Question 850009
Sides are these:
x
y
z.
{{{y=2+2x}}};
{{{z=3x}}};
{{{x+2+2x+3x=38}}}.


The perimeter equation is ready for solving for x, the requested "first side" of the triangle.
{{{6x+2=38}}}
{{{6x=36}}}
{{{highlight(x=6)}}}


IS THIS A RIGHT TRIANGLE AS DESCRIBED?
x=6; y=2+2*6=14; z=3*6=18.
Test in Pythagorean Theorem:
{{{6^2+14^2=18^2}}}
{{{36+196=324}}}
{{{232=324}}} FALSE!!!!
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This cannot be a right triangle.  A triangle, yes; but NOT a right triangle.