IntriangleABC, altitude AD = 18, median BE = 9√5 and medianCF = 15. Find BC.

Latika Leekha
8 years ago
In the given triangle ABC, let AH be the median for BC. Let G be the centroid i.e. G is the point of intersection of the medians AD, BE and CF.
Let GH be the height of $\Delta$ GBC and FK be the height of $\Delta$FBC relative to the side BC.
Using the property of median that it divides in the ratio 1:2, we get
CG = 2/3 CF = 2/3. 15 = 10.
BG = 2/3 BE = 2/3 . 9√5 = 6√5.
This means $\Delta$BFK is ismilar to $\Delta$ BAD.
Now BF= ½ BA (Since CF is the median)
Hence, FK = ½ AD = ½ . 18 = 9.
On similar lines we get, $\Delta$CGH is similar to $\Delta$CFK.
So, CG/CF = GH/FK
Hence, 10/15 = GH/FK
This means, GH = 2/3 FK = 2/3. 9 = 6.
Now, usinh pythagoras theorem in $\Delta$GHB, we get
GH2 + HB2 = GB2.
Hence, 62 + HB2 = (6√5)2.
Hence, this gives, HB2 = 180-36 = 144
Hence, HB = 12.
Again using Pythagoras theorem in $\Delta$GHC, we get
GH2 + HC2 = GC2.
62 + HC2 = 102.
Hence, HC = 8.
Now, BC = HB + HC = 12 + 8 = 20.
Hence, the length of BC is 20.