Question

Walls of two buildings on either side of a street are parellel to each other. A ladder 5.8 m long is placed on the street such that its top just reaches the window of a building at the height of 4 m. On turning the ladder over to the other side of the street , its top touches the window of the other building at a height 4.2 m. Find the width of the street.

Answer 1

Let the length of the ladder be 5.8 m.

According to Pythagoras theorem, in ∆EAB

$$\begin{gathered} {\mathrm{EA}}^2+{\mathrm{AB}}^2={\mathrm{EB}}^2 \\ \Rightarrow {\left(4.2\right)}^2+{\mathrm{AB}}^2={\left(5.8\right)}^2 \\ \Rightarrow 17.64+{\mathrm{AB}}^2=33.64 \\ \Rightarrow {\mathrm{AB}}^2=33.64-17.64 \\ \Rightarrow {\mathrm{AB}}^2=16 \\ \Rightarrow \mathrm{AB}=4\mathrm{m}...\left(1\right) \end{gathered}$$

In ∆DCB

$$\begin{gathered} {\mathrm{DC}}^2+{\mathrm{CB}}^2={\mathrm{DB}}^2 \\ \Rightarrow {\left(4\right)}^2+{\mathrm{CB}}^2={\left(5.8\right)}^2 \\ \Rightarrow 16+{\mathrm{CB}}^2=33.64 \\ \Rightarrow {\mathrm{CB}}^2=33.64-16 \\ \Rightarrow {\mathrm{CB}}^2=17.64 \\ \Rightarrow \mathrm{CB}=4.2\mathrm{m}...\left(2\right) \end{gathered}$$

From (1) and (2), we get

$$\begin{gathered} \mathrm{AB}+\mathrm{BC}=\left(4+4.2\right)\mathrm{m} \\ =8.2\mathrm{m} \end{gathered}$$

Hence, the width of the street is 8.2 m.