If $(a , b , c)$ is a Pythagorean triple, then so is $(ka , kb , kc)$

1.1k Views Asked by At

From trigonometry text:

Show that if $(a , b , c)$ is a Pythagorean triple then so is $(ka , kb , kc)$ for any integer $k > 0$. How would you interpret this geometrically?

Can someone please give a visual proof if possible?

1

There are 1 best solutions below

0
On BEST ANSWER

Edit: The formula answer makes it easy:

For a,b,c; $c^2 = a^2 + b^2 $. To verify ka,kb, and kc: $(ka)^2 + (kb)^2 = (kc)^2 $

$$(kc)^2 = k^2(c^2) = k^2(a^2 + b^2) = k^2a^2 + k^2b^2 = (ka)^2 + (kb)^2 $$

Here's a simple visual answer. You were correct, if the sides are proportional, then the triangles are similar. So, here are two 3/4/5 triangles, one in metric and one in inches:

Metric Triangles

Imperial Units