Medians and centroids
Median Centroid Right Triangle Example Example involving properties of medians
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- So we're told that AE is equal to twelve that's this side right over
- here and EC is equal to 18 EC is equal to 18
- And then they've drawn a bunch of median's 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
- So that ED is equal to DC CB is equal to BA
- AF is equal to F here that F B and D are the mid points
- And the G then would be the centroid
- Where the medians intersect
- And so the first thing they ask us is what is the area of B G C?
- So B C G 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 to
- Divide a triangle into six
- 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 twelve
- So this is going to twelve which I'm gonna space
- This entire distance right over here is 18 they tell us that
- So the area the area of AEC, AEC is going to be
- For the one half times the base which is 18 times a height
- Which is 12 which is equals to 9 times 12 which is 108
- That's the area of the entire right triangle triangle AEC
- If we want the area of BGC
- If we lower this little altitude right over here
- The ones that are bounded by the medians
- Then we just have to divide this by 6
- Coz they all have equal area we've that in previous videos
- 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 we get ten and then 48
- This would be 18 it would be 18
- And that's right coz it would be a 108 is the same as 18 times 6
- So we get our first part the area of that right over there is 18
- And if we want we can say hey the area of any of these triangles
- These the ones that are bounded by the medians
- This is gonna be 18, this is gonna 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, the centroid is
- Always two thirds along the way of the median
- 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 can just take two thirds of that
- And that will give us the length of AG
- And lucky for us this is a right triangle and we know that
- We know that F and D are the mid points
- So for example we know that this AE is 12 that was given
- We know that ED is half of this 18
- So ED right over here I'm using a new color ED is going to be 9
- ED is going to be 9 so then we could just use
- A Pythagorean Theorem to figure out what AD is
- AD is the hypotenuse of a of this right triangle AD
- So we're looking at triangle AED right now
- So we know that let me write this down
- We know that 12 squared 12 squared plus 9 squared
- Plus 9 squared is going to be equal to AD squared
- Is going to be equal to AD, AD squared
- 12 squared is 144, 144 plus 81 plus, plus 81
- And so this is going to be equaled to AD squared
- So this is what this is 225
- So we have 255 is equal to AD squared is equal to AD squared
- And 225 you may or may not recognize is 15 squared
- So AD, AD is equal to 15
- You wan to take the principal root the positive root
- Coz we're talking about distances or length 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 two thirds of AD
- AG is equal to two thirds of AD
- We've proved that in the previous video
- That the centroid is two thirds along the way
- Of any of these medians or we could do it for any of the medians
- So it's equals to two thirds times 15 which is equal to
- Which is equal to 10
- So AG right over here, AG right over here is equal to
- Is equal to 10 so we did this second part
- Now this third part what is the area of F of FGH
- So let me 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 one way that we can thing about finding what HG is
- Is to remind ourselves, 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
- Both of them have a base of 6
- So this is 6 and this is 6 over here
- And 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 you can say if we just take this triangle up here
- So we're talking about the area, area of triangle AFG
- So we know it's one half times its base
- Which is 6 times its height which is GH times GH
- That's as one half 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 three times GH is equal to 18
- GH if we divide both sides of this by
- If both sides of this by 3 GH is equals to 6
- So that is one way to do it GH is equals to 6
- You could also made a similarity argument
- And you could say look this is this triangle up here
- Is similar to this larger triangle over here
- This is this hypotenuse is two thirds of the length
- Of this entire thing
- So this is going to be two thirds of this 9
- So that's another way that you could've 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
- If we know what AH is because we know A to F is 6
- So FH is gonna be AH minus AF
- So let's figure out what AH is
- Well once again we can make a similarity argument
- And if you wanna do it formally we see
- That both of the larger this this larger triangle
- And this smaller triangle both have a 90 degree there
- They both have this angle in common
- So they have two angles in common
- They're definitely similar triangles
- And so we know the ratio of H we're gonna 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 10,10 to the ratio of AD
- which we already figured out was 15
- So one way to think about it is H is gonna be two thirds of 12
- We'll we can just work through the math
- Just using the similar triangles
- So this right hand side over here is just 2 over 3
- And so AH multiplying both sides by 12
- Is equal to two thirds times 12 which is just if you do it
- It is just H so AH here is 8, AH is 8, AF is 6
- So FH right over here, FH right over here is gonna be 2
- So now we have enough information
- to figure out the area of FHG
- it's going to be one half let me write over here
- it's going to be one half times the base
- I'll just use FH as the base here although
- Or I could just do it either way I can use well FH as the base
- One half 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 where going back
- We figured out most of them
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