Grade 10Magical Mathematics[Interesting Approach]

In

triangleABC, altitude AD = 18, median BE = 9√5
and
median

CF = 15. Find BC
.

ajinkya bhole

16 Years agoGrade 10

12 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.

Hence, FK/AD = BF/BA.

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.