Im giving 30 points pls help

Step-by-step explanation:
Pathagory and Theorem
a^2+b^2=c^2
(x-3)^2+(x-4)^2=6^2
expand:
2x^2-14x+25=36
2x^2-14x-11=0
x=(\sqrt{71}+7)/2
perimeter=(x-3)+(x-4)+6=2x-1
insert the value for x into 2x-1
[tex]\sqrt{71}[/tex]+6=perimeter
Hope that helps :)
9514 1404 393
Answer:
2x -1
Step-by-step explanation:
The perimeter is the sum of the side lengths.
P = 6 + (x -4) + (x -3)
P = 2x -1 . . . . . . collect terms
The perimeter is 2x -1.
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Additional comment
There is nothing in the problem statement that suggests this is a right triangle. Without knowing at least one angle, we cannot solve for a numerical value for the perimeter. The problem title suggests we're to leave the result in polynomial form.