Given, a ladder 10 m long reaches a window 8 m above the ground.
Let BA be the wall and AC be the ladder,
Therefore, by Pythagoras theorem,
\(AC^2 = AB^2 + BC^2\)
\(10^2 = 8^2 + BC^2\)
\(BC^2 = 100 – 64\)
\(BC^2 = 36\)
BC = 6m
Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.