A 50-ft diagonal brace on a bridge connects a support of the center of the bridge to a side support on the?

Question

A 50-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 10 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 1

Let,

height = h ft,



distance = h+10 ft.



h^2 +(h+10)^2 = 50^2

2h^2 +20h -2400 = 0

OR,

h^2 +10h -1200 = 0

(h+40)(h-30) = 0,

Hence,

horizontal-distance = 40 ft, >================< ANSWER

and

vertical-distance = 30 ft. >================< ANSWER.

Answer 2

Note that the horizontal distance, the vertical distance, and the diagonal brace make a right triangle.



Let x denote the vertical height. Then, since the horizontal distance that the brace spans is 10 ft more than the vertical distance, the horizontal distance is x + 10.



By the Pythagorean Theorem:

x^2 + (x + 10)^2 = 50^2

==> x^2 + x^2 + 20x + 100 = 2500

==> 2x^2 + 20x - 2400 = 0, by setting the right side equal to 0

==> 2(x + 40)(x - 30) = 0, by factoring

==> x = -40 and x = 30, by the zero-product property.



Since the vertical distance cannot be negative, x = 30.



Therefore, the horizontal distance spanned is 30 + 10 = 40 ft and the vertical distance spanned is 30 ft.



I hope this helps!

Answer 3

(H + 10)^2 + H^2 = 50^2 gives H^2 + 10H -- 1200 = 0

whence (H + 40)(H -- 30) = 0 giving H = 30 ft, -- 40 ft discarded

Height spannes = 30 ft and Distance spanned = 2*(30 + 10) ft = 80 ft ANSWER