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November 25, 2014

November 25, 2014

Posted by **anonymous** on Tuesday, June 25, 2013 at 5:22am.

- math -
**MathMatet**, Tuesday, June 25, 2013 at 9:20amSince D, E bisect BC and AC respectively, we have

AE=139/2=69.5

BD=178/2=89

Let medians AD and BE meet at K.

Let KD=x, DE=y

Then

BK=2y, AK=2x

since medians intersect each other at third points.

Consider right triangle AKE right angled at K:

(2x)²+y²=69.5² (Pythagoras)

=>

4x²+y²=69.5²....(1)

Consider right triangle BKD,

x²+4y²=89².....(2)

4(1)-(2) to eliminate y:

15x²=4*69.5²-89²

=>

x²=760

Substitute x² in (2):

4y²=89²-760=7161

Finally, consider right triangle AKB:

AB²

=4x²+4y²

=3040+7161

=10201

=>

AB=101

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