As below,

I know how to find the length. but how to prove?

thanks.

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- Oct 16th 2011, 05:27 AMBabyMiloVector Triangle
As below,

I know how to find the length. but how to prove?

thanks. - Oct 16th 2011, 05:58 AMPlatoRe: Vector Triangle
- Oct 16th 2011, 06:07 AMBabyMiloRe: Vector Triangle
$\displaystyle \sqrt 21$

$\displaystyle \sqrt 17$

$\displaystyle \sqrt 38$ - Oct 16th 2011, 07:59 AMPlatoRe: Vector Triangle
- Oct 16th 2011, 08:20 AMBabyMiloRe: Vector Triangle
- Oct 16th 2011, 08:36 AMPlatoRe: Vector Triangle
- Oct 16th 2011, 10:36 AMHallsofIvyRe: Vector Triangle
Three line segments, of lengths a, b, and c, can form a triangle as long as no one length is greater than the sum of the other two (the

**shortest**distance between two points is a straight line). But if $\displaystyle c^2= a^2+ b^2$ it is certainly true that the longest side is c and c< a+ b. A right**triangle**is a**triangle**as Plato said.

(To see that $\displaystyle c^2= a^2+ b^2$ leads to $\displaystyle c< a+ b$ [as long as neither of a or b is 0], add the positive number 2ab to both sides of $\displaystyle c^2= a^2+ b^2$. You get $\displaystyle c^2+ 2ab= a^2+ 2ab+ b^2= (a+ b)^2$. Since $\displaystyle (a+ b)^2$ is larger than $\displaystyle c^2$, a+ b is larger than c.)