### A ladder $40 \space m$ long reaches a window which is $24 \space m$ above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window $32 \space m$ high. Find the width of the street. A 24 m 32 m C B E D 40 m 40 m

$56 \space m$

Step by Step Explanation:
1. Let $AB$ be the street and let $C$ be the foot of the ladder. Let $D$ and $E$ be the given windows such that $AD = 24 \space m$ and $BE = 32 \space m$.

Then, $CD$ and $CE$ are the two positions of the ladder.

Clearly, $\angle CAD = 90^\circ$, $\angle CBE = 90^\circ$ and $CD = CE = 40 \space m.$

From right $\Delta CAD$, we have \begin{aligned} CD^2 =& AC^2 + AD^2 &&[\text{ By Pythagoras' theorem }] \\ \implies AC^2 =& CD^2 - AD^2 \\ =& [(40)^2 - (24)^2] \space m^2 \\ =& [ 1600 - 576 ] \space m^2 \\ =& 1024 \space m^2 \\ \implies AC =& \sqrt{ 1024 } = 32 \space m. \end{aligned}
2. From right $\Delta CBE$ , we have \begin{aligned} CE^2 =& CB^2 + BE^2 &&[\text{ By Pythagoras' theorem }] \\ \implies CB^2 =& CE^2 - BE^2 \\ =& [ (40)^2 - (32)^2 ] \space m^2 \\ =& [ 1600 - 1024 ] \space m^2 \\ =& 576 \space m^2 \\ \implies CB =& \sqrt{ 576 } = 24 \space m. \end{aligned}
3. Therefore, width of the street = $AC + CB$ = $32 \space m + 24 \space m = 56 \space m$.