The distance between city [tex]$A$[/tex] and city [tex]$B$[/tex] is 22 miles. The distance between city [tex]$B$[/tex] and city [tex]$C$[/tex] is 54 miles. The distance between city [tex]$A$[/tex] and city [tex]$C$[/tex] is 51 miles. What type of triangle is created by the three cities?

A. An acute triangle, because [tex]$22^2 + 54^2 \ \textgreater \ 51^2$[/tex]
B. An acute triangle, because [tex]$22^2 + 51^2 \ \textgreater \ 54^2$[/tex]
C. An obtuse triangle, because [tex]$22^2 + 54^2 \ \textgreater \ 51^2$[/tex]
D. An obtuse triangle, because [tex]$22^2 + 51^2 \ \textgreater \ 54^2$[/tex]



Answer :

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]