Question

A ladder 17 m long reaches a window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building.

Answer 1

Let AB denote the height of the window above the ground.
AC denotes the length of the wall.
BC denote the distance of the foot of the ladder from the building.
$AB=15m$
$AC=17m$

Triangle ABC forms a right angled triangle.
∴ $A{C}^2=A{B}^2+B{C}^2$ [Pythagoras theroem]
⇒$(17{)}^2=(15{)}^2+(BC{)}^2$
⇒ $289=225+(BC{)}^2$
⇒ $(BC{)}^2=49$
⇒ $BC=\sqrt{49}$
⇒ $BC=7$

Hence, the distance of the foot of the ladder from the building is 7 m