Ladder and Box problem for Trigonometry

A 7 ft ladder leans against a wall. There is a point near the bottom of the ladder that is 1 ft from the ground and 1 ft from the wall. Find the exact or approximate distance from the top of the ladder to the ground

Please show ALL working so I can follow and understand what you're doing.


I tried solving it and set it up as a quadratic equation:

x^4 + 2x^3 - 47x^2 + 2x +1 = 0

I tried plugging it into my calculator and graphing it but I can't seem to do maximum properly on it. Please help

Let P be the point on the ladder 1ft from the wall and ground. Let N be the point on the wall closest to P, let O be where the ladder meets the wall. Then by Pythagoras, ON^2=OP^2-NP^2=36-1=35 and ON=sqrt(35).

Let R be where the ladder touches the ground and let Q be where the wall meets the ground. The by similar triangles, OQ/OR=ON/OP=sqrt(35)/6. But OR=7, so OQ=7sqrt(35)/6 is the required distance.