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Medians $\overline{AD}$ and $\overline{BE}$ of $\triangle ABC$ are perpendicular. If $AD= 15$ and $BE = 20$, then what is the area of $\triangle ABC$?

 

Edit: I got the answer never mind

michaelcai  Aug 24, 2017
edited by michaelcai  Aug 24, 2017
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Even though Michaelcai has the answer, it might be instructive to see what it is.....

 

Let the base of the triangle be AB  and let AD and BE intersect at right angles at F

 

 AF and FB  will be the legs right triangle AFB with AF = (2/3)AD and FB   = (2/3)(BE)

And AB will form the hypotenuse of AFB  and will  = sqrt [ (2/3)AD^2 + (2/3)BE^2 ] =

sqrt [ 10^2 + (40/3)^2 ] = 50/3

 

And  BD  will form a hypotenuse of triangle BFD  with BF = (2/3)BE  and FD  = (1/3)AD

And since D is a median on BC, BC  = 2BD =   2 sqrt [ (40/3)^2 + (5)^2 ] = 2 * 5 sqrt [ 73 ] / 3 =

sqrt (7300)/3

 

And AE will form a hypotenuse of triangle AFE  with AF  =  (2/3)AD and FE = (1/3)(BE)

And since E is a median  on  AC, AC = 2*sqrt [ 10^2 + (20/3)^2 ] = 2 * 10sqrt (13) / 3  =

sqrt(5200)/3

 

Now....since we know all the sides, we could use Heron's Formula to find the area, but all those roots would make this messy...!!!

 

Instead...lets use the Law of Cosines to find apex angle ACB....so we have that

 

AB^2  = BC^2 + AC^2  - 2 (BC)(AC)cos ACB

 

[50/3]^2  = 7300/9 + 5200/9  - 2 (sqrt(7300)*sqrt(5200)/9 * cos ACB

 

[ 2500 - 7300 - 5200] /9  =  -2 (sqrt(7300) *sqrt(5200) /9 * cos ACB

 

-10000  = -2 [ sqrt(7300)*sqrt (5200) ] * cosACB

 

arccos  [ 5000  /  [ sqrt(7300)*sqrt(5200] ] = ACB  ≈ 35.75°

 

So....the area of ABC  =  (1/2)(AC)(BC) sinACB  =

(1/2)sqrt(5200)/3 * sqrt (7300)/3 sin (35.75)   =  200 units^2

 

Here's a pic :

 

BTW....we can confirm that the answer is correct because C  = ( 4/3, 24)

 

So...the height of ABC  = 24 and the base  = 50/3

 

So.....the area  = (1/2) (24) (50/3)  = 12 * 50 / 3  = 600 / 3  = 200 units^2

cool cool cool

CPhill  Aug 25, 2017

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