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

$4 (3 + \sqrt{2}) = 17.657$ $4(3+sqrt2)=17.657$ , nearly.

$- - \to X Y = < 7, 4 > - < 3, 0 > = < 4, 4 >$ $vec(XY)=<7, 4> - <3, 0> = <4, 4>$ .

So, $X Y = \sqrt{4^{2} + 4^{2}} = 4 \sqrt{2}$ $XY=sqrt(4^2+4^2)=4sqrt2$ .

$- - \to Y Z = < 10, 0 > - < 7, 4 > = < 3, - 4 >$ $vec(YZ)=<10, 0> - <7, 4> =<3, -4>$

So, $Y Z = \sqrt{3^{2} + {(- 4)}^{2}} = 5$ $YZ= sqrt(3^2+(-4)^2)=5$

$- - \to Z X = < 3, 0 > - < 10, 0 > = < - 47 - 4 >$ $vec(ZX)= <3, 0> - <10, 0> = <-47 -4>$

So, $Z X = \sqrt{7^{2} + 0} = 7 .$ $ZX=sqrt(7^2+0)=7.$

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

$= 4 (3 + \sqrt{2}) = 17.657$ $=4(3+sqrt2)=17.657$ , nearly.