A ladder ^@ 40 \space m^@ long reaches a window which is ^@24 \ | Math Question and Answer

Answer:

$56 \space m$

Step by Step Explanation:

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

You can reuse this answer
Creative Commons License

Chat with us