To determine the type of triangle created by the three cities A, B, and C with given distances AB = 22 miles, BC = 54 miles, and AC = 51 miles, we need to consider the relationship between the squares of these side lengths. The type of triangle (acute, obtuse, or right) can be determined by comparing the sums of the squares of the sides.
Step-by-Step Solution:
1. Calculate the squares of the side lengths:
- [tex]\(AB^2 = 22^2\)[/tex]
- [tex]\(BC^2 = 54^2\)[/tex]
- [tex]\(AC^2 = 51^2\)[/tex]
2. Compute the values:
- [tex]\(AB^2 = 484\)[/tex]
- [tex]\(BC^2 = 2916\)[/tex]
- [tex]\(AC^2 = 2601\)[/tex]
3. Compare the sums of pairs of these squares:
- Comparison 1: [tex]\(AB^2 + BC^2\)[/tex] with [tex]\(AC^2\)[/tex]
- [tex]\(484 + 2916 = 3400\)[/tex]
- Compare with [tex]\(AC^2 = 2601\)[/tex]
- Since [tex]\(3400 > 2601\)[/tex], this indicates [tex]\(AB^2 + BC^2 > AC^2\)[/tex], suggesting it could be an acute angle at C.
- Comparison 2: [tex]\(AB^2 + AC^2\)[/tex] with [tex]\(BC^2\)[/tex]
- [tex]\(484 + 2601 = 3085\)[/tex]
- Compare with [tex]\(BC^2 = 2916\)[/tex]
- Since [tex]\(3085 > 2916\)[/tex], this indicates [tex]\(AB^2 + AC^2 > BC^2\)[/tex], suggesting it could be an acute angle at B.
- Comparison 3: [tex]\(AC^2 + BC^2\)[/tex] with [tex]\(AB^2\)[/tex]
- [tex]\(2601 + 2916 = 5517\)[/tex]
- Compare with [tex]\(AB^2 = 484\)[/tex]
- Since [tex]\(5517 > 484\)[/tex], this indicates [tex]\(AC^2 + BC^2 > AB^2\)[/tex], suggesting it could be an acute angle at A.
Each pairwise comparison shows that the sum of the squares of two side lengths is greater than the square of the third side. This confirms that:
- The triangle is acute because [tex]\(AB^2 + BC^2 > AC^2\)[/tex], [tex]\(AB^2 + AC^2 > BC^2\)[/tex], and [tex]\(AC^2 + BC^2 > AB^2\)[/tex].
Thus, the type of triangle formed by cities A, B, and C is indeed an acute triangle.
