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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.

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#### Solution

Let the length of the ladder be 5.8 m.

According to Pythagoras theorem, in ∆EAB

\[{EA}^2 + {AB}^2 = {EB}^2 \]

\[ \Rightarrow \left( 4 . 2 \right)^2 + {AB}^2 = \left( 5 . 8 \right)^2 \]

\[ \Rightarrow 17 . 64 + {AB}^2 = 33 . 64\]

\[ \Rightarrow {AB}^2 = 33 . 64 - 17 . 64\]

\[ \Rightarrow {AB}^2 = 16\]

\[ \Rightarrow AB = 4 m . . . \left( 1 \right)\]

In ∆DCB

\[{DC}^2 + {CB}^2 = {DB}^2 \]

\[ \Rightarrow \left( 4 \right)^2 + {CB}^2 = \left( 5 . 8 \right)^2 \]

\[ \Rightarrow 16 + {CB}^2 = 33 . 64\]

\[ \Rightarrow {CB}^2 = 33 . 64 - 16\]

\[ \Rightarrow {CB}^2 = 17 . 64\]

\[ \Rightarrow CB = 4 . 2 m . . . \left( 2 \right)\]

From (1) and (2), we get

\[AB + BC = \left( 4 + 4 . 2 \right) m\]

\[ = 8 . 2 m\]

Hence, the width of the street is 8.2 m.

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