Step-by-step explanation:
2 similar triangles (the angles are the same, and the lengths of the corresponding sides have the same multiplication factor from one triangle to the other) :
1. support post (14ft), post to back (b), the long roof line
2. front to post (f), support post (14ft), dotted line
so, because the multiplication factor is the same for the sides, the ratio
"support post" / f
must be the same as
b / "support post"
14 / f = b / 14
b = 14²/f = 196/f
and then we know that
f + b = 53
by putting the first into the second equation we get
f + 196/f = 53
f² + 196 = 53f
f² - 53f + 196 = 0
the solution of such a quadratic equation in general is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
x = f
a = 1
b = -53
c = 196
f = (53 ± sqrt(53² - 4×1×196))/(2×1) =
= (53 ± sqrt(2809 - 784))/2 = (53 ± sqrt(2025))/2 =
= (53 ± 45)/2
f1 = (53 + 45)/2 = 98/2 = 49
f2 = (53 - 45)/2 = 8/2 = 4
as the distance "front to post" is shorter than "back to post", f2 = 4 must be or solution.
therefore, the support post is positioned 4ft from the front of the house.
