SOLUTION: A 75-ft diagonal brace on a bridge connects a support of the center of the bridge to a side support on the bridge. The horizontal distance that it spans is 15 ft longer than the he

Question 551512: A 75-ft diagonal brace on a bridge connects a support of the center of the bridge to a side support on the bridge. The horizontal distance that it spans is 15 ft longer than the height that it reaches on the side of the bridge. Find the horizontal and vertical distances spanned by this brace.
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
let height be x ft
horizontal distance = x+5 ft
hypotenuse = 75
Pythagoras theorem states that

Hypotenuse ^2= leg1^2+leg2^2

Hypotenuse = 75 ft
leg1 = x ft
leg2 = x+5

75^2=x^2+(x+5)^2
75^2=x^2+x^2+10x+25
2x^2+10x-5600=0
/2
x^2+5x-2800=0
Find the roots of the equation by quadratic formula
20
a= 1 ,b= 5 ,c= -2800

b^2-4ac= 25 + 11200
b^2-4ac= 11225

x1=( -5 + 105.95 )/2
x1= 50.47
x2=( -5 -105.95 ) / 2
x2= -55.47
height = 50.47
horizontal length = (50.47+5)=55.47 ft

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