Question

in yardley, the library is due south of the courthouse and due west of the community swimming pool. if the distance between the library and the courthouse is 3.1 miles and the distance between the courthouse and the city pool is 5.9 miles, how far is the library from the community pool? if necessary, round to the nearest tenth.

Explanation:

Step1: Identify the right - triangle

The positions of the library, courthouse, and community pool form a right - triangle. Let the distance between the library and the courthouse be $a = 3.1$ miles, the distance between the courthouse and the community pool be $c = 5.9$ miles, and the distance between the library and the community pool be $b$. According to the Pythagorean theorem $a^{2}+b^{2}=c^{2}$.

Step2: Rearrange the Pythagorean theorem

We can rewrite the Pythagorean theorem to solve for $b$: $b=\sqrt{c^{2}-a^{2}}$.

Step3: Substitute the values

Substitute $a = 3.1$ and $c = 5.9$ into the formula: $b=\sqrt{5.9^{2}-3.1^{2}}=\sqrt{(5.9 + 3.1)(5.9 - 3.1)}$ (using the difference - of - squares formula $x^{2}-y^{2}=(x + y)(x - y)$). First, calculate $(5.9 + 3.1)=9$ and $(5.9 - 3.1)=2.8$. Then $b=\sqrt{9\times2.8}=\sqrt{25.2}$.

Step4: Calculate the value of $b$

$\sqrt{25.2}\approx5.0$ (rounded to the nearest tenth).

Answer:

$5.0$