Question

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. what type of triangle is created by the three cities? an acute triangle, because (22^2 + 54^2 > 51^2) an acute triangle, because (22^2 + 51^2 > 54^2) an obtuse triangle, because (22^2 + 54^2 > 51^2) an obtuse triangle, because (22^2 + 51^2 > 54^2)

Explanation:

Step1: Recall the triangle inequality theorem for types of triangles

For a triangle with side lengths \(a\), \(b\), and \(c\) (where \(c\) is the longest side), we use the following:

• If \(a^{2}+b^{2}>c^{2}\), the triangle is acute.
• If \(a^{2}+b^{2}=c^{2}\), the triangle is right.
• If \(a^{2}+b^{2}

Here, the side lengths are \(a = 22\), \(b=51\), \(c = 54\) (since \(54\) is the longest side).

Step2: Calculate \(a^{2}+b^{2}\) and \(c^{2}\)

First, calculate \(22^{2}\), \(51^{2}\) and \(54^{2}\):
\(22^{2}=22\times22 = 484\)
\(51^{2}=51\times51=2601\)
\(54^{2}=54\times54 = 2916\)

Then, calculate \(a^{2}+b^{2}\):
\(22^{2}+51^{2}=484 + 2601=3085\)

Now, compare \(a^{2}+b^{2}\) and \(c^{2}\):
\(3085>2916\) (since \(3085 = 22^{2}+51^{2}\) and \(2916=54^{2}\))

Since \(22^{2}+51^{2}>54^{2}\), by the triangle inequality theorem for acute triangles, the triangle is acute.

Answer:

an acute triangle, because \(22^{2}+51^{2}>54^{2}\) (corresponding to the option "an acute triangle, because \(22^{2}+51^{2}>54^{2}\)")