Q52. - Algebra 1

Question

A ladder is leaning on a building. The base of the ladder is $\mathbf{10}\mathit{f}\mathit{e}\mathit{e}\mathit{t}$ from the building, and the ladder reaches up $\mathbf{15}\mathit{f}\mathit{e}\mathit{e}\mathit{t}$ on the building. How long is the ladder?

Step-by-Step Solution

Answer

The ladder is $\mathbf{5}\sqrt{\mathbf{13}}$ feet long

Step1. Given

The base of the ladder $=10feet$

The ladder reaches up to a height of $15feet$

Step2. Draw the rough sketch for the given problem.

The rough sketch for the given problem is:

$AC$ is the building and $CE$ is the ground.

$BC$ is the height up to which ladder can reach on the building.

$CD$ is the distance of the base of the ladder from the building.

$BD$ is the ladder.

Step3. Determine the length of the ladder.

In the triangle $BCD$, $\angle BCD$ is a right angle. Therefore, the triangle $BCD$ is right-angled.

According to the Pythagorean theorem, it can be noticed that:

${\left(BD\right)}^2={\left(BC\right)}^2+{\left(CD\right)}^2$

Therefore,

$\begin{array}{l}{\left(BD\right)}^2={\left(BC\right)}^2+{\left(CD\right)}^2\\ {\left(BD\right)}^2={\left(15\right)}^2+{\left(10\right)}^2\\ {\left(BD\right)}^2=225+100\\ {\left(BD\right)}^2=325\\ BD=\sqrt{325}\\ BD=\sqrt{25\times 13}\\ BD=\sqrt{5^2\times 13}\\ BD=5\sqrt{13}\end{array}$

Therefore, the ladder is $5\sqrt{13}$ feet long.