How do you find the perimeter of a triangle with vertices X(3, 0), Y(7, 4), and Z(10, 0)?

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Answer 1 · 2017-01-20T08:27:29

$4(3+sqrt2)=17.657$ , nearly.

$vec(XY)=<7, 4> - <3, 0> = <4, 4>$ .

So, $XY=sqrt(4^2+4^2)=4sqrt2$ .

$vec(YZ)=<10, 0> - <7, 4> =<3, -4>$

So, $YZ= sqrt(3^2+(-4)^2)=5$

$vec(ZX)= <3, 0> - <10, 0> = <-47 -4>$

So, $ZX=sqrt(7^2+0)=7.$

The perimeter = XY +YZ + ZX = 4sqrt2+5+7#

$=4(3+sqrt2)=17.657$ , nearly.