A 10-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 2 ft longer that the height that it reaches on the side of the bridge. Find the horizontal and vertical distances spanned by this brace.

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A 10-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 2 ft longer that the height that it reaches on the side of the bridge. Find the horizontal and vertical distances spanned by this brace.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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This is what i have so far (x + 2)^2 + x^2 = 100 X^2 +4 + 2 (x) (2) +x^2 = 100 Using, (a + b)^2 = a^2 + b^2 +2ab 2x^2 +4x -96 = 0 So x^2 + 2x -48 = 0 Its 8 feet and 6 feet. X is 6 ft. Using this formula, Ax^2 +bx +c = 0 x = -b ± √(b^2- 4ac)/2a 8=-b±√(b^2-4ac)/2a
from here I have no clue.
or am i wrong.

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Lol Look at the pythagorean thereom Ill send a pic with it Solve for y [4y^2 + y^2] = (5^1/2)*y = 10 y = 10/(5)^1/2 = 4.5 ft (vertical) and 9.0ft (horizontal)
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I am sorry that is a mistake...I made the horizontal distance 2 times the vertical new solution in a minute
ok
u there.

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