work through the math just using the similar triangles. So this right-hand side over here is just 2/3. And so AH-- multiplying both sides by 12-- is equal to 2/3 times 12, which is just 8. So AH here is 8. AH is 8. AF is 6. So FH right over here is going to be 2. And so now we have enough information to figure out the area of FHG. So let me write it over here. It's going to be 1/2 times the base. I'll just use FH as the base here, although I could do it either way. Well, I'll use FH as the base. 1/2 times 2 times the height-- times 6-- which is equal to 6. And we are done. And you could keep going. You could figure out the length of pretty much all of these segments here using some of these techniques or any of these areas. Well, we've actually figured out most of them.