Answer :
To determine the type of triangle formed by the three cities, given the distances between them, we need to examine their respective squared distances and apply the properties of triangles, particularly using the Pythagorean theorem for right triangles and its extension to acute and obtuse triangles. Here is a step-by-step solution:
1. Calculate the squares of the given distances:
- Distance between city A and city B (AB) = 22 miles
- Distance between city B and city C (BC) = 54 miles
- Distance between city A and city C (AC) = 51 miles
[tex]\[ AB^2 = 22^2 = 484 \][/tex]
[tex]\[ BC^2 = 54^2 = 2916 \][/tex]
[tex]\[ AC^2 = 51^2 = 2601 \][/tex]
2. Compare the sums of the squared distances to determine the type of triangle:
- For a right triangle: One combination of the squares must equal another, i.e., [tex]\(AB^2 + BC^2 = AC^2\)[/tex] or any permutations.
- For an acute triangle: The sum of the squares of any two sides must be greater than the square of the third side for all combinations.
- For an obtuse triangle: The sum of the squares of any two sides must be less than the square of the third side for at least one combination.
3. Evaluate the conditions step by step:
- Check if the triangle is right:
[tex]\[ AB^2 + BC^2 = 484 + 2916 = 3400 \neq AC^2 (2601) \][/tex]
[tex]\[ AB^2 + AC^2 = 484 + 2601 = 3085 \neq BC^2 (2916) \][/tex]
[tex]\[ BC^2 + AC^2 = 2916 + 2601 = 5517 \neq AB^2 (484) \][/tex]
Since none of these combinations are equal, the triangle is not a right triangle.
- Check if the triangle is acute:
[tex]\[ AB^2 + AC^2 = 484 + 2601 = 3085 > BC^2 (2916) \][/tex]
[tex]\[ AB^2 + BC^2 = 3400 > AC^2 (2601) \][/tex]
[tex]\[ BC^2 + AC^2 = 5517 > AB^2 (484) \][/tex]
Since all these conditions are satisfied, the triangle is an acute triangle.
Based on the evaluated conditions, an acute triangle is formed because:
[tex]\[ 22^2 + 51^2 > 54^2 \quad (\text{i.e., } 484 + 2601 > 2916) \][/tex]
Therefore, the correct choice is:
- An acute triangle, because [tex]\(22^2 + 51^2 > 54^2\)[/tex].
1. Calculate the squares of the given distances:
- Distance between city A and city B (AB) = 22 miles
- Distance between city B and city C (BC) = 54 miles
- Distance between city A and city C (AC) = 51 miles
[tex]\[ AB^2 = 22^2 = 484 \][/tex]
[tex]\[ BC^2 = 54^2 = 2916 \][/tex]
[tex]\[ AC^2 = 51^2 = 2601 \][/tex]
2. Compare the sums of the squared distances to determine the type of triangle:
- For a right triangle: One combination of the squares must equal another, i.e., [tex]\(AB^2 + BC^2 = AC^2\)[/tex] or any permutations.
- For an acute triangle: The sum of the squares of any two sides must be greater than the square of the third side for all combinations.
- For an obtuse triangle: The sum of the squares of any two sides must be less than the square of the third side for at least one combination.
3. Evaluate the conditions step by step:
- Check if the triangle is right:
[tex]\[ AB^2 + BC^2 = 484 + 2916 = 3400 \neq AC^2 (2601) \][/tex]
[tex]\[ AB^2 + AC^2 = 484 + 2601 = 3085 \neq BC^2 (2916) \][/tex]
[tex]\[ BC^2 + AC^2 = 2916 + 2601 = 5517 \neq AB^2 (484) \][/tex]
Since none of these combinations are equal, the triangle is not a right triangle.
- Check if the triangle is acute:
[tex]\[ AB^2 + AC^2 = 484 + 2601 = 3085 > BC^2 (2916) \][/tex]
[tex]\[ AB^2 + BC^2 = 3400 > AC^2 (2601) \][/tex]
[tex]\[ BC^2 + AC^2 = 5517 > AB^2 (484) \][/tex]
Since all these conditions are satisfied, the triangle is an acute triangle.
Based on the evaluated conditions, an acute triangle is formed because:
[tex]\[ 22^2 + 51^2 > 54^2 \quad (\text{i.e., } 484 + 2601 > 2916) \][/tex]
Therefore, the correct choice is:
- An acute triangle, because [tex]\(22^2 + 51^2 > 54^2\)[/tex].
