To determine the type of triangle formed by the three cities \( A \), \( B \), and \( C \) with distances given as follows:
- The distance between city \( A \) and city \( B \) is 22 miles.
- The distance between city \( B \) and city \( C \) is 54 miles.
- The distance between city \( A \) and city \( C \) is 51 miles.
We need to analyze the type of triangle by comparing the squares of these distances. Let’s denote:
- \( AB = 22 \) miles
- \( BC = 54 \) miles
- \( AC = 51 \) miles
We will calculate and compare the squares of these distances.
1. Calculating the squared distances:
[tex]\[
AB^2 = 22^2 = 484
\][/tex]
[tex]\[
BC^2 = 54^2 = 2916
\][/tex]
[tex]\[
AC^2 = 51^2 = 2601
\][/tex]
2. Comparing the squared sums:
- To determine if the triangle is acute, we compare the sum of the squares of two sides with the square of the third side.
First, we check if:
[tex]\[
AB^2 + BC^2 > AC^2 \quad \text{(484 + 2916 > 2601)}
\][/tex]
Indeed:
[tex]\[
484 + 2916 = 3400 > 2601
\][/tex]
Therefore, \( AB^2 + BC^2 > AC^2 \).
Since \( AB^2 + BC^2 > AC^2 \), this means that the given triangle is acute.
Given these calculations, we conclude that the triangle formed by cities \( A \), \( B \), and \( C \) is an acute triangle because \( 22^2 + 54^2 > 51^2 \). Thus, the correct answer is:
an acute triangle, because [tex]\( 22^2 + 54^2 > 51^2 \)[/tex]
