<PRE><B>Question 1007682</B>


Using Pythagoras theorem
x^2+(3x+3)^2 = 625


x^2+9x^2+18x +9 =625

10x^2+18x -616=0

/2
5x^2+9x-308=0

5x^2-35x+44x-308=0

5x(x-7)+44(x-7)=0

(5x+4)(x-7)=0

5x+4 = 0 OR x-7=0

x= -4/5 Or 7

x cannot be negative

Therefore x =7 

Check
x^2= 49
(3x+3)^2 = 576

576 + 49 = 625

x= 7
(3x+3)= 24

Hypotenuse = 25

Perimeter = 7+24+25 = 56

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