Question 151487
good so far...


the height and width are two legs of a right triangle with the diagonal cable as the hypotenuse


by Pythagoras __ h^2+w^2=(sqrt(194)/2)^2 __ substituting __ (w+4)^2+w^2=194/4 __ 2w^2+8w+16=194/4


multiplying by 4 __ 8w^2+32w+64=194 __ subtracting 194 __ 8w^2+32w-130=0 __ dividing by 2 __ 4w^2+16w-65=0


factoring __ (2w+13)(2w-5)=0


2w+13=0 __ w=-13/2 __ negative value not realistic


2w-5=0 __ w=5/2 __ substituting h=(5/2)+4 __ h=6.5