Question

In given figure a ladder $25m$ long reaches a window of a building $20m$ above the ground. Determine the distance of the foot of the ladder from the building.

Answer 1

Suppose that AB is the ladder, B is the window and CB is the building. Then, triangle ABC is a right triangle with right-angle at C.

By pythagoras theoram, we have

$\therefore$ ${AB}^2=A{C}^2+{BC}^2$

$\Rightarrow$ ${25}^2={AC}^2+{20}^2$

$\Rightarrow$ ${AC}^2=625-400=225$

$\Rightarrow$ $AC=\sqrt{225}m=15m$

Hence, the foot of the ladder is at a distance $15m$ from the building.