Question

Question 15
Foot of a 10m long ladder leaning against a vertical wall is 6m away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches.

Answer 1

Let AB be a vertical wall and AC = 10m is a ladder. The top of the ladder reaches to A and distance of ladder from the base of the wall BC is 6m.

In right angled $\Delta ABC$,
$A{C}^2=A{B}^2+B{C}^2$ [ by Pythagoras theorem]
$\Rightarrow (10{)}^2=A{B}^2+(6{)}^2$
$\Rightarrow 100=A{B}^2+36$
$\Rightarrow A{B}^2=100-36=64$
$\therefore AB=\sqrt{64}=8m$
Hence, the height of the point on the wall where the top of the ladder reaches is 8 m.