A $16.5\ ft.$ long ladder is leaned against a wall. The ladder reaches the wall to a height of $13.2\ ft.$ Find the distance between the wall and the foot of the ladder.

Given: A $16.5\ ft.$ long ladder is leaned against a wall. The ladder reaches the wall to a height of $13.2\ ft.$

To do: To find the distance between the wall and the foot of the ladder.

Solution:

 Let $AB$ be the ladder and $AC$ be the wall. $BC$ will be the distance between the wall and the foot of the ladder.

As given, $AB=16.5\ ft.$

$AC=13.2\ ft.$

In $vartriangle ABC$, On using the Pythagoras theorem:

$AB^2=AC^2+BC^2$

$\Rightarrow ( 16.5)^2=( 13.2)^2+BC^2$

$\Rightarrow 272.25=174.24+BC^2$

$\Rightarrow BC^2=272.25-174.24$

$\Rightarrow BC^2=98.01$

$\Rightarrow BC=\sqrt{98.01}$

$\Rightarrow BC=9.9\ ft.$

Thus, the distance between the wall and the foot of the ladder is $9.9\ ft$.

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Updated on: 10-Oct-2022

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