+618 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
+128474
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
