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The sum of the x-coordinates of the vertices of a triangle in the Cartesian plane equals \(\sqrt13\). Let equal the sum of the x-coordinates of the midpoints of the sides of the triangle. Find \(S^2\).

tertre  Mar 22, 2018
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The sum of the x-coordinates of the vertices of a triangle in the Cartesian plane equals\(\sqrt{13}\)

Let S equal the sum of the x-coordinates of the midpoints of the sides of the triangle. Find S^2

 

Let the x vaues of the vertices be a,b,and c

\(a+b+c=\sqrt{13}\)

 

the x values of the midpoints of the sides are 

    \(\frac{a+b}{2},\:\;\frac{b+c}{2},\;\;\frac{a+c}{2}\\ sum=\frac{a+b}{2}+\frac{b+c}{2}+\frac{a+c}{2}\\ sum=\frac{a+b+b+c+a+c}{2}\\ sum=\frac{2(a+b+c)}{2}\\ S=a+b+c = \sqrt{13}\\ S^2=13\)

Melody  Mar 22, 2018

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