A ladder ^@ 41 \space m^@ long just reaches the top of a buildi | Math Question and Answer

Answer:

$9 \space m$

Step by Step Explanation:

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 $9 \space m$ .

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