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In the diagram below, D  is the midpoint of the base of isosceles triangle ABC. Points G1  and G2 are the centroids of triangle ABC  and triangle ACD, respectively. We know AD=8  and G1G2=4. What is the perimeter of triangle ABC?
 

 Nov 17, 2019
 #1
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From the diagram, BD = DC = G_1 G_2 = 4.  Then from the Pythaogrean Theorem, AB = sqrt(4^2 + 8^2) = 4*sqrt(5), so the peimeter is AB + AC + BC = 4*sqrt(5) + 4*sqrt(5) + 8 = 8*sqrt(5) + 8.

 Nov 17, 2019
 #2
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Draw   median AE   to bisect  BD

Let the intersection of G1G2  and AD   =  F

 

Since G1G2 is parallel  to BC, then   triangles  A G1 F   and  AED   are similar

 

But since   AE is a median.....then  AG1  is 2/3  of AE

 

So....  AE  = 3/2  of AG1

 

So....ED  is (3/2)  of    G1F  =  (3/2) (2)  = 3

 

And, by symmetry, ED  = (1/4) BC  

 

So BC   =  4ED  =  4 * 3  = 12

 

So  (1/2) BC  = BD  = 6 

 

And by the Pythagorean Theorem

 

AD^2  + BD^2  =  AB^2

 

8^2 + 6^2  = AB^2

 

100  =  AB^2

 

10  = AB  = AC

 

So....the perimeter  of ABC  =  BC + AB + AC  =  12 + 10 + 10  =  32

 

 

cool cool cool

 Nov 17, 2019

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