Practice Set 2.1 | Q 10 | Page 39
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.
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.
Explanation:-
The problem is to find the width of a street when the length of a ladder is given. Let us assume the length of the ladder is 5.8 m. We are also given that the ladder is resting against a wall, which makes an angle of 90 degrees with the ground. Using the Pythagoras theorem, we can find the length of the sides of the right-angled triangle formed by the ladder, the ground, and the wall.
Let us consider ∆EAB. Using the Pythagoras theorem, we can write:
[ {EA}^2 + {AB}^2 = {EB}^2 ]
Plugging in the given values, we get:
[ left( 4 . 2 right)^2 + {AB}^2 = left( 5 . 8 right)^2 ]
Simplifying this equation, we get:
[ 17 . 64 + {AB}^2 = 33 . 64]
[ {AB}^2 = 33 . 64 – 17 . 64]
[ {AB}^2 = 16]
[ AB = 4 text{ m} dots (1) ]
Now let us consider ∆DCB. Using the Pythagoras theorem, we can write:
[ {DC}^2 + {CB}^2 = {DB}^2 ]
Plugging in the given values, we get:
[ left( 4 right)^2 + {CB}^2 = left( 5 . 8 right)^2 ]
Simplifying this equation, we get:
[ 16 + {CB}^2 = 33 . 64]
[ {CB}^2 = 33 . 64 – 16]
[ {CB}^2 = 17 . 64]
[ CB = 4 . 2 text{ m} dots (2) ]
From equations (1) and (2), we can add the values of AB and CB to get the width of the street:
[ AB + BC = left( 4 + 4 . 2 right) m]
[ = 8 . 2 m]
Therefore, the width of the street is 8.2 m.
Chapter 2 – Pythagoras Theorem- Text Book Solution
Practice Set 2.1 | Q 10 | Page 39