Question:
Foot of a 10 m long ladder leaning against a vertical wall is 6 m away from the base of the wall. Find the height of the point on the wall where the top of
the ladder reaches.
Solution:
Let AB be a vertical wall and AC = 10 m is a ladder. The top of the ladder reaches to A and distance of ladder from the base of the wall BC is 6 m.
In right angled $\triangle A B C, \quad A C^{2}=A B^{2}+B C^{2} \quad$ [by Pythagoras theorem]
$\Rightarrow \quad \cdot \quad(10)^{2}=A B^{2}+(6)^{2}$
$\Rightarrow \quad 100=A B^{2}+36$
$\Rightarrow \quad A B^{2}=100-36=64$
$\therefore \quad A B=\sqrt{64}=8 \mathrm{~cm}$
Hence, the height of the point on the wall where the top of the ladder reaches is 8 cm.