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


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Answer:

^@ 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^@.

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