Chapter 9.6, Problem 21PSC

To determine

To calculate: The distance between the foot of ladder from bottom of the building.

Explanation of Solution

Given information :

A 650 cm ladder is placed against a building

It touches the building at 520 cm

Again, ladder is moved to reach a point 80 cm higher up

Formula used :

Pythagoras theorem

Proof :

Let AC be ladder and AB be the height equals to 520 cm and AD is the height where ladder reached after moving up further.

In the figure below,

AC= CD = 650 cm, AB = 520 cm and AD = 80 cm

  

Now in triangle, DBC, we have

  $\begin{array}{l}{\text{DC}}^2={\text{DB}}^2+{\text{BC}}^2\\ {\left(600\right)}^2={\left(650\right)}^2+{\text{BC}}^2\\ \Rightarrow {\text{BC}}^2={\left(650\right)}^2-{\left(600\right)}^2=1250\times 50\\ \Rightarrow \text{BC}=\sqrt{62500}=250\text{ cm}\end{array}$

  $\therefore$ Distance of foot of ladder from building = 250 cm