### A ladder $41 \space m$ long just reaches the top of a building $40 \space m$ high from the ground. Find the distance of the foot of the ladder from the building. A B C 40 cm 41 cm

$9 \space m$
1. Let $AB$ be the building and $CB$ be the ladder.
Then, $AB = 40 \space m$, $CB = 41 \space m$ and $\angle CAB = 90^\circ$.
2. By pythagoras' theorem, we have \begin{aligned} CB^2 =& AB^2 + AC^2 \\ \implies AC^2 =& CB^2 - AB^2 = [ (41)^2 - (40)^2 ] \space m^2 \\ =& ( 1681 - 1600 ) \space m^2 = 81 \space m^2 \\ \implies AC =& \sqrt{ 81 } \space m^2 = 9 \space m \end{aligned} Hence, the distance of the foot of the ladder from the building is $9 \space m$.