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.
Answer:
^@9 \space m^@
- Let ^@AB^@ be the building and ^@CB^@ be the ladder.
Then, ^@ AB = 40 \space m^@, ^@CB = 41 \space m^@ and ^@\angle CAB = 90^\circ^@. - 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^@.