Question

A ladder of 17m long reaches a wall 15m high.How far is the foot of the ladder from the wall.

Answer 1

Hint: We should know that the wall is straight and makes an angle of with respect to the ground.That means here a right-angle triangle will be formed.And the given problem will be solved using pythagoras theorem i.e., $A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}$,where AB,BC,AC are the lengths of sides of $\Delta ABC$ respectively.

Complete answer:
From the problem we have given that,
The length of wall = 15m
The length of the ladder = 17m
As the wall and ground makes an angle , here the ladder will act as a hypotenuse.
Therefore here a right-angled triangle is formed.
Now denote the length of wall = AB=15m
Also the length of ladder = AC =17m
And the distance between the foot of ladder to the wall = BC = ?

Now from the pythagorean principle,
$A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}$⟶equation(1)
And substituting all the values in equation(1) then we get as follows
${{17}^{2}}={{15}^{2}}+B{{C}^{2}}$
Now,it is simplified as follows
$\begin{align}
  & B{{C}^{2}}={{17}^{2}}-{{15}^{2}} \\
 & B{{C}^{2}}=289-225 \\
 & B{{C}^{2}}=64 \\
\end{align}$
$\begin{align}
  & BC=\sqrt{64} \\
 & BC=\sqrt{{{8}^{2}}} \\
 & BC=8m \\
\end{align}$
Therefore, the distance between the foot of the ladder from the wall is “8m”.

Note:The must be careful with the units that is we have to see whether all the parameters are with same or not.And also here the pythagorean theorem is only applicable for right-angled triangle.And also here we have to consider the ground as base which is uniform.