the ladder goes over a 6 foot high fence that is 2 feet away from the building
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To answer this question we need to know the length of the ladder. The longer the ladder the further we can move it away from the fence to reach the building. Let's call the length of the ladder L and the distance from the building to the bottom of the ladder we'll call x. We have a right-angled triangle with hypotenuse L and we'll call the height from the ground where the ladder meets the building H. The fence creates another right-angled triangle similar to the larger one. Now for some equations. Similar triangles means we can write (x-2)/6=x/H. That is, (distance from ladder to fence)/(height of fence)=(distance from ladder to building)/(the height of the ladder from the ground on the building. Cross multiply: Hx-2H=6x, so x=2H/(H-6). From Pythagoras' theorem, L^2=H^2+x^2. Substitute the value we found for x and we get L^2=H^2+4H^2/(H-6)^2. Multiply through by (H-6)^2: L^2(H-6)^2=H^2(H -6)^2+4H^2⇒L^2H^2-12L^2H+36L^2=H^4-12H^3+36H^2+4H^2=H^4-12H^3+40H^2. So 36L^2=H^4-12H^3+(40-L^2)H^2+12L^2H. Remember that L is to be regarded as a constant, even though unknown. The polynomial H^4-12H^3+(40-L^2)H^2+12L^2H-36L^2=0 has to be solved to find H in terms of L.

However, we are asked to find the minimum height H. This means that we take the lowest value root. I used a calculator to find the roots of the polynomial for different values of L. When L=15 ft, H=7.0686 ft; L=20 ft, H=6.7125 ft; L=25 ft, H=6.5423 ft. Other lengths of ladder seem impractical (or can be interpolated). 

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