The distance between city A and city B is 22 miles. The distance between city B 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 :

Let's analyze the given distances to classify the triangle formed by the cities A, B, and C.

We are given:
- The distance between city A and city B (AB) = 22 miles.
- The distance between city B and city C (BC) = 54 miles.
- The distance between city A and city C (AC) = 51 miles.

To classify the triangle based on its angles, we'll use the properties of the squares of the sides and the Pythagorean theorem:

1. If [tex]\( a^2 + b^2 > c^2 \)[/tex], where [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] are the sides of the triangle, the triangle is acute.
2. If [tex]\( a^2 + b^2 < c^2 \)[/tex], the triangle is obtuse.
3. If [tex]\( a^2 + b^2 = c^2 \)[/tex], the triangle is right.

Calculating the squares of each side:
- [tex]\( AB^2 = 22^2 = 484 \)[/tex]
- [tex]\( BC^2 = 54^2 = 2916 \)[/tex]
- [tex]\( AC^2 = 51^2 = 2601 \)[/tex]

Now, let's compare the sums of the squares of two sides to the square of the third side to determine the type of triangle:

Step 1: Check [tex]\( AB^2 + BC^2 \)[/tex] and compare it to [tex]\( AC^2 \)[/tex]:
- [tex]\( AB^2 + BC^2 = 484 + 2916 = 3400 \)[/tex]
- [tex]\( AC^2 = 2601 \)[/tex]

Since [tex]\( 484 + 2916 > 2601 \)[/tex]:
- [tex]\( 22^2 + 54^2 > 51^2 \)[/tex]
This comparison indicates that the triangle is obtuse.

Therefore, the triangle formed by the cities A, B, and C is an obtuse triangle. The correct statement is:

"An obtuse triangle, because [tex]\(22^2+54^2 > 51^2\)[/tex]"